Which is the best order to practise Class 12 Maths NCERT solutions?

Follow the NCERT chapter order: Relations and Functions, Inverse Trigonometric Functions, Matrices, Determinants, Continuity and Differentiability, Application of Derivatives, Integrals, Application of Integrals, Differential Equations, Vector Algebra, Three Dimensional Geometry, Linear Programming and Probability — the same order used on this page.

How do I score full marks in Class 12 Maths board solutions?

Write every method step — state the rule or formula, substitute values, simplify, and box the final answer. The CBSE marking scheme awards method marks even when the final number is wrong.

Are these NCERT solutions enough for JEE Main preparation?

NCERT exercises build the fundamentals — calculus, matrices and vectors — that JEE Main tests heavily. Use these solved problems to master the standard methods, then practise JEE-level problems for speed.

Which Class 12 maths chapters carry the most board marks?

Calculus: continuity and differentiability, application of derivatives, integrals and differential equations together dominate the board weightage, followed by matrices and determinants, vectors and three-dimensional geometry.

Free PDF: Class 12 Maths NCERT So… | Narayan Gurukul Academy

01

How to Use These NCERT Solutions

Each chapter below opens with the key idea and then walks through representative NCERT exercise questions from start to finish — the step where the marks are won or lost. Follow each line of working with a pencil before checking your own attempt.

Board pattern

Marks in the CBSE paper are awarded for method steps, not just the final answer. Practise writing every line: state the formula, substitute, simplify, then box the answer.

Pair with the revision notes

For theory, definitions and exam pointers chapter by chapter, use the Class 12 Maths Notes hub. These solutions complement that hub — same NCERT order, worked problems instead of theory.

02

Chapter 1 — Relations and Functions

This chapter is the analytical backbone of Class 12 Maths. Everything here is tested again inside later chapters — inverse trigonometry, matrices and probability all lean on one-one/onto reasoning, invertible maps and binary operations. The exercises below carry every question of the NCERT textbook with short, exam-pattern working. Attempt each line with a pencil before opening its solution.

Board pattern

Relations questions award marks chiefly for stating each property (reflexive, symmetric, transitive) and disproving it with a concrete counterexample. Always write the test — e.g. "(1,1) ∉ R, so R is not reflexive". A bare "yes" or "no" scores zero.

$Ex 1.1 — Types of relations: reflexive, symmetric, transitive, equivalence (16 Q)$
$Ex 1.2 — One-one, onto, bijective functions (12 Q)$
$Ex 1.3 — Composition of functions and inverses (14 Q)$
$Ex 1.4 — Binary operations: closure, commutativity, associativity, identity (13 Q)$
$Misc — Mixed problems on relations, functions and operations (19 Q)$

03

Exercise 1.1 — Types of Relations

16Exercise questions

Step-by-step solution

1(i) Test reflexivity: need (x,x) for all x, i.e. 3x − x = 2x = 0, false for x ≥ 1. E.g. (1,1) ∉ R.
2(i) Test symmetry: (1,3) ∈ R since 3·1 − 3 = 0, but (3,1) requires 9 − 1 = 8 ≠ 0, so not symmetric.
3(i) Test transitivity: (1,3) ∈ R and (3,9) ∈ R, but (1,9) needs 3·1 − 9 = −6 ≠ 0.
4(iii) Reflexive and transitive hold; not symmetric: (1,2) ∈ R but (2,1) ∉ R.
5(iv) x − y ∈ ℤ: reflexive, symmetric and transitive — an equivalence relation.
6(v) (a) and (b) are equivalence relations; (c),(d),(e) have none of the three properties.

Final answer

(i) none; (ii) none (R = {(1,6),(2,7),(3,8)}); (iii) reflexive & transitive, not symmetric; (iv) equivalence relation; (v) equivalence for (a),(b); none for (c),(d),(e).

Step-by-step solution

1Write R explicitly: x < 4 gives R = {(1,6),(2,7),(3,8)}.
2Not reflexive ((1,1) ∉ R), not symmetric ((1,6) ∈ R but (6,1) ∉ R).
3Not transitive: no chain — (1,6),(6,·) ∉ R, so transitivity cannot hold.

Final answer

None of the three properties.

Final answer

R = {(1,2),(2,3),(3,4),(4,5),(5,6)} — not reflexive, not symmetric, not transitive.

Step-by-step solution

1Reflexive: a ≤ a for every a, so (a,a) ∈ R.
2Transitive: a ≤ b and b ≤ c imply a ≤ c, so (a,c) ∈ R.
3Not symmetric: (1,2) ∈ R but (2,1) ∉ R since 2 ≰ 1.

Final answer

Reflexive and transitive, not symmetric.

Final answer

None. Not reflexive (a = ½ fails since ½ ≤ ⅛ is false); not symmetric ((1,2) ∈ R, (2,1) ∉ R); not transitive ((28,4) ∈ R and (4,3) ∈ R, but (28,3) ∉ R since 28 ≰ 27).

Final answer

Symmetric: (1,2) ∈ R ⇒ (2,1) ∈ R and vice versa. Not reflexive ((1,1),(2,2),(3,3) ∉ R). Not transitive ((1,2) ∈ R and (2,1) ∈ R, but (1,1) ∉ R).

Step-by-step solution

1Reflexive: every book has the same number of pages as itself.
2Symmetric: if x,y have equal page counts then y,x also do.
3Transitive: x ≡ y and y ≡ z in page count ⇒ x ≡ z.

Final answer

Equivalence relation — its equivalence classes are the books grouped by page count.

Step-by-step solution

1Reflexive: |a − a| = 0, which is even.
2Symmetric: |a − b| even ⇔ |b − a| even.
3Transitive: |a − b| and |b − c| even ⇒ a,b same parity and b,c same parity ⇒ a,c same parity ⇒ |a − c| even.
4Parity classes: {1,3,5} (odd) and {2,4} (even) — elements within a class are related; across classes |a − b| is odd, not related.

Final answer

Equivalence relation with classes {1,3,5} and {2,4}; no cross-class relation.

Step-by-step solution

1(i) Reflexive: |a − a| = 0 is a multiple of 4. Symmetric and transitive follow from |a − b| mod 4 behaviour (congruence mod 4).
2Elements related to 1 (mod 4): {1, 5, 9}.
3(ii) Equality is trivially reflexive, symmetric, transitive; elements related to 1: just {1}.

Final answer

(i) {1, 5, 9}; (ii) {1}.

Final answer

On A = {1,2,3}: (i) R = {(1,2),(2,1)}; (ii) R = {(1,2)}; (iii) R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}; (iv) R = {(1,1),(2,2),(3,3),(1,2)}; (v) R = {(1,1),(2,2),(1,2),(2,1)}.

Final answer

Equivalence: equality of distances is reflexive, symmetric, transitive. Class of P = {Q : OQ = OP} = the circle centred at O passing through P.

Final answer

Similarity is an equivalence relation. T₁ and T₃ are related: (6,8,10) = 2 × (3,4,5). T₂ is related to neither.

Step-by-step solution

1Let A = {1,2,3} with R₁ = equality relation and R₂ = the full relation A × A.
2Both are equivalence relations, and R₁ ∪ R₂ = A × A, which is also an equivalence relation — not a counterexample.
3Take R₁ = {(a,b) : a ≡ b mod 2}, R₂ = {(a,b) : a ≡ b mod 3} on integers. R₁ ∪ R₂ need not be transitive: (0,2) ∈ R₁, (2,3) ∈ ? 2 ≡ 3 mod 1 only — neither, so use (0,3) ∈ (mod 3 of R₂?) — simpler: R₁ ∪ R₂ fails reflexivity? No. The standard counterexample: integers with R₁ (a − b even) and R₂ (a − b a multiple of 3). Then (0, 4) ∈ R₁ and (4, 6) ∈ R₂, but (0, 6) ∈ neither class since 6 − 0 is even (∈ R₁ actually).

Final answer

Union need not be transitive, hence not an equivalence relation in general.

Final answer

Parallel lines treating a line as parallel to itself gives reflexivity; symmetry and transitivity clear. Related lines: every line of the family y = 2x + c, c ∈ ℝ.

Step-by-step solution

1Reflexive: (1,1),(2,2),(3,3),(4,4) all present ✓.
2Symmetric? (1,2) ∈ R but (2,1) ∉ R — not symmetric.
3Transitive: check (1,3) and (3,2) give (1,2) ✓; (1,2) & (2,2) → (1,2) ✓; all chains hold.

Final answer

Option (B) — reflexive and transitive but not symmetric.

Step-by-step solution

1Membership needs a = b − 2 with b > 6.
2(A): (2,4) → b = 4 not > 6, no.
3(B): (3,8) → 3 = 8 − 2? No, 3 ≠ 6.
4(C): (6,8) → 6 = 8 − 2 ✓ and 8 > 6 ✓.
5(D): (8,7) → 8 = 7 − 2? No.

Final answer

Option (C) — (6,8) ∈ R.

04

Exercise 1.2 — One-One, Onto and Bijective Functions

12Exercise questions

Step-by-step solution

1One-one: f(x₁) = f(x₂) ⇒ 1/x₁ = 1/x₂ ⇒ x₁ = x₂.
2Onto: for y ≠ 0, x = 1/y ∈ ℝ* satisfies f(x) = y.
3With domain N: same one-one proof, but not onto — e.g. y = 1.5 ∈ ℝ* has no preimage among rationals of form 1/n with n ∈ N (only reciprocals of pure integers hit).

Final answer

Bijective ℝ* → ℝ*; not true for domain N (onto fails).

Final answer

(i) one-one, not onto (e.g. 2 has no preimage). (ii) not one-one (f(1)=f(−1)), not onto. (iii) not one-one, not onto. (iv) one-one, not onto (x³ = 2 has no natural solution). (v) bijective.

Step-by-step solution

1Not one-one: f(1.2) = 1 = f(1.8).
2Not onto: no x has f(x) = 0.5 since [x] is always an integer.

Final answer

Neither one-one nor onto.

Final answer

Not one-one: |1| = |−1|. Not onto: no x gives |x| = −1.

Final answer

Not one-one: f(2) = f(3) = 1. Not onto: no x gives ½.

Step-by-step solution

1Distinct domain elements 1, 2, 3 map to distinct values 4, 5, 6.
2Hence f is injective (it is not onto — 7 is never the image).

Final answer

f is one-one; not onto (7 unreached).

Final answer

(i) Bijective — linear with slope −4 ≠ 0, onto since x = (3 − y)/4 for any y. (ii) Neither: not one-one (f(1) = f(−1) = 2), not onto (values ≥ 1).

Step-by-step solution

1One-one: f(a₁,b₁) = f(a₂,b₂) ⇒ (b₁,a₁) = (b₂,a₂) ⇒ a₁=a₂, b₁=b₂.
2Onto: (b,a) ∈ B × A is the image of (a,b) ∈ A × B.

Final answer

f is a bijection.

Step-by-step solution

1Not one-one: f(1) = 1 and f(2) = 1, so two different inputs share an image.
2Onto: for any m ∈ N, the even input n = 2m gives f(2m) = m, so every natural is hit.

Final answer

Not bijective — f is onto but not one-one.

Step-by-step solution

1One-one: (x₁−2)/(x₁−3) = (x₂−2)/(x₂−3) ⇒ cross-multiply ⇒ x₁ = x₂.
2Onto: y = (x−2)/(x−3) ⇒ y(x−3) = x−2 ⇒ x(1−y) = 3y − 2 ⇒ x = (3y−2)/(1−y). For y ≠ 1, x ≠ 3 (would need 3y−2 = 3−3y i.e. 6y = 5 — possible! so check: when x = 3, y = (1)/(0)? undefined — exclude; 1 − y = 0 only if y=1, excluded) — x ∈ A for all y ∈ B.

Final answer

f is bijective (one-one and onto).

Step-by-step solution

1f(1) = f(−1) = 1 → not one-one.
2x⁴ ≥ 0 always → not onto.

Final answer

Option (D).

Final answer

Option (A) — f is a bijection: f(x₁)=f(x₂) ⇒ x₁=x₂ and x = y/3 covers ℝ.

05

Exercise 1.3 — Composition of Functions and Inverses

14Exercise questions

Step-by-step solution

1gof(x) = g(f(x)) computed pointwise:
2x = 1: f(1) = 2, g(2) = 3 → gof(1) = 3.
3x = 3: f(3) = 5, g(5) = 1 → gof(3) = 1.
4x = 4: f(4) = 1, g(1) = 3 → gof(4) = 3.

Final answer

gof = {(1,3),(3,1),(4,3)}.

Step-by-step solution

1(f+g)oh(x) = (f+g)(h(x)) = f(h(x)) + g(h(x)) = foh(x) + goh(x).
2(f·g)oh(x) = (f·g)(h(x)) = f(h(x))·g(h(x)) = (foh)(x)·(goh)(x).

Final answer

Both distributivity identities; verified pointwise.

Step-by-step solution

1(i) gof(x) = g(|x|) = |5|x| − 2|; fog(x) = f(|5x − 2|) = |5x − 2| (already absolute).
2(ii) gof(x) = (8x³)^{1/3} = 2x; fog(x) = 8(x^{1/3})³ = 8x.

Final answer

(i) gof = |5|x|−2|, fog = |5x−2|. (ii) gof(x) = 2x, fog(x) = 8x.

Step-by-step solution

1fof(x) = f(f(x)) = (4·f(x)+3)/(6·f(x)−4).
2Substitute f(x) = (4x+3)/(6x−4): numerator = (16x+12)/(6x−4) + 3 = (16x+12+18x−12)/(6x−4) = 34x/(6x−4).
3Denominator = 6(4x+3)/(6x−4) − 4 = (24x+18−24x+16)/(6x−4) = 34/(6x−4).
4Ratio fof(x) = (34x)/(34) = x.

Final answer

f = f⁻¹ (f is its own inverse).

Final answer

(i) No — not one-one. (ii) No — not one-one (5 and 7 both map to 4). (iii) Yes — h is bijective.

Step-by-step solution

1One-one: x₁/(x₁+2) = x₂/(x₂+2) ⇒ x₁(x₂+2) = x₂(x₁+2) ⇒ x₁ = x₂.
2Solve y = x/(x+2) ⇒ y(x+2) = x ⇒ x(y−1) = −2y ⇒ x = 2y/(1−y).
3f⁻¹(y) = 2y/(1−y), domain y ≠ 1 (and the range of f excludes 1).

Final answer

f⁻¹(y) = 2y/(1 − y).

Step-by-step solution

1One-one: 4x₁+3 = 4x₂+3 ⇒ x₁ = x₂.
2Onto: for y ∈ R, x = (y−3)/4 gives f(x) = y.
3So f⁻¹(y) = (y−3)/4 and f⁻¹(x) = (x−3)/4 as a map R → R.

Final answer

f⁻¹(x) = (x − 3)/4.

Step-by-step solution

1Restricted domain makes f one-one: x² increasing on ℝ⁺.
2x² + 4 = y ⇒ x = √(y−4) ≥ 0, so onto [4,∞).

Final answer

f⁻¹(y) = √(y − 4), verified.

Step-by-step solution

1Complete the square: 9x² + 6x − 5 = 9(x + 1/3)² − 6.
2y = 9(x+1/3)² − 6 ⇒ (x+1/3)² = (y+6)/9 ⇒ x + 1/3 = √(y+6)/3.
3x = (√(y+6) − 1)/3, valid since x ≥ 0 needs y ≥ −5.

Final answer

f⁻¹(y) = (√(y+6) − 1)/3.

Step-by-step solution

1Suppose g₁ and g₂ are two inverses of f.
2For all y ∈ Y: f∘g₁(y) = y and f∘g₂(y) = y.
3Hence f(g₁(y)) = f(g₂(y)); since f is one-one, g₁(y) = g₂(y) for all y, so g₁ = g₂.

Final answer

The inverse is unique.

Final answer

f⁻¹(a) = 1, f⁻¹(b) = 2, f⁻¹(c) = 3; applying the same reversal twice returns f.

Final answer

f⁻¹: Y → X is itself bijective, and (f⁻¹)⁻¹ exists; following f∘f⁻¹ = I_Y and f⁻¹∘f = I_X shows (f⁻¹)⁻¹ = f.

Step-by-step solution

1fof(x) = (3 − (f(x))³)^{1/3} = (3 − (3 − x³))^{1/3} = (x³)^{1/3} = x.

Final answer

Option (C) — x.

Step-by-step solution

1y = 4x/(3x+4) ⇒ y(3x+4) = 4x ⇒ 3xy + 4y = 4x ⇒ x(4 − 3y) = 4y.
2x = 4y/(4 − 3y).

Final answer

Option (B) — g(y) = 4y/(4 − 3y).

06

Exercise 1.4 — Binary Operations

12Exercise questions

Step-by-step solution

1(i) a − b can be ≤ 0 (e.g. 1 − 2 = −1 ∉ ℤ⁺) → not a binary operation.
2(ii) product of positive integers is positive → binary operation.
3(iii) ab² ∈ ℝ always → binary operation.
4(iv) |a − b| = 0 possible (a = b) → 0 ∉ ℤ⁺ → not a binary operation.
5(v) result a ∈ ℤ⁺ always → binary operation.

Final answer

Binary operations: (ii), (iii), (v). Not: (i), (iv).

Final answer

(i) neither; (ii) commutative, not associative; (iii) both (associative since a(bc)/2·1/2 = abc/4 from either side); (iv) commutative, not associative; (v) neither; (vi) neither.

Final answer

Table rows/columns 1..5 with entry min(row, col): column j of row i is min(i,j); e.g. row 1 → 1,1,1,1,1; row 2 → 1,2,2,2,2; row 3 → 1,2,3,3,3; row 4 → 1,2,3,4,4; row 5 → 1,2,3,4,5.

Step-by-step solution

1a∗b = max{a,b} for this table.
2(2∗3)∗4 = max(2,3)=3, max(3,4)=4.
32∗(3∗4) = max(3,4)=4, max(2,4)=4.
4max is commutative and associative.

Final answer

(2∗3)∗4 = 4, 2∗(3∗4) = 4; ∗ is commutative and associative.

Step-by-step solution

1Compare a = 2, b = 4: max(2,4) = 4 but HCF(2,4) = 2.
2Values differ, so the operations are different.

Final answer

No — different operations, e.g. 2∗4 = 4 yet 2∗′4 = 2.

Step-by-step solution

1LCM(a,b) = LCM(b,a) → commutative.
2LCM is associative: LCM(LCM(a,b),c) = LCM(a,LCM(b,c)).
3Identity e needs LCM(a,e) = a for all a ⇒ e = 1.
4Invertible a needs LCM(a,b) = 1 ⇒ a = b = 1.

Final answer

Commutative, associative; identity 1; only 1 is invertible.

Final answer

Yes — for all a,b ∈ ℚ, a − b ∈ ℚ, so the operation is closed on ℚ.

Step-by-step solution

12∗3 = 2 + 4·9 = 38; then (2∗3)∗4 = 38 + 4·16 = 102.
23∗4 = 3 + 4·16 = 67; then 2∗(3∗4) = 2 + 4·67² = 2 + 17956 = 17958.
3102 ≠ 17958, so ∗ is not associative — the differing results are exactly the failure of associativity.

Final answer

(2∗3)∗4 = 102, 2∗(3∗4) = 17958; not equal — ∗ is not associative.

Step-by-step solution

1a∗b = (a+b)/4; symmetric in a,b → commutative.
2(a∗b)∗c = (a/4+b/4)/4 + c/4 = (a+b)/16 + c/4; a∗(b∗c) = a/4 + (b/4+c/4)/4 = a/4 + (b+c)/16.
3These are not equal in general, so ∗ is NOT associative — check with a=1,b=0,c=1: LHS = 1/16 + 1/4 = 5/16; RHS = 1/4 + 1/16 = 5/16. Equal? The two expressions differ only by grouping of (a+b)/16 vs (b+c)/16 — not equal for a ≠ c. So associative fails.

Final answer

∗ is commutative but not associative; identity and inverses exist only in the (a+b)/4 reading isn't an operation with identity here — identity would need a∗e = a ⇒ (a+e)/4 = a ⇒ e = 3a, which depends on a; so no identity in ℚ.

Step-by-step solution

13^{a+b} ∈ ℤ⁺ for all a,b → binary operation.
2Commutative: 3^{a+b} = 3^{b+a}. Yes.
3Associative: (a∗b)∗c = 3^{(3^{a+b})+c} while a∗(b∗c) = 3^{a + 3^{b+c}}; these differ (exponent shapes differ), so not associative.

Final answer

Binary operation, commutative, not associative.

Step-by-step solution

1Table row a, column b contains a (left operand).
2Not commutative: 1∗2 = 1 ≠ 2 = 2∗1.
3Associative: (a∗b)∗c = a∗c = a; a∗(b∗c) = a∗b = a. Equal ✓.

Final answer

∗ is associative but not commutative.

Step-by-step solution

1(a+c,b+d) = (c+a,d+b) → commutative.
2Associative: both groupings give (a+c+e, b+d+f).
3Identity (e₁,e₂): (a,b)∗(e₁,e₂) = (a,b) ⇒ e₁ = 0, but 0 ∉ N — so NO identity in N × N.
4(In ℤ × ℤ, identity is (0,0) and every element is invertible with inverse (−a,−b).)

Final answer

Commutative and associative on N × N; identity (0,0) does not exist in N × N, so no invertible elements there.

07

Miscellaneous Exercise on Chapter 1

19Exercise questions

Step-by-step solution

1Need g(f(x)) = x and f(g(x)) = x.
2Swap f's formula: x = 10y + 7 ⇒ y = (x − 7)/10.
3So g(x) = (x − 7)/10; verify gof(x) = (10x+7−7)/10 = x.

Final answer

g(x) = (x − 7)/10.

Step-by-step solution

1Odd n = 2k+1 ↦ 2k; even n = 2k ↦ 2k+1. The map swaps each adjacent pair (1,0),(2,3),(4,5),…
2Composing twice restores n, so f⁻¹ = f.

Final answer

f⁻¹ = f (f is its own inverse).

Step-by-step solution

1f(f(x)) = (x²−3x+2)² − 3(x²−3x+2) + 2.
2Expand: (x²−3x+2)² = x⁴ − 6x³ + 13x² − 12x + 4.
3Subtract 3(x²−3x+2) = 3x² − 9x + 6 and add 2: x⁴ − 6x³ + 10x² − 3x + 0.

Final answer

f(f(x)) = x⁴ − 6x³ + 10x² − 3x.

Step-by-step solution

1One-one: split x ≥ 0 and x < 0; f is strictly increasing in each branch, and the two branches meet continuously at 0. Formal: assume f(x₁)=f(x₂) and cross-multiply case-wise — identical signs force x₁ = x₂; different signs give values of opposite sign, impossible.
2Onto: for y ∈ (−1,1), x = y/(1−|y|) gives f(x) = y.

Final answer

f is a bijection from ℝ onto (−1,1); f⁻¹(y) = y/(1−|y|).

Step-by-step solution

1x₁³ = x₂³ ⇒ x₁ = x₂ (cube function is strictly increasing).

Final answer

f is injective (and surjective, hence bijective).

Step-by-step solution

1Take f(x) = x (injective), g(x) = |x|.
2gof(x) = |x|, injective on N.
3g is not injective on Z since g(1) = g(−1) = 1.

Final answer

f(x) = x, g(x) = |x|.

Step-by-step solution

1Take f(x) = x + 1 (not onto since 1 is missed).
2g(x) = 1 if x = 1, and x − 1 if x > 1.
3gof(x) = x for all x (since x+1 > 1), so onto.

Final answer

f(x) = x + 1, g(x) = 1 (x=1), x − 1 (x>1).

Step-by-step solution

1A ⊂ A is false (strict subset of itself), so not reflexive.
2Not reflexive ⇒ not an equivalence relation.

Final answer

No — R fails reflexivity.

Step-by-step solution

1A ∗ X = A ∩ X = A for all A → X is the identity.
2A is invertible iff some B has A ∗ B = X, i.e. A ∩ B = X ⇒ A = X.

Final answer

Identity = X; only X is invertible.

Step-by-step solution

1An onto map from an n-set to itself is a bijection, i.e. a permutation of the codomain.
2Number of permutations of n objects = n!.

Final answer

n!

Step-by-step solution

1(i) F is bijective: F⁻¹ = {(3,a),(2,b),(1,c)}.
2(ii) F is not one-one ((b,1),(c,1)) → not invertible.

Final answer

(i) F⁻¹ = {(3,a),(2,b),(1,c)}; (ii) no inverse exists.

Step-by-step solution

1∗: |a−b| = |b−a| → commutative. Not associative: (1∗2)∗3 = |1−2|=1, 1∗3 = 2; 1∗(2∗3) = 1∗1 = 0. Not equal.
2o: a o b = a → (a o b) o c = a o c = a; a o (b o c) = a. Associative. Not commutative: a o b = a ≠ b generally.
3a∗(b o c) = a∗b = |a−b|; (a∗b) o (a∗c) = |a−b|. Equal → ∗ distributes over o.
4o over ∗: a o (b∗c) = a; (a o b) ∗ (a o c) = a ∗ a = 0. Equal only if a = 0 → fails generally.

Final answer

∗ commutative, not associative; o associative, not commutative; ∗ distributes over o; o does NOT distribute over ∗.

Step-by-step solution

1A ∗ φ = (A − φ) ∪ (φ − A) = A ∪ φ = A.
2A ∗ A = (A − A) ∪ (A − A) = φ ∪ φ = φ.
3So each A is its own inverse.

Final answer

Identity φ; A⁻¹ = A for every A (symmetric difference is self-inverse).

Step-by-step solution

1a ∗ 0 = a + 0 = a (since a < 6) → 0 is identity.
2a ∗ (6−a) = a + (6−a) − 6 = 0 for a ≥ 1 (and 0 for a = 0).
3Hence every element is invertible; inverse of a is 6 − a.

Final answer

Identity 0; a⁻¹ = 6 − a for a ≠ 0.

Step-by-step solution

1Compute f: f(−1) = 2, f(0) = 0, f(1) = 0, f(2) = 2 — all in B.
2Compute g: g(−1) = 2⌊−0.5⌋ − 1 = 2(−1) − 1 = −3; g(0) = −1; g(1) = −1; g(2) = 2(1) − 1 = 1.
3Two values differ (e.g. x = −1: f = 2 vs g = −3) → not equal functions.

Final answer

No — f ≠ g (e.g. f(−1) = 2 but g(−1) = −3).

Step-by-step solution

1Reflexivity forces (1,1),(2,2),(3,3); symmetry plus (1,2),(1,3) forces (2,1),(3,1).
2Need not transitive: skip (2,3),(3,2) or add just one of them — the unique minimal relation R = {(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)} is not transitive ((2? nothing pairs (2,3)) — (1,2),(2,1) ok; check (1,3),(3,1): (1,1) present — wait transitive requires (1,2) and (2,1) → (1,1) ✓; (1,3),(3,1) → (1,1) ✓; but (2,1),(1,3) → (2,3) MISSING → not transitive ✓). Adding (2,3),(3,2) makes it transitive. So exactly 1 relation.

Final answer

Option (A) — 1.

Step-by-step solution

1Equivalence relations = partitions of A. Containing (1,2) means 1 and 2 are in the same block.
2Partitions with 1,2 together: {{1,2,3}} and {{1,2},{3}} — exactly 2.

Final answer

Option (B) — 2.

Step-by-step solution

1x² + 1 = 17 ⇒ x² = 16 ⇒ x = ±4.
2x² + 1 = −3 ⇒ x² = −4, impossible over ℝ.

Final answer

Option (C) — preimages {4,−4} and φ.

Step-by-step solution

1f and g are bijections with f⁻¹ = {(a,1),(b,2),(c,3)}, g⁻¹ = {(apple,a),(ball,b),(cat,c)}.
2gof(1) = apple, gof(2) = ball, gof(3) = cat; so (gof)⁻¹ = {(apple,1),(ball,2),(cat,3)}.
3f⁻¹∘g⁻¹(apple) = f⁻¹(a) = 1; similarly 2,3 — matches (gof)⁻¹.

Final answer

(gof)⁻¹ = {(apple,1),(ball,2),(cat,3)} = f⁻¹ ∘ g⁻¹. ✓

08

Chapter 2 — Inverse Trigonometric Functions

This chapter defines the inverse of each of the six trigonometric functions, fixes their principal value branches, and builds the identity toolkit used throughout calculus. Every identity here — the triple-angle forms, the tan⁻¹ addition formulas and the half-angle simplifications — reappears inside integration and coordinate geometry later in the course. Attempt each line with a pencil before opening its solution.

Board pattern

Principal-value and simplest-form questions carry 2–3 marks; identity proofs and equations carry 3–4 marks. Always begin by naming the principal branch — e.g. "sin⁻¹ : [−1, 1] → [−π/2, π/2]" — and justify that your answer lies in it. For simplifications introduce a substitution (x = tan θ, x = a sin θ, …), reduce the inner expression, and cite the branch before stripping the outer inverse.

$Ex 2.1 — Principal values of inverse trigonometric functions (14 Q)$
$Ex 2.2 — Identities, simplest forms, values and MCQs (21 Q)$
$Misc — Mixed problems: proofs, equations and multiple choice (17 Q)$

09

Exercise 2.1 — Principal Values

14Exercise questions

Step-by-step solution

1Let $sin^{- 1} (- \frac{1}{2}) = y$ , so $sin y = - \frac{1}{2}$ .
2The principal branch of $sin^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$ .
3Since $sin (- \frac{π}{6}) = - \frac{1}{2}$ , $y = - \frac{π}{6}$ .

Final answer

- \frac{π}{6}

Step-by-step solution

1Let $cos^{- 1} (\frac{3}{2}) = y$ , so $cos y = \frac{3}{2}$ .
2The principal branch of $cos^{- 1}$ is $[0, π]$ .
3Since $cos \frac{π}{6} = \frac{3}{2}$ , $y = \frac{π}{6}$ .

Final answer

\frac{π}{6}

Step-by-step solution

1Let $cosec^{- 1} (2) = y$ , so $sin y = \frac{1}{2}$ .
2The principal branch of $cosec^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}] ∖ {0}$ .
3Since $sin \frac{π}{6} = \frac{1}{2}$ , $y = \frac{π}{6}$ .

Final answer

\frac{π}{6}

Step-by-step solution

1Let $tan^{- 1} (- 3) = y$ , so $tan y = - 3$ .
2The principal branch of $tan^{- 1}$ is $(- \frac{π}{2}, \frac{π}{2})$ .
3Since $tan (- \frac{π}{3}) = - 3$ , $y = - \frac{π}{3}$ .

Final answer

- \frac{π}{3}

Step-by-step solution

1Let $cos^{- 1} (- \frac{1}{2}) = y$ , so $cos y = - \frac{1}{2}$ .
2In $[0, π]$ , the angle with cosine $- 1/2$ is $\frac{2 π}{3}$ .

Final answer

\frac{2 π}{3}

Step-by-step solution

1Let $tan^{- 1} (- 1) = y$ , so $tan y = - 1$ .
2In $(- \frac{π}{2}, \frac{π}{2})$ , the angle with tangent −1 is $- \frac{π}{4}$ .

Final answer

- \frac{π}{4}

Step-by-step solution

1Let $sec^{- 1} (\frac{2}{3}) = y$ , so $cos y = \frac{3}{2}$ .
2The principal branch of $sec^{- 1}$ is $[0, π] ∖ {\frac{π}{2}}$ .
3Since $cos \frac{π}{6} = \frac{3}{2}$ , $y = \frac{π}{6}$ .

Final answer

\frac{π}{6}

Step-by-step solution

1Let $cot^{- 1} (3) = y$ , so $cot y = 3$ .
2The principal branch of $cot^{- 1}$ is $(0, π)$ .
3Since $cot \frac{π}{6} = 3$ , $y = \frac{π}{6}$ .

Final answer

\frac{π}{6}

Step-by-step solution

1Let $cos^{- 1} (- \frac{1}{2}) = y$ , so $cos y = - \frac{1}{2}$ .
2In $[0, π]$ , the angle with cosine $- \frac{1}{2}$ is $\frac{3 π}{4}$ .

Final answer

\frac{3 π}{4}

Step-by-step solution

1Let $cosec^{- 1} (- 2) = y$ , so $sin y = - \frac{1}{2}$ .
2In $[- \frac{π}{2}, \frac{π}{2}] ∖ {0}$ , the angle with sine $- \frac{1}{2}$ is $- \frac{π}{4}$ .

Final answer

- \frac{π}{4}

Step-by-step solution

1 $tan^{- 1} (1) = \frac{π}{4}$ .
2 $cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ and $sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ .
3Sum: $\frac{π}{4} + \frac{2 π}{3} - \frac{π}{6} = \frac{3 π + 8 π - 2 π}{12} = \frac{9 π}{12} = \frac{3 π}{4}$ .

Final answer

\frac{3 π}{4}

Step-by-step solution

1 $cos^{- 1} (\frac{1}{2}) = \frac{π}{3}$ .
2 $sin^{- 1} (\frac{1}{2}) = \frac{π}{6}$ .
3Sum: $\frac{π}{3} + 2 \cdot \frac{π}{6} = \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3}$ .

Final answer

\frac{2 π}{3}

Step-by-step solution

1The range of the principal value branch of $sin^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$ .
2So $y = sin^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}]$ .

Final answer

- \frac{π}{2} \leq y \leq \frac{π}{2}

Option (B) —

Step-by-step solution

1 $tan^{- 1} 3 = \frac{π}{3}$ .
2 $sec^{- 1} (- 2) = \frac{2 π}{3}$ (principal branch $[0, π] ∖ {π /2}$ ).
3 $\frac{π}{3} - \frac{2 π}{3} = - \frac{π}{3}$ .

Final answer

- \frac{π}{3}

Option (B) —

10

Exercise 2.2 — Identities, Simplest Forms and Values

21Exercise questions

Step-by-step solution

1Let $x = sin θ$ , so $θ = sin^{- 1} x$ and $θ \in [- \frac{π}{6}, \frac{π}{6}]$ .
2Then $3 x - 4 x^{3} = 3 sin θ - 4 sin^{3} θ = sin 3 θ$ .
3Since $3 θ \in [- \frac{π}{2}, \frac{π}{2}]$ , $sin^{- 1} (sin 3 θ) = 3 θ = 3 sin^{- 1} x$ .

Final answer

3 sin^{- 1} x = sin^{- 1} (3 x - 4 x^{3})

Step-by-step solution

1Let $x = cos θ$ , so $θ = cos^{- 1} x$ and $θ \in [0, \frac{π}{3}]$ .
2Then $4 x^{3} - 3 x = 4 cos^{3} θ - 3 cos θ = cos 3 θ$ .
3Since $3 θ \in [0, π]$ , $cos^{- 1} (cos 3 θ) = 3 θ = 3 cos^{- 1} x$ .

Final answer

3 cos^{- 1} x = cos^{- 1} (4 x^{3} - 3 x)

Step-by-step solution

1Use $tan^{- 1} a + tan^{- 1} b = tan^{- 1} (\frac{a + b}{1 - ab})$ with $a = \frac{2}{11}$ , $b = \frac{7}{24}$ .
2 $\frac{a + b}{1 - ab} = \frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \cdot \frac{7}{24}} = \frac{\frac{48 + 77}{264}}{\frac{264 - 14}{264}} = \frac{125}{250} = \frac{1}{2}$ .
3Both angles are in $(0, \frac{π}{2})$ , so the sum formula applies and the result follows.

Final answer

tan^{- 1} \frac{2}{11} + tan^{- 1} \frac{7}{24} = tan^{- 1} \frac{1}{2}

Step-by-step solution

1First, $2 tan^{- 1} \frac{1}{2} = tan^{- 1} (\frac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}}) = tan^{- 1} \frac{4}{3}$ .
2Now $tan^{- 1} \frac{4}{3} + tan^{- 1} \frac{1}{7} = tan^{- 1} (\frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \cdot \frac{1}{7}}) = tan^{- 1} (\frac{\frac{28 + 3}{21}}{\frac{21 - 4}{21}}) = tan^{- 1} \frac{31}{17}$ .

Final answer

2 tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{7} = tan^{- 1} \frac{31}{17}

Step-by-step solution

1Let $x = tan θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ .
2Then $1 + x^{2} = sec θ$ and $\frac{s e c θ - 1}{t a n θ} = \frac{1 - c o s θ}{s i n θ} = tan \frac{θ}{2}$ .
3So the expression is $tan^{- 1} (tan \frac{θ}{2}) = \frac{θ}{2} = \frac{1}{2} tan^{- 1} x$

Final answer

\frac{1}{2} tan^{- 1} x

Step-by-step solution

1Let $x = cosec θ$ with $θ \in (0, \frac{π}{2})$ for $x > 1$ .
2Then $x^{2} - 1 = cosec^{2} θ - 1 = cot θ$ , giving $\frac{1}{x ^{2} - 1} = tan θ$ .
3For $x > 1$ : value $= θ = cosec^{- 1} x$ .
4For $- 1 > x$ : $θ \in (- \frac{π}{2}, 0)$ , so the value is $- cosec^{- 1} x$ .

Final answer

For x > 1: \text{cosec}^{-1}x; for x < −1: -\text{cosec}^{-1}x

Step-by-step solution

1 $\frac{1 - c o s x}{1 + c o s x} = \frac{2 s i n ^{2} \frac{x}{2}}{2 c o s ^{2} \frac{x}{2}} = tan^{2} \frac{x}{2}$ .
2For $x \in (π, 2 π)$ , $\frac{x}{2} \in (\frac{π}{2}, π)$ , so $tan \frac{x}{2} < 0$ .
3 $\frac{1 - c o s x}{1 + c o s x} = - tan \frac{x}{2} = tan (π - \frac{x}{2})$ .
4Hence the expression $= π - \frac{x}{2}$

Final answer

π - \frac{x}{2}

Step-by-step solution

1Divide numerator and denominator by $cos x$ :
2 $\frac{c o s x - s i n x}{c o s x + s i n x} = \frac{1 - t a n x}{1 + t a n x} = tan (\frac{π}{4} - x)$ .
3For $0 < x < \frac{3 π}{4}$ , $\frac{π}{4} - x \in (- \frac{π}{2}, \frac{π}{4})$ , giving $\frac{π}{4} - x$ .
4For $\frac{3 π}{4} < x < π$ , the value is $\frac{5 π}{4} - x$ .

Final answer

π/4 − x for 0 < x < 3π/4; 5π/4 − x for 3π/4 < x < π.

Step-by-step solution

1Let $x = a sin θ$ , so $θ = sin^{- 1} \frac{x}{a}$ and $∣ θ ∣ < \frac{π}{2}$ .
2 $\frac{x}{a ^{2} - x ^{2}} = \frac{a s i n θ}{a c o s θ} = tan θ$ .
3So the expression $= tan^{- 1} (tan θ) = θ = sin^{- 1} \frac{x}{a}$

Final answer

sin^{- 1} \frac{x}{a}

Step-by-step solution

1Let $x = a tan θ$ , so $θ = tan^{- 1} \frac{x}{a}$ and $∣ θ ∣ \leq \frac{π}{6}$ .
2 $\frac{3 a ^{2} x - x ^{3}}{a ^{3} - 3 a x ^{2}} = \frac{3 t a n θ - t a n ^{3} θ}{1 - 3 t a n ^{2} θ} = tan 3 θ$ .
3Since $3 θ \in [- \frac{π}{2}, \frac{π}{2}]$ , the expression $= 3 θ = 3 tan^{- 1} \frac{x}{a}$

Final answer

3 tan^{- 1} \frac{x}{a}

Step-by-step solution

1 $sin^{- 1} \frac{1}{2} = \frac{π}{6}$ .
2 $2 cos (2 \cdot \frac{π}{6}) = 2 cos \frac{π}{3} = 2 \cdot \frac{1}{2} = 1$ .
3 $tan^{- 1} (1) = \frac{π}{4}$

Final answer

\frac{π}{4}

Step-by-step solution

1 $tan^{- 1} a + cot^{- 1} a = \frac{π}{2}$ for all $a \in R$ .
2 $cot \frac{π}{2} = 0$

Final answer

0

Step-by-step solution

1Let $x = tan α$ ; then $\frac{2 x}{1 + x ^{2}} = sin 2 α$ , so $sin^{- 1} \frac{2 x}{1 + x ^{2}} = 2 tan^{- 1} x$ .
2Let $y = tan β$ ; then $\frac{1 - y ^{2}}{1 + y ^{2}} = cos 2 β$ , so $cos^{- 1} \frac{1 - y ^{2}}{1 + y ^{2}} = 2 tan^{- 1} y$ .
3Half the sum: $tan^{- 1} x + tan^{- 1} y = tan^{- 1} (\frac{x + y}{1 - x y})$ since $x y < 1$ .
4Taking tan $= \frac{x + y}{1 - x y}$

Final answer

\frac{x + y}{1 - x y}

Step-by-step solution

1Since the sine of the angle is 1, the angle is $\frac{π}{2}$ .
2 $sin^{- 1} \frac{1}{5} + cos^{- 1} x = \frac{π}{2}$ .
3 $cos^{- 1} x = \frac{π}{2} - sin^{- 1} \frac{1}{5} = cos^{- 1} \frac{1}{5}$ .

Final answer

x = \frac{1}{5}

Step-by-step solution

1 $tan^{- 1} a + tan^{- 1} b = tan^{- 1} (\frac{a + b}{1 - ab})$ with $a = \frac{x - 1}{x - 2}$ , $b = \frac{x + 1}{x + 2}$ .
2 $a + b = \frac{2 x ^{2} - 4}{x ^{2} - 4}$ and $1 - ab = \frac{- 3}{x ^{2} - 4}$ .
3So $\frac{a + b}{1 - ab} = - \frac{2 x ^{2} - 4}{3} = tan \frac{π}{4} = 1$ ⟹ $2 x^{2} - 4 = - 3$ ⟹ $x^{2} = \frac{1}{2}$ .

Final answer

x = \pm \frac{1}{2}

Step-by-step solution

1 $\frac{2 π}{3} \in / [- \frac{π}{2}, \frac{π}{2}]$ .
2 $sin \frac{2 π}{3} = sin \frac{π}{3} = \frac{3}{2}$ .
3 $sin^{- 1} (\frac{3}{2}) = \frac{π}{3}$

Final answer

\frac{π}{3}

Step-by-step solution

1 $\frac{3 π}{4} \in / (- \frac{π}{2}, \frac{π}{2})$ .
2 $tan \frac{3 π}{4} = - 1$ .
3 $tan^{- 1} (- 1) = - \frac{π}{4}$

Final answer

- \frac{π}{4}

Step-by-step solution

1Let $sin^{- 1} \frac{3}{5} = α$ , so $tan α = \frac{3}{4}$ .
2Let $cot^{- 1} \frac{3}{2} = β$ , so $tan β = \frac{2}{3}$ .
3 $tan (α + β) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \cdot \frac{2}{3}} = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{6}$

Final answer

\frac{17}{6}

Step-by-step solution

1 $cos \frac{7 π}{6} = - \frac{3}{2}$ .
2The principal angle in $[0, π]$ with this cosine is $\frac{5 π}{6}$ .

Final answer

\frac{5 π}{6}

Option (B) —

Step-by-step solution

1 $sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ .
2 $\frac{π}{3} - (- \frac{π}{6}) = \frac{π}{2}$ .
3 $sin \frac{π}{2} = 1$

Final answer

Option (D) — 1.

Step-by-step solution

1 $tan^{- 1} 3 = \frac{π}{3}$ .
2 $cot^{- 1} (- 3) = \frac{5 π}{6}$ (principal branch $(0, π)$ ).
3 $\frac{π}{3} - \frac{5 π}{6} = - \frac{π}{2}$

Final answer

- \frac{π}{2}

Option (B) —

11

Miscellaneous Exercise on Chapter 2

17Exercise questions

Step-by-step solution

1 $cos \frac{13 π}{6} = cos (2 π + \frac{π}{6}) = cos \frac{π}{6}$ .
2 $cos^{- 1} (cos \frac{π}{6}) = \frac{π}{6}$

Final answer

\frac{π}{6}

Step-by-step solution

1 $tan \frac{7 π}{6} = tan (π + \frac{π}{6}) = tan \frac{π}{6}$ .
2 $tan^{- 1} (tan \frac{π}{6}) = \frac{π}{6}$

Final answer

\frac{π}{6}

Step-by-step solution

1Let $sin^{- 1} \frac{3}{5} = θ$ , so $tan θ = \frac{3}{4}$ .
2 $2 tan^{- 1} \frac{3}{4} = tan^{- 1} (\frac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}}) = tan^{- 1} (\frac{\frac{3}{2}}{\frac{7}{16}}) = tan^{- 1} \frac{24}{7}$

Final answer

2 sin^{- 1} \frac{3}{5} = tan^{- 1} \frac{24}{7}

Step-by-step solution

1Convert to tangent form: $sin^{- 1} \frac{8}{17} = tan^{- 1} \frac{8}{15}$ and $sin^{- 1} \frac{3}{5} = tan^{- 1} \frac{3}{4}$ .
2 $tan^{- 1} \frac{8}{15} + tan^{- 1} \frac{3}{4} = tan^{- 1} (\frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \cdot \frac{3}{4}}) = tan^{- 1} (\frac{\frac{77}{60}}{\frac{36}{60}}) = tan^{- 1} \frac{77}{36}$

Final answer

tan^{- 1} \frac{77}{36}

Proved — LHS reduces to

Step-by-step solution

1 $cos^{- 1} \frac{4}{5} = tan^{- 1} \frac{3}{4}$ and $cos^{- 1} \frac{12}{13} = tan^{- 1} \frac{5}{12}$ .
2 $tan^{- 1} \frac{3}{4} + tan^{- 1} \frac{5}{12} = tan^{- 1} (\frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \cdot \frac{5}{12}}) = tan^{- 1} \frac{56}{33}$ .
3 $tan^{- 1} \frac{56}{33} = cos^{- 1} \frac{33}{65}$ since hypotenuse $5 6^{2} + 3 3^{2} = 65$ .

Final answer

cos^{- 1} \frac{33}{65}

Proved — LHS reduces to

Step-by-step solution

1 $cos^{- 1} \frac{12}{13} = tan^{- 1} \frac{5}{12}$ and $sin^{- 1} \frac{3}{5} = tan^{- 1} \frac{3}{4}$ .
2 $tan^{- 1} \frac{5}{12} + tan^{- 1} \frac{3}{4} = tan^{- 1} (\frac{\frac{5}{12} + \frac{3}{4}}{1 - \frac{5}{12} \cdot \frac{3}{4}}) = tan^{- 1} \frac{56}{33}$ .
3 $tan^{- 1} \frac{56}{33} = sin^{- 1} \frac{56}{65}$ since hypotenuse $5 6^{2} + 3 3^{2} = 65$ .

Final answer

sin^{- 1} \frac{56}{65}

Proved — LHS reduces to

Step-by-step solution

1 $sin^{- 1} \frac{5}{13} = tan^{- 1} \frac{5}{12}$ and $cos^{- 1} \frac{3}{5} = tan^{- 1} \frac{4}{3}$ .
2 $tan^{- 1} \frac{5}{12} + tan^{- 1} \frac{4}{3} = tan^{- 1} (\frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \cdot \frac{4}{3}}) = tan^{- 1} (\frac{\frac{21}{12}}{\frac{16}{36}}) = tan^{- 1} \frac{63}{16}$

Final answer

tan^{- 1} \frac{63}{16}

Proved — RHS reduces to

Step-by-step solution

1 $tan^{- 1} \frac{1}{5} + tan^{- 1} \frac{1}{7} = tan^{- 1} (\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{35}}) = tan^{- 1} \frac{6}{17}$ .
2 $tan^{- 1} \frac{1}{3} + tan^{- 1} \frac{1}{8} = tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{24}}) = tan^{- 1} \frac{11}{23}$ .
3 $tan^{- 1} \frac{6}{17} + tan^{- 1} \frac{11}{23} = tan^{- 1} (\frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \cdot \frac{11}{23}}) = tan^{- 1} (\frac{\frac{325}{391}}{\frac{325}{391}}) = tan^{- 1} (1) = \frac{π}{4}$

Final answer

\frac{π}{4}

Proved — the four-angle sum equals

Step-by-step solution

1Let $x = tan^{2} θ$ , so $θ = tan^{- 1} x$ and $θ \in [0, \frac{π}{4}]$ .
2 $\frac{1 - x}{1 + x} = \frac{1 - t a n ^{2} θ}{1 + t a n ^{2} θ} = cos 2 θ$ .
3 $\frac{1}{2} cos^{- 1} (cos 2 θ) = \frac{1}{2} \cdot 2 θ = θ = tan^{- 1} x$

Final answer

Proved.

Step-by-step solution

1 $1 + sin x = (cos \frac{x}{2} + sin \frac{x}{2})^{2}$ and $1 - sin x = (cos \frac{x}{2} - sin \frac{x}{2})^{2}$ .
2For $x \in (0, \frac{π}{4})$ , both square roots are positive.
3 $\frac{( c o s \frac{x}{2} + s i n \frac{x}{2} ) + ( c o s \frac{x}{2} - s i n \frac{x}{2} )}{( c o s \frac{x}{2} + s i n \frac{x}{2} ) - ( c o s \frac{x}{2} - s i n \frac{x}{2} )} = \frac{2 c o s \frac{x}{2}}{2 s i n \frac{x}{2}} = cot \frac{x}{2}$ .
4 $cot^{- 1} (cot \frac{x}{2}) = \frac{x}{2}$

Final answer

Proved.

Step-by-step solution

1Let $x = cos θ$ , so $θ = cos^{- 1} x \in [0, \frac{3 π}{4}]$ .
2 $1 + x = 2 cos \frac{θ}{2}$ and $1 - x = 2 sin \frac{θ}{2}$ .
3 $\frac{1 + x - 1 - x}{1 + x + 1 - x} = \frac{c o s \frac{θ}{2} - s i n \frac{θ}{2}}{c o s \frac{θ}{2} + s i n \frac{θ}{2}} = \frac{1 - t a n \frac{θ}{2}}{1 + t a n \frac{θ}{2}} = tan (\frac{π}{4} - \frac{θ}{2})$ .
4Thus the expression $= \frac{π}{4} - \frac{θ}{2} = \frac{π}{4} - \frac{1}{2} cos^{- 1} x$

Final answer

Proved.

Step-by-step solution

1It suffices to show $sin^{- 1} \frac{1}{3} + sin^{- 1} \frac{2 2}{3} = \frac{π}{2}$ .
2Let $A = sin^{- 1} \frac{2 2}{3}$ , so $sin A = \frac{2 2}{3}$ and $cos A = \frac{1}{3}$ .
3Hence $A = cos^{- 1} \frac{1}{3} = \frac{π}{2} - sin^{- 1} \frac{1}{3}$ .
4 $\frac{9 π}{8} - \frac{9}{4} sin^{- 1} \frac{1}{3} = \frac{9}{4} (\frac{π}{2} - sin^{- 1} \frac{1}{3}) = \frac{9}{4} sin^{- 1} \frac{2 2}{3}$

Final answer

Proved.

Step-by-step solution

1 $2 tan^{- 1} (cos x) = tan^{- 1} (\frac{2 c o s x}{1 - c o s ^{2} x}) = tan^{- 1} (\frac{2 c o s x}{s i n ^{2} x})$ .
2Equate arguments: $\frac{2 c o s x}{s i n ^{2} x} = \frac{2}{s i n x}$ .
3 $cos x = sin x$ ⟹ $tan x = 1$ ⟹ $x = \frac{π}{4}$

Final answer

x = \frac{π}{4}

Step-by-step solution

1 $\frac{1 - x}{1 + x} = \frac{1 - t a n θ}{1 + t a n θ} = tan (\frac{π}{4} - θ)$ where $θ = tan^{- 1} x$ .
2 $tan^{- 1} (tan (\frac{π}{4} - θ)) = \frac{π}{4} - θ = \frac{θ}{2}$ .
3 $\frac{π}{4} = \frac{3 θ}{2}$ ⟹ $θ = \frac{π}{6}$ ⟹ $x = \frac{1}{3}$

Final answer

x = \frac{1}{3}

Step-by-step solution

1Let $tan^{- 1} x = α$ , so $tan α = x$ and $sin α = \frac{x}{1 + x ^{2}}$ .
2 $sin (tan^{- 1} x) = sin α = \frac{x}{1 + x ^{2}}$

Final answer

\frac{x}{1 + x ^{2}}

Option (D) —

Step-by-step solution

1Let $x = sin θ$ ; then $sin^{- 1} (1 - sin θ) - 2 θ = \frac{π}{2}$ ⟹ $1 - sin θ = cos 2 θ$ .
2 $1 - sin θ = 1 - 2 sin^{2} θ$ ⟹ $sin θ (2 sin θ - 1) = 0$ ⟹ $x = 0$ or $x = \frac{1}{2}$ .
3Check: $x = \frac{1}{2}$ gives $\frac{π}{6} - \frac{π}{3} = - \frac{π}{6} \neq = \frac{π}{2}$ ; so $x = 0$ only.

Final answer

x = 0

Option (C) —

Step-by-step solution

1 $tan^{- 1} \frac{x}{y} - tan^{- 1} \frac{x - y}{x + y} = tan^{- 1} (\frac{\frac{x}{y} - \frac{x - y}{x + y}}{1 + \frac{x}{y} \cdot \frac{x - y}{x + y}})$ .
2Numerator: $\frac{x ^{2} + y ^{2}}{y ( x + y )}$ ; denominator: $\frac{x ^{2} + y ^{2}}{y ( x + y )}$ .
3Ratio = 1, so the difference $= tan^{- 1} (1) = \frac{π}{4}$

Final answer

\frac{π}{4}

Option (C) —

12

Chapter 3 — Matrices

Matrices are rectangular arrays of numbers arranged in rows and columns, enclosed in brackets. They provide a compact way to represent and solve systems of linear equations, and are indispensable in physics, engineering, economics and computer science. This chapter builds the vocabulary — order, equality, transpose, symmetry — and the arithmetic (addition, scalar multiplication, product) you will use throughout determinants, linear transformations and beyond.

Board pattern

Marks are awarded for showing every intermediate step: state the order before multiplying, write each element computation, and verify both sides of a matrix identity separately. For transpose questions, always write A′ explicitly before comparing. A bare numeric answer without the working earns at most 1 mark.

$Ex 3.1 — Order of matrices, elements, construction, equality, identity and zero matrices (10 Q)$
$Ex 3.2 — Addition, subtraction, scalar multiplication, matrix equations, properties (28 Q)$
$Ex 3.3 — Transpose, symmetric and skew-symmetric matrices, decomposition (19 Q)$
$Ex 3.4 — Elementary row and column operations, inverse by row reduction (4 Q)$
$Misc — Mixed problems on matrices (15 Q)$

13

Exercise 3.1 — Basics of Matrices

10Exercise questions

Step-by-step solution

1A has 3 rows and 4 columns.
2Number of elements = 3 × 4 = 12.
3a₁₃ = 19, a₂₁ = 35, a₃₃ = −5, a₂₄ = 12, a₂₃ = 5/3.

Final answer

Order: 3 × 4; elements: 12; a₁₃ = 19, a₂₁ = 35, a₃₃ = −5, a₂₄ = 12, a₂₃ = 5/3.

Step-by-step solution

124 = 1×24, 2×12, 3×8, 4×6, 6×4, 8×3, 12×2, 24×1 — eight possible orders.
213 is prime, so 13 = 1×13 or 13×1 — two possible orders.

Final answer

24 elements: orders 1×24, 2×12, 3×8, 4×6, 6×4, 8×3, 12×2, 24×1 (8 orders). 13 elements: orders 1×13, 13×1.

Step-by-step solution

118 = 1×18, 2×9, 3×6, 6×3, 9×2, 18×1 — six possible orders.
25 is prime, so 5 = 1×5 or 5×1 — two possible orders.

Final answer

18 elements: orders 1×18, 2×9, 3×6, 6×3, 9×2, 18×1. 5 elements: orders 1×5, 5×1.

Step-by-step solution

1a₁₁ = (1+1)²/2 = 2, a₁₂ = (1+2)²/2 = 9/2, a₂₁ = (2+1)²/2 = 9/2, a₂₂ = (2+2)²/2 = 8.

Final answer

A = (2 \frac{9}{2} \frac{9}{2} 8)

Step-by-step solution

1Compute each element:
2a₁₁ = 4/2 = 2, a₁₂ = 5/2, a₁₃ = 3, a₁₄ = 7/2.
3a₂₁ = 7/2, a₂₂ = 4, a₂₃ = 9/2, a₂₄ = 5.
4a₃₁ = 5, a₃₂ = 11/2, a₃₃ = 6, a₃₄ = 13/2.

Final answer

A = 2 \frac{7}{2} 5 \frac{5}{2} 4 \frac{11}{2} 3 \frac{9}{2} 6 \frac{7}{2} 5 \frac{13}{2}

Step-by-step solution

1Compare corresponding elements: 4 = y and x = 2.

Final answer

x = 2, y = 4.

Step-by-step solution

1Compare: a + b = 6, c + d = 8. No further constraint from the given equations, so infinitely many solutions exist.
2For example a = 2, b = 4, c = 3, d = 5 is one solution.

Final answer

a + b = 6 and c + d = 8 (infinitely many solutions, e.g. a=2, b=4, c=3, d=5).

Step-by-step solution

1Compare (1,1): 2a + b = 3.
2Compare (1,2): 3a − b = 1.
3Add: 5a = 4, so a = 4/5.
4Then b = 3 − 2(4/5) = 7/5.

Final answer

a = 4/5, b = 7/5.

Step-by-step solution

1Compare entries: x + y = 3, z = 2, 5 + w = 7, y + z = 6.
2From z = 2 and y + z = 6: y = 4.
3From x + y = 3: x = −1.
4From 5 + w = 7: w = 2.

Final answer

x = −1, y = 4, z = 2, w = 2.

Step-by-step solution

1Compare the four entries: a − b = −1, 2a − c = 5, 2a − b = 0, 3c + d = 13.
2Subtract (1,1) from (2,1): (2a − b) − (a − b) = 0 − (−1), so a = 1.
3From a − b = −1: b = a + 1 = 2.
4From 2a − c = 5: c = 2(1) − 5 = −3.
5From 3c + d = 13: d = 13 − 3(−3) = 22.

Final answer

a = 1, b = 2, c = −3, d = 22.

14

Exercise 3.2 — Matrix Algebra

28Exercise questions

Step-by-step solution

1(i) A + B: add corresponding entries.
2(ii) A − B: subtract corresponding entries.
3(iii) 3A − C: compute 3A first, then subtract C.
4(iv) 3A + 4B: compute 3A and 4B, then add.

Final answer

(i) A + B = (3177); (ii) A - B = (15 1 - 3); (iii) 3 A - C = (8672); (iv) 3 A + 4 B = (1012426)

Step-by-step solution

1(i) Multiply row by column for each entry.
2(ii) A 3×1 times 1×3 gives a 3×3 matrix.
3(iii) Multiply each row of the first matrix by each column of the second.
4(iv) A 2×3 times 3×1 gives a 2×1 matrix.

Final answer

(i) (a^{2} + b^{2} 0 0 a^{2} + b^{2}); (ii) 2463694812; (iii) (- 3 2 01 - 2 1); (iv) (2032)

Step-by-step solution

1(i) Compute A + B and B + A separately; both give the same result.
2(ii) Compute A − B and B − A; they are negatives of each other, hence not equal unless zero.

Final answer

(i) A + B = B + A = (5555); (ii) A - B = (- 3 1 - 1 3) \neq = B - A = (3 - 1 1 - 3)

Step-by-step solution

1Compute A²: multiply A by itself.
2Compute B²: multiply B by itself.
3Subtract: A² − B².

Final answer

A^{2} = (16016), B^{2} = (6 - 2 - 3 7), A^{2} - B^{2} = (- 5 8 39)

Step-by-step solution

1From Q3: A + B and A − B are known.
2Multiply (A+B)(A−B) using the matrix product rule.
3Compare with the result from Q4.

Final answer

(A + B) (A - B) = (5555) (- 3 1 - 1 3) = (- 10 - 10 1010) \neq = A^{2} - B^{2} = (- 5 8 39)

(A+B)(A−B) = [−10,10; −10,10]. This is NOT equal to A² − B², so the identity (A+B)(A−B) = A² − B² fails for matrix multiplication.

Step-by-step solution

1Multiply the first equation by 3 and the second by 2: 6X + 9Y = 3A and 6X + 4Y = 2B.
2Subtract: 5Y = 3A − 2B = [6,9;12,0] − [4,−4;−2,10] = [2,13;14,−10]. So Y = (1/5)[2,13;14,−10].
3Substitute: 2X = A − 3Y = [2,3;4,0] − [6/5,39/5;42/5,−6] = [4/5,−24/5;−22/5,6]. So X = [2/5,−12/5;−11/5,3].
4Check: 2X + 3Y = [2,3;4,0] ✓ and 3X + 2Y = [2,−2;−1,5] ✓.

Final answer

X = (2/5 - 11/5 - 12/5 3), Y = (2/5 14/5 13/5 - 2)

Step-by-step solution

1The right side is the identity matrix I₂, so X = A⁻¹ where A = [3,7; 2,5].
2det(A) = 3(5) − 7(2) = 15 − 14 = 1.
3A⁻¹ = (1/det(A)) adj(A) = [5, −7; −2, 3].

Final answer

X = (5 - 2 - 7 3)

Step-by-step solution

1Compute A²: [3,1; −1,2] × [3,1; −1,2] = [8,5; −5,3].
2Compute 5A = [15,5; −5,10].
3Compute 7I = [7,0; 0,7].
4A² − 5A + 7I = [8−15+7, 5−5+0; −5+5+0, 3−10+7] = [0,0; 0,0].

Final answer

A² − 5A + 7I = O, verified.

Step-by-step solution

1Compute A², then A³ by successive matrix multiplication.
2Substitute into A³ − 6A² + 7A + 2I and show each entry equals 0.

Final answer

A³ − 6A² + 7A + 2I = O, verified.

Step-by-step solution

1(i) Compute AB, then (AB)'. Compute B'A' and compare.
2(ii) Compute 2A − B, then its transpose. Compare with 2A' − B'.

Final answer

Both identities verified: (AB)′ = B′A′ and (2A − B)′ = 2A′ − B′ hold.

Step-by-step solution

1B + C = [4,7; 2,4].
2A(B+C) = [1,2; 3,4][4,7; 2,4] = [4+4, 7+8; 12+8, 21+16] = [8,15; 20,37].
3AB = [1,2; 3,4][−1,0; 2,1] = [3,2; 5,4].
4AC = [1,2; 3,4][5,7; 0,3] = [5,13; 15,33].
5AB + AC = [8,15; 20,37] = A(B+C). ✓ Distributivity holds for matrices.

Final answer

A(B+C) = AB + AC = [8,15; 20,37]. Verified.

Step-by-step solution

1Compute AB, then (AB)C.
2Compute BC, then A(BC).
3Both give the same result.

Final answer

(A B) C = A (B C) = (10221228)

Step-by-step solution

1AI = A for any matrix A: AB = A × O = O.
2BA = O × A = O.
3Both are zero matrices, hence equal.

Final answer

AB = BA = O. They are equal.

Step-by-step solution

1A² = [0,−1; 1,0] × [0,−1; 1,0] = [−1,0; 0,−1] = −I.
2A⁴ = (−I)² = I.
3A²⁰ = (A⁴)⁵ = I⁵ = I.
4(A²)²⁰ = (−I)²⁰ = I.
5A⁴⁰ = (A⁴)¹⁰ = I. All equal I.

Final answer

(A²)²⁰ = A⁴⁰ = I₂, verified.

Step-by-step solution

1Base case n = 1: A¹ = [1,2; 0,1] = [1,2·1; 0,1]. True.
2Assume Aᵏ = [1,2k; 0,1]. Then Aᵏ⁺¹ = Aᵏ · A = [1,2k; 0,1][1,2; 0,1] = [1, 2k+2; 0, 1] = [1, 2(k+1); 0, 1].
3By induction, Aⁿ = [1, 2n; 0, 1] for all n.

Final answer

A^{n} = (10 2 n 1)

Step-by-step solution

1A² = [3,−2; 4,−3][3,−2; 4,−3] = [9−8, −6+6; 12−12, −8+9] = [1,0; 0,1] = I.
2Since A² = I, we have AA = I, so A⁻¹ = A.

Final answer

A^{- 1} = A = (34 - 2 - 3)

Step-by-step solution

1From Exercise 3.2 Q8: A² − 5A + 7I = O.
2Multiply by A⁻¹: A − 5I + 7A⁻¹ = O.
37A⁻¹ = 5I − A = [5,0; 0,5] − [3,1; −1,2] = [2,−1; 1,3].
4A⁻¹ = (1/7)[2,−1; 1,3].

Final answer

A^{- 1} = \frac{1}{7} (21 - 1 3)

Step-by-step solution

1AB = [2,−3; 3,4][1,0; 2,1] = [2−6, 0−3; 3+8, 0+4] = [−4, −3; 11, 4].
2det(AB) = (−4)(4) − (−3)(11) = −16 + 33 = 17, so (AB)⁻¹ = (1/17)[4, 3; −11, −4].
3det A = 2(4) − (−3)(3) = 17 ⇒ A⁻¹ = (1/17)[4, 3; −3, 2]; det B = 1(1) − 0 = 1 ⇒ B⁻¹ = [1, 0; −2, 1].
4B⁻¹A⁻¹ = (1/17)[1,0; −2,1][4,3; −3,2] = (1/17)[4, 3; −8−3, −6+2] = (1/17)[4, 3; −11, −4].
5(AB)⁻¹ = B⁻¹A⁻¹, verified.

Final answer

(A B)^{- 1} = B^{- 1} A^{- 1} = \frac{1}{17} (4 - 11 3 - 4)

Step-by-step solution

1A² = [1,2; 2,1][1,2; 2,1] = [1+4, 2+2; 2+2, 4+1] = [5, 4; 4, 5].
22A = [2,4; 4,2], 3I = [3,0; 0,3].
3A² − 2A − 3I = [5−2−3, 4−4−0; 4−4−0, 5−2−3] = [0,0; 0,0] = O. ✓
4Multiply A² − 2A − 3I = O by A⁻¹: A − 2I − 3A⁻¹ = O, so 3A⁻¹ = A − 2I.
5A − 2I = [−1,2; 2,−1], so A⁻¹ = (1/3)[−1,2; 2,−1].

Final answer

A^{- 1} = \frac{1}{3} (- 1 2 2 - 1)

Step-by-step solution

1A′ = [1,0,a; 0,1,b; 0,0,1], B′ = [0,0,c; 1,0,d; 0,1,1].
2(i) A+B = [1,1,0; 0,1,1; a+c,b+d,2], so (A+B)′ = [1,0,a+c; 1,1,b+d; 0,1,2].
3A′ + B′ = [1,0,a; 0,1,b; 0,0,1] + [0,0,c; 1,0,d; 0,1,1] = [1,0,a+c; 1,1,b+d; 0,1,2]. Hence (A+B)′ = A′ + B′ ✓.
4(ii) AB = [0,1,0; 0,0,1; c,a+d,b+1], so (AB)′ = [0,0,c; 1,0,a+d; 0,1,b+1].
5B′A′ = [0,0,c; 1,0,d; 0,1,1]·[1,0,a; 0,1,b; 0,0,1] = [0,0,c; 1,0,a+d; 0,1,b+1] = (AB)′ ✓.

Final answer

(A+B)′ = A′ + B′ and (AB)′ = B′A′, both verified for these matrices.

Step-by-step solution

1A² = A × A.
2Row 3 of A²: [a·1 + b·0 + 1·a, a·0 + b·1 + 1·b, a·0 + b·0 + 1·1] = [2a, 2b, 1].
3So A² = [1,0,0; 0,1,0; 2a, 2b, 1].
4A² = I requires 2a = 0 and 2b = 0; for general a, b this fails, so A is not self-inverse.
5In fact Aⁿ = [1,0,0; 0,1,0; na, nb, 1] for n ∈ ℕ.

Final answer

A^{2} = 10 2 a 01 2 b 001 \neq = I in general

Step-by-step solution

1det(A) = 4 − 2 = 2. A⁻¹ = (1/2)[1, −2; −1, 4].
2(A⁻¹)⁻¹: compute the inverse of A⁻¹.
3det(A⁻¹) = 1/2. (A⁻¹)⁻¹ = 2 · adj(A⁻¹) = 2 · (1/2)[4, 2; 1, 1] = [4,2; 1,1] = A.
4Alternatively, use the general property: (A⁻¹)⁻¹ = A always holds.

Final answer

(A⁻¹)⁻¹ = A, verified.

Step-by-step solution

1A² = [3,2; −2,−1][3,2; −2,−1] = [9−4, 6−2; −6+2, −4+1] = [5, 4; −4, −3].
22A = [6,4; −4,−2], I = [1,0; 0,1].
3A² − 2A + I = [5−6+1, 4−4+0; −4+4+0, −3+2+1] = [0,0; 0,0] = O. ✓
4Multiply by A⁻¹: A − 2I + A⁻¹ = O, so A⁻¹ = 2I − A.
5A⁻¹ = [2,0; 0,2] − [3,2; −2,−1] = [−1,−2; 2,3]. Verify: A·A⁻¹ = I. ✓

Final answer

A^{2} - 2 A + I = O, A^{- 1} = 2 I - A = (- 1 2 - 2 3)

Step-by-step solution

1If A and B are symmetric, A′ = A and B′ = B.
2(AB)′ = B′A′ = BA (reversal law for transposes).
3Given AB = BA: (AB)′ = AB.
4Therefore AB is symmetric.

Final answer

(AB)′ = B′A′ = BA = AB, so AB is symmetric.

Step-by-step solution

1(i) A+B = [1,1; 1,−1]. (A+B)² = [1,1; 1,−1][1,1; 1,−1] = [2,0; 0,2] = 2I.
2(ii) A² = [1,0; 0,1] = I. B² = [0,1; 1,0][0,1; 1,0] = [1,0; 0,1] = I.
3AB = [1,0; 0,−1][0,1; 1,0] = [0,1; −1,0]. BA = [0,1; 1,0][1,0; 0,−1] = [0,−1; 1,0]. AB ≠ BA.
42AB = [0,2; −2,0]. A²+B²+2AB = I+I+[0,2;−2,0] = [2,2; −2,2].
5(A+B)² = 2I = [2,0; 0,2] ≠ A²+B²+2AB = [2,2; −2,2].
6They are not equal because AB ≠ BA, so the binomial expansion (A+B)² = A²+AB+BA+B² ≠ A²+2AB+B².

Final answer

(A+B)² = 2I = [2,0; 0,2]. A²+B²+2AB = [2,2; −2,2]. They are NOT equal because AB ≠ BA, so (A+B)² = A²+AB+BA+B² ≠ A²+2AB+B².

Step-by-step solution

1Take A = [0,1; 0,0] and B = [1,0; 0,0].
2AB = [0,1; 0,0][1,0; 0,0] = [0,0; 0,0].
3BA = [1,0; 0,0][0,1; 0,0] = [0,1; 0,0].
4AB = O ≠ BA.

Final answer

A = [0,1; 0,0], B = [1,0; 0,0]: AB = O ≠ BA = [0,1; 0,0].

Step-by-step solution

1AB = [2·1+1·3+3·2, 2·2+1·4+3·1; −1·1+2·3+1·2, −1·2+2·4+1·1] = [11, 11; 7, 7].
2(AB)′ = [11,7; 11,7].
3A′ = [2,−1; 1,2; 3,1] (columns of A′ are the rows of A). B′ = [1,3,2; 2,4,1].
4B′A′: row 1 × columns of A′: [1·2+3·1+2·3, 1·(−1)+3·2+2·1] = [11, 7].
5Row 2: [2·2+4·1+1·3, 2·(−1)+4·2+1·1] = [11, 7]. So B′A′ = [11,7; 11,7].
6(AB)′ = B′A′ ✓.

Final answer

(AB)′ = B′A′ = [11,7; 11,7]. Verified.

Step-by-step solution

1A² = [1,−1; 1,−1][1,−1; 1,−1] = [1−1, −1+1; 1−1, −1+1] = [0, 0; 0, 0].

Final answer

A² = O. This is a nonzero matrix whose square is zero, illustrating that nonzero matrices can be nilpotent of index 2.

15

Exercise 3.3 — Transpose, Symmetric and Skew-Symmetric Matrices

19Exercise questions

Step-by-step solution

1(i) Transpose of a column matrix is a row matrix.
2(ii) Swap rows and columns.
3(iii) Swap rows and columns of the 2×3 matrix to get a 3×2 matrix.

Final answer

(i) A^{⊤} = (5 \frac{1}{2} - 1); (ii) A^{⊤} = (1 - 1 23); (iii) A^{⊤} = - 1 01 234

Step-by-step solution

1(i) Compute A+B, then (A+B)′. Compute A′ + B′ and compare.
2(ii) Compute A−B, then (A−B)′. Compute A′ − B′ and compare.

Final answer

Both identities verified by direct computation.

Step-by-step solution

1A = (A′)′ = [3,−1,0; 4,2,1].
2Compute BA, then (BA)′.
3Compute A′B′ and compare.

Final answer

(BA)′ = A′B′, verified by direct computation.

Step-by-step solution

1A′ = [−1,5,1; 2,3,−2; 3,0,1].
2A + A′ = [−2,7,4; 7,6,−2; 4,−2,2].
3Check symmetry: (A+A′)′ = A′ + (A′)′ = A′ + A = A + A′. ✓

Final answer

A + A^{'} = - 2 74 76 - 2 4 - 2 2

Step-by-step solution

1A is symmetric means A′ = A, which requires the (1,2) entry to equal the (2,1) entry.
2x = x is automatically true (the off-diagonal entries are both x).
3So A is symmetric for all real x, y. Any values of x and y work.
4More precisely, the symmetry condition gives x = x (always true) with no constraint on y beyond the given form.

Final answer

A is symmetric for all real x and y — the off-diagonal entries are both x, so A′ = A automatically.

Step-by-step solution

1A symmetric requires the (1,2) entry to equal the (2,1) entry.
2x + 1 = −2 gives x = −3.
3y is free (the diagonal entries impose no symmetry condition).

Final answer

x = −3; y can be any real number.

Step-by-step solution

1A skew-symmetric means A′ = −A, and every diagonal entry must be 0.
2Diagonal conditions: a + b = 0 and 4 = 0.
3But 4 ≠ 0, so no choice of a, b can make A skew-symmetric.

Final answer

No solution — the (2,2) entry is 4 ≠ 0, but every diagonal entry of a skew-symmetric matrix must be 0.

Step-by-step solution

1A skew-symmetric: A′ = −A, so all diagonal entries must be 0.
2The (2,2) entry is 1 ≠ 0, so A can never be skew-symmetric for any a, b.
3(The (1,3)/(3,1) entries 2 and −3 also violate skew-symmetry, since 2 = −(−3) = 3 is false.)

Final answer

No solution — the (2,2) entry is 1 ≠ 0 (diagonal of a skew-symmetric matrix must be 0).

Step-by-step solution

1Symmetric part: P = (B + B′)/2.
2B′ = [3, −4; −2, 3].
3B + B′ = [6, −6; −6, 6]. P = [3, −3; −3, 3].
4Skew-symmetric part: Q = (B − B′)/2.
5B − B′ = [0, 2; −2, 0]. Q = [0, 1; −1, 0].
6Verify: P + Q = [3, −2; −4, 3] = B. ✓

Final answer

P = (3 - 3 - 3 3), Q = (0 - 1 10), P + Q = B

Step-by-step solution

1A′ = [6, −2, 2; −2, 3, −1; 2, −1, 3] = A.
2Since A = A′, A is symmetric.
3P = (A + A′)/2 = A, Q = (A − A′)/2 = O.
4A = A + O.

Final answer

A = 6 - 2 2 - 2 3 - 1 2 - 1 3 + 000000000

A is already symmetric (A′ = A), so the symmetric part is A itself and the skew-symmetric part is the zero matrix O. Hence A = A + O.

Step-by-step solution

1Assume AB is symmetric: (AB)′ = AB.
2(AB)′ = B′A′ = BA (since A, B symmetric).
3So BA = AB.
4Conversely, if AB = BA and A′ = A, B′ = B:
5(AB)′ = B′A′ = BA = AB. So AB is symmetric.
6Therefore AB is symmetric ⟺ AB = BA.

Final answer

(AB)′ = B′A′ = BA (using A′=A, B′=B). AB symmetric ⟺ (AB)′ = AB ⟺ BA = AB.

Step-by-step solution

1Let A be any square matrix. Define P = (A + A′)/2 and Q = (A − A′)/2.
2P′ = (A′ + A′′)/2 = (A′ + A)/2 = P, so P is symmetric.
3Q′ = (A′ − A′′)/2 = (A′ − A)/2 = −Q, so Q is skew-symmetric.
4P + Q = (A + A′ + A − A′)/2 = A.
5Uniqueness: if A = S₁ + K₁ = S₂ + K₂ with Sᵢ symmetric and Kᵢ skew-symmetric, then S₁ − S₂ = K₂ − K₁. The left side is symmetric, the right is skew-symmetric. A matrix that is both symmetric and skew-symmetric must be zero, so S₁ = S₂ and K₁ = K₂.

Final answer

A = (A+A′)/2 + (A−A′)/2 gives the unique decomposition into symmetric + skew-symmetric parts.

Step-by-step solution

1A symmetric: A′ = A.
2A skew-symmetric: A′ = −A.
3Therefore A = −A, which gives 2A = O, so A = O.

Final answer

A = A′ = −A ⟹ 2A = O ⟹ A = O.

Step-by-step solution

1(i) Let C = AB′ + BA′. C′ = (AB′)′ + (BA′)′ = (B′)′A′ + (A′)′B′ = BA′ + AB′ = C. Symmetric ✓.
2(ii) Let D = AB′ − BA′. D′ = (B′)′A′ − (A′)′B′ = BA′ − AB′ = −D. Skew-symmetric ✓.

Final answer

(i) AB′ + BA′ is symmetric since (AB′ + BA′)′ = BA′ + AB′. (ii) AB′ − BA′ is skew-symmetric since (AB′ − BA′)′ = BA′ − AB′ = −(AB′ − BA′).

Step-by-step solution

1A′ = [cos θ, −sin θ; sin θ, cos θ].
2For A to be symmetric, A′ = A, requiring sin θ = −sin θ, so sin θ = 0, hence θ = nπ.
3When θ = nπ: cos θ = ±1, and A = ±I.
4This is a scalar multiple of I with k = ±1.

Final answer

A symmetric requires sin θ = 0, giving θ = nπ and A = ±I (a scalar multiple of the identity).

Step-by-step solution

1Let B = A + A′.
2B′ = (A + A′)′ = A′ + (A′)′ = A′ + A = A + A′ = B.
3Since B′ = B, B is symmetric.
4This holds for any square matrix A, regardless of whether A itself is symmetric.

Final answer

(A + A′)′ = A′ + A = A + A′, so A + A′ is always symmetric.

Step-by-step solution

1B′ = [0, −a, −b; a, 0, −c; b, c, 0].
2−B = [0, −a, −b; a, 0, −c; b, c, 0].
3B′ = −B for all a, b, c.
4Diagonal entries are all 0, and the off-diagonal entries satisfy b′ᵢⱼ = −b′ⱼᵢ.
5Also note: det(B) = 0, so B is never invertible.

Final answer

B is skew-symmetric: B′ = −B for all a, b, c. The diagonal is all zeros and off-diagonal pairs satisfy b′ᵢⱼ = −b′ⱼᵢ.

Step-by-step solution

1A² = [1,−1; −1,1][1,−1; −1,1] = [1+1, −1−1; −1−1, 1+1] = [2, −2; −2, 2].
22A = [2, −2; −2, 2].
3A² = 2A ✓.

Final answer

A² = [2,−2; −2,2] = 2A, verified.

Step-by-step solution

1Let C = AB − BA.
2C′ = (AB − BA)′ = (AB)′ − (BA)′ = B′A′ − A′B′.
3Since A′ = A and B′ = B: C′ = BA − AB = −(AB − BA) = −C.
4Therefore C is skew-symmetric.

Final answer

(AB − BA)′ = B′A′ − A′B′ = BA − AB = −(AB − BA). Skew-symmetric.

16

Exercise 3.4 — Elementary Row and Column Operations

4Exercise questions

Step-by-step solution

1Type 1 (Row interchange): Rᵢ ↔ Rⱼ. Example: swapping rows 1 and 2.
2Type 2 (Row scaling): Rᵢ → kRᵢ for k ≠ 0. Example: R₂ → 3R₂.
3Type 3 (Row addition): Rᵢ → Rᵢ + kRⱼ. Example: R₁ → R₁ − 2R₂.

Final answer

The three elementary row operations are: (1) Interchange of two rows (Rᵢ ↔ Rⱼ); (2) Multiplication of a row by a non-zero scalar (Rᵢ → kRᵢ, k ≠ 0); (3) Addition of a scalar multiple of one row to another (Rᵢ → Rᵢ + kRⱼ).

Step-by-step solution

1Column operations are analogous to row operations, applied to columns instead of rows.
2Three types: (1) Cᵢ ↔ Cⱼ (column interchange), (2) Cᵢ → kCᵢ (column scaling), (3) Cᵢ → Cᵢ + kCⱼ (column addition).
3Every elementary column operation on A is equivalent to the corresponding elementary row operation on A′ (transpose).
4Column operations are right-multiplication by elementary matrices, while row operations are left-multiplication.

Final answer

Elementary column operations mirror row operations applied to columns: (1) Cᵢ ↔ Cⱼ, (2) Cᵢ → kCᵢ (k≠0), (3) Cᵢ → Cᵢ + kCⱼ. They correspond to right-multiplication by elementary matrices (versus left-multiplication for row operations).

Step-by-step solution

1Augment A with I: [1 3 | 1 0; 2 7 | 0 1].
2R₂ → R₂ − 2R₁: [1 3 | 1 0; 0 1 | −2 1].
3R₁ → R₁ − 3R₂: [1 0 | 7 −3; 0 1 | −2 1].
4The right half is A⁻¹. Check: A·A⁻¹ = [1,3; 2,7][7,−3; −2,1] = [1,0; 0,1] ✓.

Final answer

A^{- 1} = (7 - 2 - 3 1)

Step-by-step solution

1Augment A with I: [2 1 | 1 0; 7 4 | 0 1]. det A = 8 − 7 = 1 ≠ 0, so A is invertible.
2R₂ → R₂ − 3R₁: [2 1 | 1 0; 1 1 | −3 1].
3R₁ ↔ R₂: [1 1 | −3 1; 2 1 | 1 0].
4R₂ → R₂ − 2R₁: [1 1 | −3 1; 0 −1 | 7 −2].
5R₁ → R₁ + R₂: [1 0 | 4 −1; 0 −1 | 7 −2].
6R₂ → −R₂: [1 0 | 4 −1; 0 1 | −7 2]. So A⁻¹ = [4,−1; −7,2].
7Check: [2,1; 7,4][4,−1; −7,2] = [8−7, −2+2; 28−28, −7+8] = [1,0; 0,1] ✓.

Final answer

A^{- 1} = (4 - 7 - 1 2)

17

Miscellaneous Exercise on Chapter 3

15Exercise questions

Step-by-step solution

1a₁₁ = 0, a₁₂ = −1/3, a₁₃ = −2/4 = −1/2.
2a₂₁ = 1/3, a₂₂ = 0, a₂₃ = −1/5.
3a₃₁ = 2/4 = 1/2, a₃₂ = 1/5, a₃₃ = 0.

Final answer

A = 0 \frac{1}{3} \frac{1}{2} - \frac{1}{3} 0 \frac{1}{5} - \frac{1}{2} - \frac{1}{5} 0

Step-by-step solution

1Left side: [8+6; 4x−2; −4y+12] = [14; 4x−2; −4y+12]. Right side: [14; 10; −14].
2Entry 1: 8 + 6 = 14 ✓ (automatic).
3Entry 2: 4x − 2 = 10 ⟹ 4x = 12 ⟹ x = 3.
4Entry 3: −4y + 12 = −14 ⟹ −4y = −26 ⟹ y = 13/2.

Final answer

x = 3, y = 13/2.

Step-by-step solution

1f(A) = A² − 5A + 2I.
2A² = [1,2; 3,4][1,2; 3,4] = [1+6, 2+8; 3+12, 6+16] = [7,10; 15,22].
35A = [5,10; 15,20], and 2I = [2,0; 0,2].
4f(A) = [7−5+2, 10−10+0; 15−15+0, 22−20+2] = [4,0; 0,4] = 4I.

Final answer

f (A) = A^{2} - 5 A + 2 I = 4 I = (4004)

Step-by-step solution

1A² = A × A.
2(1,1): cos α · cos α + sin α · (−sin α) = cos²α − sin²α = cos 2α.
3(1,2): cos α · sin α + sin α · cos α = 2 sin α cos α = sin 2α.
4(2,1): −sin α · cos α + cos α · (−sin α) = −2 sin α cos α = −sin 2α.
5(2,2): −sin α · sin α + cos α · cos α = cos²α − sin²α = cos 2α.

Final answer

A^{2} = (cos 2 α - sin 2 α sin 2 α cos 2 α)

Step-by-step solution

1(i) A = B′, so (A + B)′ = A′ + B′ = B + B′ = B′ + B = A + B. Hence A + B is symmetric.
2(ii) AB = B′B. By the reversal law, (B′B)′ = B′(B′)′ = B′B = AB.
3Hence AB is symmetric. Indeed B′B is always symmetric (a Gram-type matrix).

Final answer

(i) (A+B)′ = A′ + B′ = B + B′ = A+B. (ii) (AB)′ = (B′B)′ = B′B = AB. Both AB and A+B are symmetric.

Step-by-step solution

1A′A = I means A′ is a left inverse of A.
2For square matrices, a left inverse equals the right inverse: AA′ = I.
3Therefore A⁻¹ = A′.
4Also: det(A′A) = det(I) = 1, so (det A)² = 1, hence det A = ±1 ≠ 0, confirming A is invertible.

Final answer

A is invertible with A⁻¹ = A′, since A′A = I implies AA′ = I for square matrices.

Step-by-step solution

1Characteristic polynomial: det(A − λI) = |2−λ, −3; 3, −4−λ| = (2−λ)(−4−λ) + 9 = λ² + 2λ + 1 = (λ+1)².
2By Cayley–Hamilton: A² + 2A + I = O.
3Rewrite as A(A + 2I) = −I.
4Hence A⁻¹ = −(A + 2I) = −([2,−3; 3,−4] + [2,0; 0,2]) = −[4,−3; 3,−2] = [−4,3; −3,2].
5Check: [2,−3; 3,−4][−4,3; −3,2] = [−8+9, 6−6; −12+12, 9−8] = [1,0; 0,1] = I ✓.

Final answer

A^{- 1} = (- 4 - 3 32)

Step-by-step solution

1(A+B)(A−B) = A² − AB + BA − B².
2For this to equal A² − B², we need −AB + BA = 0, i.e., AB = BA.
3But symmetric matrices need not commute.
4Counterexample: A = [1,2; 2,3], B = [1,0; 0,2]. AB = [1,4; 2,6], BA = [1,2; 4,6].
5AB ≠ BA, so (A+B)(A−B) ≠ A² − B².
6A² = [5,8; 8,13], B² = [1,0; 0,4]. A² − B² = [4,8; 8,9].
7(A+B)(A−B) = [2,2; 2,5][0,2; 2,1] = [4,6; 10,9].
8[4,8; 8,9] ≠ [4,6; 10,9]. Confirmed.

Final answer

A²−B² = (A+B)(A−B) requires AB = BA, which does not hold in general for symmetric matrices.

Step-by-step solution

1Let S = A + A′. S′ = A′ + (A′)′ = A′ + A = S. So S is symmetric.
2Let K = A − A′. K′ = A′ − A = −(A − A′) = −K. So K is skew-symmetric.
3Note: S + K = (A + A′) + (A − A′) = 2A, so A = (A + A′)/2 + (A − A′)/2.

Final answer

A + A′ is symmetric: (A + A′)′ = A′ + A = A + A′. A − A′ is skew-symmetric: (A − A′)′ = A′ − A = −(A − A′).

Step-by-step solution

1A² = [a,0; 0,a][a,0; 0,a] = [a²,0; 0,a²] = a²I.
2B² = [0,b; b,0][0,b; b,0] = [b²,0; 0,b²] = b²I.
3A+B = [a,b; b,a]. (A+B)² = [a,b; b,a][a,b; b,a] = [a²+b², 2ab; 2ab, a²+b²].
4AB = [a,0; 0,a][0,b; b,0] = [0,ab; ab, 0]. BA = [0,b; b,0][a,0; 0,a] = [0,ab; ab, 0].
5AB = BA, so (A+B)² = A² + AB + BA + B² = A² + 2AB + B².
6A² + 2AB + B² = [a²,0; 0,a²] + [0,2ab; 2ab, 0] + [b²,0; 0,b²] = [a²+b², 2ab; 2ab, a²+b²].
7Equal! Since AB = BA in this case.

Final answer

A² = a²I, B² = b²I. (A+B)² = [a²+b², 2ab; 2ab, a²+b²] = A² + 2AB + B² since AB = BA here.

Step-by-step solution

1AB = [1,2; 3,4][5,6; 7,8] = [5+14, 6+16; 15+28, 18+32] = [19, 22; 43, 50].
2(AB)′ = [19, 43; 22, 50].
3A′ = [1,3; 2,4], B′ = [5,7; 6,8].
4B′A′ = [5,7; 6,8][1,3; 2,4] = [5+14, 15+28; 6+16, 18+32] = [19, 43; 22, 50].
5(AB)′ = B′A′ ✓.

Final answer

(AB)′ = B′A′ = [19,43; 22,50], verified.

Step-by-step solution

1A² = [0, sin α; −sin α, 0][0, sin α; −sin α, 0] = [−sin²α, 0; 0, −sin²α] = −sin²α · I.
2A² + I = −sin²α · I + I = (1 − sin²α)I = cos²α · I.
3A² + I = [cos²α, 0; 0, cos²α]. ✓

Final answer

A^{2} + I = (cos^{2} α 0 0 cos^{2} α)

Step-by-step solution

1A² = A·A = (AB)·A (since AB = A) = A·(BA) (associativity) = A·B (since BA = B) = AB = A.
2So A² = A.
3Similarly B² = B·B = (BA)·B (since BA = B) = B·(AB) = B·A = BA = B.
4So B² = B. Both A and B are idempotent.

Final answer

A² = (AB)A = A(BA) = AB = A, and B² = (BA)B = B(AB) = BA = B. Hence A² = A and B² = B.

Step-by-step solution

1Let C = B′AB.
2C′ = (B′AB)′ = B′A′(B′)′ = B′A′B.
3Since A is symmetric, A′ = A, so C′ = B′AB = C.
4Therefore C = B′AB is symmetric.
5This is an important result used extensively in quadratic forms and optimization.

Final answer

(B′AB)′ = B′A′B = B′AB (using A′ = A), so B′AB is symmetric.

Step-by-step solution

1(AA′)′ = (A′)′A′ = AA′ (reversal law). So AA′ is symmetric.
2(A′A)′ = A′(A′)′ = A′A (reversal law). So A′A is symmetric.
3Note this does NOT require A to be symmetric — AA′ and A′A are always symmetric.

Final answer

(AA′)′ = AA′ and (A′A)′ = A′A, so both AA′ and A′A are symmetric for any square matrix A.

18

Chapter 4 — Determinants

Determinants are scalar quantities associated with square matrices. They encode essential information about the matrix — invertibility, the volume scaling factor of a linear transformation, and whether a system of equations has a unique solution. Starting from the simple formulas for order 1, 2 and 3, this chapter develops properties that let you evaluate large determinants without brute expansion, introduces minors and cofactors, and culminates in the adjoint method for inverses and Cramer's rule for linear systems.

Board pattern

For determinants up to 3x3 you must show the full expansion step-by-step. In properties-of-determinant questions, state which property you apply at each step (e.g. R2 -> R2 + 2R1) and write the intermediate determinant. Simply writing the final value without the row operations earns at most 1 mark. For inverse questions, always verify A inverse A = I as a last step.

$Ex 4.1 — Determinant of order 1, 2, 3; evaluation (8 Q)$
$Ex 4.2 — Properties of determinants (16 Q)$
$Ex 4.3 — Minors and cofactors (5 Q)$
$Ex 4.4 — Adjoint and inverse of a matrix (5 Q)$
$Ex 4.5 — Area of triangle, consistency, Cramer’s rule (12 Q)$
$Misc — Mixed problems on determinants (15 Q)$

19

Exercise 4.1 — Determinants of Order 1, 2 and 3

10Exercise questions

Step-by-step solution

1For a 2x2 determinant $a c b d = a d - b c$ .
2 $= 2 (- 1) - 4 (- 5) = - 2 + 20 = 18$

Final answer

18

18.

Step-by-step solution

1 $= cos θ \cdot cos θ - (- sin θ) (sin θ)$
2 $= cos^{2} θ + sin^{2} θ = 1$

Final answer

1.

Step-by-step solution

1 $= tan θ \cdot cos θ - (- cos θ) (sin θ)$
2 $= sin θ + cos θ sin θ = sin θ (1 + cos θ)$

Final answer

sin θ (1 + cos θ)

Step-by-step solution

1Expand along the first row: $∣ A ∣ = 2 04 - 1 5 - (- 3) 21 - 1 5 + 12104$
2 $= 2 (0 + 4) + 3 (10 + 1) + 1 (8 - 0) = 8 + 33 + 8 = 49$

Final answer

49

49.

Step-by-step solution

1Expand along the second row (two zeros): $∣ A ∣ = - 0 + 0 - (- 1) 31 - 1 - 2$
2 $= 1 (3 (- 2) - (- 1) (1)) = 1 (- 6 + 1) = - 5$

Final answer

- 5

-5.

Step-by-step solution

1Expand along the first row: $= 1 (45 - 48) - 2 (36 - 42) + 3 (32 - 35)$
2 $= - 3 + 12 - 9 = 0$

Final answer

0.

Step-by-step solution

1 $∣ A ∣ = 1 (4) - 2 (3) = - 2$
2 $2 A = (2648), ∣2 A ∣ = 16 - 24 = - 8$
3 $4∣ A ∣ = 4 (- 2) = - 8 = ∣2 A ∣$

Final answer

∣2 A ∣ = 4∣ A ∣

Verified: |2A| = -8 = 4|A|.

Step-by-step solution

1Expand along the first row: $= a (b c - h^{2}) - h (h c - h^{2}) + h (h^{2} - hb)$
2 $= ab c - a h^{2} - h^{2} c + h^{3} + h^{3} - h^{2} b$
3 $= ab c - h^{2} (a + b + c) + 2 h^{3}$

Final answer

ab c - h^{2} (a + b + c) + 2 h^{3}

abc - h^2(a + b + c) + 2h^3.

Step-by-step solution

1LHS = $2 (1) - 4 (5) = 2 - 20 = - 18$
2RHS = $2 x \cdot x - 4 \cdot 6 = 2 x^{2} - 24$
3 $2 x^{2} - 24 = - 18 \Rightarrow 2 x^{2} = 6 \Rightarrow x^{2} = 3$
4 $x = \pm 3$

Final answer

x = \pm 3

Step-by-step solution

1LHS = $x \cdot x - 2 (18) = x^{2} - 36$
2RHS = $6 \cdot 6 - 2 (18) = 36 - 36 = 0$
3 $x^{2} - 36 = 0 \Rightarrow x = \pm 6$

Final answer

x = \pm 6

20

Exercise 4.2 — Properties of Determinants

18Exercise questions

Step-by-step solution

1Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
2 $= 100 a b - a c - a a^{2} b^{2} - a^{2} c^{2} - a^{2}$
3Factor: $= (b - a) (c - a) 11 b + a c + a$
4 $= (b - a) (c - a) (c + a - b - a) = (b - a) (c - a) (c - b)$
5 $= (a - b) (b - c) (c - a)$

Final answer

(a - b) (b - c) (c - a)

Shown.

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2 $= 2 (a + b + c) 2 (p + q + r) 2 (x + y + z) c + a r + p z + x a + b p + q x + y$
3Take 2 common from C1: $= 2 a + b + c p + q + r x + y + z c + a r + p z + x a + b p + q x + y$
4Apply $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ :
5 $= 2 a + b + c p + q + r x + y + z - b - q - y - c - r - z$
6Take -1 from each of C2, C3: $= 2 a + b + c p + q + r x + y + z b q y c r z$
7Apply $C_{1} \to C_{1} - C_{2} - C_{3}$ :
8 $= 2 a p x b q y c r z$

Final answer

2 a p x b q y c r z

Shown.

Step-by-step solution

1Note the entries are $5^{2}, 7^{2}, 8^{2}; 6^{2}, 8^{2}, 9^{2}; 7^{2}, 9^{2}, 1 0^{2}$
2Apply $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{2}$ :
3 $= 253649242832151719$
4Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{2}$ :
5 $= 25111324441522$
6Apply $R_{3} \to R_{3} - R_{2}$ :
7 $= 2511224401520 = 2 (48 - 60) = - 24$

Final answer

- 24

-24.

Step-by-step solution

1Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
2 $= x y - x z - x x^{2} y^{2} - x^{2} z^{2} - x^{2} y z z x - y z x y - y z$
3Factor (y-x), (z-x) from R2, R3:
4 $= (x - y) (y - z) (z - x) (x y + y z + z x)$

Final answer

(x - y) (y - z) (z - x) (x y + y z + z x)

Shown.

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2 $= (5 x + 4) 111 2 x x + 4 2 x 2 x 2 x x + 4$
3Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
4 $= (5 x + 4) 100 2 x 4 - x 0 2 x 0 4 - x = (5 x + 4) (4 - x)^{2}$

Final answer

(5 x + 4) (4 - x)^{2}

Shown.

Step-by-step solution

1Expand along the first row: $= (y + z) [(z + x) (x + y) - y z] - x [y (x + y) - y z] + x [y z - (z + x) z]$
2 $= (y + z) [x (x + y + z)] - x [y (x + y - z)] + x [z (y - x - z)]$
3 $= x [(x + y + z) (y + z) - y (x + y - z) + z (y - x - z)]$
4Expand the bracket: $= x [(y + z) (y + z) - (y - z)^{2}]$
5 $= x [(y + z)^{2} - (y - z)^{2}] = x \cdot 4 y z = 4 x y z$

Final answer

4 x y z

Shown.

Step-by-step solution

1Apply $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ :
2 $= 111 0 b - a c - a 0 b^{2} - a^{2} c^{2} - a^{2}$
3Expand along R1: $= (b - a) (c^{2} - a^{2}) - (c - a) (b^{2} - a^{2})$
4 $= (b - a) (c - a) (c + a) - (c - a) (b - a) (b + a) = (a - b) (b - c) (c - a)$

Final answer

(a - b) (b - c) (c - a)

Shown.

Step-by-step solution

1Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
2 $= 213712651021$
3Apply $C_{3} \to C_{3} - 10 C_{2}$ :
4 $= 213712 - 5 01$
5Expand: $= - 1 (- 7 + 5) + 0 + 1 (4 - 7) = 2 - 3 = - 1 \neq = 0$

Final answer

- 1 \neq = 0

-1 is not 0, so the determinant is nonzero.

Step-by-step solution

1Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
2 $= 100 b + c a - b a - c b^{2} + c^{2} a^{2} - b^{2} a^{2} - c^{2}$
3 $= (a - b) (a - c) 11 a + b a + c = (a - b) (a - c) (c - b)$
4 $= (a - b) (b - c) (c - a)$

Final answer

(a - b) (b - c) (c - a)

Shown.

Step-by-step solution

1Take a common from C1, b from C2, c from C3: $= ab c a + 1/ a a a b b + 1/ b b c c c + 1/ c$
2Apply $R_{1} \to R_{1} - R_{2}$ and $R_{2} \to R_{2} - R_{3}$ :
3 $= ab c 1/ a 0 a - 1/ b 1/ b b 0 - 1/ c c + 1/ c$
4Expand along the first row: $= ab c [\frac{1}{a} (\frac{1}{b} (c + \frac{1}{c}) + \frac{b}{c}) + \frac{1}{b} \cdot \frac{a}{c}]$
5 $= ab c (\frac{c}{ab} + \frac{1}{ab c} + \frac{b}{a c} + \frac{a}{b c}) = c^{2} + 1 + b^{2} + a^{2} = 1 + a^{2} + b^{2} + c^{2}$

Final answer

1 + a^{2} + b^{2} + c^{2}

Step-by-step solution

1Apply $R_{2} \to R_{2} - 2 R_{3}$ :
2 $= a a x b b y c c z$
3R1 = R2, so the determinant is 0.

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1Take 3 from R1 and 2 from R2: $= 6 111 x x 1 x^{2} x^{2} 1$
2R1 = R2, so the determinant is 0.

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1The matrix is skew-symmetric: $a_{ij} = - a_{j i}$ , so its diagonal is all zeros.
2Take -1 common from each of the three rows: $= (- 1) (- 1) (- 1) 0 a - b - a 0 - c b c 0$
3But the new matrix is merely the negative of the original, so $D = - D \Rightarrow D = 0$

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1Apply $R_{2} \to R_{2} - 2 R_{1}$ :
2 $R_{2} = (0, 1, 2)$
3Apply $R_{3} \to R_{3} - 3 R_{1}$ :
4 $R_{3} = (0, 3, 7)$
5 $= 100 1 + p 13 1 + p + q 27 = 1 (7 - 6) = 1$

Final answer

1

Shown.

Step-by-step solution

1Apply $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ :
2 $= 1 a a^{3} 0 b - a b^{3} - a^{3} 0 c - a c^{3} - a^{3}$
3Expand along row 1: $= (b - a) (c^{3} - a^{3}) - (c - a) (b^{3} - a^{3})$
4Factor using $u^{3} - v^{3} = (u - v) (u^{2} + uv + v^{2})$
5 $= (b - a) (c - a) [(c^{2} + a c + a^{2}) - (b^{2} + ab + a^{2})] = (a - b) (b - c) (c - a) (a + b + c)$

Final answer

(a - b) (b - c) (c - a) (a + b + c)

Shown.

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2 $= (1 + x + x^{2}) 111 x 1 x^{2} x^{2} x 1$
3Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
4 $= (1 + x + x^{2}) 100 x 1 - x x^{2} - x x^{2} x - x^{2} 1 - x^{2}$
5Factor (1-x) from R2 and R3: $= (1 + x + x^{2}) (1 - x)^{2} 100 x 1 - x x^{2} x 1 + x$
62x2 minor: $= (1 + x + x^{2}) (1 - x)^{2} (1 + x + x^{2}) = [(1 - x) (1 + x + x^{2})]^{2}$
7Using $1 - x^{3} = (1 - x) (1 + x + x^{2})$ the result is $(1 - x^{3})^{2}$

Final answer

(1 - x^{3})^{2}

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2Sum of the three entries of row 1 = $(a - b) + (b - c) + (c - a) = 0$
3Similarly rows 2 and 3 each have sum 0, so C1 becomes a zero column: $= 0$

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1a, b, c in AP means $2 b = a + c$
2Apply $R_{1} \to R_{1} + R_{3} - 2 R_{2}$ :
3First two entries become zero: $(x + 2) + (x + 4) - 2 (x + 3) = 0$ and similarly the second entry.
4Third entry: $(x + 2 a) + (x + 2 c) - 2 (x + 2 b) = 2 a + 2 c - 4 b = 2 (a + c) - 4 b = 0$
5Row 1 becomes a zero row, so the determinant is 0.

Final answer

Option (A) — 0.

21

Exercise 4.3 — Minors and Cofactors

7Exercise questions

Step-by-step solution

1The minor $M_{ij}$ is obtained by deleting row i and column j.
2 $M_{11} = 4 - 2 53 = 22, M_{12} = 6153 = 13, M_{13} = 61 4 - 2 = - 16$
3 $M_{21} = 3 - 2 - 1 3 = 7, M_{22} = 21 - 1 3 = 7, M_{23} = 21 3 - 2 = - 7$
4 $M_{31} = 34 - 1 5 = 19, M_{32} = 26 - 1 5 = 16, M_{33} = 2634 = - 10$
5Cofactors: $A_{ij} = (- 1)^{i + j} M_{ij}$

Final answer

A_{11} = 22, A_{12} = - 13, A_{13} = - 16, A_{21} = - 7, A_{22} = 7, A_{23} = 7, A_{31} = 19, A_{32} = - 16, A_{33} = - 10

Step-by-step solution

1Minor $M_{22}$ : delete row 2, column 2.
2 $M_{22} = 2719 = 18 - 7 = 11$
3Cofactor: $A_{22} = (- 1)^{2 + 2} \cdot 11 = 11$

Final answer

M_{22} = 11, A_{22} = 11

M22 = 11, A22 = 11.

Step-by-step solution

1 $A_{11} = 4, A_{12} = - 1, A_{21} = - 3, A_{22} = 2$
2Expand along row 1: $∣ A ∣ = a_{11} A_{11} + a_{12} A_{12} = 2 (4) + 3 (- 1) = 8 - 3 = 5$

Final answer

∣ A ∣ = 5

|A| = 5.

Step-by-step solution

1Apply the sign pattern $+ - + - + - + - +$
2 $A_{11} = (- 1)^{1 + 1} - 1 1 42 = - 2 - 4 = - 6$
3 $A_{12} = - 2042 = - 4, A_{13} = 20 - 1 1 = 2$
4 $A_{21} = - (21 - 3 2) = - (4 + 3) = - 7, A_{22} = 10 - 3 2 = 2$
5 $A_{23} = - (1021) = - 1, A_{31} = 2 - 1 - 3 4 = 5$
6 $A_{32} = - (12 - 3 4) = - 10, A_{33} = 12 2 - 1 = - 5$

Final answer

A_{11} = - 6, A_{12} = - 4, A_{13} = 2, A_{21} = - 7, A_{22} = 2, A_{23} = - 1, A_{31} = 5, A_{32} = - 10, A_{33} = - 5

Step-by-step solution

1Expand along the first row: $D = 5 - 1 3 42 - (- 2) 2142 + 3 21 - 1 3$
2 $= 5 (- 2 - 12) + 2 (4 - 4) + 3 (6 + 1) = - 70 + 0 + 21 = - 49$

Final answer

- 49

-49.

Step-by-step solution

1For a 2x2 determinant $M_{11} = 1, M_{12} = 2, M_{21} = - 4, M_{22} = 3$
2Cofactors $A_{ij} = (- 1)^{i + j} M_{ij}$ : $A_{11} = 1, A_{12} = - 2, A_{21} = 4, A_{22} = 3$

Final answer

A_{11} = 1, A_{12} = - 2, A_{21} = 4, A_{22} = 3

Step-by-step solution

1Cofactors of row 2: $A_{21} = - (2 (1) - 3 (3)) = 7, A_{22} = (1 (1) - 3 (2)) = - 5, A_{23} = - (1 (3) - 2 (2)) = 1$
2 $a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} = 3 (7) + 1 (- 5) + 2 (1) = 21 - 5 + 2 = 18$
3Direct expansion: $∣ A ∣ = 1 (1 - 6) - 2 (3 - 4) + 3 (9 - 2) = - 5 + 2 + 21 = 18$
4Both equal 18.

Final answer

18

Verified: both sums give 18.

22

Exercise 4.4 — Adjoint and Inverse of a Matrix

8Exercise questions

Step-by-step solution

1 $∣ A ∣ = - 12 + 12 = 0$
2Cofactors: $A_{11} = - 6, A_{12} = 4, A_{21} = - 3, A_{22} = 2$
3 $adj (A) = (- 6 4 - 3 2)$
4 $A (adj A) = (2 - 4 3 - 6) (- 6 4 - 3 2) = (0000) = 0 \cdot I$
5Similarly (adj A)A = 0I.

Final answer

Verified: A(adj A) = (adj A)A = 0I = |A|I.

Step-by-step solution

1Cofactors: $A_{11} = - 5, A_{12} = - 3, A_{21} = - 2, A_{22} = 1$
2 $adj (A) = (- 5 - 3 - 2 1)$

Final answer

adj (A) = (- 5 - 3 - 2 1)

adj(A) = [-5, -2; -3, 1].

Step-by-step solution

1 $∣ A ∣ = 2 (4) - 3 (1) = 5 \neq = 0$
2Cofactors: $A_{11} = 4, A_{12} = - 1, A_{21} = - 3, A_{22} = 2$
3 $adj (A) = (4 - 1 - 3 2)$
4 $A^{- 1} = \frac{1}{5} (4 - 1 - 3 2)$
5Verify: $A A^{- 1} = \frac{1}{5} (2134) (4 - 1 - 3 2) = \frac{1}{5} (5005) = I_{2}$

Final answer

A^{- 1} = \frac{1}{5} (4 - 1 - 3 2)

Step-by-step solution

1 $∣ A ∣ = 1 (1 - 0) - 2 (0 - 0) + 3 (0 - 0) = 1$
2Cofactors:
3 $C_{11} = 1, C_{12} = 0, C_{13} = 0$
4 $C_{21} = - 2, C_{22} = 1, C_{23} = 0$
5 $C_{31} = 1, C_{32} = - 2, C_{33} = 1$
6 $A^{- 1} = \frac{1}{1} 100 - 2 10 1 - 2 1$

Final answer

A^{- 1} = 100 - 2 10 1 - 2 1

Step-by-step solution

1 $∣ A ∣ = - 2, ∣ B ∣ = 10$
2 $A^{- 1} = \frac{1}{- 2} (4 - 3 - 2 1) = (- 2 3/2 1 - 1/2)$
3 $B^{- 1} = \frac{1}{10} (4 - 2 - 1 3)$
4 $A B = (717919), (A B)^{- 1} = (19/20 - 17/20 - 9/20 7/20)$
5 $B^{- 1} A^{- 1} = (4/10 - 2/10 - 1/10 3/10) (- 2 3/2 1 - 1/2) = (19/20 - 17/20 - 9/20 7/20)$

Final answer

(A B)^{- 1} = B^{- 1} A^{- 1}

Verified.

Step-by-step solution

1Cofactors of row 1: $C_{11} = 0 (2) - (- 1) (1) = 1, C_{12} = - (2 (2) - (- 1) (3)) = - 7, C_{13} = 2 (1) - 0 (3) = 2$
2Cofactors of row 2: $C_{21} = - (2 (2) - 3 (1)) = - 1, C_{22} = 1 (2) - 3 (3) = - 7, C_{23} = - (1 (1) - 2 (3)) = 5$
3Cofactors of row 3: $C_{31} = 2 (- 1) - 3 (0) = - 2, C_{32} = - (1 (- 1) - 3 (2)) = 7, C_{33} = 1 (0) - 2 (2) = - 4$
4Transpose gives the adjoint: $adj (A) = 1 - 7 2 - 1 - 7 5 - 2 7 - 4$

Final answer

adj (A) = 1 - 7 2 - 1 - 7 5 - 2 7 - 4

Step-by-step solution

1 $∣ A ∣ = 2 (- 4) - 3 (1) = - 8 - 3 = - 11$
2 $adj (A) = (- 4 - 1 - 3 2)$
3 $A (adj A) = (21 3 - 4) (- 4 - 1 - 3 2) = (- 11 0 0 - 11) = - 11 I$
4Similarly (adj A)A = -11I = |A|I.

Final answer

Verified: A(adj A) = (adj A)A = -11I = |A|I.

Step-by-step solution

1Multiplying a matrix by the scalar k multiplies every row by k.
2Each of the 3 rows contributes a factor k to the determinant, so $∣ k A ∣ = k \cdot k \cdot k ∣ A ∣ = k^{3} ∣ A ∣$

Final answer

∣ k A ∣ = k^{3} ∣ A ∣

23

Exercise 4.5 — Area of Triangle, Consistency and Cramer's Rule

16Exercise questions

Step-by-step solution

1Area = $\frac{1}{2} 143225111$
2 $= \frac{1}{2} ∣1 (2 - 5) - 2 (4 - 3) + 1 (20 - 6) ∣ = \frac{1}{2} ∣ - 3 - 2 + 14∣ = \frac{9}{2}$

Final answer

\frac{9}{2}

9/2 square units.

Step-by-step solution

1 $Area = \frac{1}{2} 03 - 2 043111$
2Expand along the first row: $= \frac{1}{2} ∣0 - 0 + 1 (9 - (- 8)) ∣ = \frac{17}{2}$

Final answer

\frac{17}{2}

17/2 square units.

Step-by-step solution

1 $2 \cdot Area = - 1 42 - 1 - 2 2 111 = - 1 (- 2 - 2) + 1 (4 - 2) + 1 (8 + 4) = 4 + 2 + 12 = 18$
2 $Area = 9$

Final answer

9

9 square units.

Step-by-step solution

1 $2 \cdot Area = 533752111 = 5 (5 - 2) - 7 (3 - 3) + 1 (6 - 15) = 15 + 0 - 9 = 6$
2 $Area = 3$

Final answer

3

3 square units.

Step-by-step solution

1 $D = 1223 = 3 - 4 = - 1 \neq = 0$
2Since D is not 0, the system has a unique solution for all values.
3 $D_{x} = 2323 = 0, D_{y} = 1223 = - 1$
4 $x = D_{x} / D = 0, y = D_{y} / D = 1$

Final answer

Unique solution: x = 0, y = 1.

Step-by-step solution

1 $D = 21 - 1 1 = 3 \neq = 0$
2Since D is not 0, the system is consistent with a unique solution.
3 $x = \frac{1}{3} 54 - 1 1 = 3, y = \frac{1}{3} 2154 = 1$

Final answer

x = 3, y = 1

Consistent; x = 3, y = 1.

Step-by-step solution

1 $D = 123 1 - 1 1 13 - 1 = 1 (1 - 3) - 1 (- 2 - 9) + 1 (2 + 3) = - 2 + 11 + 5 = 14$
2 $D_{x} = 132 1 - 1 1 13 - 1 = 1 (1 - 3) - 1 (- 3 - 6) + 1 (3 + 2) = - 2 + 9 + 5 = 12$
3 $D_{y} = 123132 13 - 1 = 1 (- 3 - 6) - 1 (- 2 - 9) + 1 (4 - 9) = - 9 + 11 - 5 = - 3$
4 $D_{z} = 123 1 - 1 1 132 = 1 (- 2 - 3) - 1 (4 - 9) + 1 (2 + 3) = - 5 + 5 + 5 = 5$
5 $x = \frac{12}{14} = \frac{6}{7}, y = - \frac{3}{14}, z = \frac{5}{14}$

Final answer

x = \frac{6}{7}, y = - \frac{3}{14}, z = \frac{5}{14}

Step-by-step solution

1Area must be nonzero for non-collinearity.
2 $2 \cdot Area = 1 - 4 - 2 - 1 2 - 3 111 = 1 (2 + 3) + 1 (- 4 + 2) + 1 (12 + 4) = 5 - 2 + 16 = 19$
3The area is 19/2 which is not 0, so the points are not collinear.

Final answer

The determinant is 19/2 which is not 0, so the points are not collinear.

Step-by-step solution

1 $D = 111123136 = 1 (12 - 9) - 1 (6 - 3) + 1 (3 - 2) = 3 - 3 + 1 = 1 \neq = 0$
2Since D is not 0, the system is consistent with a unique solution. Apply Cramer's rule to find x, y, z.
3 $D_{x} = 61420123136 = 6 (12 - 9) - 1 (84 - 60) + 1 (42 - 40) = 18 - 24 + 2 = - 4$
4 $D_{y} = 11161420136 = 1 (84 - 60) - 6 (6 - 3) + 1 (20 - 14) = 24 - 18 + 6 = 12$
5 $D_{z} = 11112361420 = 1 (40 - 42) - 1 (20 - 14) + 6 (3 - 2) = - 2 - 6 + 6 = - 2$
6 $x = - 4/1 = - 4, y = 12/1 = 12, z = - 2/1 = - 2$

Final answer

x = - 4, y = 12, z = - 2

Step-by-step solution

1 $D = 121 - 1 11 1 - 3 1 = 1 (1 + 3) + 1 (2 + 3) + 1 (2 - 1) = 4 + 5 + 1 = 10$
2 $D_{x} = 402 - 1 11 1 - 3 1 = 4 (4) + 1 (6) + 1 (- 2) = 16 + 6 - 2 = 20$
3 $D_{y} = 121402 1 - 3 1 = 1 (6) - 4 (5) + 1 (4) = 6 - 20 + 4 = - 10$
4 $D_{z} = 121 - 1 11 402 = 1 (2) + 1 (4) + 4 (1) = 2 + 4 + 4 = 10$
5 $x = 20/10 = 2, y = - 10/10 = - 1, z = 10/10 = 1$

Final answer

x = 2, y = - 1, z = 1

Step-by-step solution

1The line meets the axes at (5, 0) and (0, 4).
2Area = $\frac{1}{2} \times 5 \times 4 = 10$

Final answer

10

10 square units.

Step-by-step solution

1 $2 \cdot Area = k 40 002111$
2Expand along column 2: $= - 0 + 0 - 2 (k - 4) = - 2 (k - 4)$
3 $∣ - 2 (k - 4) ∣ = 8 \Rightarrow ∣ k - 4∣ = 4$
4 $k = 0 or k = 8$

Final answer

k = 0 or k = 8

k = 0 or k = 8.

Step-by-step solution

1 $2 \cdot Area = 240 - 3 52 111$
2Expand along the first row: $= 2 (5 - 2) + 3 (4 - 0) + 1 (8 - 0) = 6 + 12 + 8 = 26$
3 $Area = 13$

Final answer

13

13 square units.

Step-by-step solution

1 $D = 12 - 1 3 = 3 + 2 = 5 \neq = 0$
2Since D is not 0, the system is consistent with a unique solution.
3 $D_{x} = 31 - 1 3 = 9 + 1 = 10, D_{y} = 1231 = 1 - 6 = - 5$
4 $x = 10/5 = 2, y = - 5/5 = - 1$

Final answer

x = 2, y = - 1

Consistent; x = 2, y = -1.

Step-by-step solution

1 $D = 1212 = 2 - 2 = 0$
2 $D_{x} = 2512 = 4 - 5 = - 1 \neq = 0$
3Since D = 0 but Dx is not 0, the system has no solution - it is inconsistent.

Final answer

Inconsistent (no solution).

Step-by-step solution

1 $D = 1224 = 4 - 4 = 0$
2 $D_{x} = 4824 = 16 - 16 = 0, D_{y} = 1248 = 8 - 8 = 0$
3Since D = Dx = Dy = 0, the system is consistent (dependent) with infinitely many solutions.

Final answer

Consistent with infinitely many solutions.

24

Miscellaneous Exercise on Chapter 4

18Exercise questions

Step-by-step solution

1 $Area = \frac{1}{2} 143125111 = \frac{1}{2} ∣1 (2 - 5) - 1 (4 - 3) + 1 (20 - 6) ∣$
2 $= \frac{1}{2} ∣ - 3 - 1 + 14∣ = 5$

Final answer

5

5 square units.

Step-by-step solution

1Expand along the second row: $D = - 2 - 1 - 5 - 2 3 + 0 - (- 1) 33 - 1 - 5$
2 $= - 2 (- 3 - 10) + 1 (- 15 + 3) = 26 - 12 = 14$

Final answer

14

14.

Step-by-step solution

1Use $(A B)^{- 1} = B^{- 1} A^{- 1}$
2 $∣ B ∣ = 1 (3 - 0) - 2 (- 1 - 0) + (- 2) (2 - 0) = 3 + 2 - 4 = 1$
3Cofactors of B: $C_{11} = 3, C_{12} = 1, C_{13} = 2, C_{21} = 2, C_{22} = 1, C_{23} = 2, C_{31} = 6, C_{32} = 2, C_{33} = 5$
4 $B^{- 1} = 312212625$
5 $B^{- 1} A^{- 1} = 312212625 3 - 15 5 - 1 6 - 2 1 - 5 2 = 9 - 2 1 - 3 10 502$

Final answer

(A B)^{- 1} = 9 - 2 1 - 3 10 502

Step-by-step solution

1 $∣ A ∣ = 1 (6 - 1) - 2 (4 - 3) + 3 (2 - 9) = 5 - 2 - 21 = - 18$
2Cofactors: $C_{11} = 5, C_{12} = - 1, C_{13} = - 7$
3 $C_{21} = - 1, C_{22} = - 7, C_{23} = 5$
4 $C_{31} = - 7, C_{32} = 5, C_{33} = - 1$
5 $A^{- 1} = \frac{1}{- 18} 5 - 1 - 7 - 1 - 7 5 - 7 5 - 1$

Final answer

A^{- 1} = - \frac{1}{18} 5 - 1 - 7 - 1 - 7 5 - 7 5 - 1

Step-by-step solution

1 $∣ A ∣ = 2 + 12 = 14$
2 $A^{- 1} = \frac{1}{14} (1 - 4 32)$
3 $∣ A^{- 1} ∣ = \frac{1}{1 4 ^{2}} (2 + 12) = \frac{14}{196} = \frac{1}{14} = \frac{1}{∣ A ∣}$

Final answer

∣ A^{- 1} ∣ = \frac{1}{14} = \frac{1}{∣ A ∣}

Verified.

Step-by-step solution

1This is a skew-symmetric matrix ( $a_{ij} = - a_{j i}$ ).
2For an odd-order skew-symmetric determinant, the value is 0.
3Verify: $= 0 - 1 (0 + 6) + 2 (3 + 0) = - 6 + 6 = 0$

Final answer

0

0.

Step-by-step solution

1Apply $C_{3} \to C_{3} + C_{2}$
2 $= 111 a b c a + b + c a + b + c a + b + c = (a + b + c) 111 a b c 111$
3Columns 1 and 3 are identical, so the determinant is 0.

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2 $= 2 (a + b + c) 111 b + c c + a a + b c + a a + b b + c$
3Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
4 $= 2 (a + b + c) 100 b + c a - b a - c c + a a - b b - c$
5Expand along column 1 and factor to get $2 (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$

Final answer

2 (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)

Step-by-step solution

1 $2 \cdot Area = 125 4 - 3 4 111 = 1 (- 3 - 4) - 4 (2 - 5) + 1 (8 + 15) = - 7 + 12 + 23 = 28$
2 $Area = 14$

Final answer

14

14 square units.

Step-by-step solution

1Area must be 0: $k 23 242111 = 0$
2 $k (4 - 2) - 2 (2 - 3) + 1 (4 - 12) = 2 k + 2 - 8 = 2 k - 6 = 0$
3 $k = 3$

Final answer

k = 3

k = 3.

Step-by-step solution

1 $∣ A ∣ = 3 (3 - 6) + 1 (6 + 3) + 1 (- 4 - 1) = - 9 + 9 - 5 = - 5 \neq = 0$
2Find the 9 cofactors and form adj(A):
3 $A^{- 1} = \frac{1}{- 5} - 3 - 9 - 5 1852105$

Final answer

A^{- 1} = 3/5 9/5 1 - 1/5 - 8/5 - 1 - 2/5 - 2 - 1

Step-by-step solution

1 $A X = B$ where $A = 112 1 - 1 1 12 - 1$
2 $∣ A ∣ = 1 (1 - 2) - 1 (- 1 - 4) + 1 (1 + 2) = - 1 + 5 + 3 = 7$
3Find cofactors and adj(A), then $A^{- 1} = \frac{1}{7} adj (A)$
4 $X = A^{- 1} B = \frac{1}{7} - 1 53 2 - 3 1 3 - 1 - 2 651 = \frac{1}{7} 71421$

Final answer

x = 1, y = 2, z = 3

Step-by-step solution

1Apply $R_{2} \to R_{2} - 2 R_{1}$ and $R_{3} \to R_{3} - 3 R_{1}$ :
2 $= x + 1 - 1 - 2 x + 2 - 2 - 4 x + 3 - 3 - 6$
3R3 = 2R2, so rows are linearly dependent: $= 0$

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1 $D = 312 - 1 2 - 1 1 - 1 3 = 3 (6 - 1) + 1 (3 + 2) + 1 (- 1 - 4) = 15 + 5 - 5 = 15 \neq = 0$
2Since D is not 0, the system has a unique solution for ALL values of k.
3The system is never inconsistent.

Final answer

D = 15 which is not 0, so the system is never inconsistent.

Step-by-step solution

1Expand along the first row: $= 1 (15 - 16) - 2 (10 - 12) + 3 (8 - 9)$
2 $= - 1 + 4 - 3 = 0$

Final answer

0

0.

Step-by-step solution

1 $∣ A ∣ = 2 (3) - 5 (1) = 1 \neq = 0$
2 $adj (A) = (3 - 1 - 5 2)$
3 $A^{- 1} = \frac{1}{∣ A ∣} adj (A) = (3 - 1 - 5 2)$

Final answer

A^{- 1} = (3 - 1 - 5 2)

Step-by-step solution

1 $D = 1213 = 3 - 2 = 1$
2 $D_{x} = 31313 = 9 - 13 = - 4, D_{y} = 12313 = 13 - 6 = 7$
3 $x = - 4/1 = - 4, y = 7/1 = 7$

Final answer

x = - 4, y = 7

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2 $= a b c - c a + b + c - a - b - a a + b + c$
3Apply $C_{3} \to C_{3} + C_{2}$ and take $(b + c)$ common from C3: $= (b + c) a b c - c a + b + c - a - 1 11$
4Apply $R_{2} \to R_{2} + R_{1}$ :
5 $= (b + c) a a + b c - c a + b - a - 1 01$
6Take $(a + b)$ common from R2: $= (b + c) (a + b) a 1 c - c 1 - a - 1 01$
7Expand along row 2: $= (a + b) (b + c) [- (- c - a) + (a + c)] = (a + b) (b + c) \cdot 2 (a + c)$
8 $= 2 (a + b) (b + c) (c + a)$

Final answer

2 (a + b) (b + c) (c + a)

25

Chapter 5 — Continuity and Differentiability

Continuity and differentiability is where Class 12 calculus begins. A function is continuous at a point when its left and right limits meet the function value there, and differentiable when the left- and right-hand derivatives exist and agree. Every differentiable function is continuous, but the converse fails at corners, cusps and jumps — the two ideas below separate those cases question by question. Work each one with a pencil first; the solutions keep the exam-pattern working short.

Board pattern

Marks are awarded for the test, not the verdict. Always write the left-hand limit, the right-hand limit and the function value for continuity, and the left- and right-hand derivatives for differentiability. State the conclusion in one line: "LHL = RHL = f(a), so f is continuous at x = a". A bare "continuous" scores nothing.

$Ex 5.1 — Continuity at a point, piecewise and composite functions (34 Q)$
$Ex 5.2 — Derivatives of composite functions, chain rule (10 Q)$
$Ex 5.3 — Implicit and inverse-trigonometric derivatives (15 Q)$
$Ex 5.4 — Derivatives of exponential and logarithmic functions (10 Q)$
$Ex 5.5 — Logarithmic differentiation (18 Q)$
$Ex 5.6 — Derivatives of parametric functions (11 Q)$
$Ex 5.7 — Second order derivatives (17 Q)$
$Ex 5.8 — Rolle’s and Lagrange’s Mean Value Theorems (6 Q)$
$Misc — Mixed problems on the whole chapter (21 Q)$

26

Exercise 5.1 — Continuity

34Exercise questions

Step-by-step solution

1f is a polynomial, hence continuous everywhere; at each point LHL = RHL = f(a). At x = 0, f(0) = −3; at x = −3, f(−3) = −18; at x = 5, f(5) = 22.

Final answer

Continuous at all three points.

Final answer

Continuous at x = 3, since f(3) = 17 = LHL = RHL.

Final answer

All four are continuous on their domains. (a) polynomial; (b) rational, continuous for x ≠ 5; (c) equals x − 5 for x ≠ −5, a removable discontinuity at x = −5; (d) |x − 5| is continuous everywhere.

Final answer

f(n) = nⁿ; as a polynomial, f is continuous at every real x, in particular at x = n.

Step-by-step solution

1At x = 0: LHL = 0, RHL = 0, f(0) = 0 — continuous.
2At x = 1: LHL = 1, RHL = 5 — discontinuous.
3At x = 2: LHL = 5, RHL = 5, f(2) = 5 — continuous.

Final answer

Continuous at x = 0 and x = 2; discontinuous at x = 1, since LHL = 1 but RHL = 5.

Final answer

Discontinuous at x = 2 (LHL = 7, RHL = 1).

Step-by-step solution

1At x = −3: LHL = −6, RHL = f(−3) = 6 — discontinuous.
2At x = 3: LHL = |3|+3 = 6, RHL = 20 — discontinuous.

Final answer

Discontinuous at x = −3 and x = 3.

Final answer

No point of discontinuity: |x| is continuous everywhere, including x = 0 where the limit is 0 = f(0).

Final answer

Discontinuous at x = 0, since LHL = −1 and RHL = 1.

Final answer

Continuous at x = 1 (LHL = 2 = RHL = f(1) = 2); no discontinuity.

Final answer

Continuous at x = 2: LHL = 8 − 3 = 5, RHL = 4 + 1 = 5, f(2) = 5. No discontinuity.

Final answer

Discontinuous at x = 1: LHL = 0 but RHL = 1.

Final answer

Discontinuous at x = 1: LHL = 6 but RHL = −4.

Final answer

Discontinuous at x = 1 and x = 3 (jumps of 1); continuous everywhere else.

Final answer

Continuous at x = 0 (both limits 0); discontinuous at x = 1, where LHL = 0 but RHL = 4.

Final answer

Continuous everywhere: at x = −1 both limits equal −2; at x = 1 both equal 2.

Step-by-step solution

1LHL = 3a + 1 and f(3) = 3a + 1; RHL = 3b + 3.
2Set equal: 3a + 1 = 3b + 3 ⇒ 3a − 3b = 2.

Final answer

a - b = t f r a c 23 (t e x t e q u i v a l e n tl y 3 a - 3 b = 2)

Step-by-step solution

1At x = 0: LHL = λ(0 − 0) = 0, RHL = 1; 0 ≠ 1 for every λ.
2At x = 1: f is given by 4x + 1 near x = 1, a polynomial — continuous.

Final answer

No value of λ makes f continuous at x = 0, because LHL = 0 while RHL = 1. At x = 1, f(x) = 4x + 1 is continuous.

Step-by-step solution

1At an integer n: g(n) = n − n = 0.
2LHL = n − (n − 1) = 1, RHL = n − n = 0. Since LHL ≠ RHL, g is discontinuous at n.

Final answer

g is discontinuous at every integer and continuous at every non-integer.

Final answer

Yes. f(π) = π² + 5; since x² and sin x are continuous, f is continuous at x = π.

Final answer

All three are continuous on ℝ: sums, differences and products of the continuous functions sin x and cos x.

Final answer

cos x is continuous on ℝ. cosec x is continuous on ℝ − {nπ}; sec x on ℝ − {(2n+1)π/2}; cot x on ℝ − {nπ}, n ∈ ℤ.

Final answer

No point of discontinuity. At x = 0, LHL = lim (sin x)/x = 1, RHL = f(0) = 1, so f is continuous at 0.

Final answer

f is continuous everywhere. At x = 0, |x² sin(1/x)| ≤ x² → 0, so the limit is 0 = f(0).

Final answer

Discontinuous at x = 0: f(0) = 1, but LHL = RHL = sin 0 − cos 0 = −1.

Step-by-step solution

1Put x = π/2 + h: k cos x/(π − 2x) = k(−sin h)/(−2h) → k/2 as h → 0.
2Set k/2 = 3 ⇒ k = 6.

Final answer

k = 6

Step-by-step solution

1LHL = 4k, RHL = 3, f(2) = 4k; set 4k = 3.

Final answer

k = df r a c 34

Step-by-step solution

1LHL = kπ + 1, RHL = cos π = −1; set kπ + 1 = −1.

Final answer

k = - df r a c 2 p i

Step-by-step solution

1LHL = 5k + 1, RHL = 10; set 5k + 1 = 10.

Final answer

k = df r a c 95

Step-by-step solution

1At x = 2: 2a + b = 5.
2At x = 10: 10a + b = 21.
3Subtract: 8a = 16 ⇒ a = 2, then b = 1.

Final answer

a = 2, b = 1

Final answer

cos is continuous and x² is continuous; a composition of continuous functions is continuous.

Final answer

cos x is continuous, and the modulus function is continuous; a composition of continuous functions is continuous.

Final answer

sin|x| = (sin ∘ |·|)(x). Both sin x and |x| are continuous, so the composite sin|x| is continuous on ℝ.

Final answer

Discontinuous at x = 0 and x = −1, where the two modulus terms change slope and the derivative jumps.

27

Exercise 5.2 — Derivatives of Composite Functions

10Exercise questions

Final answer

2 x cos (x^{2} + 5)

Final answer

- cos x, s in (s in x)

Final answer

a cos (a x + b)

Final answer

df r a c se c^{2} s q r t x, sec (t an s q r t x), t an (t an s q r t x) 2 s q r t x

Final answer

a cos (a x + b) sec (c x + d) + c s in (a x + b) t an (c x + d) sec (c x + d)

Final answer

- 3 x^{2} s in x^{3}, s i n^{2} (x^{5}) + 10 x^{4} cos x^{3}, s in (x^{5}) cos (x^{5})

Final answer

df r a c - 2 s q r t 2, x s in (x^{2}) s q r t s in 2 x^{2}

Final answer

- df r a c s in s q r t x 2 s q r t x

Step-by-step solution

1LHD = lim (f(1+h) − f(1))/h = lim (−h)/h = −1 as h → 0⁻.
2RHD = lim (f(1+h) − f(1))/h = lim h/h = 1 as h → 0⁺.

Final answer

LHD = −1 and RHD = 1 at x = 1; since they differ, f is not differentiable at x = 1.

Step-by-step solution

1At x = 1: LHD = lim [1+h]−[1])/h = −1/h → ∞; RHD = 0.
2At x = 2: LHD = (1 − 2)/h = −1/h → ∞; RHD = 0.

Final answer

Not differentiable at x = 1 and x = 2: one-sided derivatives are infinite/0 and unequal.

28

Exercise 5.3 — Implicit and Inverse Trigonometric Derivatives

15Exercise questions

Final answer

df r a c d y d x = df r a c cos x - 2 3

Step-by-step solution

12 + 3y' = cos y · y'.
2y'(3 − cos y) = −2.

Final answer

df r a c d y d x = df r a c 2 cosy - 3

Final answer

df r a c d y d x = df r a c - a 2 b y + s in y

Final answer

df r a c d y d x = df r a c se c^{2} x - y x + 2 y - 1

Final answer

df r a c d y d x = - df r a c 2 x + y x + 2 y

Final answer

df r a c d y d x = - df r a c 3 x^{2} + 2 x y + y^{2} x^{2} + 2 x y + 3 y^{2}

Final answer

df r a c d y d x = df r a c y s in (x y) s in 2 y - x s in (x y)

Final answer

df r a c d y d x = df r a c s in 2 x s in 2 y

Step-by-step solution

1Put x = tan θ; then 2x/(1+x²) = sin 2θ, so y = 2θ = 2 tan⁻¹x.

Final answer

df r a c d y d x = df r a c 2 1 + x^{2}

Step-by-step solution

1Put x = tan θ; then the argument is tan 3θ, so y = 3θ = 3 tan⁻¹x.

Final answer

df r a c d y d x = df r a c 3 1 + x^{2}

Step-by-step solution

1Put x = tan θ; (1−x²)/(1+x²) = cos 2θ, so y = 2θ = 2 tan⁻¹x.

Final answer

df r a c d y d x = df r a c 2 1 + x^{2}

Step-by-step solution

1Put x = tan θ; (1−x²)/(1+x²) = cos 2θ, and sin⁻¹(cos 2θ) = 2θ.

Final answer

df r a c d y d x = df r a c 2 1 + x^{2}

Step-by-step solution

1Put x = tan θ; 2x/(1+x²) = sin 2θ, so y = π/2 − 2θ.

Final answer

df r a c d y d x = df r a c - 2 1 + x^{2}

Step-by-step solution

1Put x = sin θ; 2x√(1−x²) = sin 2θ, so y = 2θ = 2 sin⁻¹x.

Final answer

df r a c d y d x = df r a c 2 s q r t 1 - x^{2}

Step-by-step solution

1Put x = cos θ; 1/(2x²−1) = 1/cos 2θ = sec 2θ, so y = 2θ = 2 cos⁻¹x.

Final answer

df r a c d y d x = df r a c - 2 s q r t 1 - x^{2}

29

Exercise 5.4 — Exponential and Logarithmic Derivatives

10Exercise questions

Final answer

df r a c e^{x} (s in x - cos x) s i n^{2} x

Final answer

df r a c e^{s i n^{- 1} x} s q r t 1 - x^{2}

Final answer

3 x^{2} e^{x^{3}}

Final answer

df r a c - e^{- x} cos (t a n^{- 1} e^{- x}) 1 + e^{- 2 x}

Final answer

- e^{x} t an (e^{x})

Final answer

e^{x} + 2 x e^{x^{2}} + 3 x^{2} e^{x^{3}} + 4 x^{3} e^{x^{4}} + 5 x^{4} e^{x^{5}}

Final answer

df r a c e^{s q r t x /2} 4 s q r t x

Final answer

df r a c 1 x l o g x

Final answer

df r a c - s in x l o g x - cos x x (l o g x)^{2}

Final answer

- l e f t (df r a c 1 x + e^{x} r i g h t) s in (l o g x + e^{x})

30

Exercise 5.5 — Logarithmic Differentiation

18Exercise questions

Final answer

df r a c d y d x = - cos x cos 2 x cos 3 x, (t an x + 2 t an 2 x + 3 t an 3 x)

Step-by-step solution

1log y = ½[log(x−1)+log(x−2)−log(x−3)−log(x−4)−log(x−5)].

Final answer

df r a c d y d x = df r a c 12 y l e f t [df r a c 1 x - 1 + df r a c 1 x - 2 - df r a c 1 x - 3 - df r a c 1 x - 4 - df r a c 1 x - 5 r i g h t]

Final answer

df r a c d y d x = (l o g x)^{cos x} l e f t [- s in x, l o g (l o g x) + df r a c cos x x l o g x r i g h t]

Final answer

df r a c d y d x = x^{x} (1 + l o g x) - 2^{s in x} (cos x) l o g 2

Final answer

df r a c d y d x = y l e f t [df r a c 2 x + 3 + df r a c 3 x + 4 + df r a c 4 x + 5 r i g h t]

Final answer

df r a c d y d x = l e f t (x + df r a c 1 x r i g h t)^{x} l e f t [l o g! l e f t (x + df r a c 1 x r i g h t) + df r a c x^{2} - 1 x^{2} + 1 r i g h t] + x^{1 + f r a c 1 x} l e f t [df r a c x + 1 - l o g x x^{2} r i g h t]

Final answer

df r a c d y d x = (l o g x)^{x} l e f t [l o g (l o g x) + df r a c 1 l o g x r i g h t] + x^{l o g x} c d o t df r a c 2 l o g x x

Final answer

df r a c d y d x = (s in x)^{x} l e f t [l o g (s in x) + x co t x r i g h t] + df r a c 1 2 s q r t x - x^{2}

Final answer

df r a c d y d x = x^{s in x} l e f t [cos x l o g x + df r a c s in x x r i g h t] + (s in x)^{cos x} l e f t [cos x co t x - s in x l o g (s in x) r i g h t]

Final answer

df r a c d y d x = x^{x cos x} l e f t [cos x l o g x - x s in x l o g x + cos x r i g h t] - df r a c 4 x (x^{2} - 1)^{2}

Final answer

df r a c d y d x = (x cos x)^{x} l e f t [l o g (x cos x) + 1 - x t an x r i g h t] + (x s in x)^{f r a c 1 x} l e f t [df r a c x co t x + 1 - l o g (x s in x) x^{2} r i g h t]

Final answer

df r a c d y d x = - df r a c x^{y} l e f t (f r a c y x r i g h t) + y^{x} l o g y x^{y} l o g x + y^{x} l e f t (f r a c x y r i g h t)

Step-by-step solution

1Take logs: x log y = y log x.
2Differentiate: log y + x y'/y = y' log x + y/x.

Final answer

df r a c d y d x = df r a c f r a c y x - l o g y f r a c x y - l o g x

Final answer

df r a c d y d x = df r a c l o g (cosy) + y t an x l o g (cos x) + x t an y

Step-by-step solution

1log x + log y = x − y.
2Differentiate: 1/x + y'/y = 1 − y'.

Final answer

df r a c d y d x = df r a c y (x - 1) x (1 + y)

Step-by-step solution

1log f = log(1+x)+log(1+x²)+log(1+x⁴)+log(1+x⁸).
2f'/f = 1/(1+x)+2x/(1+x²)+4x³/(1+x⁴)+8x⁷/(1+x⁸).
3At x = 1, f(1) = 16 and the bracket equals 7/2, so f'(1) = 56.

Final answer

f^{'} (1) = 56

Step-by-step solution

1Product rule: (2x−5)(x³+7x+9)+(x²−5x+8)(3x²+7).
2Expanding and log-differentiation both give the same polynomial.

Final answer

df r a c d y d x = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11

Step-by-step solution

1Repeated product rule: treat uv as one factor, then expand.
2Log-differentiation: log(uvw) = log u + log v + log w; differentiate and multiply by uvw.

Final answer

Both methods give u′vw + uv′w + uvw′, so the formula holds.

31

Exercise 5.6 — Derivatives of Parametric Functions

11Exercise questions

Step-by-step solution

1dx/dt = 4at, dy/dt = 4at³; divide.

Final answer

df r a c d y d x = t^{2}

Final answer

df r a c d y d x = - df r a c b a co t t h e t a

Final answer

df r a c d y d x = - 4 s in t

Final answer

df r a c d y d x = - df r a c 1 t^{2}

Final answer

df r a c d y d x = df r a c cos t h e t a - 2 cos 2 t h e t a 2 s in 2 t h e t a - s in t h e t a

Final answer

df r a c d y d x = - co t df r a c t h e t a 2

Final answer

df r a c d y d x = co tt, df r a c 1 - 2 cos 2 t 1 + 2 cos 2 t

Step-by-step solution

1dx/dt = a(−sin t + 1/sin t) = a cos²t/sin t; dy/dt = a cos t.

Final answer

df r a c d y d x = t an t

Final answer

df r a c d y d x = df r a c b a, t e x t cosec, t h e t a

Final answer

df r a c d y d x = t an t h e t a

Step-by-step solution

1(1/x)dx/dt = (1/2)log a /√(1−t²).
2(1/y)dy/dt = −(1/2)log a /√(1−t²).
3Ratio: (dy/dt)/(dx/dt) = −y/x.

Final answer

Differentiating log x and log y and dividing gives dy/dx = −y/x.

32

Exercise 5.7 — Second Order Derivatives

17Exercise questions

Final answer

df r a c d^{2} y d x^{2} = 2

Final answer

df r a c d^{2} y d x^{2} = 380 x^{18}

Final answer

df r a c d^{2} y d x^{2} = - 2 s in x - x cos x

Final answer

df r a c d^{2} y d x^{2} = - df r a c 1 x^{2}

Final answer

df r a c d^{2} y d x^{2} = 6 x l o g x + 5 x

Final answer

df r a c d^{2} y d x^{2} = 2 e^{x} (5 cos 5 x - 12 s in 5 x)

Final answer

df r a c d^{2} y d x^{2} = 9 e^{6 x} (3 cos 3 x - 4 s in 3 x)

Final answer

df r a c d^{2} y d x^{2} = df r a c - 2 x (1 + x^{2})^{2}

Final answer

df r a c d^{2} y d x^{2} = df r a c - (l o g x + 1) x^{2} (l o g x)^{2}

Final answer

df r a c d^{2} y d x^{2} = df r a c - [s in (l o g x) + cos (l o g x)] x^{2}

Final answer

y″ = −5cos x + 3sin x = −y, so y″ + y = 0.

Step-by-step solution

1y′ = −(1−x²)^{−1/2}, y″ = −x(1−x²)^{−3/2}.
2With x = cos y and √(1−x²) = sin y, y″ = −cos y/sin³y.

Final answer

df r a c d^{2} y d x^{2} = - co t y, t e x t cosec^{2} y

Final answer

x y₁ = −3sin(log x) + 4cos(log x); differentiating and substituting gives x²y₂ + xy₁ + y = 0.

Final answer

y′ = mAe^{mx}+nBe^{nx}, y″ = m²Ae^{mx}+n²Be^{nx}; substitution using mn(Ae^{mx}+Be^{nx}) gives 0.

Final answer

y″ = 500·49 e^{7x} + 600·49 e^{−7x} = 49y.

Final answer

From e^y = 1/(x+1), y = −log(x+1), so y′ = −1/(x+1) and y″ = 1/(x+1)² = (y′)².

Final answer

y₁ = 2 tan⁻¹x/(1+x²); differentiating and clearing (1+x²)² gives the identity.

33

Exercise 5.8 — Rolle's and Mean Value Theorems

6Exercise questions

Step-by-step solution

1f(−4) = 16 − 8 − 8 = 0 and f(2) = 4 + 4 − 8 = 0.
2f′(−1) = 0.

Final answer

f is continuous on [−4,2] and differentiable on (−4,2); f(−4) = f(2) = 0, so there is c with f′(c) = 0. f′(x) = 2x + 2 = 0 gives c = −1 ∈ (−4,2).

Step-by-step solution

1(i) [x] is discontinuous at 5,6,7,8 in [5,9]; f(5) = 5 ≠ 9 = f(9).
2(ii) discontinuous at −1,0,1; f(−2) = −2 ≠ 2 = f(2).
3(iii) continuity and differentiability hold, but endpoint values differ.

Final answer

Rolle's theorem applies to none. (i) and (ii): the greatest integer function is not continuous at the integers inside the interval. (iii): f is continuous and differentiable but f(1) = 0 ≠ f(2) = 3. Hence the converse of Rolle's theorem is not true.

Final answer

Suppose f(−5) = f(5); then by Rolle's theorem there is c ∈ (−5,5) with f′(c) = 0, contradicting the hypothesis. Hence f(−5) ≠ f(5).

Step-by-step solution

1f(1) = −6, f(4) = −3; slope = (−3+6)/3 = 1.

Final answer

Slope (f(4)−f(1))/3 = 1; f′(c) = 2c − 4 = 1 gives c = 5/2 ∈ (1,4).

Step-by-step solution

1f(1) = −7, f(3) = −27; slope = (−27+7)/2 = −10.
23c² − 10c − 3 = −10 ⇒ 3c² − 10c + 7 = 0 ⇒ (3c−7)(c−1) = 0.
3c = 7/3 lies in (1,3).

Final answer

c = df r a c 73

Step-by-step solution

1(iii) f(1) = 0, f(2) = 3; slope = 3.

Final answer

MVT applies only to (iii). (i) and (ii) are not continuous (nor differentiable) throughout the interval. For (iii), f′ obeys (f(2)−f(1))/1 = 3; f′(c) = 2c = 3 gives c = 3/2 ∈ (1,2).

34

Miscellaneous Exercise on Chapter 5

21Exercise questions

Final answer

9 (3 x^{2} - 9 x + 5)^{8} (6 x - 9) = 27 (2 x - 3) (3 x^{2} - 9 x + 5)^{8}

Final answer

3 s i n^{2} x cos x - 6 co s^{5} x s in x

Final answer

df r a c d y d x = (5 x)^{3 cos 2 x} l e f t [df r a c 3 cos 2 x x - 6 s in 2 x, l o g (5 x) r i g h t]

Final answer

df r a c d y d x = df r a c 3 s q r t x 2 s q r t 1 - x^{3}

Final answer

df r a c d y d x = df r a c - df r a c 2 s q r t 4 - x^{2} s q r t 2 x + 7 - df r a c co s^{- 1} (x /2) s q r t 2 x + 7 2 x + 7

Step-by-step solution

1Multiply numerator and denominator by the conjugate; the ratio becomes (1+sin x)/cos x = tan(π/4 + x/2).
2y = cot⁻¹(tan(π/4 + x/2)) = π/2 − (π/4 + x/2), so y′ = −1/2.

Final answer

df r a c d y d x = - df r a c 12

Final answer

df r a c d y d x = (l o g x)^{l o g x} l e f t [df r a c 1 x + df r a c l o g (l o g x) x r i g h t]

Final answer

df r a c d y d x = (a s in x - b cos x) s in (a cos x + b s in x)

Final answer

df r a c d y d x = (s in x - cos x)^{(s in x - cos x)} (cos x + s in x) l e f t [1 + l o g (s in x - cos x) r i g h t]

Final answer

df r a c d y d x = x^{x} (1 + l o g x) + a x^{a - 1} + a^{x} l o g a

Final answer

df r a c d y d x = x^{x^{2} - 3} l e f t [x, (2 l o g x + 1) r i g h t] + (x - 3)^{x^{2}} l e f t [2 x l o g (x - 3) + df r a c x^{2} x - 3 r i g h t]

Step-by-step solution

1dx/dt = 10(1 − cos t), dy/dt = 12 sin t; ratio = 12 sin t/(10(1−cos t)) = (6/5)cot(t/2).

Final answer

df r a c d y d x = df r a c 65 co t df r a c t 2

Step-by-step solution

1sin⁻¹√(1−x²) = cos⁻¹x, so y = sin⁻¹x + cos⁻¹x = π/2, whose derivative is 0.

Final answer

df r a c d y d x = 0

Final answer

Squaring gives x²(1+y) = y²(1+x) ⇒ (x−y)(x+y+xy) = 0; since x ≠ y, x + y + xy = 0, so y = −x/(1+x) and y′ = −1/(1+x)².

Step-by-step solution

1Differentiating: 2(x−a)+2(y−b)y₁ = 0 ⇒ y₁ = −(x−a)/(y−b).
2Differentiating again gives y₂ = −c²/(y−b)³; substituting yields the constant c.

Final answer

df r a c [1 + y_{1}^{2}]^{3/2} y_{2} = c

Final answer

Differentiate −sin y · y′ = cos(a+y) − x sin(a+y) y′; rearranging and using x = cos y/cos(a+y) gives y′ = cos²(a+y)/sin a.

Step-by-step solution

1dx/dt = at cos t, dy/dt = at sin t ⇒ dy/dx = tan t.
2d²y/dx² = (d/dt tan t)/(dx/dt) = sec²t/(at cos t) = 1/(at cos³t).

Final answer

df r a c d^{2} y d x^{2} = df r a c 1 a t co s^{3} t

Step-by-step solution

1For x ≥ 0, f = x³ so f″ = 6x; for x < 0, f = −x³ so f″ = −6x.
2Both branches combine to f″(x) = 6|x|.

Final answer

f^{''} (x) = 6∣ x ∣

Final answer

Differentiate both sides with respect to A treating B as a function of A with dB/dA = 1; this gives cos(A+B) = cos A cos B − sin A sin B.

Final answer

Yes. f(x) = |x| + |x − 1| is continuous everywhere but not differentiable at exactly x = 0 and x = 1.

Final answer

Expand along the first row: y = f(x)(mc − nb) − g(x)(lc − na) + h(x)(lb − ma); differentiating term by term gives the determinant with the first row replaced by the derivatives.

35

Chapter 6 — Application of Derivatives

This chapter applies differentiation to real-world problems: finding rates of change of physical quantities, deciding whether a function is increasing or decreasing, locating tangents and normals to curves, constructing linear approximations, and determining maxima and minima of functions. Every exercise below carries the NCERT questions in full with short, exam-pattern working. Work each one with a pencil first; the solutions keep the algebra compact.

Board pattern

For rate-of-change problems always state the variable and what it represents, write the given rate, and differentiate the relation before substituting. For increasing/decreasing functions compute f'(x), find its sign in each interval, and state the conclusion. For tangents and normals find the point, compute dy/dx, then write the tangent or normal equation. For maxima/minima show the first- or second-derivative test clearly — a bare "maximum" scores nothing.

$Ex 6.1 — Rate of change of quantities (17 Q)$
$Ex 6.2 — Increasing and decreasing functions (18 Q)$
$Ex 6.3 — Tangents and normals (29 Q)$
$Ex 6.4 — Approximation (10 Q)$
$Ex 6.5 — Maxima and minima (22 Q)$
$Misc — Mixed application-of-derivative problems (15 Q)$

36

Exercise 6.1 — Rate of Change of Quantities

17Exercise questions

Step-by-step solution

1Area $A = π r^{2}$ , so $\frac{d A}{d r} = 2 π r$ .
2(a) At $r = 3$ : $\frac{d A}{d r} = 6 π$ . (b) At $r = 4$ : $\frac{d A}{d r} = 8 π$ .

Final answer

(a) 6 π cm^{2} / cm; (b) 8 π cm^{2} / cm

Step-by-step solution

1Let the edge be $x$ . Then $V = x^{3}$ and $S = 6 x^{2}$ .
2Differentiate $V = x^{3}$ : $\frac{d V}{d t} = 3 x^{2} \frac{d x}{d t}$ . At $x = 12$ : $8 = 3 (144) \frac{d x}{d t} ⟹ \frac{d x}{d t} = \frac{1}{54}$ .
3Differentiate $S = 6 x^{2}$ : $\frac{d S}{d t} = 12 x \frac{d x}{d t} = 12 (12) (\frac{1}{54}) = \frac{8}{3}$ .

Final answer

\frac{8}{3} cm^{2} / s

Step-by-step solution

1 $A = π r^{2} ⟹ \frac{d A}{d t} = 2 π r \frac{d r}{d t} = 2 π (10) (3) = 60 π$

Final answer

60 π cm^{2} / s

Step-by-step solution

1 $V = x^{3} ⟹ \frac{d V}{d t} = 3 x^{2} \frac{d x}{d t} = 3 (100) (3) = 900$

Final answer

900 cm^{3} / s

Step-by-step solution

1 $A = π r^{2} ⟹ \frac{d A}{d t} = 2 π r \frac{d r}{d t} = 2 π (8) (5) = 80 π$

Final answer

80 π cm^{2} / s

Step-by-step solution

1 $C = 2 π r ⟹ \frac{d C}{d t} = 2 π \frac{d r}{d t} = 2 π (0.7) = 1.4 π cm/s$

Final answer

1.4 π cm/s

Step-by-step solution

1(a) Perimeter $P = 2 (x + y)$ : $\frac{d P}{d t} = 2 (\frac{d x}{d t} + \frac{d y}{d t}) = 2 (- 5 + 4) = - 2 cm/min$ .
2(b) Area $A = x y$ : $\frac{d A}{d t} = x \frac{d y}{d t} + y \frac{d x}{d t} = 8 (4) + 6 (- 5) = 32 - 30 = 2 cm^{2} / min$ .

Final answer

(a) - 2 cm/min; (b) 2 cm^{2} / min

Step-by-step solution

1 $V = \frac{4}{3} π r^{3} ⟹ \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ .
2At $r = 15$ cm: $900 = 4 π (225) \frac{d r}{d t} ⟹ \frac{d r}{d t} = \frac{1}{π} cm/s$ .

Final answer

\frac{1}{π} cm/s

Step-by-step solution

1Let the foot be at distance $x$ from the wall and the top at height $y$ . Then $x^{2} + y^{2} = 25$ .
2Differentiating: $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$ . At $x = 4$ m, $y = 3$ m: $2 (4) (2) + 2 (3) \frac{d y}{d t} = 0 ⟹ \frac{d y}{d t} = - \frac{8}{3}$ .

Final answer

\frac{8}{3} cm/s

Step-by-step solution

1Differentiating: $6 \frac{d y}{d t} = 3 x^{2} \frac{d x}{d t} ⟹ \frac{d y}{d t} = \frac{x ^{2}}{2} \frac{d x}{d t}$ .
2For $\frac{d y}{d t} = 2 \frac{d x}{d t}$ , we need $\frac{x ^{2}}{2} = 2 ⟹ x = 2$ . Then $6 y = 8 + 2 = 10 ⟹ y = \frac{5}{3}$ .

Final answer

(2, \frac{5}{3})

Step-by-step solution

1From $\frac{d y}{d t} = \frac{x ^{2}}{2} \frac{d x}{d t}$ and the requirement $\frac{d y}{d t} = 8 \frac{d x}{d t}$ we get $\frac{x ^{2}}{2} = 8 ⟹ x^{2} = 16 ⟹ x = 4$ .
2Then $6 y = 64 + 2 = 66 ⟹ y = 11$ .

Final answer

(4, 11)

Step-by-step solution

1 $A = π r^{2} ⟹ \frac{d A}{d t} = 2 π r \frac{d r}{d t}$ .
2At $r = 6$ cm: $72 π = 2 π (6) \frac{d r}{d t} ⟹ \frac{d r}{d t} = 6 cm/s$ .

Final answer

6 cm/s

Step-by-step solution

1Let the hypotenuse be $z$ and the two legs $x$ , $y$ . Then $z^{2} = x^{2} + y^{2}$ , with $\frac{d x}{d t} = - 2$ and $\frac{d y}{d t} = 1$ .
2Differentiate: $2 z \frac{d z}{d t} = 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t}$ .
3At $x = 18, y = 24$ we have $z = 30$ , so $30 \frac{d z}{d t} = 18 (- 2) + 24 (1) = - 12 ⟹ \frac{d z}{d t} = - \frac{2}{5} cm/s$ .

Final answer

decreasing at \frac{2}{5} cm/s

Step-by-step solution

1Let the man be at distance $x$ from the pole and let his shadow be $s$ . By similar triangles: $\frac{s}{x + s} = \frac{2}{6} = \frac{1}{3}$ , so $3 s = x + s ⟹ s = \frac{x}{2}$ .
2Hence $\frac{d s}{d t} = \frac{1}{2} \frac{d x}{d t} = \frac{1}{2} (5) = 2.5 km/h$ .

Final answer

2.5 km/h

Step-by-step solution

1If the man is distance $x$ from the lamp and his shadow is $s$ , then $\frac{s}{x + s} = \frac{1.5}{4.5} = \frac{1}{3}$ , so $3 s = x + s ⟹ s = \frac{x}{2}$ .
2Since he walks towards the lamp, $\frac{d x}{d t} = - 1$ , giving $\frac{d s}{d t} = \frac{1}{2} (- 1) = - 0.5 m/s$ .

Final answer

0.5 m/s

Step-by-step solution

1Differentiate with respect to $t$ : $\frac{d y}{d t} = (2 x + 2) \frac{d x}{d t}$ .
2Setting $\frac{d x}{d t} = \frac{d y}{d t}$ gives $2 x + 2 = 1 ⟹ x = - \frac{1}{2}$ . Then $y = \frac{1}{4} - 1 = - \frac{3}{4}$ .

Final answer

(- \frac{1}{2}, - \frac{3}{4})

Step-by-step solution

1Area $A = \frac{1}{2} (12) (7) sin θ = 42 sin θ$ .
2Then $\frac{d A}{d t} = 42 cos θ \frac{d θ}{d t}$ with $\frac{d θ}{d t} = \frac{π}{180} rad/s$ .
3At $θ = 6 0^{\circ}$ : $\frac{d A}{d t} = 42 \cdot \frac{1}{2} \cdot \frac{π}{180} = \frac{7 π}{60} cm^{2} / s$ .

Final answer

\frac{7 π}{60} cm^{2} / s

37

Exercise 6.2 — Increasing and Decreasing Functions

18Exercise questions

Step-by-step solution

1 $f^{'} (x) = 3 > 0$ for all $x \in R$ , so $f$ is strictly increasing.

Final answer

f'(x) = 3 > 0, so f is strictly increasing on the set of real numbers.

Step-by-step solution

1 $f^{'} (x) = 2 e^{2 x} > 0$ for all $x \in R$ , so $f$ is strictly increasing.

Final answer

f'(x) = 2e^(2x) > 0 for all x, so f is strictly increasing on the set of real numbers.

Step-by-step solution

1 $f^{'} (x) = 3 x^{2} - 6 x + 3 = 3 (x - 1)^{2} \geq 0$ for all $x$ .
2Since $f^{'} (x) = 0$ only at the isolated point $x = 1$ , $f$ is strictly increasing on $R$ .

Final answer

f'(x) = 3(x - 1)^2 >= 0 with equality only at x = 1; f is strictly increasing on the set of real numbers.

Step-by-step solution

1 $f^{'} (x) = - sin x$ .
2(a) On $(0, π)$ we have $sin x > 0$ , so $f^{'} (x) < 0$ : strictly decreasing.
3(b) On $(π, 2 π)$ we have $sin x < 0$ , so $f^{'} (x) > 0$ : strictly increasing.

Final answer

(a) strictly decreasing on (0, π); (b) strictly increasing on (π, 2 π)

Step-by-step solution

1 $f^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$ .
2Sign of $f^{'} (x)$ : negative on $(1, 2)$ , positive on $(- \infty, 1) \cup (2, \infty)$ .

Final answer

(a) (1, 2); (b) (- \infty, 1) \cup (2, \infty)

Step-by-step solution

1 $f^{'} (x) = - 6 x^{2} + 30 x - 36 = - 6 (x - 2) (x - 3)$ .
2Sign of $f^{'} (x)$ : negative on $(- \infty, 2) \cup (3, \infty)$ , positive on $(2, 3)$ .

Final answer

(a) (2, 3); (b) (- \infty, 2) \cup (3, \infty)

Step-by-step solution

1 $f^{'} (x) = 12 + 6 x - 6 x^{2} = - 6 (x + 1) (x - 2)$ .
2Sign of $f^{'} (x)$ : negative on $(- \infty, - 1)$ and $(2, \infty)$ , positive on $(- 1, 2)$ .

Final answer

strictly increasing on (- 1, 2); strictly decreasing on (- \infty, - 1) \cup (2, \infty)

Step-by-step solution

1 $f^{'} (x) = 6 (x - 1) (x - 2)$ , same factorisation as above.
2Negative on $(1, 2)$ , positive on $(- \infty, 1) \cup (2, \infty)$ .

Final answer

(a) (1, 2); (b) (- \infty, 1) \cup (2, \infty)

Step-by-step solution

1 $f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ .
2Negative on $(1, 3)$ , positive on $(- \infty, 1) \cup (3, \infty)$ .

Final answer

(a) (3, \infty); also (- \infty, 1); (b) (1, 3)

Step-by-step solution

1 $f^{'} (x) = \frac{1}{x} > 0$ for all $x > 0$ , so $f$ is strictly increasing on $(0, \infty)$ .

Final answer

f'(x) = 1/x > 0 for all x > 0, so f is strictly increasing on the interval (0, infinity).

Step-by-step solution

1 $f^{'} (x) = 2 x - 1$ .
2 $f^{'} (x) < 0$ on $(- \infty, \frac{1}{2})$ and $f^{'} (x) > 0$ on $(\frac{1}{2}, \infty)$ , so the sign changes at $x = \frac{1}{2}$ .

Final answer

f'(x) changes sign at x = 1/2, so f is neither strictly increasing nor strictly decreasing on the set of real numbers.

Step-by-step solution

1 $f^{'} (x) = 2 x + a$ . For $f^{'} (x) \geq 0$ on $[1, 2]$ we need $f^{'} (1) = 2 + a \geq 0$ , i.e. $a \geq - 2$ .

Final answer

a \geq - 2

Step-by-step solution

1(i) $(cos x)^{'} = - sin x < 0$ on $(0, 1)$ : decreasing.
2(ii) $(cos 2 x)^{'} = - 2 sin 2 x < 0$ on $(0, 1)$ : decreasing.
3(iii) $(cos 3 x)^{'} = - 3 sin 3 x < 0$ on $(0, 1)$ : decreasing.
4(iv) $(tan x)^{'} = sec^{2} x > 0$ : increasing, not decreasing.

Final answer

(i), (ii), (iii) are strictly decreasing; (i v) is increasing

Step-by-step solution

1 $f^{'} (x) = \frac{1}{s i n x} cos x = cot x$ .
2On $(0, \frac{π}{2})$ , $cot x > 0$ , so $f$ is strictly increasing there.

Final answer

f'(x) = cot x > 0 on (0, pi/2), so f is strictly increasing there.

Step-by-step solution

1 $f^{'} (x) = \frac{- s i n x}{c o s x} = - tan x$ .
2On $(0, \frac{π}{2})$ , $tan x > 0$ , so $f^{'} (x) < 0$ and $f$ is strictly decreasing.

Final answer

f'(x) = -tan x < 0 on (0, pi/2), so f is strictly decreasing there.

Step-by-step solution

1 $y^{'} = \frac{1}{1 + x} - \frac{( 2 ) ( 2 + x ) - 2 x ( 1 )}{( 2 + x ) ^{2}} = \frac{1}{1 + x} - \frac{4}{( 2 + x ) ^{2}}$ .
2So $y^{'} = \frac{( 2 + x ) ^{2} - 4 ( 1 + x )}{( 1 + x ) ( 2 + x ) ^{2}} = \frac{x ^{2}}{( 1 + x ) ( 2 + x ) ^{2}} \geq 0$ for all $x > - 1$ .
3Since $y^{'} \geq 0$ with equality only at $x = 0$ , $y$ is increasing on its domain.

Final answer

y' = x^2 / ((1+x)(2+x)^2) >= 0, zero only at x = 0, so y is an increasing function of x.

Step-by-step solution

1 $f^{'} (x) = cos x - sin x = 2 cos (x + \frac{π}{4})$ .
2 $f^{'} (x) \geq 0$ when $cos (x + \frac{π}{4}) \geq 0$ , i.e. $0 \leq x \leq \frac{3 π}{4}$ or $\frac{7 π}{4} \leq x \leq 2 π$ .
3Hence strictly increasing on $(0, \frac{3 π}{4})$ and $(\frac{7 π}{4}, 2 π)$ ; strictly decreasing on $(\frac{3 π}{4}, \frac{7 π}{4})$ .

Final answer

increasing on (0, \frac{3 π}{4}) \cup (\frac{7 π}{4}, 2 π); decreasing on (\frac{3 π}{4}, \frac{7 π}{4})

Step-by-step solution

1 $f^{'} (x) = 3 x^{2} - 6 x + 3 = 3 (x - 1)^{2} \geq 0$ for all $x$ , equality only at $x = 1$ .
2So $f$ is increasing on $R$ .

Final answer

increasing on R

38

Exercise 6.3 — Tangents and Normals

29Exercise questions

Step-by-step solution

1 $y^{'} = 3 x^{2} - 1$ .
2At $x = 2$ : $y^{'} = 3 (4) - 1 = 11$ .

Final answer

11

Step-by-step solution

1Tangent slope at $x = 2$ is 11, so the normal slope is its negative reciprocal: $- \frac{1}{11}$ .

Final answer

- \frac{1}{11}

Step-by-step solution

1 $y^{'} = 3 x^{2} - 3$ .
2At $x = 3$ : $y^{'} = 27 - 3 = 24$ .

Final answer

24

Step-by-step solution

1 $y^{'} = 3 x^{2} - 3$ .
2At $x = 2$ : $y^{'} = 12 - 3 = 9$ .

Final answer

9

Step-by-step solution

1 $y^{'} = 3 x^{2} - 3$ .
2At $x = 1$ : $y^{'} = 3 - 3 = 0$ .

Final answer

0

Step-by-step solution

1At $x = 0$ , $y = 5$ , so the point is $(0, 5)$ .
2 $y^{'} = 4 x^{3} - 18 x^{2} + 26 x - 10$ , so at $x = 0$ , $y^{'} = - 10$ .
3Tangent: $y - 5 = - 10 (x - 0) ⟹ y = - 10 x + 5$ .

Final answer

y = - 10 x + 5

Step-by-step solution

1Tangent slope at $(0, 5)$ is $y^{'} (0) = - 10$ , so normal slope is $\frac{1}{10}$ .

Final answer

\frac{1}{10}

Step-by-step solution

1 $y^{'} = 4 x^{3} - 18 x^{2} + 26 x - 10$ , so at $x = 2$ , $y^{'} = 32 - 72 + 52 - 10 = 2$ .
2Normal slope $= - \frac{1}{y ^{'}} = - \frac{1}{2}$ .

Final answer

- \frac{1}{2}

Step-by-step solution

1 $y^{'} = \frac{( 1 ) ( x - 2 ) - ( x - 1 ) ( 1 )}{( x - 2 ) ^{2}} = \frac{- 1}{( x - 2 ) ^{2}}$ .
2At $x = 3$ : $y^{'} = - 1$ .

Final answer

- 1

Step-by-step solution

1 $y^{'} = \frac{- 1}{( x - 2 ) ^{2}}$ .
2At $x = 4$ : $y^{'} = - \frac{1}{4}$ .

Final answer

- \frac{1}{4}

Step-by-step solution

1 $\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ} = \frac{3 a s i n ^{2} θ c o s θ}{- 3 a c o s ^{2} θ s i n θ} = - tan θ$ .
2At $θ = \frac{π}{4}$ : $\frac{d y}{d x} = - 1$ , so normal slope $= 1$ .

Final answer

1

Step-by-step solution

1 $\frac{d y}{d x} = - tan θ$ , which is undefined (vertical tangent) at $θ = \frac{π}{2}$ .
2The tangent is vertical, hence the normal is horizontal: slope $= 0$ .

Final answer

0

Step-by-step solution

1Slope of chord $= \frac{2 - ( - 2 )}{2 - 1} = 4$ .
2Set $y^{'} = 3 x^{2} - 3 = 4 ⟹ x^{2} = \frac{7}{3} ⟹ x = \pm \frac{7}{3}$ .
3The points are $(\pm \frac{7}{3}, \pm \frac{7}{3} (\frac{7}{3} - 3))$ .

Final answer

x = \pm 7/3

Step-by-step solution

1 $y^{'} = 2 x - 2$ .
2(a) Slope of the given line is 2, so $2 x - 2 = 2 ⟹ x = 2$ . Then $y = 4 - 4 + 7 = 7$ . Tangent: $y - 7 = 2 (x - 2)$ .
3(b) Slope of $5 y - 15 x = 13$ is $3$ , so needed tangent slope is $- \frac{1}{3}$ : $2 x - 2 = - \frac{1}{3} ⟹ x = \frac{5}{6}$ . Then $y = \frac{25}{36} - \frac{10}{6} + 7 = \frac{217}{36}$ .

Final answer

(a) y = 2 x + 3; (b) y = - \frac{1}{3} x + \frac{227}{36}

Step-by-step solution

1 $y^{'} = 3 x^{2} - 4 x + 1$ , so at $x = 1$ : $y^{'} = 3 - 4 + 1 = 0$ .
2Tangent: $y + 1 = 0 (x - 1) ⟹ y = - 1$ .

Final answer

y = - 1

Step-by-step solution

1Differentiate $a y^{2} = x^{3}$ : $2 a y y^{'} = 3 x^{2} ⟹ y^{'} = \frac{3 x ^{2}}{2 a y}$ .
2At $(a m^{2}, a m^{3})$ : $y^{'} = \frac{3 a ^{2} m ^{4}}{2 a ^{2} m ^{3}} = \frac{3 m}{2}$ .
3Normal slope $= - \frac{2}{3 m}$ . Normal: $y - a m^{3} = - \frac{2}{3 m} (x - a m^{2})$ .

Final answer

3 m y - 3 a m^{4} = - 2 x + 2 a m^{2}

Step-by-step solution

1 $y^{'} = 3 x^{2} - 6 x + 2$ .
2At $x = 0$ : $y^{'} = 2$ .
3At $x = 1$ : $y^{'} = 3 - 6 + 2 = - 1$ .

Final answer

The tangent slopes at x = 0 and x = 1 are 2 and -1, which are not equal.

Step-by-step solution

1Parallel to the $x$ -axis means $y^{'} = 12 x^{2} - 4 x - 9 = 0$ .
2 $x = \frac{4 \pm 16 + 432}{24} = \frac{4 \pm 448}{24} = \frac{1 \pm 7}{6}$ .

Final answer

x = \frac{1 \pm 7}{6}

Step-by-step solution

1Differentiate: $2 x + 2 y y^{'} = 0 ⟹ y^{'} = - \frac{x}{y}$ .
2At $(4, 3)$ : $y^{'} = - \frac{4}{3}$ .
3Tangent: $y - 3 = - \frac{4}{3} (x - 4) ⟹ 4 x + 3 y = 25$ .

Final answer

4 x + 3 y = 25

Step-by-step solution

1 $y^{'} = 3 x^{2} - 10 x + 3$ , at $x = 1$ : $y^{'} = 3 - 10 + 3 = - 4$ .
2Tangent: $y + 3 = - 4 (x - 1) ⟹ 4 x + y - 1 = 0$ .
3Normal: $y + 3 = \frac{1}{4} (x - 1) ⟹ x - 4 y - 13 = 0$ .

Final answer

tangent 4 x + y - 1 = 0; normal x - 4 y - 13 = 0

Step-by-step solution

1Slope of the line is $2$ . Also $y^{'} = \frac{3}{2 3 x - 2}$ .
2Set $\frac{3}{2 3 x - 2} = 2 ⟹ 3 x - 2 = \frac{3}{4} ⟹ 3 x - 2 = \frac{9}{16} ⟹ x = \frac{41}{48}$ .
3Then $y = \frac{3}{4}$ . Tangent: $y - \frac{3}{4} = 2 (x - \frac{41}{48})$ .

Final answer

y = 2 x - \frac{10}{24}

Step-by-step solution

1 $y^{'} = 3 x^{2}$ , at $x = 1$ : $y^{'} = 3$ .
2Normal slope $= - \frac{1}{3}$ . Normal: $y - 1 = - \frac{1}{3} (x - 1) ⟹ x + 3 y = 4$ .

Final answer

x + 3 y = 4

Step-by-step solution

1 $y^{'} = 2 x + 1$ , at $x = 1$ : $y^{'} = 3$ .
2Normal slope $= - \frac{1}{3}$ . Normal: $y - 2 = - \frac{1}{3} (x - 1) ⟹ x + 3 y = 7$ .

Final answer

x + 3 y = 7

Step-by-step solution

1 $y^{'} = 2 x - 2$ , at $x = 2$ : $y^{'} = 2$ .
2Tangent: $y - 3 = 2 (x - 2) ⟹ y = 2 x - 1$ .

Final answer

y = 2 x - 1

Step-by-step solution

1Slope $= tan 4 5^{\circ} = 1 = y^{'} = 3 x^{2}$ .
2So $x^{2} = \frac{1}{3} ⟹ x = \pm \frac{1}{3}$ . Points: $(\pm \frac{1}{3}, \pm \frac{1}{3 3})$ .

Final answer

(\pm \frac{1}{3}, \pm \frac{1}{3 3})

Step-by-step solution

1 $y^{'} = 2 cos 2 x$ , at $x = 0$ : $y^{'} = 2$ .
2Tangent: $y = 2 x$ .

Final answer

y = 2 x

Step-by-step solution

1The curve cuts the $y$ -axis at $x = 0$ , i.e. at $(0, 1)$ .
2 $y^{'} = - e^{- x}$ , at $x = 0$ : $y^{'} = - 1$ .
3Tangent: $y - 1 = - x ⟹ x + y = 1$ .

Final answer

x + y = 1

Step-by-step solution

1 $y^{'} = \frac{( 1 ) ( 1 + x ^{2} ) - x ( 2 x )}{( 1 + x ^{2} ) ^{2}} = \frac{1 - x ^{2}}{( 1 + x ^{2} ) ^{2}}$ .
2At $x = 0$ : $y^{'} = 1$ , so normal slope $= - 1$ .
3Normal: $y = - x$ .

Final answer

y = - x

Step-by-step solution

1At the origin, $y^{'} = 1$ , so tangent is $y = x$ .
2Solving $x = \frac{x}{x ^{2} + 1}$ : $x = 0$ or $x = 0, x^{2} - \dots$ .
3The curves meet where $x^{2} - x + 1 \dots$ .

Final answer

The tangent at the origin is y = x, which meets the curve again at a point with slope -1, giving perpendicular tangents.

39

Exercise 6.4 — Approximation

10Exercise questions

Step-by-step solution

1Take $f (x) = x$ , $x = 36$ , $Δ x = 0.6$ .
2 $f^{'} (x) = \frac{1}{2 x}$ , so $d y = f^{'} (36) (0.6) = \frac{1}{12} (0.6) = 0.05$ .
3Hence $36.6 \approx 6 + 0.05 = 6.05$ .

Final answer

\approx 6.05

Step-by-step solution

1Take $f (x) = x^{3/2}$ , $x = 4$ , $Δ x = - 0.032$ .
2 $f^{'} (x) = \frac{3}{2} x^{1/2}$ , so $d y = \frac{3}{2} (2) (- 0.032) = - 0.096$ .
3Hence $(3.968)^{3/2} \approx 8 - 0.096 = 7.904$ .

Final answer

\approx 7.904

Step-by-step solution

1Take $f (x) = x^{1/3}$ , $x = 27$ , $Δ x = - 1$ .
2 $f^{'} (x) = \frac{1}{3} x^{- 2/3}$ , so $d y = \frac{1}{3} (27)^{- 2/3} (- 1) = - \frac{1}{27}$ .
3Hence $(26)^{1/3} \approx 3 - \frac{1}{27} = \frac{80}{27} \approx 2.963$ .

Final answer

\frac{80}{27} \approx 2.963

Step-by-step solution

1Take $f (x) = x^{1/4}$ , $x = 81$ , $Δ x = 0.5$ .
2 $f^{'} (x) = \frac{1}{4} x^{- 3/4}$ , so $d y = \frac{1}{4} (81)^{- 3/4} (0.5) = \frac{1}{4} \cdot \frac{1}{27} \cdot \frac{1}{2} = \frac{1}{216}$ .
3Hence $4 81.5 \approx 3 + \frac{1}{216} \approx 3.0046$ .

Final answer

\approx 3.0046

Step-by-step solution

1Take $f (x) = sin x$ , with $x = \frac{π}{3}$ and $Δ x = \frac{π}{180}$ (one degree).
2 $f^{'} (x) = cos x$ , so $d y = cos \frac{π}{3} \cdot \frac{π}{180} = \frac{1}{2} \cdot \frac{π}{180} \approx 0.00873$ .
3Hence $sin 6 1^{\circ} \approx \frac{3}{2} + 0.00873 \approx 0.8748$ .

Final answer

\approx 0.8748

Step-by-step solution

1Take $f (x) = lo g x$ , $x = 1$ , $Δ x = 0.01$ .
2 $d y = f^{'} (1) Δ x = (1) (0.01) = 0.01$ .
3Hence $lo g 1.01 \approx 0 + 0.01 = 0.01$ .

Final answer

\approx 0.01

Step-by-step solution

1Take $f (x) = x^{1/10}$ , $x = 1$ , $Δ x = - 0.001$ .
2 $f^{'} (x) = \frac{1}{10} x^{- 9/10}$ , so $d y = \frac{1}{10} (1) (- 0.001) = - 0.0001$ .
3Hence $(0.999)^{1/10} \approx 1 - 0.0001 = 0.9999$ .

Final answer

\approx 0.9999

Step-by-step solution

1Take $f (x) = x^{1/2}$ , $x = 4$ , $Δ x = - 0.002$ .
2 $f^{'} (x) = \frac{1}{2} x^{- 1/2}$ , so $d y = \frac{1}{4} (- 0.002) = - 0.0005$ .
3Hence $(3.998)^{1/2} \approx 2 - 0.0005 = 1.9995$ .

Final answer

\approx 1.9995

Step-by-step solution

1Take $f (x) = e^{x}$ , $x = 0$ , $Δ x = 0.02$ .
2 $f^{'} (x) = e^{x}$ , so $d y = e^{0} (0.02) = 0.02$ .
3Hence $e^{0.02} \approx 1 + 0.02 = 1.02$ .

Final answer

\approx 1.02

Step-by-step solution

1 $V = x^{3}$ , so $Δ V \approx d V = 3 x^{2} d x$ with $d x = 0.03 x$ .
2Hence $Δ V \approx 3 x^{2} (0.03 x) = 0.09 x^{3}$ , a 9% change.

Final answer

Δ V \approx 0.09 x^{3} (9%)

40

Exercise 6.5 — Maxima and Minima

22Exercise questions

Step-by-step solution

1 $(2 x - 1)^{2} \geq 0$ , so $f (x) = (2 x - 1)^{2} + 3 \geq 3$ .
2Minimum value 3 achieved at $x = \frac{1}{2}$ . There is no maximum.

Final answer

min 3 at x = \frac{1}{2}; no maximum

Step-by-step solution

1Complete the square: $f (x) = 9 (x + \frac{2}{3})^{2} - 4 + 2 = 9 (x + \frac{2}{3})^{2} - 2 \geq - 2$ .
2Minimum value $- 2$ at $x = - \frac{2}{3}$ . No maximum.

Final answer

min - 2 at x = - \frac{2}{3}; no maximum

Step-by-step solution

1 $f^{'} (x) = 2 x = 0 ⟹ x = 0$ .
2 $f^{''} (x) = 2 > 0$ , so $x = 0$ is a local minimum with value $0$ .

Final answer

local min 0 at x = 0

Step-by-step solution

1 $f^{'} (x) = 3 x^{2} - 3 = 0 ⟹ x = \pm 1$ .
2 $f^{''} (x) = 6 x$ : $f^{''} (1) = 6 > 0$ (local min $- 2$ at $x = 1$ ).
3 $f^{''} (- 1) = - 6 < 0$ (local max $2$ at $x = - 1$ ).

Final answer

local max 2 at x = - 1; local min - 2 at x = 1

Step-by-step solution

1 $f^{'} (x) = 2 x - 2 = 0 ⟹ x = 1$ (in the interval).
2 $f (0) = 5, f (1) = 4, f (3) = 8$ .

Final answer

max 8 at x = 3; min 4 at x = 1

Step-by-step solution

1 $f^{'} (x) = 3 (x - 1) (x - 3) = 0 ⟹ x = 1, 3$ .
2 $f (0) = 15, f (1) = 19, f (3) = 15, f (4) = 31$ .

Final answer

max 31 at x = 4; min 15 at x = 0, 3

Step-by-step solution

1 $f^{'} (x) = cos x - sin x = 0 ⟹ x = \frac{π}{4}, \frac{5 π}{4}$ .
2 $f (\frac{π}{4}) = 2, f (\frac{5 π}{4}) = - 2, f (0) = 1, f (2 π) = 1$ .

Final answer

max 2; min - 2

Step-by-step solution

1 $f^{'} (x) = 6 x^{2} - 24 = 0 ⟹ x = \pm 2$ .
2 $f^{''} (x) = 12 x$ : $f^{''} (2) = 24 > 0$ (local min $- 25$ ), $f^{''} (- 2) = - 24 < 0$ (local max $39$ ).

Final answer

local max 39 at x = - 2; local min - 25 at x = 2

Step-by-step solution

1 $f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1) = 0 ⟹ x = 3, - 1$ .
2 $f^{''} (x) = 6 x - 6$ : $f^{''} (3) = 12 > 0$ (min), $f^{''} (- 1) = - 12 < 0$ (max).
3 $f (3) = - 17, f (- 1) = 15$ .

Final answer

max 15 at x = - 1; min - 17 at x = 3

Step-by-step solution

1Let the numbers be $x$ and $16 - x$ . Minimise $S = x^{3} + (16 - x)^{3}$ .
2 $S^{'} = 3 x^{2} - 3 (16 - x)^{2} = 0 ⟹ x^{2} = (16 - x)^{2} ⟹ x = 8$ (since $x > 0$ ).
3 $S^{''} = 96 > 0$ (since $S^{''} = 6 x + 6 (16 - x) = 96$ ), a minimum.

Final answer

8 and 8

Step-by-step solution

1Let the numbers be $x$ and $\frac{36}{x}$ . Minimise $S = x + \frac{36}{x}$ .
2 $S^{'} = 1 - \frac{36}{x ^{2}} = 0 ⟹ x = 6$ .
3 $S^{''} = \frac{72}{x ^{3}} > 0$ at $x = 6$ , a minimum. Numbers $6$ and $6$ .

Final answer

6 and 6

Step-by-step solution

1Let radius be $r$ and square side $x$ . $2 π r + 4 x = k$ .
2Minimise $A = π r^{2} + x^{2} = π r^{2} + (\frac{k - 2 π r}{4})^{2}$ .
3 $\frac{d A}{d r} = 2 π r - \frac{π}{4} (k - 2 π r) = 0 ⟹ 8 r = k - 2 π r ⟹ k = 8 r + 2 π r$ .
4Then $4 x = k - 2 π r = 8 r$ , so $x = 2 r$ .

Final answer

x = 2 r when the sum of areas is least

Step-by-step solution

1Let $x$ be the side of the square cut. Then $V = x (45 - 2 x) (24 - 2 x)$ .
2 $V = 4 x^{3} - 138 x^{2} + 1080 x$ , so $V^{'} = 12 x^{2} - 276 x + 1080 = 12 (x - 5) (x - 18)$ .
3Reject $x = 18$ (24 - 2x would be negative). Check $x = 5$ : $V^{''} = 24 x - 276 = - 156 < 0$ , a maximum.

Final answer

x = 5 cm

Step-by-step solution

1Let the width be $x$ and height of rectangle $h$ . Perimeter $3 x + 2 h = 12$ .
2Area $A = x h + \frac{3}{4} x^{2} = x (6 - \frac{3 x}{2}) + \frac{3}{4} x^{2}$ .
3 $A^{'} = 6 - 3 x + \frac{3}{2} x = 0 ⟹ x = \frac{12}{6 - 3}$ .

Final answer

x = \frac{12}{6 - 3}

Step-by-step solution

1Fixed volume $V = \frac{1}{3} π r^{2} h$ , curved surface $S = π r r^{2} + h^{2}$ .
2Substitute $h = \frac{3 V}{π r ^{2}}$ to get $S^{2} = π^{2} r^{2} (r^{2} + \frac{9 V ^{2}}{π ^{2} r ^{4}})$ .
3Minimising $S$ (via $S^{2}$ ) gives $π^{2} (2 r^{2} - \frac{9 V ^{2}}{π ^{2} r ^{4}}) = 0 ⟹ π r^{2} 2 = 3 V / r \dots h = 2 r$ .

Final answer

h = 2 r

Step-by-step solution

1Let one vertex be at angle $θ$ . The sides are $2 r cos θ$ and $2 r sin θ$ .
2Area $A = 4 r^{2} sin θ cos θ = 2 r^{2} sin 2 θ$ , maximised when $sin 2 θ = 1$ , i.e. $θ = \frac{π}{4}$ .
3Then both sides equal $r 2$ — a square. Max area $2 r^{2}$ .

Final answer

square of side r 2, area 2 r^{2}

Step-by-step solution

1 $P^{'} (x) = - 72 - 36 x = 0 ⟹ x = - 2$ .
2 $P^{''} (x) = - 36 < 0$ , so a maximum at $x = - 2$ : $P (- 2) = 41 + 144 - 72 = 113$ .

Final answer

113

Step-by-step solution

1 $f^{'} (x) = - 2 sin 2 x + 1 = 0 ⟹ sin 2 x = \frac{1}{2} ⟹ x = \frac{π}{12}$ .
2 $f (0) = 1, f (\frac{π}{12}) = \frac{3}{2} + \frac{π}{12}, f (\frac{π}{2}) = - 1 + \frac{π}{2}$ .

Final answer

max = 1 at x = 0;

Step-by-step solution

1Let base side be $x$ and height $h$ . $x^{2} h = 500$ .
2Surface area $S = x^{2} + 4 x h = x^{2} + \frac{2000}{x}$ .
3 $S^{'} = 2 x - \frac{2000}{x ^{2}} = 0 ⟹ x^{3} = 1000 ⟹ x = 10$ . Then $h = \frac{500}{100} = 5$ .

Final answer

base 10 m \times 10 m, height 5 m

Step-by-step solution

1Distance squared $D = x^{2} + (x^{2} - c)^{2}$ .
2 $D^{'} = 2 x + 4 x (x^{2} - c) = 2 x (1 + 2 x^{2} - 2 c) = 0 ⟹ x = 0$ or $x^{2} = c - \frac{1}{2}$ .
3If $c \leq \frac{1}{2}$ , minimum at $x = 0$ , distance $c$ . If $c > \frac{1}{2}$ , use $x^{2} = c - \frac{1}{2}$ . Distance $c - \frac{1}{4}$ .

Final answer

distance = {c, c - \frac{1}{4}, c \leq \frac{1}{2} c > \frac{1}{2}

Step-by-step solution

1 $f^{'} (x) = 6 x^{2} + 24 > 0$ throughout $[0, 3]$ , so $f$ is increasing.
2Maximum at $x = 3$ : $f (3) = 54 + 72 + 7 = 133$ .

Final answer

133

Step-by-step solution

1Total surface area $2 π r^{2} + 2 π r h = A$ .
2Volume $V = π r^{2} h = π r^{2} (\frac{A - 2 π r ^{2}}{2 π r}) = \frac{A r - 2 π r ^{3}}{2}$ .
3 $V^{'} = \frac{A}{2} - 3 π r^{2} = 0 ⟹ r^{2} = \frac{A}{6 π} ⟹ h = 2 r$ .
4Max volume $V = \frac{1}{3} π r^{2} (2 r) \dots = \frac{A ^{3/2}}{6 6 π}$ .

Final answer

V_{m a x} = \frac{A ^{3/2}}{6 6 π}

41

Miscellaneous Exercise — Mixed Problems on Application of Derivatives

15Exercise questions

Step-by-step solution

1Time of fall: $490 = \frac{1}{2} (9.8) t^{2} ⟹ t^{2} = 100 ⟹ t = 10 s$ .
2(a) Distance in 10th second: $s_{10} - s_{9} = \frac{1}{2} g (1 0^{2} - 9^{2}) = \frac{1}{2} (9.8) (19) = 93.1 m$ .
3(b) Striking velocity: $v = g t = 9.8 (10) = 98 m/s$ .

Final answer

(a) 93.1 m; (b) 98 m/s

Step-by-step solution

1 $V = \frac{4}{3} π r^{3} ⟹ \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} = 3$ .
2At $r = 2$ : $4 π (4) \frac{d r}{d t} = 3 ⟹ \frac{d r}{d t} = \frac{3}{16 π}$ .
3Surface area $S = 4 π r^{2}$ : $\frac{d S}{d t} = 8 π r \frac{d r}{d t} = 8 π (2) (\frac{3}{16 π}) = 3 cm^{2} / s$ .

Final answer

3 cm^{2} / s

Step-by-step solution

1 $f^{'} (x) = \frac{x ^{2}}{( 1 + x ) ( 2 + x ) ^{2}} \geq 0$ for all $x > 0$ .
2Since $f^{'} (x) > 0$ for $x > 0$ , $f$ is increasing on $(0, \infty)$ .

Final answer

f'(x) = x^2 / ((1+x)(2+x)^2) > 0 for x > 0, so f is increasing.

Step-by-step solution

1It cuts the $x$ -axis where $y = 0$ , i.e. $x = 7$ .
2 $y^{'} = \frac{( 1 ) ( x ^{2} - 5 x + 6 ) - ( x - 7 ) ( 2 x - 5 )}{( x ^{2} - 5 x + 6 ) ^{2}}$ , giving $y^{'} (7) = \frac{( 1 ) ( 20 ) - 0}{2 0 ^{2}} = \frac{1}{20}$ .
3Tangent: $y = \frac{1}{20} (x - 7) ⟹ x - 20 y - 7 = 0$ .

Final answer

x - 20 y - 7 = 0

Step-by-step solution

1 $y^{'} = 18 x - 4$ . Parallel to $y = x$ means $18 x - 4 = 1 ⟹ x = \frac{5}{18}$ .
2Then $y = 9 (\frac{25}{324}) - \frac{20}{18} + 5 = \frac{25}{36} - \frac{10}{9} + 5 = \frac{133}{36}$ .

Final answer

(\frac{5}{18}, \frac{133}{36})

Step-by-step solution

1 $y^{'} = 12 x^{2} - 10 x^{4}$ .
2Tangent at $(a, f (a))$ through origin: $\frac{f ( a )}{a} = f^{'} (a)$ .
3So $4 a^{2} - 2 a^{4} = 12 a^{2} - 10 a^{4} ⟹ 8 a^{4} - 8 a^{2} = 0 ⟹ a^{2} (a^{2} - 1) = 0$ , giving $a = 0, \pm 1$ .
4Points: $(0, 0), (1, 2), (- 1, - 2)$ .

Final answer

(0, 0), (1, 2), (- 1, - 2)

Step-by-step solution

1 $y = \frac{x ^{2}}{4}$ , so $y^{'} = \frac{x}{2}$ ; normal slope $= - \frac{2}{x}$ .
2Normal at $(t, t^{2} /4)$ : $y - \frac{t ^{2}}{4} = - \frac{2}{t} (x - t)$ .
3Through $(1, 2)$ : $2 - \frac{t ^{2}}{4} = - \frac{2}{t} (1 - t) ⟹ t^{3} - 8 t + 8 = 0$ .
4t = 2 is a root, giving normal $x + y = 3$ .

Final answer

x + y = 3

Step-by-step solution

1Fix $V = \frac{1}{3} π r^{2} h$ . Curved surface $S = π r r^{2} + h^{2}$ .
2Minimising $S^{2} (\propto r^{2} (r^{2} + h^{2}))$ subject to $r^{2} h = const$ gives $r^{2} (6 h) = 2 r (3 r^{2} + 3 h^{2}) \dots$ .
3The extremum yields $r^{2} = 2 h^{2}$ .

Final answer

r^{2} = 2 h^{2}

Step-by-step solution

1Take vertex $(a cos θ, b sin θ)$ . Then area $A = \frac{1}{2} (2 a cos θ) (b sin θ) \dots$ .
2Optimising $sin θ cos θ$ over $θ$ gives maximum $\frac{3 3}{4} ab$ .

Final answer

\frac{3 3}{4} ab

Step-by-step solution

1 $f^{'} (x) = \frac{6}{5} x^{3} - \frac{12}{5} x^{2} - 6 x + \frac{36}{5} = \frac{6}{5} (x - 2) (x - 3) (x + 1)$ .
2Sign of $f^{'} (x)$ : negative on $(- \infty, - 1) \cup (2, 3)$ , positive on $(- 1, 2) \cup (3, \infty)$ .

Final answer

inc on (- 1, 2) \cup (3, \infty); dec on (- \infty, - 1) \cup (2, 3)

Step-by-step solution

1 $y^{'} = 2 x - 1 = 0 ⟹ x = \frac{1}{2}$ .
2 $y = \frac{1}{4} - \frac{1}{2} - 2 = - \frac{9}{4}$ .

Final answer

(\frac{1}{2}, - \frac{9}{4})

Step-by-step solution

1 $x^{2} + y^{2} = 2500$ : $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$ .
2At $x = 30$ , $y = 40$ : $30 (2) + 40 \frac{d y}{d t} = 0 ⟹ \frac{d y}{d t} = - \frac{3}{2} m/s$ .

Final answer

\frac{3}{2} m/s downwards

Step-by-step solution

1 $V = x^{3} ⟹ Δ V \approx 3 x^{2} Δ x = 3 (25) (0.2) = 15 cm^{3}$ .

Final answer

15 cm^{3}

Step-by-step solution

1 $y^{'} = 3 x^{2} - 12 x + 12 = 3 (x - 2)^{2}$ , $y^{''} = 6 x - 12 = 6 (x - 2)$ .
2At $x = 2$ : $y^{''} = 0$ and $y^{''}$ changes sign (negative before, positive after), so $x = 2$ is an inflection point.

Final answer

y'' changes sign at x = 2, confirming a point of inflection.

Step-by-step solution

1 $f^{'} (x) = 2 cos x - 2 sin 2 x = 2 cos x (1 - 2 sin x) = 0$ .
2So $cos x = 0$ or $sin x = \frac{1}{2}$ .
3Checking values: max at $x = \frac{π}{6}, \frac{5 π}{6}$ with $f = \frac{3}{2}$ .

Final answer

\frac{3}{2}

42

Chapter 7 — Integrals

Integration is one of the two fundamental operations of calculus, the other being differentiation. Since differentiation and integration are inverse processes, every differentiation formula gives a corresponding integration formula. This chapter begins with antiderivatives, progresses through techniques — substitution, trigonometric identities, partial fractions, and integration by parts — and culminates in definite integrals, the Fundamental Theorem of Calculus, and properties that make evaluation tractable. The exercises below carry every question of the NCERT textbook with short, exam-pattern working.

Board pattern

Integration questions award marks for the technique chosen, intermediate steps, and the final answer with an arbitrary constant C. Always write the substitution explicitly (e.g. Let u = x^2), show the transformed integral, integrate, and back-substitute. For definite integrals, show the antiderivative evaluated at both limits.

$Ex 7.1 — Integration as inverse of differentiation: basic antiderivatives (22 Q)$
$Ex 7.2 — Integration by substitution (22 Q)$
$Ex 7.3 — Integration using trigonometric identities (14 Q)$
$Ex 7.4 — Integration of particular functions: partial fractions I (11 Q)$
$Ex 7.5 — Integration by partial fractions (continued) (15 Q)$
$Ex 7.6 — Integration by parts (15 Q)$
$Ex 7.7 — Definite integrals: fundamental theorem (8 Q)$
$Ex 7.8 — Evaluation of definite integrals by substitution (8 Q)$
$Ex 7.9 — Some properties of definite integrals (9 Q)$
$Ex 7.10 — More definite integral problems (8 Q)$
$Ex 7.11 — Properties and evaluation (10 Q)$
$Misc — Mixed problems on integration (20 Q)$

43

Exercise 7.1 — Integration as Inverse of Differentiation

18Exercise questions

Final answer

(i) - \frac{1}{2} cos 2 x + C; (ii) \frac{1}{3} sin 3 x + C; (iii) \frac{1}{2} e^{2 x} + C; (i v) - \frac{1}{π} cos π x + C; (v) \frac{1}{π} sin π x + C

Final answer

(i) \frac{1}{4} sin 4 x + C; (ii) ln ∣ cos x ∣ + C

Final answer

(i) \frac{4}{3} e^{3 x} + x + C; (ii) \frac{x ^{3}}{3} - x + C; (iii) \frac{a}{3} x^{3} + \frac{b}{2} x^{2} + c x + C

Step-by-step solution

1(i) Use $sin^{2} θ = \frac{1 - c o s 2 θ}{2}$ :
2Let $θ = 2 x + 5$ : $\int \frac{1 - c o s ( 4 x + 10 )}{2} d x = \frac{x}{2} - \frac{s i n ( 4 x + 10 )}{8} + C$
3(ii) Use $2 sin A cos B = sin (A + B) + sin (A - B)$

Final answer

(i) \frac{x}{2} - \frac{s i n ( 4 x + 10 )}{8} + C; (ii) - \frac{1}{14} cos 7 x + \frac{1}{2} cos x + C

Step-by-step solution

1Multiply numerator and denominator by conjugate, then separate and integrate.

Final answer

(i) x - tan x + sec x + C; (ii) tan x - x + tan \frac{x}{2} + C

Step-by-step solution

1Perform polynomial long division, then integrate term by term.

Final answer

(i) \frac{x ^{3}}{3} + C; (ii) \frac{x ^{2}}{2} + 3 x - 2 ln ∣ x + 2∣ + C; (iii) \frac{x ^{2}}{2} - 2 x + 3 ln ∣ x + 1∣ + C

Final answer

\frac{x ^{2}}{2} + 3 sin x + 2 cos x - 2 ln ∣ x ∣ + C

Final answer

\frac{x ^{3}}{3} - x + ln ∣ x ∣ + C

Step-by-step solution

1Note $sin^{2} x + cos^{2} x = 1$ and $2 sin x cos x = sin 2 x$
2So $1 - sin 2 x = (sin x - cos x)^{2}$ , hence
3 $\frac{s i n x + c o s x}{∣ s i n x - c o s x ∣}$
4For $sin x > cos x$ : $\frac{s i n x + c o s x}{s i n x - c o s x}$ ; integrate by letting $u = sin x - cos x$

Final answer

- ln ∣ sin x - cos x ∣ + C

Step-by-step solution

1Let $u = 1 + sin x$ , $d u = cos x d x$

Final answer

2 1 + sin x + C

Final answer

sec x + tan x + C

Step-by-step solution

1Split: $\frac{1}{c o s ^{2} x} - \frac{1}{s i n ^{2} x} = sec^{2} x - csc^{2} x$

Final answer

tan x + cot x + C

Step-by-step solution

1Rewrite as $1 - (csc^{2} x - 1) = 2 - csc^{2} x$

Final answer

2 x + cot x + C

Step-by-step solution

1Use $tan^{2} x = sec^{2} x - 1$

Final answer

tan x - x + C

Step-by-step solution

1This is $\frac{1/ c o s ^{2} x}{1/ s i n ^{2} x} = tan^{2} x$

Final answer

tan x - x + C

Step-by-step solution

1Rewrite as $\frac{c o s x}{s i n x}$

Final answer

ln ∣ sin x ∣ + C

Final answer

ln ∣ csc x - cot x ∣ + C

Step-by-step solution

1Standard form: multiply by $\frac{s e c x + t a n x}{s e c x + t a n x}$

Final answer

ln ∣ sec x + tan x ∣ + C

44

Exercise 7.2 — Integration by Substitution

20Exercise questions

Step-by-step solution

1Let $u = x + 1$ , so $x = u - 1$ , $d x = d u$
2 $\int \frac{u - 1}{u} d u = u - ln ∣ u ∣ + C = x + 1 - ln ∣ x + 1∣ + C$

Final answer

x - ln ∣ x + 1∣ + C

Step-by-step solution

1Multiply by $\frac{x + x - 1}{x + x - 1}$ :
2 $\int (x + x - 1) d x = \frac{2}{3} x^{3/2} + \frac{2}{3} (x - 1)^{3/2} + C$

Final answer

\frac{2}{3} (x^{3/2} + (x - 1)^{3/2}) + C

Step-by-step solution

1Partial fractions: $\frac{1}{x ( x ^{2} + 1 )} = \frac{1}{x} - \frac{x}{x ^{2} + 1}$

Final answer

ln ∣ x ∣ - \frac{1}{2} ln (x^{2} + 1) + C

Step-by-step solution

1Write $sin^{3} x = sin x (1 - cos^{2} x)$ and let $u = cos x$

Final answer

- cos x + \frac{cos ^{3} x}{3} + C

Step-by-step solution

1Let $u = cos x$ , $d u = - sin x d x$

Final answer

- cos (cos x) + C

Step-by-step solution

1Let $u = x - 1$ , so $x = u^{2} + 1$ , $d x = 2 u d u$
2 $\int \frac{( u ^{2} + 1 ) ^{2}}{u} \cdot 2 u d u = 2 \int (u^{2} + 1)^{2} d u$

Final answer

\frac{2}{5} (x - 1)^{5/2} + \frac{4}{3} (x - 1)^{3/2} + 2 x - 1 + C

Step-by-step solution

1Let $u = sin x$ , $d u = cos x d x$
2Partial fractions: $\frac{1}{( 1 + u ) ( 2 + u )} = \frac{1}{1 + u} - \frac{1}{2 + u}$

Final answer

ln \frac{1 + sin x}{2 + sin x} + C

Step-by-step solution

1Let $u = x^{2} + 1$ , $d u = 2 x d x$

Final answer

x^{2} + 1 + C

Step-by-step solution

1Let $u = x^{2} + 1$ , $x^{2} = u - 1$ , $x d x = \frac{d u}{2}$

Final answer

\frac{1}{3} (x^{2} + 1)^{3/2} - (x^{2} + 1)^{1/2} + C

Step-by-step solution

1Use $cos 2 x = 2 cos^{2} x - 1$

Final answer

2 sin x - ln ∣ sec x + tan x ∣ + C

Step-by-step solution

1For $x \in (0, π)$ : $sin^{- 1} (cos x) = \frac{π}{2} - x$

Final answer

\frac{π}{2} x - \frac{x ^{2}}{2} + C

Step-by-step solution

1Let $u = 1 - x^{2}$ , $x^{2} = 1 - u$ , $x d x = - \frac{d u}{2}$

Final answer

- \frac{1}{3} (1 - x^{2})^{3/2} + (1 - x^{2})^{1/2} + C

Step-by-step solution

1Let $u = cos^{- 1} x$ , $d u = - \frac{d x}{1 - x ^{2}}$

Final answer

- \frac{( cos ^{- 1} x ) ^{2}}{2} + C

Step-by-step solution

1For $x > 0$ : $tan^{- 1} (1/ x) = \frac{π}{2} - tan^{- 1} x$

Final answer

\frac{π}{2} x - x tan^{- 1} x + \frac{1}{2} ln (1 + x^{2}) + C

Step-by-step solution

1Let $u = 1 + x^{2}$ , $u^{2} = 1 + x^{2}$ , $u d u = x d x$
2Then $\int \frac{u}{x} \cdot \frac{u d u}{x} = \int \frac{u ^{2}}{u ^{2} - 1} d u$

Final answer

1 + x^{2} - ln \frac{1 + x ^{2} + 1}{x} + C

Step-by-step solution

1Simplify: $\frac{x ^{2} - 1}{x ( x ^{2} + 1 )} = \frac{x}{x ^{2} + 1} - \frac{1}{x ( x ^{2} + 1 )}$

Final answer

ln ∣ x ∣ - \frac{1}{2} ln (x^{2} + 1) + C

Step-by-step solution

1Let $u = 1 + sin x$ , $d u = cos x d x$
2Rewrite integrand as $\frac{s i n x}{c o s x 1 + s i n x}$ and use substitution.

Final answer

2 (\frac{1}{1 + sin x} - ln ∣ 1 + sin x + 1∣) + C

Step-by-step solution

1Let $u = 1 + cos x$ , $d u = - sin x d x$

Final answer

\frac{1}{1 + cos x} + C

Step-by-step solution

1Write $sin x = A (sin x + cos x) + B (cos x - sin x)$ to find constants.

Final answer

\frac{x}{2} - \frac{1}{4} ln ∣1 + sin 2 x ∣ + C

Step-by-step solution

1Let $u = sin^{- 1} (x^{2})$ , $d u = \frac{2 x}{1 - x ^{4}} d x$

Final answer

\frac{( sin ^{- 1} x ^{2} ) ^{2}}{4} + C

45

Exercise 7.3 — Integration Using Trigonometric Identities

12Exercise questions

Step-by-step solution

1Use $sin^{2} x = \frac{1 - c o s 2 x}{2}$

Final answer

\frac{x}{2} - \frac{sin 2 x}{4} + C

Step-by-step solution

1Use $cos^{2} x = \frac{1 + c o s 2 x}{2}$

Final answer

\frac{x}{2} + \frac{sin 2 x}{4} + C

Step-by-step solution

1Write $sin^{3} x = sin x (1 - cos^{2} x)$

Final answer

- cos x + \frac{cos ^{3} x}{3} + C

Step-by-step solution

1Write $cos^{3} x = cos x (1 - sin^{2} x)$

Final answer

sin x - \frac{sin ^{3} x}{3} + C

Step-by-step solution

1Use $sin^{2} x = \frac{1 - c o s 2 x}{2}$ then square it.

Final answer

\frac{3 x}{8} - \frac{sin 2 x}{4} + \frac{sin 4 x}{32} + C

Step-by-step solution

1Use $sin^{2} x = 1 - cos^{2} x = (1 - cos x) (1 + cos x)$

Final answer

cos x + C

Step-by-step solution

1Use $tan^{2} x = sec^{2} x - 1$ and $sec^{2} x = 1 + tan^{2} x$

Final answer

\frac{tan ^{3} x}{3} - tan x + x + C

Step-by-step solution

1Split into $\frac{s i n x}{c o s ^{2} x} + \frac{c o s x}{s i n ^{2} x} = sec x tan x + csc x cot x$

Final answer

sec x - csc x + C

Step-by-step solution

1Rewrite as $\frac{c o s x}{s i n ^{2} x} = csc x cot x$

Final answer

- csc x + C

Step-by-step solution

1Use $sin^{2} x cos^{2} x = \frac{s i n ^{2} 2 x}{4} = \frac{1 - c o s 4 x}{8}$

Final answer

\frac{x}{8} - \frac{sin 4 x}{32} + C

Step-by-step solution

1Factor $cos 2 x = (cos x + sin x) (cos x - sin x)$

Final answer

\frac{sin x - cos x}{cos x + sin x} + C

Step-by-step solution

1Let $u = tan x$ , $d u = sec^{2} x d x$

Final answer

\frac{tan ^{5} x}{5} + \frac{tan ^{3} x}{3} + C

46

Exercise 7.4 — Integration of Particular Functions

11Exercise questions

Step-by-step solution

1Factor: $x^{2} + 3 x - 10 = (x + 5) (x - 2)$
2Partial fractions: $\frac{3 x + 5}{( x + 5 ) ( x - 2 )} = \frac{A}{x + 5} + \frac{B}{x - 2}$ , find $A = - 1, B = 2$

Final answer

- ln ∣ x + 5∣ + 2 ln ∣ x - 2∣ + C

Step-by-step solution

1Long division: $1 + \frac{- 4 x - 4}{( x + 1 ) ( x + 3 )}$
2Partial fractions on the remainder.

Final answer

x + \frac{2}{x + 1} - \frac{6}{x + 3} + C wait, partial fractions give: x - 2 ln ∣ x + 1∣ + ln ∣ x + 3∣ + C

Step-by-step solution

1Factor and use partial fractions: $\frac{x}{( x ^{2} + 1 ) ( x - 1 )} = \frac{A}{x - 1} + \frac{B x + C}{x ^{2} + 1}$ , $A = \frac{1}{2}, B = - \frac{1}{2}, C = \frac{1}{2}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - \frac{1}{4} ln (x^{2} + 1) + \frac{1}{2} tan^{- 1} x + C

Step-by-step solution

1Partial fractions: $\frac{1}{( x - 1 ) ( x ^{2} + 1 )} = \frac{1/2}{x - 1} + \frac{- x /2 + 1/2}{x ^{2} + 1}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - \frac{1}{4} ln (x^{2} + 1) + \frac{1}{2} tan^{- 1} x + C

Step-by-step solution

1Partial fractions: $\frac{x}{( x - 1 ) ( x - 2 ) ( x - 3 )} = \frac{1/2}{x - 1} - \frac{2}{x - 2} + \frac{3/2}{x - 3}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - 2 ln ∣ x - 2∣ + \frac{3}{2} ln ∣ x - 3∣ + C

Step-by-step solution

1Partial fractions: $\frac{2 x + 3}{( x + 1 ) ( x + 2 )} = \frac{1}{x + 1} + \frac{1}{x + 2}$

Final answer

ln ∣ x + 1∣ + ln ∣ x + 2∣ + C

Step-by-step solution

1Write $\frac{1 - x ^{2}}{x ( 1 + x ^{2} )} = \frac{1}{x} - \frac{2 x}{1 + x ^{2}}$

Final answer

ln ∣ x ∣ - ln (1 + x^{2}) + C

Step-by-step solution

1Long division: $x + \frac{- x - 1}{( x + 1 ) ( x + 2 )} = x + \frac{- 1}{x + 2}$

Final answer

\frac{x ^{2}}{2} - ln ∣ x + 2∣ + C

Step-by-step solution

1Long division: $5 + \frac{- 2 x - 4}{x ^{2} + x + 1}$
2Complete the square for the remainder integral.

Final answer

5 x - ln (x^{2} + x + 1) + \frac{4}{3} tan^{- 1} \frac{2 x + 1}{3} + C

Step-by-step solution

1Partial fractions: $\frac{1}{x ( x ^{2} + 1 )} = \frac{1}{x} - \frac{x}{x ^{2} + 1}$

Final answer

ln ∣ x ∣ - \frac{1}{2} ln (x^{2} + 1) + C

Step-by-step solution

1Partial fractions: $\frac{1}{( x ^{2} + 1 ) ( x ^{2} + 4 )} = \frac{1}{3} (\frac{1}{x ^{2} + 1} - \frac{1}{x ^{2} + 4})$

Final answer

\frac{1}{3} tan^{- 1} x - \frac{1}{6} tan^{- 1} \frac{x}{2} + C

47

Exercise 7.5 — Integration by Partial Fractions

13Exercise questions

Step-by-step solution

1Partial fractions: $\frac{x}{( x + 1 ) ( x + 2 )} = \frac{- 1}{x + 1} + \frac{2}{x + 2}$

Final answer

- ln ∣ x + 1∣ + 2 ln ∣ x + 2∣ + C

Step-by-step solution

1Factor: $x^{2} - 9 = (x - 3) (x + 3)$
2 $\frac{1}{( x - 3 ) ( x + 3 )} = \frac{1}{6} (\frac{1}{x - 3} - \frac{1}{x + 3})$

Final answer

\frac{1}{6} ln \frac{x - 3}{x + 3} + C

Step-by-step solution

1Partial fractions: $\frac{3 x - 1}{( x - 1 ) ( x - 2 ) ( x - 3 )} = \frac{1}{x - 1} - \frac{5}{x - 2} + \frac{4}{x - 3}$

Final answer

ln ∣ x - 1∣ - 5 ln ∣ x - 2∣ + 4 ln ∣ x - 3∣ + C

Step-by-step solution

1Partial fractions: $\frac{x ^{2} + 1}{( x - 1 ) ^{2} ( x + 3 )} = \frac{1}{x + 3} + \frac{3}{( x - 1 ) ^{2}}$

Final answer

ln ∣ x + 3∣ - \frac{3}{x - 1} + C

Step-by-step solution

1Partial fractions: $\frac{1 - x ^{2}}{x ( 1 + x ^{2} )} = \frac{1}{x} - \frac{2 x}{1 + x ^{2}}$

Final answer

ln ∣ x ∣ - ln (1 + x^{2}) + C

Step-by-step solution

1Partial fractions: $\frac{x}{( x - 1 ) ( x - 2 ) ( x - 3 )} = \frac{1/2}{x - 1} - \frac{2}{x - 2} + \frac{3/2}{x - 3}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - 2 ln ∣ x - 2∣ + \frac{3}{2} ln ∣ x - 3∣ + C

Step-by-step solution

1Partial fractions: $\frac{x}{( x + 1 ) ( x ^{2} + 1 )} = \frac{- 1/2}{x + 1} + \frac{x /2 + 1/2}{x ^{2} + 1}$

Final answer

- \frac{1}{2} ln ∣ x + 1∣ + \frac{1}{4} ln (x^{2} + 1) + \frac{1}{2} tan^{- 1} x + C

Step-by-step solution

1Partial fractions: $\frac{x ^{2} - 3 x + 2}{x ( x - 1 ) ^{2}} = \frac{2}{x} - \frac{1}{x - 1} - \frac{1}{( x - 1 ) ^{2}}$

Final answer

2 ln ∣ x ∣ - ln ∣ x - 1∣ + \frac{1}{x - 1} + C

Step-by-step solution

1Long division: $1 + \frac{5 x - 8}{x ^{2} + 4}$
2Split: $\frac{5 x}{x ^{2} + 4} - \frac{8}{x ^{2} + 4}$

Final answer

x + \frac{5}{2} ln (x^{2} + 4) - 4 tan^{- 1} \frac{x}{2} + C

Step-by-step solution

1Factor: $x^{4} - 1 = (x^{2} - 1) (x^{2} + 1) = (x - 1) (x + 1) (x^{2} + 1)$
2Partial fractions.

Final answer

\frac{1}{4} ln \frac{x - 1}{x + 1} - \frac{1}{2} tan^{- 1} x + C

Step-by-step solution

1Partial fractions: $\frac{2 x - 1}{( x - 1 ) ( x - 2 ) ( x - 3 )} = \frac{1/2}{x - 1} - \frac{3}{x - 2} + \frac{5/2}{x - 3}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - 3 ln ∣ x - 2∣ + \frac{5}{2} ln ∣ x - 3∣ + C

Step-by-step solution

1Partial fractions in $u = x^{2}$ : $\frac{u}{( u + 1 ) ( u + 4 )} = \frac{- 1/3}{u + 1} + \frac{4/3}{u + 4}$

Final answer

- \frac{1}{3} tan^{- 1} x + \frac{4}{3} \cdot \frac{1}{2} tan^{- 1} \frac{x}{2} + C = - \frac{1}{3} tan^{- 1} x + \frac{2}{3} tan^{- 1} \frac{x}{2} + C

Step-by-step solution

1Long division: $1 + \frac{2}{x ^{2} - 1}$
2Partial fractions: $\frac{2}{( x - 1 ) ( x + 1 )} = \frac{1}{x - 1} - \frac{1}{x + 1}$

Final answer

x + ln \frac{x - 1}{x + 1} + C

48

Exercise 7.6 — Integration by Parts

12Exercise questions

Step-by-step solution

1By parts: $u = x, d v = sin x d x$ → $d u = d x, v = - cos x$

Final answer

- x cos x + sin x + C

Step-by-step solution

1By parts: $u = x, d v = sin 3 x d x$

Final answer

- \frac{x cos 3 x}{3} + \frac{sin 3 x}{9} + C

Step-by-step solution

1By parts twice.

Final answer

x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + C

Step-by-step solution

1By parts: $u = x, d v = e^{x} d x$

Final answer

x e^{x} - e^{x} + C

Step-by-step solution

1By parts: $u = lo g x, d v = x d x$

Final answer

\frac{x ^{2}}{2} lo g x - \frac{x ^{2}}{4} + C

Step-by-step solution

1By parts: $u = lo g x, d v = x^{2} d x$

Final answer

\frac{x ^{3}}{3} lo g x - \frac{x ^{3}}{9} + C

Step-by-step solution

1By parts: $u = sin^{- 1} x, d v = x d x$

Final answer

\frac{x ^{2}}{2} sin^{- 1} x - \frac{1}{4} (x 1 - x^{2} + sin^{- 1} x) + C

Step-by-step solution

1By parts: $u = cos^{- 1} x, d v = x d x$

Final answer

\frac{x ^{2}}{2} cos^{- 1} x + \frac{1}{4} (x 1 - x^{2} - cos^{- 1} x) + C

Step-by-step solution

1Recognise $\frac{d}{d x} (\frac{e ^{x}}{x}) = \frac{e ^{x}}{x} - \frac{e ^{x}}{x ^{2}}$

Final answer

\frac{e ^{x}}{x} + C

Step-by-step solution

1Recognise $\frac{d}{d x} (\frac{e ^{x}}{x}) = \frac{x e ^{x} - e ^{x}}{x ^{2}}$

Final answer

- \frac{e ^{x}}{x} + C

Step-by-step solution

1Rewrite as $\int e^{x} (\frac{x - 1}{x + 1} + \frac{2}{( x + 1 ) ^{2}}) d x$ or recognise the standard form.

Final answer

\frac{e ^{x} ( x - 1 )}{x + 1} + C

Step-by-step solution

1Simplify the rational function, then use the formula $\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + C$

Final answer

e^{x} (x - 2) + C (after simplification)

49

Exercise 7.7 — Definite Integrals

8Exercise questions

Step-by-step solution

1Use $cos^{2} x = \frac{1 + c o s 2 x}{2}$ :
2 $[\frac{x}{2} + \frac{s i n 2 x}{4}]_{0}^{π /2} = \frac{π}{4}$

Final answer

\frac{π}{4}

Step-by-step solution

1By symmetry with $cos^{2} x$ , the answer is the same.

Final answer

\frac{π}{4}

Step-by-step solution

1Write $sin^{3} x = sin x (1 - cos^{2} x)$ , let $u = cos x$

Final answer

\frac{2}{3}

Step-by-step solution

1By symmetry with $sin^{3} x$

Final answer

\frac{2}{3}

Step-by-step solution

1Use power-reduction formulas twice.

Final answer

\frac{3 π}{8}

Step-by-step solution

1Use power-reduction: $sin^{4} x cos^{2} x = \frac{1 - c o s 2 x}{2} \cdot \frac{1 - c o s 2 x}{2} \cdot \frac{1 + c o s 2 x}{2}$

Final answer

\frac{π}{32}

Step-by-step solution

1Use $sin^{4} x = (\frac{1 - c o s 2 x}{2})^{2}$

Final answer

\frac{3 π}{4}

Step-by-step solution

1Reduction: $sin^{5} x = sin x (1 - cos^{2} x)^{2}$ , let $u = cos x$

Final answer

\frac{8}{15}

50

Exercise 7.8 — Evaluation of Definite Integrals by Substitution

8Exercise questions

Step-by-step solution

1Let $u = x + 1$ :
2 $\int_{1}^{2} \frac{u - 1}{u} d u = [u - ln u]_{1}^{2} = 1 - ln 2$

Final answer

1 - ln 2

Step-by-step solution

1Let $u = x^{2} + 1$ :
2 $\frac{1}{2} \int_{1}^{3} u^{- 1/2} d u = [u]_{1}^{3} = 3 - 1$

Final answer

3 - 1

Step-by-step solution

1Long division: $1 - \frac{4}{x ^{2} + 4}$

Final answer

2 - 2 tan^{- 1} 1 = 2 - \frac{π}{2}

Step-by-step solution

1Let $u = tan^{- 1} x$ , $d u = \frac{d x}{1 + x ^{2}}$

Final answer

\frac{π ^{2}}{32}

Step-by-step solution

1Standard integral: $\int \frac{d x}{1 - x ^{2}} = sin^{- 1} x$

Final answer

\frac{π}{2}

Step-by-step solution

1Antiderivative: $tan^{- 1} x$

Final answer

\frac{π}{4}

Step-by-step solution

1Let $u = x^{3}$ , $d u = 3 x^{2} d x$

Final answer

\frac{1}{3} tan^{- 1} 1 = \frac{π}{12}

Step-by-step solution

1Simplify: $\frac{x}{2 x} = \frac{1}{2}$

Final answer

\frac{1}{2}

51

Exercise 7.9 — Some Properties of Definite Integrals

8Exercise questions

Step-by-step solution

1Let $t = a + b - x$ : $d x = - d t$
2When $x = a, t = b$ ; when $x = b, t = a$
3 $\int_{b}^{a} f (t) (- d t) = \int_{a}^{b} f (t) d t$

Final answer

Proved.

Step-by-step solution

1Apply the property with $a = 0, b = π /2$ :
2 $I = \int_{0}^{π /2} \frac{c o s ^{n} x}{c o s ^{n} x + s i n ^{n} x} d x$ ; add: $2 I = \int_{0}^{π /2} 1 d x = \frac{π}{2}$

Final answer

\frac{π}{4}

Step-by-step solution

1Use the property with $a = 0, b = π$

Final answer

\frac{π ^{2}}{4}

Step-by-step solution

1Let $u = 1 - x$ and use the property.

Final answer

- 1

Step-by-step solution

1Use property: $\int_{0}^{π} x f (sin x) d x = \frac{π}{2} \int_{0}^{π} f (sin x) d x$

Final answer

\frac{π ^{2}}{4}

Step-by-step solution

1The integrand has period $π$ ; over a full period of an odd power of cosine, the integral is zero.

Final answer

0

Step-by-step solution

1Let $u = cos x$ :
2 $\int_{0}^{1} u^{3} (1 - u^{2}) d u = \int_{0}^{1} (u^{3} - u^{5}) d u$

Final answer

\frac{1}{4} - \frac{1}{6} = \frac{1}{12}

Step-by-step solution

1By the same property as the $sin^{n} x$ problem, the answer is always $π /4$

Final answer

\frac{π}{4}

52

Exercise 7.10 — More Definite Integral Problems

8Exercise questions

Step-by-step solution

1Use $sin^{4} x = (\frac{1 - c o s 2 x}{2})^{2}$

Final answer

\frac{3 π}{16}

Step-by-step solution

1By symmetry: same as $\int_{0}^{π /2} sin^{4} x d x$

Final answer

\frac{3 π}{16}

Step-by-step solution

1Use reduction formula.

Final answer

\frac{5 π}{32}

Step-by-step solution

1Use $sin^{3} x = sin x (1 - cos^{2} x)$

Final answer

\frac{4}{3}

Step-by-step solution

1Odd power of sine over full period: the integral is zero by symmetry.

Final answer

0

Step-by-step solution

1Odd function integrated over symmetric interval: zero.

Final answer

0

Step-by-step solution

1Use the property: add with $\frac{s i n ^{3} x}{s i n ^{3} x + c o s ^{3} x}$

Final answer

\frac{π}{4}

Step-by-step solution

1Use property: $\int_{0}^{π} x f (sin x) d x = \frac{π}{2} \int_{0}^{π} f (sin x) d x$
2 $= \frac{π}{2} \int_{0}^{π} \frac{s i n x}{1 + c o s ^{2} x} d x$
3Let $u = cos x$ : $= \frac{π}{2} \cdot \frac{π}{2} = \frac{π ^{2}}{4}$

Final answer

\frac{π ^{2}}{4}

53

Exercise 7.11 — Properties of Definite Integrals

9Exercise questions

Step-by-step solution

1Use property: $\int_{0}^{π /2} lo g sin x d x = \int_{0}^{π /2} lo g cos x d x$
2Add: $2 I = \int_{0}^{π /2} lo g (\frac{s i n 2 x}{2}) d x$
3Substitute $u = 2 x$ and simplify.

Final answer

- \frac{π}{2} ln 2

Step-by-step solution

1By the property just proved, same as the above.

Final answer

- \frac{π}{2} ln 2

Step-by-step solution

1Use $I = \int_{0}^{π} x f (sin x) d x = \frac{π}{2} \int_{0}^{π} f (sin x) d x$
2 $= \frac{π}{2} \int_{0}^{π} \frac{s i n x}{1 + c o s ^{2} x} d x = \frac{π}{2} \cdot \frac{π}{2} = \frac{π ^{2}}{4}$

Final answer

\frac{π ^{2}}{4}

Step-by-step solution

1Use the substitution property: add the integral with $cos x$ in numerator.

Final answer

\frac{π}{4}

Step-by-step solution

1Recognise: $\frac{d}{d x} (\frac{e ^{2 x}}{2 x}) = \frac{e ^{2 x}}{x} - \frac{e ^{2 x}}{2 x ^{2}}$

Final answer

\frac{e ^{4}}{4} - \frac{e ^{2}}{2}

Step-by-step solution

1Use the property $\int_{0}^{π} f (cos x) d x = \int_{0}^{π} f (- cos x) d x$ and add.

Final answer

- π ln 2

Step-by-step solution

1Periodicity and symmetry: double the integral over $[0, π]$

Final answer

\frac{π}{8}

Step-by-step solution

1Use property with $f (x) = \frac{t a n x}{s e c x + t a n x}$

Final answer

\frac{π}{2} (\frac{π}{2} - ln 2)

Step-by-step solution

1Let $I = \int_{0}^{π /2} \frac{c o s ^{α} x}{c o s ^{α} x + s i n ^{α} x} d x$
2Apply property $I = \int_{0}^{π /2} \frac{s i n ^{α} x}{s i n ^{α} x + c o s ^{α} x} d x$
3Add: $2 I = \frac{π}{2}$

Final answer

I = \frac{π}{4}

54

Miscellaneous Exercise on Chapter 7

14Exercise questions

Step-by-step solution

1Divide by $x^{2}$ : $\frac{1 + 1/ x ^{2}}{x ^{2} + 1/ x ^{2}}$
2Let $u = x - 1/ x$ : $\int \frac{d u}{u ^{2} + 2} = \frac{1}{2} tan^{- 1} \frac{u}{2}$

Final answer

\frac{1}{2} tan^{- 1} \frac{x - 1/ x}{2} + C

Step-by-step solution

1Factor: $x^{4} + 1 = (x^{2} + 2 x + 1) (x^{2} - 2 x + 1)$
2Partial fractions, then complete the square in each quadratic.

Final answer

\frac{1}{2 2} (tan^{- 1} \frac{x 2 + 1}{2} - tan^{- 1} \frac{x 2 - 1}{2}) + C

Step-by-step solution

1Divide by $x^{2}$ : $\frac{1 - 1/ x ^{2}}{x ^{2} + 1/ x ^{2}}$
2Let $u = x + 1/ x$ : $\int \frac{d u}{u ^{2} - 2} = \frac{1}{2 2} ln \frac{u - 2}{u + 2}$

Final answer

\frac{1}{2 2} ln \frac{x ^{2} - 2 x + 1}{x ^{2} + 2 x + 1} + C

Step-by-step solution

1Recognise: $\frac{d}{d x} (\frac{e ^{x}}{2 + 3 x}) = \frac{e ^{x} ( 2 + 3 x ) - 3 e ^{x}}{( 2 + 3 x ) ^{2}} = \frac{e ^{x} ( 3 x - 1 )}{( 2 + 3 x ) ^{2}}$
2Adjust coefficients to match.

Final answer

\frac{e ^{x}}{3 ( 2 + 3 x )} + C

Step-by-step solution

1Simplify: $\frac{1 + s i n x}{1 + c o s x} = \frac{( s i n \frac{x}{2} + c o s \frac{x}{2} ) ^{2}}{2 c o s ^{2} \frac{x}{2}}$

Final answer

e^{x} tan \frac{x}{2} + C

Step-by-step solution

1Simplify similarly.

Final answer

- e^{x} cot \frac{x}{2} + C

Step-by-step solution

1Multiply by $\frac{1 - s i n x}{1 - s i n x}$

Final answer

tan x - sec x + C

Step-by-step solution

1Use $1 + sin 2 x = (sin x + cos x)^{2}$

Final answer

- cos x + sin x + C

Step-by-step solution

1Long division and partial fractions.

Final answer

\frac{x ^{2}}{2} - x + \frac{1}{2} ln (x^{2} + 1) + \frac{1}{2} ln ∣ x + 1∣ + C

Step-by-step solution

1Partial fractions: $\frac{1}{x - 1} - \frac{5}{x - 2} + \frac{4}{x - 3}$

Final answer

ln ∣ x - 1∣ - 5 ln ∣ x - 2∣ + 4 ln ∣ x - 3∣ + C

Step-by-step solution

1Let $u = x^{2}$ , $d u = 2 x d x$

Final answer

\frac{e - 1}{2}

Step-by-step solution

1Use $sin x cos x = \frac{s i n 2 x}{2}$

Final answer

\frac{3 π}{256}

Step-by-step solution

1Let $u = sin^{- 1} x$ , $d u = \frac{d x}{1 - x ^{2}}$

Final answer

\frac{( sin ^{- 1} x ) ^{2}}{2} + C

Step-by-step solution

1Use $sin^{2} x = \frac{1 - c o s 2 x}{2}$

Final answer

\frac{π}{2}

55

Chapter 8 — Application of Integrals

Definite integrals are not just algebraic exercises — they measure real geometric quantities. The area under a curve, the area enclosed between two curves, and the area of regions bounded by lines and conic sections all fall within this chapter. You already know how to evaluate definite integrals; here you learn to set up the correct limits, decide which curve lies on top, and split regions at intersection points. Every question below is from the NCERT textbook and board pattern, solved step-by-step.

Board pattern

Marks are always for the working, not the answer. Always (a) sketch the region, (b) find the intersection points, (c) write the integral with correct limits and correct upper curve, and (d) evaluate step by step. For curves that cross the x-axis, split the integral and take the absolute values of negative portions. State every final area in square units, and conclude with the exact answer.

$Ex 8.1 — Area under simple curves: parabolas, ellipses, sine curves, lines (13 Q)$
$Ex 8.2 — Area enclosed between a curve and a line, and between two curves (6 Q)$
$Misc — Mixed problems on application of integrals (10 Q)$

56

Exercise 8.1 — Area Under Simple Curves

13Exercise questions

Step-by-step solution

1The parabola opens rightward with vertex at the origin; the vertical line x = 3 closes the region.
2By symmetry about the x-axis, total area = 2 × area in the first quadrant.
3In Q1, $y = 2 x$
4Area = $2 \int_{0}^{3} 2 x d x = 4 \int_{0}^{3} x^{1/2} d x = 4 [\frac{2}{3} x^{3/2}]_{0}^{3}$
5= $\frac{8}{3} \cdot 3^{3/2} = \frac{8}{3} \cdot 33 = 83$

Final answer

83 sq. units

Step-by-step solution

1In the first quadrant, $y = 3 x$
2Required area = $\int_{2}^{4} 3 x d x = 3 [\frac{2}{3} x^{3/2}]_{2}^{4}$
3= $2 (4^{3/2} - 2^{3/2}) = 2 (8 - 22) = 16 - 42$

Final answer

16 - 42 sq. units

Step-by-step solution

1In the first quadrant, $x = 2 y$
2Area = $\int_{2}^{4} 2 y d y = 2 [\frac{2}{3} y^{3/2}]_{2}^{4}$
3= $\frac{4}{3} (4^{3/2} - 2^{3/2}) = \frac{4}{3} (8 - 22) = \frac{32 - 8 2}{3}$

Final answer

\frac{32 - 8 2}{3} sq. units

Step-by-step solution

1This is a standard ellipse with $a = 4$ and $b = 3$
2By symmetry, total area = 4 × area in the first quadrant.
3In Q1, $y = \frac{3}{4} 16 - x^{2}$
4Area = $4 \int_{0}^{4} \frac{3}{4} 16 - x^{2} d x = 3 \int_{0}^{4} 16 - x^{2} d x$
5Use $\int_{0}^{a} a^{2} - x^{2} d x = \frac{π a ^{2}}{4}$ : Area = $3 \cdot \frac{π \cdot 16}{4} = 12 π$
6(Equivalently, the ellipse area formula $π ab = π \cdot 4 \cdot 3 = 12 π$ .

Final answer

12 π sq. units

Step-by-step solution

1Here $a = 2$ and $b = 3$
2Area of an ellipse is $π ab$
3Area = $π \cdot 2 \cdot 3 = 6 π$

Final answer

6 π sq. units

Step-by-step solution

1sin x is positive on (0, π) and negative on (π, 2π), so take absolute values.
2Area = $\int_{0}^{2 π} ∣ sin x ∣ d x = \int_{0}^{π} sin x d x + \int_{π}^{2 π} sin x d x$
3= $[- cos x]_{0}^{π} + [- cos x]_{π}^{2 π} = (1 + 1) + ∣ (- 1) - 1∣$
4= 2 + 2 = 4

Final answer

4 sq. units

Step-by-step solution

1In the first quadrant, $y = 2 a x$
2By symmetry, area = 2 × area in Q1.
3Area = $2 \int_{0}^{a} 2 a x d x = 4 a [\frac{2}{3} x^{3/2}]_{0}^{a}$
4= $\frac{8 a}{3} \cdot a^{3/2} = \frac{8 a ^{2}}{3}$

Final answer

\frac{8 a ^{2}}{3} sq. units

Step-by-step solution

1The line $y = \frac{x}{3}$ and the circle $x^{2} + y^{2} = 4$ meet where $x^{2} + \frac{x ^{2}}{3} = 4 \Rightarrow x = 3$ , i.e. at (√3, 1).
2On [0, √3] the top boundary is the line; on [√3, 2] it is the circle arc $y = 4 - x^{2}$
3Area = $\int_{0}^{3} \frac{x}{3} d x + \int_{3}^{2} 4 - x^{2} d x$
4First integral: $[\frac{x ^{2}}{2 3}]_{0}^{3} = \frac{3}{2}$
5Second integral: $[\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{3}^{2} = (0 + π) - (\frac{3}{2} + \frac{2 π}{3}) = \frac{π}{3} - \frac{3}{2}$
6Area = $\frac{3}{2} + \frac{π}{3} - \frac{3}{2} = \frac{π}{3}$

Final answer

\frac{π}{3} sq. units

Step-by-step solution

1Name the vertices A(−1,0), B(1,3), C(3,2).
2AB: $y = \frac{3}{2} (x + 1)$
3BC: $y = \frac{7 - x}{2}$
4AC: $y = \frac{x + 1}{2}$
5Area = $\int_{- 1}^{1} (\frac{3 ( x + 1 )}{2} - \frac{x + 1}{2}) d x + \int_{1}^{3} (\frac{7 - x}{2} - \frac{x + 1}{2}) d x$
6= $\int_{- 1}^{1} (x + 1) d x + \int_{1}^{3} (3 - x) d x = [\frac{x ^{2}}{2} + x]_{- 1}^{1} + [3 x - \frac{x ^{2}}{2}]_{1}^{3}$
7= $(\frac{1}{2} + 1) - (\frac{1}{2} - 1) + (9 - \frac{9}{2}) - (3 - \frac{1}{2}) = 2 + 2 = 4$

Final answer

4 sq. units

Step-by-step solution

1x² + 1 ≥ 1 > 0 on [0, 3], so the curve stays above the x-axis.
2Area = $\int_{0}^{3} (x^{2} + 1) d x = [\frac{x ^{3}}{3} + x]_{0}^{3} = (9 + 3) - 0 = 12$

Final answer

12 sq. units

Step-by-step solution

1sin x ≥ 0 on [0, π].
2Area = $\int_{0}^{π} sin x d x = [- cos x]_{0}^{π} = - cos π + cos 0 = 1 + 1 = 2$

Final answer

2 sq. units

Step-by-step solution

1Both circles are centred at the origin with radii 2 and 3 respectively.
2The required region is the ring between the two circles.
3Area = $π (3^{2}) - π (2^{2}) = 9 π - 4 π = 5 π$

Final answer

5 π sq. units

Step-by-step solution

1x³ is negative on (−2, 0) and positive on (0, 2), so split the integral.
2Area = $\int_{- 2}^{0} x^{3} d x + \int_{0}^{2} x^{3} d x = [\frac{x ^{4}}{4}]_{- 2}^{0} + [\frac{x ^{4}}{4}]_{0}^{2}$
3= |0 − 4| + (4 − 0) = 4 + 4 = 8

Final answer

8 sq. units

57

Exercise 8.2 — Area Enclosed Between a Curve and a Line, and Between Two Curves

6Exercise questions

Step-by-step solution

1Circle: $x^{2} + y^{2} = \frac{9}{4}$ , radius $\frac{3}{2}$ . Parabola: $y = \frac{x ^{2}}{4}$
2Intersections: $x^{2} + \frac{x ^{4}}{16} = \frac{9}{4} \Rightarrow x^{4} + 16 x^{2} - 36 = 0 \Rightarrow x^{2} = 2 \Rightarrow x = \pm 2$
3On the first quadrant piece, the top curve is the circle, the bottom curve is the parabola.
4Area = $2 \int_{0}^{2} (\frac{9}{4} - x^{2} - \frac{x ^{2}}{4}) d x$
5Using $\int a^{2} - u^{2} d u = \frac{u}{2} a^{2} - u^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{u}{a}$ with a = 3/2:
6First integral = $[\frac{x}{2} \frac{9}{4} - x^{2} + \frac{9}{8} sin^{- 1} \frac{2 x}{3}]_{0}^{2} = \frac{2}{4} + \frac{9}{8} sin^{- 1} \frac{2 2}{3}$
7Second integral = $[\frac{x ^{3}}{12}]_{0}^{2} = \frac{2}{6}$
8Area = $2 (\frac{2}{4} + \frac{9}{8} sin^{- 1} \frac{2 2}{3} - \frac{2}{6}) = \frac{2}{6} + \frac{9}{4} sin^{- 1} \frac{2 2}{3}$

Final answer

\frac{2}{6} + \frac{9}{4} sin^{- 1} \frac{2 2}{3} sq. units

Step-by-step solution

1The line is $x = 2 y + 3$ . It cuts the x-axis at (3, 0).
2Intersection with the parabola: $y = 2 y + 3 \Rightarrow y^{2} = 2 y + 3 \Rightarrow y = 3$ , giving (9, 3).
3The bounded region runs along the parabola from (0, 0) to (9, 3), then down the line to (3, 0), then back along the x-axis.
4Area = $\int_{0}^{3} x d x + \int_{3}^{9} (x - \frac{x - 3}{2}) d x$
5= $23 + [\frac{2}{3} x^{3/2} - \frac{( x - 3 ) ^{2}}{4}]_{3}^{9}$
6= $23 + (18 - 9) - (23 - 0) = 9$

Final answer

9 sq. units

Step-by-step solution

1Intersections: $(m x)^{2} = 4 a x \Rightarrow m^{2} x \cdot x = 4 a x \Rightarrow x = 0$ or $x = \frac{4 a}{m ^{2}}$
2In Q1, the parabola gives $y = 2 a x$ and the line is $y = m x$
3Area = $\int_{0}^{4 a / m^{2}} (2 a x - m x) d x = [\frac{4 a}{3} x^{3/2} - \frac{m x ^{2}}{2}]_{0}^{4 a / m^{2}}$
4= $\frac{4 a}{3} (\frac{4 a}{m ^{2}})^{3/2} - \frac{m}{2} \cdot \frac{16 a ^{2}}{m ^{4}} = \frac{32 a ^{2}}{3 m ^{3}} - \frac{8 a ^{2}}{m ^{3}} = \frac{8 a ^{2}}{3 m ^{3}}$

Final answer

\frac{8 a ^{2}}{3 m ^{3}} sq. units

Step-by-step solution

1They intersect at (0, 0) and at (4a, 4a).
2The parabola y² = 4ax gives $y = 2 a x$ ; the other is $y = \frac{x ^{2}}{4 a}$
3On (0, 4a), the first parabola lies above the second.
4Area = $\int_{0}^{4 a} (2 a x - \frac{x ^{2}}{4 a}) d x = [\frac{4 a}{3} x^{3/2} - \frac{x ^{3}}{12 a}]_{0}^{4 a}$
5= $\frac{4 a}{3} (8 a^{3/2}) - \frac{64 a ^{3}}{12 a} = \frac{32 a ^{2}}{3} - \frac{16 a ^{2}}{3} = \frac{16 a ^{2}}{3}$

Final answer

\frac{16 a ^{2}}{3} sq. units

Step-by-step solution

1Centres (0, 0) and (2, 0), each of radius 2 — the circles cut at two points.
2Intersections: $x^{2} + y^{2} = 4$ and $x^{2} - 4 x + y^{2} = 0 \Rightarrow 4 = 4 x \Rightarrow x = 1$ , so points $(1, \pm 3)$
3By symmetry the common (lens-shaped) area = 2 × the first quadrant piece.
4In Q1, the upper boundary of the lens is the arc of circle 2, $y = 4 - (x - 2)^{2}$ , on [0, 1], then the arc of circle 1, $y = 4 - x^{2}$ , on [1, 2].
5Half area = $\int_{0}^{1} 4 - (x - 2)^{2} d x + \int_{1}^{2} 4 - x^{2} d x$
6First integral (u = x − 2): $\int_{- 2}^{- 1} 4 - u^{2} d u = [\frac{u}{2} 4 - u^{2} + 2 sin^{- 1} \frac{u}{2}]_{- 2}^{- 1} = \frac{2 π}{3} - \frac{3}{2}$
7Second integral: $[\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{1}^{2} = π - (\frac{3}{2} + \frac{π}{3}) = \frac{2 π}{3} - \frac{3}{2}$
8Half area = $\frac{4 π}{3} - 3$ ; total = $2 (\frac{4 π}{3} - 3) = \frac{8 π}{3} - 23$

Final answer

\frac{8 π}{3} - 23 sq. units

Step-by-step solution

1Intersections: $x^{2} = 4 \Rightarrow x = \pm 2$
2Area = $\int_{- 2}^{2} (4 - x^{2}) d x = [4 x - \frac{x ^{3}}{3}]_{- 2}^{2} = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$

Final answer

Option (A) — 32/3.

58

Miscellaneous Exercise on Chapter 8

10Exercise questions

Step-by-step solution

1Intersections: $x^{2} = x \Rightarrow x (x - 1) = 0 \Rightarrow x = 0, 1$
2On [0, 1], the line y = x is above the parabola.
3Area = $\int_{0}^{1} (x - x^{2}) d x = [\frac{x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

Final answer

\frac{1}{6} sq. units

Step-by-step solution

1The line is x = 2 − y; the parabola is x = y².
2Intersections: $y^{2} = 2 - y \Rightarrow y^{2} + y - 2 = 0 \Rightarrow y = 1, - 2$
3Integrate along y from −2 to 1 with line minus parabola:
4Area = $\int_{- 2}^{1} (2 - y - y^{2}) d y = [2 y - \frac{y ^{2}}{2} - \frac{y ^{3}}{3}]_{- 2}^{1}$
5= $(2 - \frac{1}{2} - \frac{1}{3}) - (- 4 - 2 + \frac{8}{3}) = \frac{7}{6} + \frac{10}{3} = \frac{27}{6} = \frac{9}{2}$

Final answer

\frac{9}{2} sq. units

Step-by-step solution

1In the first quadrant, $x = \frac{y}{2}$
2Area = $\int_{1}^{4} \frac{y}{2} d y = \frac{1}{2} [\frac{2}{3} y^{3/2}]_{1}^{4} = \frac{1}{3} (8 - 1) = \frac{7}{3}$

Final answer

\frac{7}{3} sq. units

Step-by-step solution

1y = x changes sign at x = 0, so split the integral.
2Area = $\int_{- 2}^{0} x d x + \int_{0}^{3} x d x = [\frac{x ^{2}}{2}]_{- 2}^{0} + [\frac{x ^{2}}{2}]_{0}^{3}$
3= |0 − 2| + (9/2 − 0) = 2 + 9/2 = 13/2

Final answer

\frac{13}{2} sq. units

Step-by-step solution

1y = x and x + y = 6 meet at (3, 3). The line x + y = 6 meets the x-axis at (6, 0).
2Area = $\int_{0}^{3} x d x + \int_{3}^{6} (6 - x) d x = [\frac{x ^{2}}{2}]_{0}^{3} + [6 x - \frac{x ^{2}}{2}]_{3}^{6}$
3= $\frac{9}{2} + (36 - 18) - (18 - \frac{9}{2}) = \frac{9}{2} + 18 - \frac{27}{2} = 9$

Final answer

9 sq. units

Step-by-step solution

1|x| = −x on x < 0 and |x| = x on x ≥ 0.
2Area = $\int_{- 3}^{0} (- x) d x + \int_{0}^{3} x d x = [- \frac{x ^{2}}{2}]_{- 3}^{0} + [\frac{x ^{2}}{2}]_{0}^{3}$
3= $\frac{9}{2} + \frac{9}{2} = 9$

Final answer

9 sq. units

Step-by-step solution

1Parabola: $y = \frac{3 x ^{2}}{4}$ ; line: $y = \frac{3 x + 12}{2}$
2Intersections: $\frac{3 x ^{2}}{4} = \frac{3 x + 12}{2} \Rightarrow 3 x^{2} - 6 x - 24 = 0 \Rightarrow x = - 2, 4$
3On (−2, 4), the line lies above the parabola.
4Area = $\int_{- 2}^{4} (\frac{3 x + 12}{2} - \frac{3 x ^{2}}{4}) d x = [\frac{3 x ^{2}}{4} + 6 x - \frac{x ^{3}}{4}]_{- 2}^{4}$
5= $(12 + 24 - 16) - (3 - 12 + 2) = 20 + 7 = 27$

Final answer

27 sq. units

Step-by-step solution

1On [0, 3], y = x² + 2 is always above y = x.
2Area = $\int_{0}^{3} (x^{2} + 2 - x) d x = [\frac{x ^{3}}{3} + 2 x - \frac{x ^{2}}{2}]_{0}^{3}$
3= $9 + 6 - \frac{9}{2} = 15 - \frac{9}{2} = \frac{21}{2}$

Final answer

\frac{21}{2} sq. units

Step-by-step solution

1Side OA (O=(0,0), A=(2,3)): $y = \frac{3}{2} x$
2Side AB (A=(2,3), B=(4,0)): $y = \frac{3}{2} (4 - x) = 6 - \frac{3 x}{2}$
3Area = $\int_{0}^{2} \frac{3 x}{2} d x + \int_{2}^{4} (6 - \frac{3 x}{2}) d x = [\frac{3 x ^{2}}{4}]_{0}^{2} + [6 x - \frac{3 x ^{2}}{4}]_{2}^{4}$
4= $3 + (24 - 12) - (12 - 3) = 3 + 12 - 9 = 6$

Final answer

6 sq. units

Step-by-step solution

1The curve gives $x = \frac{y ^{2}}{4}$ in the first quadrant.
2Area = $\int_{0}^{3} \frac{y ^{2}}{4} d y = [\frac{y ^{3}}{12}]_{0}^{3} = \frac{27}{12} = \frac{9}{4}$

Final answer

\frac{9}{4} sq. units

59

Chapter 9 — Differential Equations

60

Exercise 9.1 — Order and Degree of a Differential Equation

12Exercise questions

Step-by-step solution

1The highest order derivative present is $\frac{d ^{4} y}{d x ^{4}}$ — order 4.
2The term $sin (\frac{d ^{3} y}{d x ^{3}})$ is transcendental in the derivative, so the equation is not a polynomial in derivatives — degree not defined.

Final answer

Order 4; degree not defined.

Step-by-step solution

1Highest order derivative present: $\frac{d y}{d x}$ — first order.
2It appears to the power 1, so degree = 1.

Final answer

Order 1; degree 1.

Step-by-step solution

1Highest order derivative present: $\frac{d ^{2} s}{d t ^{2}}$ — order 2.
2It appears with power 1, so degree = 1.

Final answer

Order 2; degree 1.

Step-by-step solution

1Highest order derivative present: $\frac{d ^{2} y}{d x ^{2}}$ — order 2.
2The term $cos (\frac{d y}{d x})$ is transcendental in the derivative, so the equation is not a polynomial in derivatives — degree not defined.

Final answer

Order 2; degree not defined.

Step-by-step solution

1Highest order derivative present: $\frac{d ^{2} y}{d x ^{2}}$ — order 2.
2It appears with power 1, so degree = 1.

Final answer

Order 2; degree 1.

Step-by-step solution

1Highest order derivative present: $y^{'''}$ = $\frac{d ^{3} y}{d x ^{3}}$ — order 3.
2Its highest power in the equation is 2, so degree = 2.

Final answer

Order 3; degree 2.

Step-by-step solution

1Highest order derivative present: $y^{'''}$ — order 3.
2It appears with power 1, so degree = 1.

Final answer

Order 3; degree 1.

Step-by-step solution

1Highest order derivative present: $y^{'}$ — order 1.
2It appears with power 1, so degree = 1.

Final answer

Order 1; degree 1.

Step-by-step solution

1Highest order derivative present: $y^{''}$ — order 2.
2It appears with power 1, so degree = 1.

Final answer

Order 2; degree 1.

Step-by-step solution

1Highest order derivative present: $y^{''}$ — order 2.
2The derivatives appear polynomially (sin y is not a function of a derivative), so degree = 1.

Final answer

Order 2; degree 1.

Step-by-step solution

1Highest order derivative present: $\frac{d ^{2} y}{d x ^{2}}$ — order 2.
2The term $sin (\frac{d y}{d x})$ makes the equation non-polynomial in the derivatives — degree not defined.

Final answer

Order 2; degree not defined.

Step-by-step solution

1Highest order derivative present: $\frac{d ^{2} y}{d x ^{2}}$ — order 2.
2Square both sides to clear the fractional power: $(\frac{d ^{2} y}{d x ^{2}})^{4} = [1 + (\frac{d y}{d x})^{2}]^{3}$
3The highest order derivative now appears with power 4, so degree = 4.

Final answer

Order 2; degree 4.

61

Exercise 9.2 — General and Particular Solutions

12Exercise questions

Step-by-step solution

1y = eˣ + 1 gives y' = eˣ and y'' = eˣ.
2 $y^{''} - y^{'} = e^{x} - e^{x} = 0$ — an identity, so it is a solution.

Final answer

Yes — the function satisfies y'' − y' = 0.

Step-by-step solution

1y' = 2x + 2.
2 $y^{'} - 2 x - 2 = (2 x + 2) - 2 x - 2 = 0$ — identity.

Final answer

Yes — it is a solution (a family of parabolas).

Step-by-step solution

1y' = −sin x.
2 $y^{'} + sin x = - sin x + sin x = 0$ — identity.

Final answer

Yes — it is a solution.

Step-by-step solution

1y² = 1 + x², so 2yy' = 2x ⇒ y' = x/y.
2 $\frac{x y}{1 + x ^{2}} = \frac{x y}{y ^{2}} = \frac{x}{y} = y^{'}$ — identity.

Final answer

Yes — it is a solution.

Step-by-step solution

1y' = A.
2 $x y^{'} = x A = A x = y$ — identity.

Final answer

Yes — it is a solution.

Step-by-step solution

1y' = sin x + x cos x, so xy' = x sin x + x² cos x = y + x² cos x.
2x² − y² = x² − x² sin²x = x² cos²x, so x√(x² − y²) = x² cos x (cos x > 0).
3Hence xy' = y + x√(x² − y²) — identity. It is a solution.

Final answer

Yes — it is a solution.

Step-by-step solution

1y² = a² − x² gives 2yy' = −2x ⇒ yy' = −x.
2 $x + y y^{'} = x - x = 0$ — identity, valid for x ∈ (−a, a).

Final answer

Yes — it is a solution on the given interval.

Step-by-step solution

1y' = −sin x − cos x and y'' = −cos x + sin x.
2 $y^{''} + y = (- cos x + sin x) + (cos x - sin x) = 0$ — identity.

Final answer

Yes — it is a solution.

Step-by-step solution

1Differentiate x + y = tan⁻¹y with respect to x: $1 + y^{'} = \frac{y ^{'}}{1 + y ^{2}}$
2Multiply by 1 + y²: (1 + y²)(1 + y') = y' ⇒ 1 + y² + y²y' = 0 — identity.

Final answer

Yes — it is a solution.

Step-by-step solution

1y' = −a sin x + b cos x, y'' = −a cos x − b sin x = −y.
2 $y^{''} + y = 0$ and the function carries two arbitrary constants — the general solution of this second order equation.

Final answer

Yes — y = a cos x + b sin x is the general solution.

Step-by-step solution

1The general solution of an equation of order n contains exactly n arbitrary constants.
2Order 4 ⇒ 4 arbitrary constants.

Final answer

Option (D) — 4.

Step-by-step solution

1A particular solution is obtained from the general solution by fixing all the arbitrary constants using the initial conditions.
2Hence it contains no arbitrary constants.

Final answer

Option (D) — 0.

62

Exercise 9.3 — Formation of Differential Equations

13Exercise questions

Step-by-step solution

1The equation has two constants a and b, so differentiate twice.
2First derivative: $\frac{1}{a} + \frac{1}{b} \frac{d y}{d x} = 0$
3Second derivative: $\frac{1}{b} \frac{d ^{2} y}{d x ^{2}} = 0 \Rightarrow \frac{d ^{2} y}{d x ^{2}} = 0$

Final answer

y^{''} = 0

Step-by-step solution

1Differentiate once: 2yy' = −2ax ⇒ yy' = −ax.
2Differentiate again: yy'' + (y')² = −a.
3Substitute −a: from yy' = −ax we have a = −yy'/x, giving yy'' + (y')² = yy'/x.
4Multiply by x: xyy'' + x(y')² − yy' = 0.

Final answer

x y y^{''} + x (y^{'})^{2} - y y^{'} = 0

Step-by-step solution

1Constant count 2 ⇒ differentiate twice.
2 $y^{'} = 3 a e^{3 x} - 2 b e^{- 2 x}, y^{''} = 9 a e^{3 x} + 4 b e^{- 2 x}$
3Eliminate a, b: note 3y + y' = 6ae³ˣ and 3y' + y'' = 6be⁻²ˣ; multiplying the first relation of the system y'' − y' − 6y = 0 — verify: substitute both terms.
4 $y^{''} - y^{'} - 6 y = (9 a e^{3 x} + 4 b e^{- 2 x}) - (3 a e^{3 x} - 2 b e^{- 2 x}) - 6 (a e^{3 x} + b e^{- 2 x}) = 0$

Final answer

y^{''} - y^{'} - 6 y = 0

Step-by-step solution

1 $y^{'} = e^{2 x} (2 a + b + 2 b x) = 2 y + b e^{2 x}, y^{''} = 4 y + 4 b e^{2 x}$
2Eliminate be²ˣ = y' − 2y: y'' = 4y + 4(y' − 2y) = 4y' − 4y.

Final answer

y^{''} - 4 y^{'} + 4 y = 0

Step-by-step solution

1 $y^{'} = e^{x} (a cos x + b sin x) + e^{x} (- a sin x + b cos x) = y + e^{x} (- a sin x + b cos x)$
2 $y^{''} = y^{'} + e^{x} (- a cos x - b sin x) + e^{x} (- a sin x + b cos x)$
3Eliminating gives y'' − 2y' + 2y = 0.

Final answer

y^{''} - 2 y^{'} + 2 y = 0

Step-by-step solution

1Constants a, b, r (3) ⇒ differentiate thrice. First: $x - a + (y - b) y^{'} = 0$
2Second: $1 + (y^{'})^{2} + (y - b) y^{''} = 0$
3 $\Rightarrow y - b = \frac{1 + ( y ^{'} ) ^{2}}{y ^{''}}$
4Third/elimination leads to $(1 + (y^{'})^{2})^{3} = r^{2} (y^{''})^{2}$ — substitute back: |1+(y')²| = r|y''|, square.

Final answer

[1 + (y^{'})^{2}]^{3} = r^{2} (y^{''})^{2}

Step-by-step solution

1Differentiate: $x y^{'} + y = a e^{x} - b e^{- x}$
2Differentiate again: $x y^{''} + 2 y^{'} = a e^{x} + b e^{- x} = x y$

Final answer

x y^{''} + 2 y^{'} - x y = 0

Step-by-step solution

1 $y^{'} = - am sin (m x) + bm cos (m x), y^{''} = - m^{2} (a cos (m x) + b sin (m x)) = - m^{2} y$

Final answer

y^{''} + m^{2} y = 0

Step-by-step solution

1 $y^{'} = - \frac{a}{x ^{2}}, y^{''} = \frac{2 a}{x ^{3}}$
2Eliminate a: from y' = −a/x², a/x³ = −y'/x, so y'' = −2y'/x ⇒ xy'' + 2y' = 0.

Final answer

x y^{''} + 2 y^{'} = 0

Step-by-step solution

1Differentiate: 2x = 4ay' ⇒ 4a = 2x/y'.
2Differentiate again: 2 = 4ay'' ⇒ substitute: 2 = (2x/y')y'' ⇒ xy'' − y' = 0.

Final answer

x y^{''} - y^{'} = 0

Step-by-step solution

1Differentiate: 2yy' = 4a.
2Differentiate again: 2(y')² + 2yy'' = 0.

Final answer

y y^{''} + (y^{'})^{2} = 0

Step-by-step solution

1For y = x: y' = 1, y'' = 0.
2Check each: (A) 0 = x²? no. (B) 1 = x²? no. (C) 0 − x²(1) + x(x) = 0 ✓. (D) 0 = x³? no.

Final answer

Option (C) — y'' − x²y' + xy = 0.

Step-by-step solution

1 $y^{'} = c_{1} e^{x} - c_{2} e^{- x}, y^{''} = c_{1} e^{x} + c_{2} e^{- x} = y$
2Hence y'' − y = 0.

Final answer

Option (B) — y'' − y = 0.

63

Exercise 9.4 — Variable Separable Method

14Exercise questions

Step-by-step solution

1Use identities: $\frac{1 - c o s x}{1 + c o s x} = \frac{2 s i n ^{2} ( x /2 )}{2 c o s ^{2} ( x /2 )} = tan^{2} (x /2) = sec^{2} (x /2) - 1$
2 $y = \int (sec^{2} (x /2) - 1) d x = 2 tan (x /2) - x + C$

Final answer

y = 2 tan (x /2) - x + C

Step-by-step solution

1 $\frac{d y}{4 - y ^{2}} = d x \Rightarrow sin^{- 1} (y /2) = x + C$

Final answer

sin^{- 1} (y /2) = x + C

Step-by-step solution

1 $\frac{d y}{y + 1} = d x \Rightarrow ln ∣ y + 1∣ = x + C$
2y(0) = 5 ⇒ ln 6 = C, so $ln \frac{y + 1}{6} = x \Rightarrow y = 6 e^{x} - 1$

Final answer

y = 6 e^{x} - 1

Step-by-step solution

1Separate: $\frac{s e c ^{2} x}{t a n x} d x + \frac{s e c ^{2} y}{t a n y} d y = 0$
2Integrate: ln|tan x| + ln|tan y| = C, so $tan x tan y = C$

Final answer

tan x tan y = C

Step-by-step solution

1 $d y = \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} d x \Rightarrow y = ln (e^{x} + e^{- x}) + C$

Final answer

y = ln (e^{x} + e^{- x}) + C

Step-by-step solution

1 $\frac{d y}{1 + y ^{2}} = (1 + x^{2}) d x \Rightarrow tan^{- 1} y = x + \frac{x ^{3}}{3} + C$

Final answer

tan^{- 1} y = x + x^{3} /3 + C

Step-by-step solution

1 $\frac{d y}{y l o g y} = \frac{d x}{x} \Rightarrow ln ∣ lo g y ∣ = ln ∣ x ∣ + C$
2Hence $lo g y = C x \Rightarrow x = C lo g y$

Final answer

x = C lo g y

Step-by-step solution

1 $y^{- 5} d y = - x^{- 5} d x \Rightarrow \frac{y ^{- 4}}{- 4} = \frac{x ^{- 4}}{4} + C$
2Hence $\frac{1}{x ^{4}} + \frac{1}{y ^{4}} = C$

Final answer

1/ x^{4} + 1/ y^{4} = C

Step-by-step solution

1 $y = \int sin^{- 1} x d x = x sin^{- 1} x + 1 - x^{2} + C$

Final answer

y = x sin^{- 1} x + 1 - x^{2} + C

Step-by-step solution

1 $y = \int tan^{- 1} x d x = x tan^{- 1} x - \frac{1}{2} ln (1 + x^{2}) + C$

Final answer

y = x tan^{- 1} x - \frac{1}{2} ln (1 + x^{2}) + C

Step-by-step solution

1 $d y = \frac{2 x ^{2} + x}{x ^{3} + x ^{2} + x + 1} d x = \frac{2 x ^{2} + x}{( x + 1 ) ( x ^{2} + 1 )} d x$
2Partial fractions: $\frac{2 x ^{2} + x}{( x + 1 ) ( x ^{2} + 1 )} = \frac{A}{x + 1} + \frac{B x + C}{x ^{2} + 1}$
3Solving: A = 1/2, B = 3/2, C = −1/2.
4 $y = \frac{1}{2} ln ∣ x + 1∣ + \frac{3}{4} ln (x^{2} + 1) - \frac{1}{2} tan^{- 1} x + C$

Final answer

y = \frac{1}{2} ln ∣ x + 1∣ + \frac{3}{4} ln (x^{2} + 1) - \frac{1}{2} tan^{- 1} x + C

Step-by-step solution

1 $d y = \frac{d x}{x ( x ^{2} - 1 )} = \frac{d x}{x ( x - 1 ) ( x + 1 )}$
2Partial fractions: $\frac{1}{x ( x - 1 ) ( x + 1 )} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$
3Solving: A = −1, B = 1/2, C = 1/2.
4 $y = - ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 1∣ + \frac{1}{2} ln ∣ x + 1∣ + C = \frac{1}{2} ln ∣ x^{2} - 1∣ - ln ∣ x ∣ + C$

Final answer

y = \frac{1}{2} ln ∣ x^{2} - 1∣ - ln ∣ x ∣ + C

Step-by-step solution

1 $\frac{y}{1 - y ^{2}} d y = - x e^{x} d x$
2 $- 1 - y^{2} = - e^{x} (x - 1) + C \Rightarrow e^{x} (x - 1) - 1 - y^{2} = C$

Final answer

e^{x} (x - 1) - 1 - y^{2} = C

Step-by-step solution

1 $\frac{d y}{y} = 2 x d x \Rightarrow ln ∣ y ∣ = x^{2} + C$
2y(0) = 1 ⇒ C = 0, so $y = e^{x^{2}}$

Final answer

y = e^{x^{2}}

64

Exercise 9.5 — Homogeneous Differential Equations

12Exercise questions

Step-by-step solution

1F(x,y) = (x² + y²)/(x² + xy) satisfies F(tx, ty) = F(x, y), so the equation is homogeneous; put y = vx, dy = v dx + x dv.
2 $(1 + v) (v d x + x d v) = (1 + v^{2}) d x \Rightarrow x (1 + v) v^{'} = 1 - v$
3 $\frac{1 + v}{1 - v} d v = \frac{d x}{x} \Rightarrow - v - 2 ln ∣1 - v ∣ = ln ∣ x ∣ + C$
4With y = vx: $\frac{y}{x} + ln \frac{( x - y ) ^{2}}{x} = C$

Final answer

y / x + lo g ∣ (x - y)^{2} / x ∣ = C

Step-by-step solution

1 $\frac{x + y}{x} has degree 0, so it is homogeneous; put y = v x : v + x v^{'} = 1 + v$
2 $x v^{'} = 1 \Rightarrow d v = \frac{d x}{x} \Rightarrow v = ln ∣ x ∣ + C$
3y = vx ⇒ $y = x ln ∣ x ∣ + C x$

Final answer

y = x lo g ∣ x ∣ + C x

Step-by-step solution

1Put y = vx: $(1 - v) (v + x v^{'}) - (1 + v) = 0 \Rightarrow x v^{'} = \frac{1 + v ^{2}}{1 - v}$
2 $\frac{1 - v}{1 + v ^{2}} d v = \frac{d x}{x} \Rightarrow tan^{- 1} v - \frac{1}{2} ln (1 + v^{2}) = ln ∣ x ∣ + C$
3Back-substitute v = y/x: $2 tan^{- 1} (y / x) = ln (x^{2} + y^{2}) + C$

Final answer

2 tan^{- 1} (y / x) = lo g (x^{2} + y^{2}) + C

Step-by-step solution

1Put y = vx, dy = v dx + x dv: $(1 - v^{2}) d x + 2 v (v d x + x d v) = 0$
2 $(1 + v^{2}) d x + 2 v x d v = 0 \Rightarrow \frac{d x}{x} + \frac{2 v}{1 + v ^{2}} d v = 0$
3 $ln ∣ x ∣ + ln (1 + v^{2}) = C \Rightarrow x (1 + v^{2}) = C$
4y = vx ⇒ $x + \frac{y ^{2}}{x} = C \Rightarrow x^{2} + y^{2} = C x$

Final answer

x^{2} + y^{2} = C x

Step-by-step solution

1Put y = vx: $v + x v^{'} = 1 + v - 2 v^{2} \Rightarrow x v^{'} = 1 - 2 v^{2}$
2 $\int \frac{d v}{1 - 2 v ^{2}} = \int \frac{d x}{x} \Rightarrow \frac{1}{2 2} ln \frac{1 + 2 v}{1 - 2 v} = ln ∣ x ∣ + C$
3y = vx: $\frac{1}{2 2} ln \frac{x + 2 y}{x - 2 y} = ln ∣ x ∣ + C$

Final answer

\frac{1}{2 2} ln \frac{x + 2 y}{x - 2 y} = ln ∣ x ∣ + C

Step-by-step solution

1Put y = vx, dy = v dx + x dv: $x (v d x + x d v) - v x d x = x 1 + v^{2} d x$
2 $x d v = 1 + v^{2} d x \Rightarrow \int \frac{d v}{1 + v ^{2}} = \int \frac{d x}{x}$
3 $ln ∣ v + 1 + v^{2} ∣ = ln ∣ x ∣ + C$
4y = vx ⇒ $y + x^{2} + y^{2} = C x^{2}$

Final answer

y + x^{2} + y^{2} = C x^{2}

Step-by-step solution

1Put y = vx, dy = v dx + x dv: $x^{2} v (cos v + v sin v) d x = x^{2} (v sin v - cos v) (v d x + x d v)$
2Cancel x² and collect dx terms: $2 v cos v d x = x (v sin v - cos v) d v$
3 $\frac{2}{x} d x = (tan v - \frac{1}{v}) d v$
4 $2 ln ∣ x ∣ = - ln ∣ cos v ∣ - ln ∣ v ∣ + C \Rightarrow x^{2} v cos v = C$
5v = y/x ⇒ $x y cos (y / x) = C$

Final answer

x y cos (y / x) = C

Step-by-step solution

1 $\frac{d y}{d x} = \frac{y}{x} - sin (y / x)$
2Put y = vx: $v + x v^{'} = v - sin v \Rightarrow x v^{'} = - sin v$
3 $csc v d v = - \frac{d x}{x} \Rightarrow ln ∣ tan (v /2) ∣ = - ln ∣ x ∣ + C$
4y = vx ⇒ $x tan (\frac{y}{2 x}) = C$

Final answer

x tan (y /2 x) = C

Step-by-step solution

1Write for x as a function of y: $\frac{d x}{d y} = \frac{x ( 2 - l o g ( y / x ))}{y}$
2Put x = uy, dx/dy = u + y du/dy: $u + y \frac{d u}{d y} = u (2 - lo g u)$
3 $y \frac{d u}{d y} = u (1 - lo g u) \Rightarrow \frac{d u}{u ( 1 - l o g u )} = \frac{d y}{y}$
4Integrate (w = 1 − log u): $- ln ∣1 - lo g u ∣ = ln ∣ y ∣ + C \Rightarrow y (1 - lo g u) = C$
5u = x/y and 1 − log(x/y) = 1 + log(y/x): $1 + lo g (\frac{y}{x}) = \frac{C}{y}$

Final answer

1 + lo g (y / x) = C / y

Step-by-step solution

1Put x = uy, dx = u dy + y du: $(1 + e^{u}) (u d y + y d u) + e^{u} (1 - u) d y = 0$
2 $(u + e^{u}) d y + y (1 + e^{u}) d u = 0 \Rightarrow \frac{1 + e ^{u}}{u + e ^{u}} d u = - \frac{d y}{y}$
3 $ln ∣ u + e^{u} ∣ = - ln ∣ y ∣ + C \Rightarrow y (u + e^{u}) = C$
4u = x/y ⇒ $x + y e^{x / y} = C$

Final answer

x + y e^{x / y} = C

Step-by-step solution

1Put y = vx: $v + x v^{'} = v + v^{2} \Rightarrow \frac{d v}{v ^{2}} = \frac{d x}{x}$
2 $- \frac{1}{v} = ln ∣ x ∣ + C$
3y(1) = 1 ⇒ v(1) = 1 ⇒ C = −1: $\frac{x}{y} = 1 - ln ∣ x ∣ \Rightarrow y = \frac{x}{1 - l n ∣ x ∣}$

Final answer

y = \frac{x}{1 - lo g ∣ x ∣}

Step-by-step solution

1Put y = vx: $cos v (v + x v^{'}) = v cos v - 1 \Rightarrow x cos v v^{'} = - 1$
2 $cos v d v = - \frac{d x}{x} \Rightarrow sin v = - ln ∣ x ∣ + C$
3y(1) = 0 ⇒ v(1) = 0 ⇒ C = 0: $sin (\frac{y}{x}) = - ln ∣ x ∣$

Final answer

sin (y / x) = - lo g ∣ x ∣

65

Exercise 9.6 — Linear Differential Equations

17Exercise questions

Step-by-step solution

1 $P = 2, Q = sin x, IF = e^{\int 2 d x} = e^{2 x}$
2 $y e^{2 x} = \int e^{2 x} sin x d x = \frac{e ^{2 x} ( 2 s i n x - c o s x )}{5} + C$

Final answer

y = \frac{2 sin x - cos x}{5} + C e^{- 2 x}

Step-by-step solution

1 $IF = e^{\int 3 d x} = e^{3 x}$
2 $y e^{3 x} = \int e^{3 x} e^{- 2 x} d x = \int e^{x} d x = e^{x} + C \Rightarrow y = e^{- 2 x} + C e^{- 3 x}$

Final answer

y = e^{- 2 x} + C e^{- 3 x}

Step-by-step solution

1 $IF = e^{\int \frac{d x}{x}} = e^{l n x} = x$
2 $y x = \int x^{3} d x = \frac{x ^{4}}{4} + C \Rightarrow y = \frac{x ^{3}}{4} + \frac{C}{x}$

Final answer

y = \frac{x ^{3}}{4} + \frac{C}{x}

Step-by-step solution

1 $IF = e^{\int s e c x d x} = sec x + tan x$
2 $y (sec x + tan x) = \int tan x (sec x + tan x) d x = sec x + tan x - x + C$

Final answer

y (sec x + tan x) = sec x + tan x - x + C

Step-by-step solution

1 $\frac{d y}{d x} + sec^{2} x y = tan x sec^{2} x, IF = e^{\int s e c^{2} x d x} = e^{t a n x}$
2 $y e^{t a n x} = \int e^{t a n x} tan x sec^{2} x d x$
3Put u = tan x: $= \int e^{u} u d u = e^{u} (u - 1) + C$
4 $y = tan x - 1 + C e^{- t a n x}$

Final answer

y = tan x - 1 + C e^{- t a n x}

Step-by-step solution

1 $\frac{d y}{d x} + \frac{2}{x} y = x lo g x, IF = e^{\int \frac{2}{x} d x} = x^{2}$
2 $y x^{2} = \int x^{3} lo g x d x = \frac{x ^{4}}{4} lo g x - \frac{x ^{4}}{16} + C$
3 $y = \frac{x ^{2}}{4} lo g x - \frac{x ^{2}}{16} + \frac{C}{x ^{2}} = \frac{x ^{2}}{16} (4 lo g x - 1) + \frac{C}{x ^{2}}$

Final answer

y = \frac{x ^{2}}{16} (4 lo g x - 1) + \frac{C}{x ^{2}}

Step-by-step solution

1 $\frac{d y}{d x} + \frac{y}{x l o g x} = \frac{2}{x}, IF = e^{\int \frac{d x}{x l o g x}} = lo g x$
2 $y lo g x = \int \frac{2}{x} lo g x d x = (lo g x)^{2} + C$
3 $y = lo g x + \frac{C}{l o g x}$

Final answer

y = lo g x + \frac{C}{lo g x}

Step-by-step solution

1Note d/dx (1+x²) = 2x, so the left side is d/dx [y(1+x²)]:
2 $y (1 + x^{2}) = \int cot x d x = ln ∣ sin x ∣ + C$

Final answer

y (1 + x^{2}) = ln ∣ sin x ∣ + C

Step-by-step solution

1 $\frac{d y}{d x} + (\frac{1}{x} + cot x) y = 1, IF = e^{\int (1/ x + c o t x) d x} = x sin x$
2 $y x sin x = \int x sin x d x = - x cos x + sin x + C$
3 $y = \frac{s i n x - x c o s x + C}{x s i n x}$

Final answer

y = \frac{sin x - x cos x + C}{x sin x}

Step-by-step solution

1 $\frac{d x}{d y} - x = y,$ linear in x $IF = e^{\int - 1 d y} = e^{- y}$
2 $x e^{- y} = \int y e^{- y} d y = - e^{- y} (y + 1) + C \Rightarrow x = - y - 1 + C e^{y}$

Final answer

x = - y - 1 + C e^{y}

Step-by-step solution

1 $\frac{d x}{d y} + \frac{1}{y} x = y,$ linear in x $IF = e^{\int \frac{d y}{y}} = y$
2 $x y = \int y^{2} d y = \frac{y ^{3}}{3} + C \Rightarrow x = \frac{y ^{2}}{3} + \frac{C}{y}$

Final answer

x = \frac{y ^{2}}{3} + \frac{C}{y}

Step-by-step solution

1 $\frac{d x}{d y} - \frac{1}{y} x = 3 y,$ linear in x $IF = e^{- \int \frac{d y}{y}} = \frac{1}{y}$
2 $\frac{x}{y} = \int 3 d y = 3 y + C \Rightarrow x = 3 y^{2} + C y$

Final answer

x = 3 y^{2} + C y

Step-by-step solution

1 $IF = e^{2 \int t a n x d x} = e^{- 2 l n c o s x} = sec^{2} x$
2 $y sec^{2} x = \int sin x sec^{2} x d x = \int tan x sec x d x = sec x + C$
3y(0) = 0 ⇒ 0 = 1 + C ⇒ C = −1: $y = cos x - cos^{2} x$

Final answer

y = cos x - cos^{2} x

Step-by-step solution

1Left side = d/dx [y(1+x²)]: $y (1 + x^{2}) = \int \frac{d x}{1 + x ^{2}} = tan^{- 1} x + C$
2y(0) = 0 ⇒ C = 0: $y = \frac{t a n ^{- 1} x}{1 + x ^{2}}$

Final answer

y = \frac{tan ^{- 1} x}{1 + x ^{2}}

Step-by-step solution

1 $IF = e^{- 3 \int c o t x d x} = e^{- 3 l n s i n x} = csc^{3} x$
2 $y csc^{3} x = \int sin 2 x csc^{3} x d x = 2 \int cot x csc x d x = - 2 csc x + C$
3y(π/2) = 2 ⇒ 2 = −2 + C ⇒ C = 4: $y csc^{3} x = - 2 csc x + 4 \Rightarrow y = 4 sin^{3} x - 2 sin^{2} x$

Final answer

y = 4 sin^{3} x - 2 sin^{2} x

Step-by-step solution

1 $IF = e^{\int c o t x d x} = sin x$
2 $y sin x = \int 4 x d x = 2 x^{2} + C$
3y(π/2) = 0 ⇒ 0 = 2(π/2)² + C ⇒ C = −π²/2: $y = (2 x^{2} - \frac{π ^{2}}{2}) csc x$

Final answer

y = (2 x^{2} - π^{2} /2) csc x

Step-by-step solution

1 $\frac{d x}{d y} + \frac{y}{1 - y ^{2}} x = \frac{a y}{1 - y ^{2}}$
2 $IF = e^{\int \frac{y}{1 - y ^{2}} d y} = e^{- \frac{1}{2} l n (1 - y^{2})} = \frac{1}{1 - y ^{2}}$

Final answer

Option (D) — 1/√(1−y²).

66

Miscellaneous Exercise — Differential Equations

13Exercise questions

Step-by-step solution

1(i) Highest derivative is y'', to power 1 — order 2, degree 1 (log x does not involve derivatives).
2(ii) Highest derivative is y', to power 3 — order 1, degree 3.
3(iii) Highest derivative is y'''' — order 4; sin y''' makes it non-polynomial in derivatives, degree not defined.

Final answer

(i) order 2, degree 1; (ii) order 1, degree 3; (iii) order 4, degree not defined.

Step-by-step solution

1 $y^{'} = - 3 e^{- 3 x}, y^{''} = 9 e^{- 3 x}$
2 $y^{''} + y^{'} - 6 y = 9 e^{- 3 x} - 3 e^{- 3 x} - 6 e^{- 3 x} = 0$ — identity.

Final answer

Yes — y = e⁻³ˣ satisfies y'' + y' − 6y = 0.

Step-by-step solution

1 $y^{'} = e^{x} (a cos x + b sin x) + e^{x} (- a sin x + b cos x) = y + e^{x} (- a sin x + b cos x)$
2 $y^{''} = y^{'} + e^{x} (- a sin x + b cos x) + e^{x} (- a cos x - b sin x) = 2 y^{'} - 2 y$
3 $y^{''} - 2 y^{'} + 2 y = 0$ — identity.

Final answer

Yes — it satisfies y'' − 2y' + 2y = 0.

Step-by-step solution

1 $\frac{s e c ^{2} y}{t a n y} d y = - \frac{3 e ^{x}}{1 - e ^{x}} d x$
2 $ln ∣ tan y ∣ = 3 ln ∣1 - e^{x} ∣ + C \Rightarrow tan y = C (1 - e^{x})^{3}$

Final answer

tan y = C (1 - e^{x})^{3}

Step-by-step solution

1 $(x + y - 1) d x = (x + y + 1) d y \Rightarrow \frac{d y}{d x} = \frac{x + y - 1}{x + y + 1}$
2Put u = x + y, du/dx = 1 + dy/dx: $\frac{d u}{d x} = \frac{2 u}{u + 1} \Rightarrow \frac{u + 1}{u} d u = 2 d x$
3 $u + ln ∣ u ∣ = 2 x + C$ with u = x + y; equivalently
4 $x - y = ln ∣ x + y ∣ + C$

Final answer

x - y = lo g ∣ x + y ∣ + C

Step-by-step solution

1 $\frac{1}{y l o g y} \frac{d y}{d x} = \frac{1}{x} (e^{x} - \frac{l o g y}{x})$ — put v = log y, y' / y = v':
2 $\frac{d v}{d x} + \frac{1}{x} v = e^{x}, IF = x$
3When v + vy/x: $v x = \int x e^{x} d x = e^{x} (x - 1) + C$
4 $lo g y = \frac{e ^{x} ( x - 1 ) + C}{x} = \frac{e ^{x} ( x - 1 )}{x} + \frac{C}{x}$

Final answer

lo g y = e^{x} (x - 1) / x + C / x

Step-by-step solution

1Homogeneous; put y = vx: $2 v + v^{2} = 2 (v + x \frac{d v}{d x})$
2 $2 x \frac{d v}{d x} = v (2 + v) - 2 v = v^{2} \Rightarrow \int \frac{d v}{v ^{2}} = \int \frac{d x}{2 x}$
3 $- \frac{1}{v} = \frac{1}{2} ln ∣ x ∣ + C$
4y(2) = 1 ⇒ v(2) = 1/2 ⇒ −2 = (1/2)ln 2 + C ⇒ C = −2 − (1/2)ln 2:
5 $\frac{x}{y} = 2 + \frac{1}{2} ln \frac{2}{∣ x ∣} \Rightarrow y = \frac{2 x}{4 + l n ( 2/ x )}$

Final answer

y = \frac{2 x}{4 + ln ( 2/ x )}

Step-by-step solution

1Put y = vx: $(1 + v^{2}) d x + 2 v (v d x + x d v) = 0$
2 $\frac{d x}{x} = - \frac{2 v}{1 + 3 v ^{2}} d v \Rightarrow ln ∣ x ∣ = - \frac{1}{3} ln ∣1 + 3 v^{2} ∣ + C$
3 $x (1 + 3 v^{2})^{1/3} = C \Rightarrow (x^{3} + 3 x y^{2})^{1/3} / x ...$ equivalently raise to power 3:
4 $x^{3} (1 + 3 v^{2}) = C^{'} \Rightarrow x^{3} + 3 x y^{2} = C^{'}$
5y(1) = 1 ⇒ 1 + 3 = C' ⇒ C' = 4: $x^{2} + 3 y^{2} = \frac{4}{x}$

Final answer

x^{2} + 3 y^{2} = 4/ x

Step-by-step solution

1Put y = vx: $(1 + v) (v + x v^{'}) + (1 - v) = 0$
2 $x (1 + v) v^{'} = - (1 + v^{2}) \Rightarrow \frac{1 + v}{1 + v ^{2}} d v = - \frac{d x}{x}$
3 $tan^{- 1} v + \frac{1}{2} ln (1 + v^{2}) = - ln ∣ x ∣ + C = ln \frac{1}{x} + C$
4y(1) = 0 ⇒ v(1) = 0 ⇒ 0 + 0 = 0 + C ⇒ C = 0:
5 $\frac{1}{2} ln (x^{2} + y^{2}) + tan^{- 1} (\frac{y}{x}) = 0$

Final answer

\frac{1}{2} lo g (x^{2} + y^{2}) + tan^{- 1} (y / x) = 0

Step-by-step solution

1Rewrite: $e^{x} \frac{d y}{d x} + e^{x} y = - 2 x \Rightarrow \frac{d}{d x} (y e^{x}) = - 2 x$
2 $y e^{x} = - x^{2} + C \Rightarrow y e^{x} + x^{2} = C$

Final answer

Option (C) — yeˣ + x² = C.

Step-by-step solution

1 $\frac{d y}{d x} = e^{x} e^{y} \Rightarrow e^{- y} d y = e^{x} d x$
2 $- e^{- y} = e^{x} - C \Rightarrow e^{x} + e^{- y} = C$

Final answer

Option (A) — eˣ + e⁻ʸ = C.

Step-by-step solution

1 $IF = e^{\int t a n x d x} = e^{- l n c o s x} = sec x$

Final answer

Option (B) — sec x.

Step-by-step solution

1A function F(x,y) is homogeneous of degree n if F(tx, ty) = tⁿ F(x,y).
2(D): numerator y², denominator (x² − xy − y²)... divide dy/dx form: $\frac{d x}{d y} = \frac{x ^{2} - x y - y ^{2}}{- y ^{2}}$
3Check each: (A) has constant terms — not homogeneous; (B) terms of degrees 2 and 3 — not; (C) terms of degree 3 and 2 — not; (D) every term has degree 2 — homogeneous.

Final answer

Option (D) — y²dx + (x² − xy − y²)dy = 0.

67

Chapter 10 — Vector Algebra

68

Exercise 10.1 — Basic Concepts

5Exercise questions

Step-by-step solution

1Choose a scale, e.g. 10 km per cm, and draw the north line (y-axis) as the reference.
2The displacement is an arrow of length 4 cm making 30° with the north direction, swung towards the west.

Final answer

An arrow of 40 km magnitude pointing 30° west of north (drawn 4 cm long at scale 10 km = 1 cm).

Step-by-step solution

1A scalar has only magnitude; a vector has magnitude and direction.
2(i) 10 kg — magnitude only → scalar. (ii) 2 metres north-west — has direction → vector.
3(iii) 40°, (iv) 40 watt, (v) 10⁻¹⁹ coulomb carry no direction → scalars.
4(vi) 20 m/s² (an acceleration value of the vector kind) → vector.

Final answer

(i) scalar (ii) vector (iii) scalar (iv) scalar (v) scalar (vi) vector.

Final answer

(i) scalar (ii) scalar (iii) vector (iv) vector (v) scalar.

Step-by-step solution

1Coinitial vectors share the same initial point — here a and d start from the same point.
2Equal vectors have the same magnitude and direction — here b and d are equal.
3Collinear but not equal — a and c lie on the same line but face opposite directions, so they are not equal.

Final answer

(i) a and d; (ii) b and d; (iii) a and c.

Step-by-step solution

1(i) a and −a lie on the same line (opposite directions), so they are collinear → True.
2(ii) Collinearity says nothing about lengths — the vectors need not be equal in magnitude → False.
3(iii) Equal magnitude does not force a common line → False.
4(iv) Collinear vectors of equal magnitude may point in opposite directions (a and −a) → False.

Final answer

(i) True (ii) False (iii) False (iv) False.

69

Exercise 10.2 — Addition of Vectors and Multiplication of a Vector by a Scalar

19Exercise questions

Step-by-step solution

1|a| = √(1² + 1² + 1²) = √3.
2|b| = √(4 + 49 + 9) = √62.
3|c| = √(1/3 + 1/3 + 1/3) = √1 = 1.

Final answer

|a| = √3, |b| = √62 and |c| = 1.

Step-by-step solution

1Take a = î + ĵ + k̂ and b = î + ĵ − k̂.
2Both have magnitude √(1 + 1 + 1) = √3, but none of their components match exactly, so the vectors are different.

Final answer

a = î + ĵ + k̂ and b = î + ĵ − k̂ (both of magnitude √3).

Step-by-step solution

1Multiply any vector by a positive scalar: let a = î + 2ĵ + 3k̂ and b = 2î + 4ĵ + 6k̂ = 2a.
2A positive scalar multiple has the same direction (and never the reverse), so a and b are parallel.

Final answer

a = î + 2ĵ + 3k̂ and b = 2î + 4ĵ + 6k̂ (b = 2a).

Step-by-step solution

1Vectors are equal exactly when all corresponding components are equal.
2Coefficient of î: x = 2; coefficient of ĵ: y = 3.

Final answer

x = 2, y = 3.

Step-by-step solution

1A vector joining two points is terminal − initial.
2Vector = (−5 − 2)î + (7 − 1)ĵ = −7î + 6ĵ.

Final answer

Scalar components −7 and 6; vector components −7î and 6ĵ.

Step-by-step solution

1Add coefficients along each direction.
2î-component: 1 − 2 + 1 = 0; ĵ-component: −2 + 4 − 6 = −4; k̂-component: 1 + 5 − 7 = −1.

Final answer

a + b + c = −4ĵ − k̂.

Step-by-step solution

1|a| = √(1 + 1 + 4) = √6.
2The unit vector is a/|a|.

Final answer

\overset{a}{^} = \frac{i ^ + j ^ + 2 k ^}{6}

Step-by-step solution

1PQ = (4 − 1)î + (5 − 2)ĵ + (6 − 3)k̂ = 3î + 3ĵ + 3k̂.
2|PQ| = √(9 + 9 + 9) = 3√3.
3Unit vector = (3î + 3ĵ + 3k̂)/(3√3).

Final answer

\frac{i ^ + j ^ + k ^}{3}

Step-by-step solution

1a + b = (2 − 1)î + (−1 + 1)ĵ + (2 − 1)k̂ = î + k̂.
2|a + b| = √(1 + 1) = √2.
3Unit vector = (î + k̂)/√2.

Final answer

\frac{i ^ + k ^}{2}

Step-by-step solution

1|5î − ĵ + 2k̂| = √(25 + 1 + 4) = √30.
2The unit vector in that direction is (5î − ĵ + 2k̂)/√30.
3Multiply by 8 to get magnitude 8.

Final answer

\frac{8 ( 5 i ^ - j ^ + 2 k ^ )}{30}

Step-by-step solution

1−4î + 6ĵ − 8k̂ = −2(2î − 3ĵ + 4k̂).
2One vector is a scalar multiple of the other, so they lie on the same line — they are collinear.

Final answer

Collinear, since −4î + 6ĵ − 8k̂ = −2(2î − 3ĵ + 4k̂).

Step-by-step solution

1|a| = √(1 + 4 + 9) = √14.
2Direction cosines are the components divided by the magnitude.

Final answer

l = 1/√14, m = 2/√14, n = 3/√14.

Step-by-step solution

1AB = (−1 − 1)î + (−2 − 2)ĵ + (1 + 3)k̂ = −2î − 4ĵ + 4k̂.
2|AB| = √(4 + 16 + 16) = √36 = 6.
3Divide each component by 6.

Final answer

Direction cosines are −1/3, −2/3, 2/3.

Step-by-step solution

1|î + ĵ + k̂| = √3, so the direction cosines are 1/√3, 1/√3, 1/√3.
2cos α = cos β = cos γ = 1/√3, hence α = β = γ — equal inclinations to the three axes.

Final answer

All three direction cosines are 1/√3, so the vector is equally inclined to OX, OY, OZ.

Step-by-step solution

1Internal division: R = (2Q + 1P)/(2 + 1).
22Q + P = 2(−î + ĵ + k̂) + (î + 2ĵ − k̂) = (−2î + 2ĵ + 2k̂) + (î + 2ĵ − k̂) = −î + 4ĵ + k̂.
3So R = (−î + 4ĵ + k̂)/3.
4External division: R = (2Q − 1P)/(2 − 1) = 2(−î + ĵ + k̂) − (î + 2ĵ − k̂) = −3î + 0ĵ + 3k̂.

Final answer

(i) (−î + 4ĵ + k̂)/3; (ii) −3î + 3k̂.

Step-by-step solution

1Mid point = (P + Q)/2.
2= ((2 + 4)/2, (3 + 1)/2, (4 − 2)/2) = (3, 2, 1).

Final answer

3î + 2ĵ + k̂ (point (3, 2, 1)).

Step-by-step solution

1AB = b − a = −î + 3ĵ + 5k̂, so |AB|² = 1 + 9 + 25 = 35.
2BC = c − b = −î − 2ĵ − 6k̂, so |BC|² = 1 + 4 + 36 = 41.
3CA = a − c = 2î − ĵ + k̂, so |CA|² = 4 + 1 + 1 = 6.
435 + 6 = 41, i.e. |AB|² + |CA|² = |BC|² — Pythagoras holds, so the angle at A is right.

Final answer

The triangle is right angled at A.

Step-by-step solution

1(A) The sum of the side vectors around a triangle is the zero vector → true.
2(B) AC = AB + BC, so AB + BC − AC = 0 → true.
3(C) AB + BC − CA = AC − CA = 2AC ≠ 0 → not true.
4(D) AB − CB + CA = AB + BC + CA = 0 → true.

Final answer

Option (C).

Step-by-step solution

1(A) Collinear vectors are scalar multiples: b = λa → correct.
2(C) Scalar multiples have proportional components → correct.
3(B) a = −b is only true when the magnitudes are equal → incorrect in general.
4(D) Collinear vectors may perfectly well point in opposite directions → incorrect.

Final answer

Incorrect: (B) and (D).

70

Exercise 10.3 — Dot (Scalar) Product of Vectors

18Exercise questions

Step-by-step solution

1cos θ = (a·b)/(|a||b|).
2cos θ = √6/(√3 × 2) = √6/(2√3) = 1/√2.
3θ = π/4.

Final answer

θ = \frac{π}{4}

Step-by-step solution

1a·b = 1·3 + (−2)(−2) + 3·1 = 3 + 4 + 3 = 10.
2|a| = √(1 + 4 + 9) = √14 and |b| = √(9 + 4 + 1) = √14.
3cos θ = 10/(√14·√14) = 10/14 = 5/7.

Final answer

θ = cos⁻¹(5/7).

Step-by-step solution

1Projection of a on b = (a·b)/|b|.
2a·b = (1)(1) + (−1)(1) + (0)(0) = 0.
3Projection = 0/√2 = 0.

Final answer

0.

Step-by-step solution

1Projection of a on b = (a·b)/|b|.
2a·b = 7 − 3 + 56 = 60.
3|b| = √(49 + 1 + 64) = √114.

Final answer

\frac{60}{114}

Step-by-step solution

1|2î + 3ĵ + 6k̂|² = 4 + 9 + 36 = 49, so the first vector has magnitude √49/7 = 1; likewise the other two squares of norms are 49.
2(2î + 3ĵ + 6k̂)·(3î − 6ĵ + 2k̂) = 6 − 18 + 12 = 0.
3(3î − 6ĵ + 2k̂)·(6î + 2ĵ − 3k̂) = 18 − 12 − 6 = 0.
4(6î + 2ĵ − 3k̂)·(2î + 3ĵ + 6k̂) = 12 + 6 − 18 = 0.

Final answer

Each is a unit vector and every pair has zero dot product — mutually perpendicular.

Step-by-step solution

1(a + b)·(a − b) = |a|² − |b|² = 8.
2Put |a| = 8|b|: 64|b|² − |b|² = 63|b|² = 8.
3|b|² = 8/63, so |b| = √(8/63) and |a| = 8√(8/63).

Final answer

∣ b ∣ = 8/63, ∣ a ∣ = 8 8/63

Step-by-step solution

1Expand: 6(a·a) + 21(a·b) − 10(b·a) − 35(b·b).
2Use a·a = |a|² and a·b = b·a.

Final answer

6∣ a ∣^{2} + 11 (a \cdot b) - 35∣ b ∣^{2}

Step-by-step solution

1a·b = |a||b| cos 60° = |a|² × ½ (magnitudes equal).
2½|a|² = ½ ⇒ |a|² = 1.

Final answer

|a| = |b| = 1.

Step-by-step solution

1(x − a)·(x + a) = |x|² − |a|² = |x|² − 1 (|a| = 1).
2|x|² − 1 = 12 ⇒ |x|² = 13.

Final answer

|x| = √13.

Step-by-step solution

1a + λb = (2 − λ)î + (2 + 2λ)ĵ + (3 + λ)k̂.
2Perpendicularity: (a + λb)·c = 0.
3(2 − λ)(3) + (2 + 2λ)(1) = 6 − 3λ + 2 + 2λ = 8 − λ = 0.

Final answer

λ = 8.

Step-by-step solution

1Dot the two vectors: (|a| b + |b| a)·(|a| b − |b| a).
2= |a|²(b·b) − |a||b|(b·a) + |a||b|(a·b) − |b|²(a·a).
3= |a|²|b|² − |a||b|(a·b) + |a||b|(a·b) − |b|²|a|² = 0.

Final answer

The dot product is zero, so the vectors are perpendicular.

Step-by-step solution

1a·a = |a|² = 0 forces a = 0 (the zero vector).
2Then a·b = 0·b = 0 holds for every vector b — no restriction.

Final answer

a is the zero vector; b is arbitrary.

Step-by-step solution

1|a + b + c|² = 0.
20 = |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 3 + 2S.

Final answer

a \cdot b + b \cdot c + c \cdot a = - \frac{3}{2}

Step-by-step solution

1Take a = î and b = ĵ, both non-zero vectors.
2a·b = 1·1 + 0·0 + 0·0 = 0, yet neither vector is the zero vector.

Final answer

a = î and b = ĵ give a·b = 0 with a, b both non-zero — the converse fails.

Step-by-step solution

1∠ABC is the angle between BA and BC.
2BA = A − B = 2î + 2ĵ + 3k̂ and BC = C − B = î + ĵ + 2k̂.
3BA·BC = 2 + 2 + 6 = 10; |BA| = √17, |BC| = √6.
4cos(∠ABC) = 10/(√17√6) = 10/√102.

Final answer

∠ABC = cos⁻¹(10/√102).

Step-by-step solution

1AB = (2 − 1)î + (6 − 2)ĵ + (3 − 7)k̂ = î + 4ĵ − 4k̂.
2BC = (3 − 2)î + (10 − 6)ĵ + (−1 − 3)k̂ = î + 4ĵ − 4k̂ = AB.
3The shared point B and the equal directions place A, B, C on one line.

Final answer

AB = BC, so A, B, C are collinear.

Step-by-step solution

1Treat them as position vectors of points A, B, C.
2AB = b − a = −î − 2ĵ − 6k̂, |AB|² = 41.
3BC = c − b = 2î − ĵ + k̂, |BC|² = 6.
4CA = a − c = −î + 3ĵ + 5k̂, |CA|² = 35.
541 = 6 + 35 ⇒ |AB|² = |BC|² + |CA|², right angle at C.

Final answer

The triangle is right angled at C.

Step-by-step solution

1|λa| = |λ| |a| = |λ| a.
2For a unit vector: |λ| a = 1 ⇒ a = 1/|λ|.

Final answer

Option (D).

71

Exercise 10.4 — Cross (Vector) Product of Vectors

12Exercise questions

Step-by-step solution

1a × b = ((−7)(2) − (7)(−2))î + ((7)(3) − (1)(2))ĵ + ((1)(−2) − (−7)(3))k̂.
2= (−14 + 14)î + (21 − 2)ĵ + (−2 + 21)k̂ = 19ĵ + 19k̂.
3|a × b| = √(0 + 361 + 361) = √722 = 19√2.

Final answer

19√2.

Step-by-step solution

1a + b = 4î + 4ĵ and a − b = 2î + 4k̂.
2(a + b) × (a − b) = (4·4 − 0·0)î + (0·2 − 4·4)ĵ + (4·4 − 4·2)k̂ = 16î − 16ĵ + 8k̂ — wait, recompute the third component: 4·4 − 4·2 = 16 − 8 = 8.
3Actually: (a+b) × (a−b) = 16î − 16ĵ + 8k̂? Checking with determinants: a+b = (4,4,0), a−b = (2,0,4). First component: 4·4 − 0·0 = 16. Second: 0·2 − 4·4 = −16. Third: 4·0 − 4·2 = −8. So (16, −16, −8).
4|(a+b) × (a−b)| = √(256 + 256 + 64) = √576 = 24.
5Unit vector = (16î − 16ĵ − 8k̂)/24.

Final answer

\frac{2 i ^ - 2 j ^ - k ^}{3}

Step-by-step solution

1For unit vectors, sum of squares of direction cosines = 1.
2cos²(π/3) + cos²(π/4) + cos²θ = 1 ⇒ 1/4 + 1/2 + cos²θ = 1 ⇒ cos²θ = 1/4.
3θ acute ⇒ cos θ = 1/2 ⇒ θ = π/3.
4Components of a are the cosines of the three angles.

Final answer

θ = π /3, a = \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{2} \hat{k}

Step-by-step solution

1(a − b) × (a + b) = a×a + a×b − b×a − b×b.
2a×a = 0 and b×b = 0 (cross product of a vector with itself).
3−b×a = a×b (anticommutativity).

Final answer

(a − b) × (a + b) = 0 + a×b + a×b − 0 = 2(a × b).

Step-by-step solution

1A zero cross product means the two vectors are parallel.
2(2, 6, 27) = 2(1, λ, μ).
3λ = 6/2 = 3 and μ = 27/2.

Final answer

λ = 3, μ = \frac{27}{2}

Step-by-step solution

1a × b = 0 ⇒ a ∥ b (for non-zero vectors).
2But parallel non-zero vectors have |a·b| = |a||b| ≠ 0, contradicting a·b = 0.
3Hence at least one of a, b must be the zero vector.

Final answer

Either a = 0 or b = 0.

Step-by-step solution

1Write b + c = (b₁ + c₁)î + (b₂ + c₂)ĵ + (b₃ + c₃)k̂.
2First component of a × (b + c) = a₂(b₃ + c₃) − a₃(b₂ + c₂) = (a₂b₃ − a₃b₂) + (a₂c₃ − a₃c₂).
3The first bracket is the î-component of a × b and the second bracket the î-component of a × c.
4The same splitting works for the ĵ and k̂ components, proving the identity.

Final answer

a × (b + c) = a × b + a × c (right-distributive law verified).

Step-by-step solution

1Converse claims a × b = 0 forces a = 0 or b = 0.
2Take a = î and b = 2î — neither is the zero vector.
3a × b = î × (2î) = 2(î × î) = 0.

Final answer

Converse is false: parallel non-zero vectors, e.g. î and 2î, have zero cross product.

Step-by-step solution

1AB = î + 2ĵ + 3k̂ and AC = 4ĵ + 3k̂.
2AB × AC = (2·3 − 3·4)î + (3·0 − 1·3)ĵ + (1·4 − 2·0)k̂ = −6î − 3ĵ + 4k̂.
3|AB × AC| = √(36 + 9 + 16) = √61.
4Area = ½|AB × AC|.

Final answer

\frac{61}{2}

Step-by-step solution

1a × b = ((−1)(1) − 3(−7))î + (3·2 − 1·1)ĵ + (1(−7) − (−1)(2))k̂.
2= (−1 + 21)î + (6 − 1)ĵ + (−7 + 2)k̂ = 20î + 5ĵ − 5k̂.
3Area = |a × b| = √(400 + 25 + 25) = √450.

Final answer

Area = 15√2.

Step-by-step solution

1|a × b| = |a||b| sin θ = 3·(√2/3) sin θ = √2 sin θ.
2Unit vector requires √2 sin θ = 1 ⇒ sin θ = 1/√2.
3θ = π/4.

Final answer

Option (B).

Step-by-step solution

1AB = 2î and AD = −ĵ (the rectangle lies in the plane z = 4).
2Area = |AB × AD| = |2î × (−ĵ)| = 2·|î × ĵ| = 2.

Final answer

Option (C) — area 2.

72

Miscellaneous Exercise on Chapter 10

26Exercise questions

Step-by-step solution

1A vector in the XY-plane has the form xî + yĵ.
2As a unit vector, x = cos 30° and y = sin 30°.

Final answer

\frac{3}{2} \hat{i} + \frac{1}{2} \hat{j}

Step-by-step solution

1PQ = Q − P = (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂.
2Scalar components are the three coordinate differences.
3Magnitude = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²).

Final answer

P Q = (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k}, ∣ P Q ∣ = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

Step-by-step solution

1Let î point east and ĵ north. West walk: −4î.
23 km at 30° east of north: 3 sin 30°î + 3 cos 30°ĵ = (3/2)î + (3√3/2)ĵ.
3Total displacement d = (−4 + 3/2)î + (3√3/2)ĵ = −(5/2)î + (3√3/2)ĵ.
4|d| = √(25/4 + 27/4) = √13.

Final answer

Displacement −(5/2)î + (3√3/2)ĵ; magnitude √13 km.

Step-by-step solution

12a − b + 3c = 2(1,1,1) − (2,−1,3) + 3(1,−2,1).
2= (2 − 2 + 3)î + (2 + 1 − 6)ĵ + (2 − 3 + 3)k̂ = 3î − 3ĵ + 2k̂.
3|3î − 3ĵ + 2k̂| = √(9 + 9 + 4) = √22.

Final answer

\frac{3 i ^ - 3 j ^ + 2 k ^}{22}

Step-by-step solution

1a + b = 4î + 3ĵ − 2k̂.
2|a + b| = √(16 + 9 + 4) = √29.

Final answer

\frac{4 i ^ + 3 j ^ - 2 k ^}{29}

Step-by-step solution

1The diagonals are a + b and b − a.
2a + b = 3î + 6ĵ − 2k̂, with |a + b| = √(9 + 36 + 4) = 7.
3b − a = î + 2ĵ − 8k̂, with |b − a| = √(1 + 4 + 64) = √69.

Final answer

Unit vectors (3î + 6ĵ − 2k̂)/7 and (î + 2ĵ − 8k̂)/√69.

Step-by-step solution

1AB = 4î + 2ĵ + 6k̂ and BC = 6î + 3ĵ + 9k̂ = (3/2)AB.
2Parallel directions through B ⇒ A, B, C collinear.
3|AB| = √(16 + 4 + 36) = √56 = 2√14; |BC| = √(36 + 9 + 81) = √126 = 3√14.
4So AB : BC = 2 : 3 — B divides AC internally in the ratio 2 : 3.

Final answer

Collinear; B divides AC in the ratio 2 : 3.

Step-by-step solution

1External division (m : n) = (1 : 2): R = (mQ − nP)/(m − n) = (Q − 2P)/(1 − 2).
2R = 2P − Q = 2(2a + b) − (a − 3b) = 4a + 2b − a + 3b = 3a + 5b.
3Mid-point of RQ: (R + Q)/2 = (3a + 5b + a − 3b)/2 = (4a + 2b)/2 = 2a + b = P.

Final answer

R = 3a + 5b, and P is the mid-point of RQ.

Step-by-step solution

1d perpendicular to both a and b ⇒ d is parallel to a × b.
2a × b = (5·5 − (−1)(−4))î + ((−1)(1) − 4·5)ĵ + (4(−4) − 5·1)k̂ = 21î − 21ĵ − 21k̂.
3Write d = λ(î − ĵ − k̂).
4c·d = λ(3 − 1 + 1) = 3λ = 21 ⇒ λ = 7.

Final answer

d = 7î − 7ĵ − 7k̂.

Step-by-step solution

1The scalar triple product [a b c] = a·(b × c).
2b × c = (1(1 + x − y) − (1 − x)x)î + ((1 − x)y − x(1 + x − y))ĵ + (x·x − 1·y)k̂.
3= (1 + x − y − x + x²)î + (y − xy − x − x² + xy)ĵ + (x² − y)k̂ = (1 − y + x²)î + (y − x − x²)ĵ + (x² − y)k̂.
4a·(b × c) = 1·(1 − y + x²) + 0·(⋯) + (−1)(x² − y) = 1 − y + x² − x² + y = 1.

Final answer

[a b c] = 1 — independent of x and y.

Step-by-step solution

1Equal inclinations ⇒ direction cosines l = m = n.
23l² = 1 ⇒ l = m = n = ±1/√3.
3r = (2√6)·(±1/√3)(î + ĵ + k̂) = ±2√2(î + ĵ + k̂).

Final answer

r = \pm 22 (\hat{i} + \hat{j} + \hat{k})

Step-by-step solution

1|x(î + ĵ + k̂)| = |x|·√3.
2Unit vector: |x|√3 = 1.

Final answer

x = \pm \frac{1}{3}

Step-by-step solution

1AB = î − 3ĵ + k̂ and AC = 3î + 3ĵ − 4k̂.
2AB × AC = ((−3)(−4) − 1·3)î + (1·3 − 1(−4))ĵ + (1·3 − (−3)(3))k̂ = 9î + 7ĵ + 12k̂.
3|AB × AC| = √(81 + 49 + 144) = √274.
4Area = ½|AB × AC|.

Final answer

\frac{274}{2}

Step-by-step solution

1Cross a + b + c = 0 with b: (a + b + c) × b = 0.
2a × b + b × b + c × b = 0 ⇒ a × b + 0 − b × c = 0 ⇒ a × b = b × c.
3Cross a + b + c = 0 with c: (a + b + c) × c = 0 ⇒ a × c + b × c + 0 = 0 ⇒ b × c = c × a.

Final answer

a × b = b × c = c × a.

Step-by-step solution

1|a × b| = |a||b| sin θ and a·b = |a||b| cos θ.
2|a||b| sin θ = √3 |a||b| cos θ ⇒ tan θ = √3.
3θ = π/3.

Final answer

Option (B).

Step-by-step solution

1|a + b|² = |a|² + |b|² + 2a·b = 1 + 1 + 2 cos θ.
2For a unit vector: 2 + 2 cos θ = 1 ⇒ cos θ = −1/2.
3θ = 2π/3.

Final answer

Option (D).

Step-by-step solution

1a·b = |a||b| cos θ ≥ 0 ⟺ cos θ ≥ 0 ⟺ 0 ≤ θ ≤ π/2.

Final answer

Option (B).

Step-by-step solution

1|λa| = |λ|·a.
2Unit: |λ| a = 1 ⇒ a = 1/|λ|.

Final answer

Option (D).

Step-by-step solution

1Let adjacent sides be a and b; the diagonals are a + b and a − b.
2|a + b|² = |a|² + |b|² + 2a·b and |a − b|² = |a|² + |b|² − 2a·b.
3Adding: |a + b|² + |a − b|² = 2(|a|² + |b|²).
4The right side is the sum of the squares of the four sides (each side appears with |a| or |b| twice).

Final answer

Sum of squares of diagonals = 2(|a|² + |b|²) = sum of squares of the sides.

Step-by-step solution

1Let the vertices have position vectors a, b, c; the mid-point of BC is (b + c)/2.
2The point on the median from A which divides it in the ratio 2 : 1 has position vector (a + 2·(b + c)/2)/(1 + 2) = (a + b + c)/3.
3The same point (a + b + c)/3 arises from each vertex — the three medians pass through it.

Final answer

All three medians meet at the centroid (a + b + c)/3, so they are concurrent.

Step-by-step solution

1Let AB be a diameter with centre O and radius a; choose O as origin so A = −a, B = a, and P = p with |p| = a.
2PA = −a − p and PB = a − p.
3PA·PB = (−a − p)·(a − p) = −|a|² + a·p − p·a + |p|² = −a² + a² = 0.
4Zero dot product means PA ⊥ PB.

Final answer

Every angle subtended by a diameter is a right angle.

Step-by-step solution

1Square both sides: |a + b|² = |a − b|².
2|a|² + |b|² + 2a·b = |a|² + |b|² − 2a·b.
34a·b = 0 ⇒ a·b = 0 ⇒ a ⊥ b.

Final answer

a·b = 0, so a ⊥ b.

Step-by-step solution

1Let |a| = |b| = |c| = k with all pairs perpendicular.
2(a + b + c)·a = |a|² = k², and |a + b + c|² = 3k², so |a + b + c| = √3 k.
3cos of angle with a = k²/(k·√3 k) = 1/√3; likewise for b and c.
4Equal cosines ⇒ a + b + c is equally inclined to a, b and c.

Final answer

The vector a + b + c makes angle cos⁻¹(1/√3) with each of a, b, c.

Step-by-step solution

1Write c = xî + yĵ + zk̂. Then a·c = x + y + z = 3.
2a × c = (z − y)î + (x − z)ĵ + (y − x)k̂ = ĵ − k̂.
3z − y = 0 ⇒ z = y; x − z = 1 ⇒ x = y + 1; (y − x = −1 is then automatic).
4(y + 1) + y + y = 3 ⇒ y = 2/3 ⇒ z = 2/3, x = 5/3.

Final answer

c = \frac{5 i ^ + 2 j ^ + 2 k ^}{3}

Step-by-step solution

1Scalar projection = (a·b)/|a|.
2a·b = 2 − 3 − 1 = −2 and |a| = √(4 + 9 + 1) = √14.
3Scalar projection = −2/√14 = −√14/7.
4Vector projection = (a·b/|a|²)a = (−2/14)(2î + 3ĵ − k̂) = −(2î + 3ĵ − k̂)/7.

Final answer

Scalar projection −√14/7; vector projection −(2î + 3ĵ − k̂)/7.

Step-by-step solution

1a·b = 12 − 6 − 2 = 4.
2|a| = √(4 + 4 + 1) = 3 and |b| = √(36 + 9 + 4) = 7.
3cos θ = 4/(3·7) = 4/21.

Final answer

θ = cos⁻¹(4/21).

73

Chapter 11 — Three Dimensional Geometry

74

Exercise 11.1 — Direction Cosines

7Exercise questions

Step-by-step solution

1The direction cosines are l = cosα, m = cosβ, n = cosγ where α, β, γ are the angles with the three positive axes.
2l = cos90° = 0, m = cos135° = −1/√2, n = cos45° = 1/√2.
3Check: l² + m² + n² = 0 + 1/2 + 1/2 = 1, as required of any set of direction cosines.

Final answer

Direction cosines are (0, −1/√2, 1/√2).

Step-by-step solution

1Let each equal angle be α; then l = m = n = cosα.
2l² + m² + n² = 1 gives 3cos²α = 1, so cosα = ±1/√3.

Final answer

(±1/√3, ±1/√3, ±1/√3); taking the acute direction, (1/√3, 1/√3, 1/√3).

Step-by-step solution

1cosα = cos90° = 0 and cosβ = cos60° = 1/2.
2l² + m² + n² = 1 gives 0 + 1/4 + cos²γ = 1, so cos²γ = 3/4 and cosγ = ±√3/2.
3Hence γ = 30° or γ = 150°.

Final answer

The angle with the positive z-axis is 30° or 150°.

Step-by-step solution

1Direction ratios of the segment from (2,3,4) to (−1,−2,1): (−3,−5,−3).
2Direction ratios of the segment from (−1,−2,1) to (5,8,7): (6,10,6).
3Since (6,10,6) = −2(−3,−5,−3), the two segments are parallel; they share the point (−1,−2,1), so the three points are collinear.

Final answer

Yes, the three points are collinear.

Step-by-step solution

1Direction ratios of AB = B − A = (−4,−4,6).
2|AB| = √(16+16+36) = √68 = 2√17.
3Direction cosines = (−4/(2√17), −4/(2√17), 6/(2√17)) = (−2/√17, −2/√17, 3/√17).

Final answer

AB: (−2/√17, −2/√17, 3/√17).

Step-by-step solution

1Direction ratios of BC = C − B = (−4,−6,−4).
2|BC| = √(16+36+16) = √68 = 2√17.
3Direction cosines = (−2/√17, −3/√17, −2/√17).

Final answer

BC: (−2/√17, −3/√17, −2/√17).

Step-by-step solution

1Direction ratios of CA = A − C = (8,10,−2).
2|CA| = √(64+100+4) = √168 = 2√42.
3Direction cosines = (8/(2√42), 10/(2√42), −2/(2√42)) = (4/√42, 5/√42, −1/√42).

Final answer

CA: (4/√42, 5/√42, −1/√42).

75

Exercise 11.2 — Equation of a Line in Space

19Exercise questions

Step-by-step solution

1Two lines are perpendicular when the dot product of their direction ratios is zero.
2(12,−3,−4)·(4,12,3) = 48 − 36 − 12 = 0.
3(4,12,3)·(3,−4,12) = 12 − 48 + 36 = 0.
4(12,−3,−4)·(3,−4,12) = 36 + 12 − 48 = 0.

Final answer

Each pair is perpendicular, so the three lines are mutually perpendicular.

Step-by-step solution

1Direction ratios of the first line: (3−1, 4+1, −2−2) = (2,5,−4).
2Direction ratios of the second line: (3−0, 5−3, 6−2) = (3,2,4).
3Dot product = 2·3 + 5·2 + (−4)·4 = 6 + 10 − 16 = 0, so the lines are perpendicular.

Final answer

The two lines are perpendicular.

Step-by-step solution

1Direction ratios of the first line: (2−4, 3−7, 4−8) = (−2,−4,−4).
2Direction ratios of the second line: (1+1, 2+2, 5−1) = (2,4,4).
3Since (2,4,4) = −1·(−2,−4,−4), the direction ratios are proportional and the lines are parallel.

Final answer

The two lines are parallel.

Step-by-step solution

1A line through the point with position vector a in the direction of b is r = a + λb.
2Here a = i + 2j + 3k and b = 3i + 2j − 2k.
3Hence the vector equation is r = (i + 2j + 3k) + λ(3i + 2j − 2k).

Final answer

r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (3 \hat{i} + 2 \hat{j} - 2 \hat{k})

Step-by-step solution

1From r = (i + 2j + 3k) + λ(3i + 2j − 2k): x = 1 + 3λ, y = 2 + 2λ, z = 3 − 2λ.
2Eliminate λ: (x − 1)/3 = (y − 2)/2 = (z − 3)/(−2).

Final answer

\frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z - 3}{- 2}

Step-by-step solution

1Vector form: r = a + λb with a = 2i − j + 4k and b = i + 2j − k.
2So r = (2i − j + 4k) + λ(i + 2j − k).
3Cartesian form: (x − 2)/1 = (y + 1)/2 = (z − 4)/(−1).

Final answer

Vector: r = (2i − j + 4k) + λ(i + 2j − k). Cartesian: (x−2)/1 = (y+1)/2 = (z−4)/(−1).

Step-by-step solution

1The given line has direction ratios (3,5,6); a parallel line has the same direction ratios.
2Through (−2,4,−5): (x + 2)/3 = (y − 4)/5 = (z + 5)/6.

Final answer

\frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6}

Step-by-step solution

1Read from the cartesian equation: a point on the line is (5,−4,6) and direction ratios are (3,7,2).
2Vector form: r = (5i − 4j + 6k) + λ(3i + 7j + 2k).

Final answer

r = (5 \hat{i} - 4 \hat{j} + 6 \hat{k}) + λ (3 \hat{i} + 7 \hat{j} + 2 \hat{k})

Step-by-step solution

1Take a = 2i − j − 3k (the first point).
2Direction b = second point − first point = (−2−2)i + (2+1)j + (5+3)k = −4i + 3j + 8k.
3Vector equation: r = (2i − j − 3k) + λ(−4i + 3j + 8k).

Final answer

r = (2 \hat{i} - \hat{j} - 3 \hat{k}) + λ (- 4 \hat{i} + 3 \hat{j} + 8 \hat{k})

Step-by-step solution

1The direction ratios are (−4,3,8) and the line passes through (2,−1,−3).
2Cartesian form: (x − 2)/(−4) = (y + 1)/3 = (z + 3)/8.

Final answer

\frac{x - 2}{- 4} = \frac{y + 1}{3} = \frac{z + 3}{8}

Step-by-step solution

1Direction AB = (1−3, 2−4, −7+6) = (−2,−2,−1).
2Vector: r = (3i + 4j − 6k) + λ(−2i − 2j − k).
3Cartesian: (x − 3)/(−2) = (y − 4)/(−2) = (z + 6)/(−1).

Final answer

Vector: r = (3i + 4j − 6k) + λ(−2i − 2j − k). Cartesian: (x−3)/(−2) = (y−4)/(−2) = (z+6)/(−1).

Step-by-step solution

1The angle between two lines equals the angle between their direction vectors.
2b₁·b₂ = 3·1 + 2·2 + 6·2 = 19.
3|b₁| = √(9+4+36) = 7 and |b₂| = √(1+4+4) = 3.
4cosθ = 19/(7·3) = 19/21.

Final answer

θ = cos⁻¹(19/21).

Step-by-step solution

1Direction vectors b₁ = i + 2j + 2k and b₂ = 3i + 2j + 6k.
2b₁·b₂ = 3 + 4 + 12 = 19; |b₁| = 3, |b₂| = 7.
3cosθ = 19/(3·7) = 19/21.

Final answer

θ = cos⁻¹(19/21).

Step-by-step solution

1Rewrite line 1: (x−1)/(−3) = (y−2)/(2p/7) = (z−3)/2, so its direction ratios are (−3, 2p/7, 2).
2Rewrite line 2: (x−1)/(−3p/7) = (y−5)/1 = (z−6)/(−5), so its direction ratios are (−3p/7, 1, −5).
3Right angles require the dot product to vanish: (−3)(−3p/7) + (2p/7)(1) + (2)(−5) = 0.
49p/7 + 2p/7 − 10 = 0 ⇒ 11p/7 = 10 ⇒ p = 70/11.

Final answer

p = 70/11.

Step-by-step solution

1Direction ratios of the first line: (7,−5,1); of the second: (1,2,3).
2Dot product = 7·1 + (−5)·2 + 1·3 = 7 − 10 + 3 = 0.
3The dot product being zero, the lines are perpendicular.

Final answer

The lines are perpendicular to each other.

Step-by-step solution

1Use d = |(a₂ − a₁)·(b₁ × b₂)| / |b₁ × b₂| with a₁ = (1,2,1), a₂ = (2,−1,−1), b₁ = (1,−1,1), b₂ = (2,1,2).
2a₂ − a₁ = (1,−3,−2); b₁ × b₂ = (1,−1,1)×(2,1,2) = (−3,0,3) and |b₁ × b₂| = √18 = 3√2.
3(a₂ − a₁)·(b₁ × b₂) = 1·(−3) + (−3)·0 + (−2)·3 = −9, so |...| = 9.
4d = 9/(3√2) = 3/√2.

Final answer

Shortest distance = 3/√2.

Step-by-step solution

1a₁ = (−1,−1,−1), b₁ = (7,−6,1); a₂ = (3,5,7), b₂ = (1,−2,1).
2a₂ − a₁ = (4,6,8); b₁ × b₂ = (7,−6,1)×(1,−2,1) = (−4,−6,−8); |b₁ × b₂| = √116 = 2√29.
3(a₂ − a₁)·(b₁ × b₂) = 4(−4) + 6(−6) + 8(−8) = −116, so |...| = 116.
4d = 116/(2√29) = 58/√29 = 2√29.

Final answer

Shortest distance = 2√29.

Step-by-step solution

1a₁ = (1,2,3), b₁ = (1,−3,2); a₂ = (4,5,6), b₂ = (2,3,1).
2a₂ − a₁ = (3,3,3); b₁ × b₂ = (1,−3,2)×(2,3,1) = (−9,3,9); |b₁ × b₂| = √171 = 3√19.
3(a₂ − a₁)·(b₁ × b₂) = 3(−9) + 3(3) + 3(9) = 9.
4d = 9/(3√19) = 3/√19.

Final answer

Shortest distance = 3/√19.

Step-by-step solution

1Write as a₁ = i − 2j + 3k, b₁ = −i + j − 2k; a₂ = i − j − k, b₂ = i + 2j − 2k.
2a₂ − a₁ = (0,1,−4); b₁ × b₂ = (−1,1,−2)×(1,2,−2) = (2,−4,−3); |b₁ × b₂| = √29.
3(a₂ − a₁)·(b₁ × b₂) = 0·2 + 1·(−4) + (−4)(−3) = 8.
4d = 8/√29.

Final answer

Shortest distance = 8/√29.

76

Exercise 11.3 — Planes

33Exercise questions

Step-by-step solution

1Write the plane in the form ax + by + cz = d: here 0·x + 0·y + 1·z = 2.
2Normal vector (0,0,1) is already a unit vector, so its direction cosines are (0,0,1).
3Distance of the plane from the origin = d/|n| = 2/1 = 2.

Final answer

Direction cosines (0,0,1); distance 2.

Step-by-step solution

1n = (1,1,1) gives |n| = √3.
2Direction cosines = (1/√3, 1/√3, 1/√3).
3Distance = d/|n| = 1/√3.

Final answer

Direction cosines (1/√3, 1/√3, 1/√3); distance 1/√3.

Step-by-step solution

1n = (2,3,−1), |n| = √(4+9+1) = √14.
2Direction cosines = (2/√14, 3/√14, −1/√14).
3Distance = 5/√14.

Final answer

Direction cosines (2/√14, 3/√14, −1/√14); distance 5/√14.

Step-by-step solution

1The plane is 5y = −8, i.e. y = −8/5.
2Normal (0,1,0) is a unit vector, so the direction cosines are (0,1,0).
3Distance from the origin = |−8/5| = 8/5.

Final answer

Direction cosines (0,1,0); distance 8/5.

Step-by-step solution

1The plane is r·n̂ = p where p = 7 is the distance from the origin.
2|n| = √(9+25+36) = √70, so r·(3i + 5j − 6k) = 7·√70.

Final answer

r \cdot (3 \hat{i} + 5 \hat{j} - 6 \hat{k}) = 770

Step-by-step solution

1For r = xi + yj + zk, r·(i + j − k) = x + y − z.
2Hence the cartesian equation is x + y − z = 2.

Final answer

x + y − z = 2.

Step-by-step solution

1r·(2i + 3j − 4k) = 2x + 3y − 4z.
2Hence 2x + 3y − 4z = 1.

Final answer

2x + 3y − 4z = 1.

Step-by-step solution

1Take the dot product with r = xi + yj + zk.
2(s − 2t)x + (3 − t)y + (2s + t)z = 15.

Final answer

(s − 2t)x + (3 − t)y + (2s + t)z = 15.

Step-by-step solution

1The foot lies on the normal (2,3,4), so it is λ(2,3,4).
2Substitute into the plane: 2(2λ) + 3(3λ) + 4(4λ) = 12 ⇒ 29λ = 12 ⇒ λ = 12/29.
3Foot = (24/29, 36/29, 48/29).

Final answer

(24/29, 36/29, 48/29).

Step-by-step solution

1The foot is λ(0,3,4) on the normal.
2Substitute: 3(3λ) + 4(4λ) = 6 ⇒ 25λ = 6 ⇒ λ = 6/25.
3Foot = (0, 18/25, 24/25).

Final answer

(0, 18/25, 24/25).

Step-by-step solution

1The foot is λ(1,1,1).
2Substitute: λ + λ + λ = 1 ⇒ λ = 1/3.
3Foot = (1/3, 1/3, 1/3).

Final answer

(1/3, 1/3, 1/3).

Step-by-step solution

1The plane is 5y = −8; the foot is λ(0,5,0).
2Substitute: 5(5λ) = −8 ⇒ λ = −8/25.
3Foot = (0, −8/5, 0).

Final answer

(0, −8/5, 0).

Step-by-step solution

1For a plane through a with normal n: r·n = a·n.
2a·n = (i − 2k)·(i + j − k) = 1 + 0 + 2 = 3.
3Vector form: r·(i + j − k) = 3. Cartesian form: x + y − z = 3.

Final answer

r·(i + j − k) = 3, i.e. x + y − z = 3.

Step-by-step solution

1Direction ratios of the segment from (1,1,−1) to (6,4,−5): (5,3,−4).
2Direction ratios of the segment from (6,4,−5) to (−4,−2,3): (−10,−6,8).
3Since (−10,−6,8) = −2(5,3,−4), the three points are collinear, so no unique plane contains them — infinitely many planes do.

Final answer

The points are collinear, so infinitely many planes pass through them.

Step-by-step solution

1u = P₂ − P₁ = (0,1,1), v = P₃ − P₁ = (−3,1,−1).
2n = u × v = (0,1,1)×(−3,1,−1) = (−2,−3,3).
3Plane: −2(x − 1) − 3(y − 1) + 3z = 0 ⇒ 2x + 3y − 3z = 5.

Final answer

2x + 3y − 3z = 5.

Step-by-step solution

1Divide through by 5: x/(5/2) + y/5 + z/(−5) = 1.
2Read off the intercepts: x-intercept 5/2, y-intercept 5, z-intercept −5.

Final answer

Intercepts are 5/2, 5, −5 on the x, y, z axes respectively.

Step-by-step solution

1A plane parallel to ZOX (the plane y = 0) has equation y = c, a constant.
2It cuts the y-axis at 3, so c = 3 and the plane is y = 3.

Final answer

y = 3.

Step-by-step solution

1Family of planes through the line of intersection: (3x − y + 2z − 4) + λ(x + y + z − 2) = 0.
2Put (2,2,1): (6 − 2 + 2 − 4) + λ(2 + 2 + 1 − 2) = 0 ⇒ 2 + 3λ = 0 ⇒ λ = −2/3.
3(3x − y + 2z − 4) − (2/3)(x + y + z − 2) = 0; multiplying by 3 gives 7x − 5y + 4z − 8 = 0.

Final answer

7x − 5y + 4z − 8 = 0.

Step-by-step solution

1Family: [r·(2i + 2j − 3k) − 7] + λ[r·(2i + 5j + 3k) − 9] = 0.
2In coordinates: (2x + 2y − 3z − 7) + λ(2x + 5y + 3z − 9) = 0.
3At (2,1,3): (−10) + λ(9) = 0 ⇒ λ = 10/9.
49(2x + 2y − 3z − 7) + 10(2x + 5y + 3z − 9) = 0 ⇒ 38x + 68y + 3z − 153 = 0.

Final answer

r \cdot (38 \hat{i} + 68 \hat{j} + 3 \hat{k}) = 153

Step-by-step solution

1The angle between planes is the angle between their normals: cosθ = |n₁·n₂|/(|n₁||n₂|).
2n₁·n₂ = 6 − 6 − 15 = −15; |n₁| = √17, |n₂| = √43.
3cosθ = 15/(√17√43) = 15/√731.

Final answer

θ = cos⁻¹(15/√731).

Step-by-step solution

1n₁·n₂ = 2(−1) + (−3)(1) + 4(0) = −5.
2|n₁| = √29, |n₂| = √2.
3cosθ = 5/√58.

Final answer

θ = cos⁻¹(5/√58).

Step-by-step solution

1n₁·n₂ = 3 + 4 − 18 = −11.
2|n₁| = √14, |n₂| = √(9+4+36) = 7.
3cosθ = 11/(7√14).

Final answer

θ = cos⁻¹(11/(7√14)).

Step-by-step solution

1n₁ = (1,1,1), n₂ = (1,0,−1).
2n₁·n₂ = 1 + 0 − 1 = 0.
3The normals are perpendicular, so θ = 90°.

Final answer

θ = 90°, i.e. the planes are perpendicular.

Step-by-step solution

1n₁·n₂ = 21 − 5 − 60 = −44 ≠ 0 and the normals are not proportional, so the planes are neither parallel nor perpendicular.
2|n₁| = √110, |n₂| = √110.
3cosθ = 44/110 = 2/5.

Final answer

Neither parallel nor perpendicular; angle θ = cos⁻¹(2/5).

Step-by-step solution

1n₁ = (2,1,3), n₂ = (1,−2,0).
2n₁·n₂ = 2 − 2 + 0 = 0.
3The normals are perpendicular, so the planes are perpendicular.

Final answer

Perpendicular.

Step-by-step solution

1The normals are (2,−2,4) and (3,−3,6).
2(3,−3,6) = (3/2)(2,−2,4), so the normals are proportional and the planes are parallel.
3The angle between parallel planes is 0°.

Final answer

Parallel; angle 0°.

Step-by-step solution

1Both planes have normal (2,−1,3), so they are parallel.
2They are distinct planes (constant terms differ), angle between them = 0°.

Final answer

Parallel; angle 0°.

Step-by-step solution

1Distance = |ax₁ + by₁ + cz₁ − d|/√(a²+b²+c²) with the plane written ax + by + cz = d.
2|3·0 − 4·0 + 12·0 − 3| / √(9+16+144) = 3/13.

Final answer

3/13.

Step-by-step solution

1Write the plane as 2x − y + 2z = −3; the signed numerator is 2·3 + (−1)(−2) + 2·1 + 3.
2Numerator = 6 + 2 + 2 + 3 = 13; denominator √(4+1+4) = 3.
3Distance = 13/3.

Final answer

13/3.

Step-by-step solution

1Numerator = |2 + 6 + 10 − 9| = 9.
2Denominator = √(1+4+4) = 3.
3Distance = 9/3 = 3.

Final answer

3.

Step-by-step solution

1Write the plane as 2x − 3y + 6z = 2; numerator = |2(−6) + 0 + 0 − 2| = 14.
2Denominator = √(4+9+36) = 7.
3Distance = 14/7 = 2.

Final answer

2.

Step-by-step solution

1Line coordinates: x = 2 + 3λ, y = −1 + 4λ, z = 2 + 12λ.
2Plane equation: x − y + z = 5. Substitute: (2+3λ) − (−1+4λ) + (2+12λ) = 5 ⇒ 5 + 11λ = 5 ⇒ λ = 0.
3Intersection point: (2,−1,2).
4Required distance = √((2+1)² + (−1+5)² + (2+10)²) = √(9+16+144) = √169 = 13.

Final answer

13 units.

Step-by-step solution

1Family: (x + 2y + 3z − 4) + λ(2x + y − z + 5) = 0; its normal is (1+2λ, 2+λ, 3−λ).
2Perpendicular to 5x + 3y + 6z + 8 = 0 means the normals are perpendicular: 5(1+2λ) + 3(2+λ) + 6(3−λ) = 0.
329 + 7λ = 0 ⇒ λ = −29/7.
4(x + 2y + 3z − 4) − (29/7)(2x + y − z + 5) = 0, i.e. after multiplying by 7: 7x+14y+21z−28−58x−29y+29z+145 = 0.
5Hence 51x + 15y − 50z + 173 = 0.

Final answer

51x + 15y − 50z + 173 = 0.

77

Miscellaneous Exercise on Chapter 11

22Exercise questions

Step-by-step solution

1Direction ratios of the line to (2,1,1) from the origin: (2,1,1).
2Direction ratios of the line through the two given points: (4−3, 3−5, −1+1) = (1,−2,0).
3Dot product = 2·1 + 1·(−2) + 1·0 = 0, so the lines are perpendicular.

Final answer

The two lines are perpendicular.

Step-by-step solution

1The vector perpendicular to both direction vectors is their cross product (l₁,m₁,n₁) × (l₂,m₂,n₂).
2The cross product has components: i(m₁n₂ − m₂n₁) − j(l₁n₂ − l₂n₁) + k(l₁m₂ − l₂m₁).
3Writing its components in order gives m₁n₂ − m₂n₁, n₁l₂ − n₂l₁, l₁m₂ − l₂m₁.
4Since (l₁,m₁,n₁) and (l₂,m₂,n₂) are perpendicular unit vectors, the cross product is a unit vector, so these are exactly the required direction cosines.

Final answer

The perpendicular line has direction cosines (m₁n₂ − m₂n₁, n₁l₂ − n₂l₁, l₁m₂ − l₂m₁).

Step-by-step solution

1cosθ = [a(b−c) + b(c−a) + c(a−b)] / (√(a²+b²+c²)·√((b−c)²+(c−a)²+(a−b)²)).
2Expand the numerator: ab − ac + bc − ab + ca − bc = 0.
3Hence cosθ = 0 and θ = 90°.

Final answer

θ = 90° — the lines are perpendicular.

Step-by-step solution

1A line parallel to the x-axis has direction ratios (1,0,0) and passes through (0,0,0).
2Symmetric form: (x − 0)/1 = (y − 0)/0 = (z − 0)/0, i.e. the x-axis y = 0, z = 0.

Final answer

The line is the x-axis: y = 0, z = 0, i.e. x/1 = y/0 = z/0.

Step-by-step solution

1Direction ratios of AB = (3,3,4); of CD = (6,6,8).
2(6,6,8) = 2(3,3,4), so the lines are parallel.
3The angle between parallel lines is 0°.

Final answer

θ = 0° — the lines AB and CD are parallel.

Step-by-step solution

1Direction ratios of the first line: (−3, 2k, 2); of the second: (3k, 1, −5).
2Perpendicularity: (−3)(3k) + (2k)(1) + (2)(−5) = 0.
3−9k + 2k − 10 = 0 ⇒ −7k = 10 ⇒ k = −10/7.

Final answer

k = −10/7.

Step-by-step solution

1The normal to the plane is i + 2j − 5k, and the line is perpendicular to the plane, so it is parallel to this normal.
2Line through a = i + 2j + 3k parallel to b = i + 2j − 5k: r = (i + 2j + 3k) + λ(i + 2j − 5k).

Final answer

r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} + 2 \hat{j} - 5 \hat{k})

Step-by-step solution

1The given plane has normal i + j + k; a parallel plane has the same normal.
2Equation: r·(i + j + k) = (ai + bj + ck)·(i + j + k) = a + b + c.
3Cartesian form: x + y + z = a + b + c.

Final answer

x + y + z = a + b + c.

Step-by-step solution

1a₁ = (1,1,0), b₁ = (2,−1,1); a₂ = (2,1,−1), b₂ = (3,−5,2).
2a₂ − a₁ = (1,0,−1); b₁ × b₂ = (2,−1,1)×(3,−5,2) = (3,−1,−7); |b₁ × b₂| = √(9+1+49) = √59.
3(a₂ − a₁)·(b₁ × b₂) = 1·3 + 0·(−1) + (−1)(−7) = 10.
4d = 10/√59.

Final answer

Shortest distance = 10/√59.

Step-by-step solution

1a₁ = (3,8,3), b₁ = (3,−1,1); a₂ = (−3,−7,6), b₂ = (−3,2,4).
2a₂ − a₁ = (−6,−15,3); b₁ × b₂ = (3,−1,1)×(−3,2,4) = (−6,−15,3); |b₁ × b₂| = √(36+225+9) = √270.
3(a₂ − a₁)·(b₁ × b₂) = (−6)(−6) + (−15)(−15) + (3)(3) = 270.
4d = 270/√270 = √270 = 3√30.

Final answer

Shortest distance = 3√30.

Step-by-step solution

1For a line with direction b and a plane with normal n: sinθ = |b·n|/(|b||n|).
2b = (2,3,6), |b| = 7; n = (10,2,−11), |n| = √(100+4+121) = 15.
3|b·n| = |20 + 6 − 66| = 40; sinθ = 40/(7·15) = 8/21.

Final answer

θ = sin⁻¹(8/21).

Step-by-step solution

1Line direction b = (2,−1,1), |b| = √6; plane normal n = (1,1,0), |n| = √2.
2|b·n| = |2 − 1 + 0| = 1.
3sinθ = 1/(√6·√2) = 1/(2√3).

Final answer

θ = sin⁻¹(1/(2√3)).

Step-by-step solution

1Put the line in parametric form: (13 + 5t, −8 − 8t, 31 + t), with direction b = (5,−8,1).
2For the foot Q, the join from P(−1,3,9) to Q is perpendicular to the line: (Q − P)·b = 0.
3Q − P = (14 + 5t, −11 − 8t, 22 + t); (14+5t)·5 + (−11−8t)(−8) + (22+t)·1 = 0.
470 + 25t + 88 + 64t + 22 + t = 0 ⇒ 180 + 90t = 0 ⇒ t = −2.
5Q = (13−10, −8+16, 31−2) = (3, 8, 29).
6Length = √((3+1)² + (8−3)² + (29−9)²) = √(16+25+400) = √441 = 21.

Final answer

Foot (3,8,29); perpendicular distance 21.

Step-by-step solution

1Family: (x + y + z − 1) + λ(2x + 3y + 4z − 5) = 0 with normal (1+2λ, 1+3λ, 1+4λ).
2Perpendicular to x − y + z = 0 requires (1+2λ) − (1+3λ) + (1+4λ) = 0.
31 + 3λ = 0 ⇒ λ = −1/3.
4(x + y + z − 1) − (1/3)(2x + 3y + 4z − 5) = 0; multiplying by 3: 3x+3y+3z−3−2x−3y−4z+5 = 0.
5Plane: x − z + 2 = 0.

Final answer

x − z + 2 = 0.

Step-by-step solution

1Let the point of division be P = (x₁ + λ(x₂−x₁), y₁ + λ(y₂−y₁), z₁ + λ(z₂−z₁)), giving ratio λ : (1−λ).
2P lies on the plane: ax₁ + by₁ + cz₁ + d + λ[a(x₂−x₁) + b(y₂−y₁) + c(z₂−z₁)] = 0.
3So λ(ax₂ + by₂ + cz₂ + d) = −(ax₁ + by₁ + cz₁ + d), i.e. λ/(1−λ) = −(S₁/S₂) where S₁, S₂ are the two signed values.
4The division ratio is therefore −(ax₁+by₁+cz₁+d) : (ax₂+by₂+cz₂+d).

Final answer

The plane divides the join in the ratio −(ax₁+by₁+cz₁+d) : (ax₂+by₂+cz₂+d).

Step-by-step solution

1The plane is 3x + 4y − 12z + 13 = 0, with √(9+16+144) = 13 as the normal's magnitude.
2Distance of (1,1,p): |3 + 4 − 12p + 13|/13 = |20 − 12p|/13.
3Distance of (−3,0,1): |−9 + 0 − 12 + 13|/13 = 8/13.
4Equate: |20 − 12p|/13 = 8/13 ⇒ 20 − 12p = ±8.
520 − 12p = 8 gives p = 1; 20 − 12p = −8 gives p = 7/3.

Final answer

p = 1 or p = 7/3.

Step-by-step solution

1Let the plane be x/a + y/b + z/c = 1, meeting the axes at (a,0,0), (0,b,0), (0,0,c).
2Centroid of ABC = (a/3, b/3, c/3) = (1,2,3), so a = 3, b = 6, c = 9.
3Plane: x/3 + y/6 + z/9 = 1, i.e. 6x + 3y + 2z = 18.

Final answer

6x + 3y + 2z = 18.

Step-by-step solution

1From l = −m − n, substitute into mn − 2nl − 2lm = 0: mn + 2n(m+n) + 2(m+n)m = 0.
2This reduces to 2m² + 5mn + 2n² = 0 = (2m + n)(m + 2n).
3Case 2m + n = 0: n = −2m, l = m ⇒ (l,m,n) ∝ (1,1,−2).
4Case m + 2n = 0: m = −2n, l = n ⇒ (l,m,n) ∝ (1,−2,1).
5Normalising: direction cosines are (1/√6, 1/√6, −2/√6) and (1/√6, −2/√6, 1/√6).

Final answer

Direction cosines are (1/√6, 1/√6, −2/√6) and (1/√6, −2/√6, 1/√6).

Step-by-step solution

1A point on the parallel line through (1,2,3) has the form (1 + 2t, 2 + 3t, 3 − 6t).
2It meets the plane when (1+2t) − (2+3t) + (3−6t) = 5 ⇒ 2 − 7t = 5 ⇒ t = −3/7.
3The direction vector has length √(4+9+36) = 7, so the required distance is |t|·7 = (3/7)·7 = 3.

Final answer

3 units.

Step-by-step solution

1Line: (5,1,6) + t(3−5, 4−1, 1−6) = (5 − 2t, 1 + 3t, 6 − 5t).
2On the YZ-plane x = 0: 5 − 2t = 0 ⇒ t = 5/2.
3y = 1 + 3(5/2) = 17/2 and z = 6 − 5(5/2) = −13/2.

Final answer

The point is (0, 17/2, −13/2).

Step-by-step solution

1The second plane can be written 2x − y + 3z + 13/3 = 0 after dividing by 3.
2Distance between parallel planes = |d₁ − d₂|/√(a²+b²+c²).
3|−4 − 13/3| / √(4+1+9) = (25/3)/√14 = 25/(3√14).

Final answer

25/(3√14).

Step-by-step solution

1u = (1,−1,1) − (1,1,1) = (0,−2,0); v = (−7,−3,−5) − (1,1,1) = (−8,−4,−6).
2n = u × v = (0,−2,0)×(−8,−4,−6) = (12,0,−16).
3Plane through (1,1,1): 12(x−1) − 16(z−1) = 0 ⇒ 12x − 16z + 4 = 0.
4Divide by 4: 3x − 4z + 1 = 0.

Final answer

3x − 4z + 1 = 0.

78

Chapter 12 — Linear Programming

Linear programming is a mathematical technique for optimising (maximising or minimising) a linear objective function subject to linear constraints. It has wide applications in business, economics, industry and military planning. In this chapter we learn to formulate real-world problems as linear programming problems (LPPs), solve them using the graphical method, and identify optimal solutions at corner points of the feasible region.

Board pattern

In board exams, 4-6 mark questions on LPP expect you to (i) define variables, (ii) write the objective function, (iii) list all constraints including non-negativity, (iv) draw the feasible region accurately, (v) find corner points, and (vi) evaluate Z at each corner point to state the optimal value. Always label axes, shade the feasible region, and verify whether it is bounded or unbounded.

$Ex 12.1 — Introduction to LPP, graphical method, corner point method (8 Q)$
$Ex 12.2 — Applications: diet, manufacturing, transportation, resource allocation (10 Q)$
$Misc — Mixed LPP problems: bounded/unbounded regions, infeasibility (8 Q)$

79

Exercise 12.1 — Linear Programming: Graphical Method

8Exercise questions

Step-by-step solution

1Constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.
2The feasible region is bounded by the lines x + y = 4, x = 0, y = 0.
3Corner points: O(0,0), A(4,0), B(0,4).
4Z at O(0,0) = 0; Z at A(4,0) = 12; Z at B(0,4) = 16.
5Maximum value is 16 at (0, 4).

Final answer

Z_{m a x} = 16 at (0, 4)

Step-by-step solution

1Plot the lines x + 2y = 8 and 3x + 2y = 12 along with x = 0, y = 0.
2Feasible region is bounded. Solve x + 2y = 8 and 3x + 2y = 12 simultaneously: subtracting gives 2x = 4, so x = 2, y = 3. Intersection at (2, 3).
3Corner points: O(0,0), A(4,0), B(2,3), C(0,4).
4Z at O = 0; Z at A = -12; Z at B = 9; Z at C = 16.
5Minimum value is -12 at (4, 0).

Final answer

Z_{m i n} = - 12 at (4, 0)

Step-by-step solution

1Plot 3x + 5y = 15 and 5x + 2y = 10 in the first quadrant.
2Find intersection: from 3x + 5y = 15 and 5x + 2y = 10. Multiply first by 2 and second by 5: 6x + 10y = 30 and 25x + 10y = 50. Subtracting: 19x = 20, x = 20/19. Then 5(20/19) + 2y = 10 gives y = 45/19.
3Corner points: O(0,0), A(2,0), B(20/19, 45/19), C(0,3).
4Z at O = 0; Z at A = 10; Z at B = 5(20/19) + 3(45/19) = 100/19 + 135/19 = 235/19 ≈ 12.37; Z at C = 9.
5Maximum value is 235/19 at (20/19, 45/19).

Final answer

Z_{m a x} = \frac{235}{19} at (\frac{20}{19}, \frac{45}{19})

Step-by-step solution

1Plot x + 3y = 3 and x + y = 1 in the first quadrant. Shade regions above both lines.
2The feasible region is unbounded (extends infinitely upward).
3Corner points: A(0,1), B(3,0).
4Z at A(0,1) = 5; Z at B(3,0) = 9.
5Since the region is unbounded, check: can Z < 5 be achieved? Draw 3x + 5y = 5; it passes through (0,1). For any point in the feasible region, 3x + 5y ≥ 5.
6Minimum value is 5 at (0, 1).

Final answer

Z_{m i n} = 5 at (0, 1)

Step-by-step solution

1Plot x + 3y = 3 and x + y = 1.
2Feasible region: below both lines in the first quadrant.
3Corner points: O(0,0), A(1,0), B(0,1).
4Z at O = 0; Z at A = 3; Z at B = 5.
5Maximum value is 5 at (0, 1).

Final answer

Z_{m a x} = 5 at (0, 1)

Step-by-step solution

1Plot x + 2y = 12 and x + y = 6.
2Feasible region: bounded by x + 2y ≤ 12 (below), x + y ≥ 6 (above), x ≥ 0, y ≥ 0.
3On x-axis (y=0): x ≤ 12 and x ≥ 6, so x in [6,12]. On y-axis (x=0): 2y ≤ 12 gives y ≤ 6, and y ≥ 6, so y = 6.
4Corner points: A(6,0), B(12,0), C(0,6).
5Z at A = 30; Z at B = 60; Z at C = 60.
6Minimum value is 30 at (6, 0); maximum value is 60 at (12, 0) and (0, 6) and along the line segment joining them.

Final answer

Z_{m i n} = 30 at (6, 0); Z_{m a x} = 60 at (12, 0) and (0, 6)

Step-by-step solution

1Plot 2x + y = 3 and x + 2y = 6.
2Find intersection: from 2x + y = 3, y = 3 - 2x. Substitute: x + 2(3 - 2x) = 6, x + 6 - 4x = 6, -3x = 0, x = 0, y = 3. Intersection at (0, 3).
3Feasible region: 2x + y ≥ 3 (above), x + 2y ≤ 6 (below), x ≥ 0, y ≥ 0.
4On x-axis: 2x ≥ 3 gives x ≥ 3/2, and x ≤ 6. So A(3/2, 0), B(6, 0).
5Corner points: A(3/2, 0), B(6, 0), C(0, 3).
6Z at A = 3/2; Z at B = 6; Z at C = 6.
7The region is unbounded (x can grow large along the x-axis). As x → ∞, Z = x + 2y → ∞, so no finite maximum.
8Minimum value is 3/2 at (3/2, 0).

Final answer

Z_{m i n} = \frac{3}{2} at (\frac{3}{2}, 0); Z has no maximum

Step-by-step solution

1Plot x₁ + x₂ = 4 and x₁ + 3x₂ = 6.
2Find intersection: from x₁ + x₂ = 4 and x₁ + 3x₂ = 6, subtracting gives 2x₂ = 2, x₂ = 1, x₁ = 3. Intersection at (3, 1).
3Feasible region is bounded. Corner points: O(0,0), A(4,0), B(3,1), C(0,2).
4Z at O = 0; Z at A = -4; Z at B = -3 + 2 = -1; Z at C = 4.
5Maximum value is 4 at (0, 2).

Final answer

Z_{m a x} = 4 at (0, 2)

80

Exercise 12.2 — Applications of Linear Programming

10Exercise questions

Step-by-step solution

1Let x = units of F₁ and y = units of F₂.
2Objective: Minimize Z = 4x + 6y.
3Constraints: 3x + 5y ≥ 80 (vitamin A), 5x + 2y ≥ 100 (minerals), x ≥ 0, y ≥ 0.
4Plot 3x + 5y = 80 and 5x + 2y = 100.
5Intersection: 3x + 5y = 80 and 5x + 2y = 100. Multiply first by 2 and second by 5: 6x + 10y = 160, 25x + 10y = 500. Subtracting: 19x = 340, x = 340/19. Then y = (80 - 3(340/19))/5 = (1520 - 1020)/95 = 500/95 = 100/19.
6Corner points of the feasible region (unbounded, above both lines): A(0, 40), B(340/19, 100/19), C(20, 0).
7Z at A = 240; Z at B = 4(340/19) + 6(100/19) = (1360 + 600)/19 = 1960/19 ≈ 103.16; Z at C = 80.
8Check unbounded region: draw 4x + 6y = 80. This does not intersect the feasible region in its interior, so Z ≥ 80 throughout.
9Minimum cost is Rs 80 at (20, 0).

Final answer

Z_{m i n} = Rs 80 at (20, 0)

Step-by-step solution

1Let x = number of tables and y = number of chairs.
2Objective: Maximize Z = 500x + 300y.
3Constraints: 2x + y ≤ 120 (carpentry), x + y ≤ 80 (finishing), x ≥ 0, y ≥ 0.
4Plot 2x + y = 120 and x + y = 80.
5Intersection: subtracting gives x = 40, y = 40.
6Corner points: O(0,0), A(60,0), B(40,40), C(0,80).
7Z at O = 0; Z at A = 30000; Z at B = 20000 + 12000 = 32000; Z at C = 24000.
8Maximum profit is Rs 32000 at (40, 40).

Final answer

Z_{m a x} = Rs 32000 at (40, 40)

Step-by-step solution

1Let x = units of Food I, y = units of Food II.
2Objective: Minimize Z = 50x + 70y.
3Constraints: 2x + y ≥ 10 (vitamin A), x + 2y ≥ 8 (vitamin B), x + 3y ≥ 7 (vitamin C), x ≥ 0, y ≥ 0.
4Plot the three lines and identify the feasible region (unbounded, above all three lines).
5Find key intersections: 2x + y = 10 and x + 2y = 8 give x = 4, y = 2. Check third: 4 + 6 = 10 ≥ 7 ✓.
6x + 2y = 8 and x + 3y = 7 give y = -1 (infeasible).
72x + y = 10 and x + 3y = 7 give x = 23/5, y = 4/5. Check second: 23/5 + 8/5 = 31/5 = 6.2 < 8, fails.
8Corner points of feasible region: A(0, 10/2) on y-axis: y ≥ 10 from first constraint (x=0 gives y ≥ 10). Actually (0, 10): 0+20=20≥8, 0+30=30≥7. But also check (0, 7/3): 0+7/3 = 2.33 < 8, fails. So on y-axis, need 2(0)+y≥10 gives y≥10. And (0,8) from second: 0+16≥8, but 0+8≥10 fails. So A(0,10).
9Also B(4,2) and on x-axis: 2x≥10 gives x≥5. C(5,0): check x+2y=5≥8 fails! So need x≥8 on x-axis. C(8,0): 16+0≥10, 8+0≥8, 8+0≥7 ✓.
10Corner points: A(0, 10), B(4, 2), C(8, 0).
11Z at A = 700; Z at B = 200 + 140 = 340; Z at C = 400.
12Minimum cost is Rs 340 at (4, 2). Check unbounded: 50x + 70y < 340 has no feasible point.

Final answer

Z_{m i n} = Rs 340 at (4, 2)

Step-by-step solution

1Let x = units of A, y = units of B.
2Objective: Maximize Z = 30x + 40y.
3Constraints: 2x + y ≤ 200 (Machine I), x + 3y ≤ 300 (Machine II), x ≥ 0, y ≥ 0.
4Plot 2x + y = 200 and x + 3y = 300.
5Intersection: from first y = 200 - 2x, substitute: x + 600 - 6x = 300, -5x = -300, x = 60, y = 80.
6Corner points: O(0,0), A(100, 0) [from 2x+y=200 on x-axis], B(60, 80), C(0, 100) [from x+3y=300 on y-axis].
7Z at O = 0; Z at A = 3000; Z at B = 1800 + 3200 = 5000; Z at C = 4000.
8Maximum profit is Rs 5000 at (60, 80).

Final answer

Z_{m a x} = Rs 5000 at (60, 80)

Step-by-step solution

1Let x = units of P, y = units of Q.
2Objective: Maximize Z = 200x + 300y.
3Constraints: x + 2y ≤ 8 (Machine A), 2x + y ≤ 10 (Machine B), x + 3y ≤ 12 (Machine C), x ≥ 0, y ≥ 0.
4Find intersections of constraint pairs.
5x + 2y = 8 and 2x + y = 10: multiply first by 2: 2x + 4y = 16, subtract second: 3y = 6, y = 2, x = 4. Check Machine C: 4 + 6 = 10 ≤ 12 ✓.
6x + 2y = 8 and x + 3y = 12: subtracting gives y = 4, x = 0. Check Machine B: 0 + 4 = 4 ≤ 10 ✓.
72x + y = 10 and x + 3y = 12: from first y = 10 - 2x, x + 30 - 6x = 12, -5x = -18, x = 18/5 = 3.6, y = 2.8. Check Machine A: 3.6 + 5.6 = 9.2 > 8, fails.
8Corner points: O(0,0), A(5,0) [from 2x+y=10], B(4,2), C(0,4) [from x+2y=8].
9Z at O = 0; Z at A = 1000; Z at B = 800 + 600 = 1400; Z at C = 1200.
10Maximum profit is Rs 1400 at (4, 2).

Final answer

Z_{m a x} = Rs 1400 at (4, 2)

Step-by-step solution

1Let a = fraction of capacity used on Machine A, b = fraction on Machine B (0 ≤ a, b ≤ 1).
2Item X produced: 120a + 60b ≥ 200. Item Y produced: 80a + 140b ≥ 160.
3Simplify: 6a + 3b ≥ 10 (from X), 4a + 7b ≥ 8 (from Y).
4Objective: Minimize Z = 2000a + 1600b.
5Constraints: 6a + 3b ≥ 10, 4a + 7b ≥ 8, 0 ≤ a ≤ 1, 0 ≤ b ≤ 1.
6Plot 6a + 3b = 10 and 4a + 7b = 8 in the a-b plane (first quadrant).
7Intersection: multiply first by 7 and second by 3: 42a + 21b = 70, 12a + 21b = 24. Subtracting: 30a = 46, a = 46/30 = 23/15 > 1. Not feasible.
8Since intersection is outside [0,1] x [0,1], the binding constraints in the feasible region are different. On b = 1: 6a + 3 ≥ 10 gives a ≥ 7/6 > 1, fails. On a = 1: 6 + 3b ≥ 10 gives b ≥ 4/3 > 1, fails.
9The problem as stated may be infeasible within the [0,1] box. In practice, the NCERT problem uses different numbers. With the given data, the factory cannot meet both demand constraints with one day of each machine. The minimum cost solution requires running Machine A for 5/3 days and Machine B for 0 days (cost Rs 3333.33), or running both for extended periods.
10For a one-day allocation: a = 1, b = 1. Check: 6+3=9 < 10, fails X demand. The demand cannot be met in one day.
11Reinterpret: let a and b be the number of days each machine runs (no upper bound). Then minimize Z = 2000a + 1600b subject to 6a + 3b ≥ 10, 4a + 7b ≥ 8, a ≥ 0, b ≥ 0.
12Corner points: A(0, 10/3), B(23/15, 8/7)... since intersection at a = 23/15 ≈ 1.53, b = (10 - 6(23/15))/3 = (150 - 138)/45 = 12/45 = 4/15 ≈ 0.27. Check 4a+7b = 92/15 + 28/15 = 120/15 = 8 ✓.
13Corner points: A(0, 10/3), B(23/15, 4/15), C(5/3, 0).
14Z at A = 16000/3 ≈ 5333; Z at B = 2000(23/15) + 1600(4/15) = 46000/15 + 6400/15 = 52400/15 ≈ 3493; Z at C = 10000/3 ≈ 3333.
15Minimum cost is Rs 10000/3 ≈ Rs 3333 at (5/3, 0) — run Machine A for 5/3 days, Machine B for 0 days.

Final answer

Z_{m i n} = Rs \frac{10 000}{3} at (\frac{5}{3}, 0)

Step-by-step solution

1Let x = pills of Brand X, y = pills of Brand Y.
2Objective: Minimize Z = 40x + 60y.
3Constraints: 20x + 30y ≥ 200 (vitamin A), 100x + 150y ≥ 500 (vitamin B), 30x + 20y ≥ 150 (vitamin C), x ≥ 0, y ≥ 0.
4Simplify: 2x + 3y ≥ 20, 2x + 3y ≥ 10, 3x + 2y ≥ 15. The second constraint is redundant (implied by the first).
5Effective: 2x + 3y ≥ 20, 3x + 2y ≥ 15.
6Plot 2x + 3y = 20 and 3x + 2y = 15.
7Intersection: multiply first by 3, second by 2: 6x + 9y = 60, 6x + 4y = 30. Subtract: 5y = 30, y = 6, x = (20 - 18)/2 = 1.
8On y-axis: 3y ≥ 20 gives y ≥ 20/3; 2y ≥ 15 gives y ≥ 7.5. So y ≥ 20/3. On x-axis: 2x ≥ 20 gives x ≥ 10; 3x ≥ 15 gives x ≥ 5. So x ≥ 10.
9Corner points: A(0, 20/3), B(1, 6), C(10, 0).
10Z at A = 60 × 20/3 = 400; Z at B = 40 + 360 = 400; Z at C = 400.
11All three corner points give Z = 400. Minimum cost is Rs 400, achieved at every point on the boundary of the feasible region (all give the same value).

Final answer

Z_{m i n} = Rs 400 at (0, \frac{20}{3}), (1, 6), and (10, 0)

Step-by-step solution

1Let x = units of T₁, y = units of T₂.
2Objective: Maximize Z = 80x + 120y.
3Constraints: 2x + y ≤ 120 (dept A), x + 3y ≤ 150 (dept B), x ≥ 0, y ≥ 0.
4Plot 2x + y = 120 and x + 3y = 150.
5Intersection: y = 120 - 2x, x + 3(120 - 2x) = 150, x + 360 - 6x = 150, -5x = -210, x = 42, y = 36.
6Corner points: O(0,0), A(60, 0), B(42, 36), C(0, 50).
7Z at O = 0; Z at A = 4800; Z at B = 3360 + 4320 = 7680; Z at C = 6000.
8Maximum profit is Rs 7680 at (42, 36).

Final answer

Z_{m a x} = Rs 7680 at (42, 36)

Step-by-step solution

1Let x = units of A, y = units of B.
2Objective: Maximize Z = 60x + 80y.
3Constraints: 3x + y ≤ 120 (M₁), 2x + 4y ≤ 160 (M₂), x ≥ 0, y ≥ 0.
4Simplify M₂: x + 2y ≤ 80.
5Plot 3x + y = 120 and x + 2y = 80.
6Intersection: from first y = 120 - 3x, x + 2(120 - 3x) = 80, x + 240 - 6x = 80, -5x = -160, x = 32, y = 24.
7Corner points: O(0,0), A(40, 0) [from 3x+y=120], B(32, 24), C(0, 40) [from x+2y=80].
8Z at O = 0; Z at A = 2400; Z at B = 1920 + 1920 = 3840; Z at C = 3200.
9Maximum profit is Rs 3840 at (32, 24).

Final answer

Z_{m a x} = Rs 3840 at (32, 24)

Step-by-step solution

1Let x = quintals transported from A to I. Then A to II = 200 - x, B to I = 150 - x, B to II = 300 - (150 - x) = 150 + x.
2Non-negativity: x ≥ 0, 200 - x ≥ 0 (≤ 200), 150 - x ≥ 0 (≤ 150), 150 + x ≥ 0 (always). So 0 ≤ x ≤ 150.
3Objective: Z = 5x + 3(200 - x) + 4(150 - x) + 2(150 + x) = 5x + 600 - 3x + 600 - 4x + 300 + 2x = 1500.
4Z = 1500 for all feasible x! The cost is constant regardless of how we distribute.
5Any schedule with 0 ≤ x ≤ 150 gives minimum (and maximum) cost Rs 1500.
6For example: x = 0 gives A→I: 0, A→II: 200, B→I: 150, B→II: 150. Cost = 0 + 600 + 600 + 300 = 1500.

Final answer

Z_{m i n} = Rs 1500 for any feasible schedule

81

Miscellaneous Exercise on Chapter 12

8Exercise questions

Step-by-step solution

1Evaluate Z at all corner points: Z(0,0) = 0; Z(0,5) = 5q; Z(6,8) = 6p + 8q; Z(6,0) = 6p; Z(3,0) = 3p.
2Since p, q > 0, the maximum must occur at one of the 'outermost' points: (6,0), (6,8), or (0,5).
3Comparing: 6p + 8q is the largest value among {6p, 6p + 8q, 5q} when p, q > 0 (since 8q > 0 adds to 6p, and 6p + 8q > 5q when 6p > -3q which is always true).
4So the maximum always occurs at (6, 8) for any p, q > 0.
5More precisely: 6p + 8q > 6p (since 8q > 0) and 6p + 8q > 5q (since 6p + 3q > 0). So the unique maximum is at (6, 8).

Final answer

The maximum of Z occurs at (6, 8) for all p, q > 0.

Step-by-step solution

1Plot 3x + 5y = 15 and 5x + 2y = 10.
2Intersection: 6x + 10y = 30 and 25x + 10y = 50. Subtracting: 19x = 20, x = 20/19, y = 45/19.
3Corner points: O(0,0), A(2,0), B(20/19, 45/19), C(0,3).
4Z at O = 0; Z at A = 14; Z at B = 7(20/19) + 11(45/19) = 140/19 + 495/19 = 635/19 ≈ 33.42; Z at C = 33.
5Maximum value is 635/19 at (20/19, 45/19).

Final answer

Z_{m a x} = \frac{635}{19} at (\frac{20}{19}, \frac{45}{19})

Step-by-step solution

1Let x = number of dolls, y = number of trucks.
2Objective: Maximize Z = 500x + 800y.
3Constraints: 2x + 4y ≤ 70 (cutting), 5x + 2y ≤ 80 (sewing), x ≥ 0, y ≥ 0.
4Simplify cutting: x + 2y ≤ 35.
5Plot x + 2y = 35 and 5x + 2y = 80.
6Intersection: subtracting first from second: 4x = 45, x = 45/4 = 11.25, y = (35 - 11.25)/2 = 23.75/2 = 11.875.
7Corner points: O(0,0), A(16, 0) [from 5x+2y=80], B(45/4, 47/4), C(0, 35/2) [from x+2y=35].
8Z at O = 0; Z at A = 8000; Z at B = 500(45/4) + 800(47/4) = 22500/4 + 37600/4 = 60100/4 = 15025; Z at C = 800(35/2) = 14000.
9Maximum profit is Rs 15025 at (45/4, 47/4).

Final answer

Z_{m a x} = Rs 15025 at (\frac{45}{4}, \frac{47}{4})

Step-by-step solution

1Evaluate Z at each corner point:
2Z(0,0) = 0; Z(5,0) = 15; Z(4,3) = 12 + 12 = 24; Z(0,5) = 20.
3Maximum value is 24 at (4, 3).

Final answer

Z_{m a x} = 24 at (4, 3)

Step-by-step solution

1Plot 3x + 5y = 15 and 5x + 2y = 10. The feasible region is unbounded (above both lines in the first quadrant).
2On x-axis: 3x ≥ 15 gives x ≥ 5, and 5x ≥ 10 gives x ≥ 2. So x ≥ 5.
3On y-axis: 5y ≥ 15 gives y ≥ 3, and 2y ≥ 10 gives y ≥ 5. So y ≥ 5.
4Intersection: 6x + 10y = 30, 25x + 10y = 50. Subtracting: 19x = 20, x = 20/19, y = 45/19.
5Corner points: A(0, 5), B(20/19, 45/19), C(5, 0).
6Z at A = 15; Z at B = 5(20/19) + 3(45/19) = 100/19 + 135/19 = 235/19 ≈ 12.37; Z at C = 25.
7Minimum is 235/19 at B(20/19, 45/19). Since the region is unbounded and Z = 5x + 3y with positive coefficients increases in all feasible directions, this is the true minimum.

Final answer

Z_{m i n} = \frac{235}{19} at (\frac{20}{19}, \frac{45}{19})

Step-by-step solution

1Let x = cans of premium, y = cans of regular.
2Objective: Maximize Z = 12x + 8y.
3Constraints: 2x + 4y ≤ 40 (blending), 3x + 2y ≤ 36 (packaging), x ≥ 0, y ≥ 0.
4Simplify blending: x + 2y ≤ 20.
5Plot x + 2y = 20 and 3x + 2y = 36.
6Intersection: subtracting first from second: 2x = 16, x = 8, y = 6.
7Corner points: O(0,0), A(12, 0) [from 3x+2y=36 on x-axis], B(8, 6), C(0, 10) [from x+2y=20 on y-axis].
8Z at O = 0; Z at A = 144; Z at B = 96 + 48 = 144; Z at C = 80.
9Maximum profit is Rs 144, achieved at both A(12, 0) and B(8, 6) and along the line segment joining them.
10Note: Z = 12x + 8y = 4(3x + 2y) = 4(36) = 144 along the entire edge 3x + 2y = 36 from (12,0) to (8,6).

Final answer

Z_{m a x} = Rs 144 at every point on the segment from (12, 0) to (8, 6)

Step-by-step solution

1Let x = acres of wheat, y = acres of barley.
2Objective: Maximize Z = 120x + 200y.
3Constraints: x + y ≤ 100 (land), 100x + 200y ≤ 14000 (investment), x ≥ 0, y ≥ 0.
4Simplify investment: x + 2y ≤ 140.
5Plot x + y = 100 and x + 2y = 140.
6Intersection: subtracting gives y = 40, x = 60.
7Corner points: O(0,0), A(100, 0), B(60, 40), C(0, 70) [from x+2y=140 on y-axis, since 140/2 = 70].
8Z at O = 0; Z at A = 12000; Z at B = 7200 + 8000 = 15200; Z at C = 14000.
9Maximum profit is Rs 15200 at (60, 40).

Final answer

Z_{m a x} = Rs 15200 at (60, 40)

Step-by-step solution

1Let x = units of Product I, y = units of Product II.
2Objective: Maximize Z = 15x + 25y.
3Constraints: x + 2y ≤ 80 (R₁), 2x + y ≤ 100 (R₂), x + 3y ≤ 120 (R₃), x ≥ 0, y ≥ 0.
4Find intersection of x + 2y = 80 and 2x + y = 100: from first x = 80 - 2y, 2(80-2y) + y = 100, 160 - 4y + y = 100, -3y = -60, y = 20, x = 40. Check R₃: 40 + 60 = 100 ≤ 120 ✓.
5x + 2y = 80 and x + 3y = 120: subtracting gives y = 40, x = 0. Check R₂: 0 + 40 = 40 ≤ 100 ✓.
62x + y = 100 and x + 3y = 120: from first y = 100 - 2x, x + 300 - 6x = 120, -5x = -180, x = 36, y = 28. Check R₁: 36 + 56 = 92 > 80, fails.
7Corner points: O(0,0), A(50, 0) [from 2x+y=100], B(40, 20), C(0, 40) [from x+3y=120 and x+2y=80 both on y-axis at y=40 and y=40, actually x+2y=80 on y-axis is y=40].
8Check A(50,0): 50+0=50≤80, 100+0=100≤100, 50+0=50≤120 ✓.
9Z at O = 0; Z at A = 750; Z at B = 600 + 500 = 1100; Z at C = 1000.
10Maximum profit is Rs 1100 at (40, 20).

Final answer

Z_{m a x} = Rs 1100 at (40, 20)

82

Chapter 13 — Probability

Probability is the branch of mathematics that quantifies uncertainty. This chapter extends the basic ideas of Class 10 and Class 11 into the territory of conditional probability, Bayes' theorem, random variables and the binomial distribution. Every question below is from the NCERT Class 12 textbook (rationalized edition); each carries short, exam-ready working.

Board pattern

Conditional probability and Bayes' theorem questions reward a clear statement of the given events, the partition, and a labelled tree diagram where applicable. Random variable questions award marks for writing the probability distribution table and using the variance formula. Always show the substitution into the formula — a bare final answer scores zero.

$Ex 13.1 — Conditional probability, multiplication theorem, independent events (22 Q)$
$Ex 13.2 — Multiplication theorem and independent events (23 Q)$
$Ex 13.3 — Bayes’ theorem (18 Q)$
$Ex 13.4 — Random variables and probability distributions (20 Q)$
$Ex 13.5 — Bernoulli trials and binomial distribution (19 Q)$
$Misc — Mixed problems on probability (28 Q)$

83

Exercise 13.1 — Conditional Probability

22Exercise questions

Step-by-step solution

1P(A|B) = P(A ∩ B)/P(B) = 0.18/0.30 = 0.6.
2P(B|A) = P(A ∩ B)/P(A) = 0.18/0.60 = 0.3.
3P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.3 − 0.18 = 0.72.

Final answer

P(A|B) = 0.6, P(B|A) = 0.3, P(A ∪ B) = 0.72.

Step-by-step solution

1Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, |S| = 8.
2(i) B = {HHH, HHT}, A ∩ B = {HHH}, so P(A|B) = 1/2.
3(ii) A = {HHT, HTH, THH, HHH}; B = {HHT, HTH, HTT, THH, THT, TTH, TTT}; A ∩ B = {HHT, HTH, THH}; P(A|B) = 3/7.

Final answer

(i) 1/2; (ii) 3/7.

Step-by-step solution

1S = {HH, HT, TH, TT}; A = {HT, TH}; B = {HH}; A ∩ B = φ.
2P(A|B) = 0.

Final answer

P (A ∣ B) = 0.

Step-by-step solution

1B = {63x : x = 1,...,6}, |B| = 6. A ∩ B = {634}, |A ∩ B| = 1.
2P(A|B) = 1/6.

Final answer

P (A ∣ B) = 1/6.

Step-by-step solution

1S = {MFS, MSF, FMS, FSM, SMF, SFM}, |S| = 6.
2A = {MFS, FMS, SMF, SFM}; B = {MFS, SFM}; A ∩ B = {MFS, SFM}; P(A|B) = 2/2 = 1.
3P(B|A) = 2/4 = 1/2.

Final answer

P (A ∣ B) = 1, P (B ∣ A) = 1/2.

Step-by-step solution

1B has 3 × 6 = 18 outcomes. A ∩ B = {(1,6),(2,5),(3,4)}: 3 outcomes.
2P(A|B) = 3/18 = 1/6.

Final answer

P (A ∣ B) = 1/6.

Step-by-step solution

1P(red) = 3/5, P(black) = 2/5.
2P(at least one red) = 1 − P(both black) = 1 − (2/5)² = 1 − 4/25 = 21/25.

Final answer

\frac{21}{25} .

Step-by-step solution

1Face cards = 12 (J,Q,K of 4 suits). Kings among face cards = 4.
2P(A|B) = 4/12 = 1/3.

Final answer

\frac{1}{3} .

Step-by-step solution

1P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.4 + 0.5 − 0.7 = 0.2.
2P(A|B) = 0.2/0.5 = 0.4; P(B|A) = 0.2/0.4 = 0.5.

Final answer

P (A \cap B) = 0.2, P (A ∣ B) = 0.4, P (B ∣ A) = 0.5.

Step-by-step solution

1P(B) = 5/13. P(A|B) = P(A ∩ B)/P(B) = 2/5, so P(A ∩ B) = (2/5)(5/13) = 2/13.
22P(A) = 5/13 ⇒ P(A) = 5/26.
3P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 5/26 + 5/13 − 2/13 = 5/26 + 3/13 = 5/26 + 6/26 = 11/26.

Final answer

\frac{11}{26} .

Step-by-step solution

1P(not A or not B) = P((A ∩ B)ᶜ) = 1 − P(A ∩ B) = 1/4, so P(A ∩ B) = 3/4.
2P(A) · P(B) = (1/2)(7/12) = 7/24 ≠ 3/4.
3P(A ∩ B) ≠ P(A)·P(B), so A and B are NOT independent.

Final answer

A and B are not independent.

Final answer

P (A \cap B) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{25} .

Step-by-step solution

1(i) P(A ∩ B) = 0.3 × 0.4 = 0.12.
2(ii) P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58.
3(iii) P(A|B) = P(A) = 0.3 (independent).
4(iv) P(B|A) = P(B) = 0.4.

Final answer

(i) 0.12; (ii) 0.58; (iii) 0.3; (iv) 0.4.

Step-by-step solution

1P(A)·P(B) = 0.5 × 0.6 = 0.3 = P(A ∩ B).

Final answer

Yes, A and B are independent.

Final answer

Mutually exclusive and exhaustive \Rightarrow P (A_{1} \cup \dots \cup A_{n}) = 1.

Step-by-step solution

1P(first ace | second ace) = P(both ace)/P(second ace).
2P(both ace) = (4/52)(3/51) = 12/2652 = 1/221.
3P(second ace) = 4/52 = 1/13 (by symmetry).
4P(first ace | second ace) = (1/221)/(1/13) = 13/221 = 1/17.

Final answer

\frac{1}{17} .

Step-by-step solution

1E = {1,2,3}, F = {1,3,5}. E ∩ F = {1,3}.
2P(E|F) = 2/3, P(F|E) = 2/3.

Final answer

P (E ∣ F) = \frac{2}{3}, P (F ∣ E) = \frac{2}{3} .

Step-by-step solution

1After drawing one white ball, 4 white and 3 black remain (7 total).
2P(second white | first white) = 4/7.

Final answer

\frac{4}{7} .

Step-by-step solution

1Numbers greater than 3: {4,5,6}. Even numbers among them: {4,6}.
2P(even | >3) = 2/3.

Final answer

\frac{2}{3} .

Step-by-step solution

1Outcomes with at least one 4: {(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(1,4),(2,4),(3,4),(5,4),(6,4)}, 11 outcomes.
2Among these, sum = 7: {(4,3),(3,4)}, 2 outcomes.
3P = 2/11.

Final answer

\frac{2}{11} .

Step-by-step solution

1Sample space: {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}.
2At least one boy: 7 outcomes (all except GGG).
3Exactly 2 boys: {BBG, BGB, GBB}, 3 outcomes.
4P = 3/7.

Final answer

\frac{3}{7} .

Step-by-step solution

1P(A|B) = 0.2/0.4 = 0.5, so P(A|B) × P(B) = 0.5 × 0.4 = 0.2.
2P(B|A) = 0.2/0.5 = 0.4, so P(B|A) × P(A) = 0.4 × 0.5 = 0.2.
3Both sides equal P(A ∩ B) = 0.2. Verified.

Final answer

Both sides equal 0.2. Verified.

84

Exercise 13.2 — Multiplication Theorem and Independent Events

23Exercise questions

Step-by-step solution

1P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 6/11 + 5/11 − 7/11 = 4/11.
2P(A|B) = (4/11)/(5/11) = 4/5.
3P(B|A) = (4/11)/(6/11) = 4/6 = 2/3.

Final answer

P (A \cap B) = \frac{4}{11}, P (A ∣ B) = \frac{4}{5}, P (B ∣ A) = \frac{2}{3} .

Step-by-step solution

1X = 0,1,2,3 with P(X = 0) = 1/8, P(X = 1) = 3/8, P(X = 2) = 3/8, P(X = 3) = 1/8.

Final answer

Distribution: 0→1/8, 1→3/8, 2→3/8, 3→1/8.

Step-by-step solution

1B = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}; |B| = 6. A ∩ B = {(3,4)}, |A ∩ B| = 1.
2P(A|B) = 1/6; P(B|A) = 1/6.
3P(A) = 1/6, so P(A|B) = P(A) and P(B|A) = P(B). Independent.

Final answer

P(A|B) = P(B|A) = \frac{1}{6};\; A \text{ and } B \text{ are independent.

Step-by-step solution

1These are independent throws: P(4 on first) = 1/6. On second throw, sum = 6 from {(1,5),(2,4),(3,3),(4,2),(5,1)}: 5/36.
2P = (1/6)(5/36) = 5/216.

Final answer

\frac{5}{216} .

Step-by-step solution

1P(both white) = (3/5)(2/4) = 6/20 = 3/10.

Final answer

\frac{3}{10} .

Step-by-step solution

1P(red) = P(first bag)·P(red|first) + P(second bag)·P(red|second)
2= (1/2)(4/8) + (1/2)(2/8) = 1/4 + 1/8 = 3/8.

Final answer

\frac{3}{8} .

Step-by-step solution

1P(F ∪ C) = P(F) + P(C) − P(F ∩ C) = 0.3 + 0.4 − 0.1 = 0.6.

Final answer

0.6.

Step-by-step solution

1(i) Perfect squares: 1,4,9,...,100 → 10 values → 10/100 = 1/10.
2(ii) Perfect cubes: 1,8,27,64 → 4 values → 4/100 = 1/25.

Final answer

(i) 1/10; (ii) 1/25.

Step-by-step solution

1P(2) = 1/6, so P(not 2) = 1 − 1/6 = 5/6.
21/6 + 5/6 = 1.

Final answer

P (not 2) = \frac{5}{6}; \frac{1}{6} + \frac{5}{6} = 1.

Step-by-step solution

1P(H) = 2P(T) and P(H) + P(T) = 1.
22P(T) + P(T) = 1 ⇒ P(T) = 1/3, P(H) = 2/3.

Final answer

P (H) = \frac{2}{3}, P (T) = \frac{1}{3} .

Step-by-step solution

1P(first king) = 4/52 = 1/13.
2After removing one king, 51 cards remain. Queen of the same suit is still present (king and queen are different cards): 1/51.
3P = (1/13)(1/51) = 1/663.

Final answer

\frac{1}{663} .

Step-by-step solution

1Total = 14 balls. P(R) = 7/14 = 1/2, P(W) = 4/14 = 2/7, P(B) = 3/14.
2(i) P(RR) = (1/2)(1/2) = 1/4.
3(ii) P(RW) + P(WR) = (1/2)(2/7) + (2/7)(1/2) = 2/7.
4(iii) 1 − P(no red) = 1 − (5/7)² = 1 − 25/49 = 24/49.

Final answer

(i) 1/4; (ii) 2/7; (iii) 24/49.

Step-by-step solution

1P(A ∪ B) = P(A) + P(B) − P(A)P(B) (independent).
20.6 = 0.2 + P(B) − 0.2·P(B) = 0.2 + 0.8P(B).
30.4 = 0.8P(B) ⇒ P(B) = 0.5.

Final answer

P (B) = 0.5.

Step-by-step solution

1After drawing one red, 2 red and 2 blue remain (4 total).
2P(second red | first red) = 2/4 = 1/2.

Final answer

\frac{1}{2} .

Step-by-step solution

1P = (1/6)(5/6)(5/6) = 25/216.

Final answer

\frac{25}{216} .

Step-by-step solution

1Let L = late. P(L) = (0.3)(0.10) + (0.2)(0.05) + (0.1)(0.12) + (0.4)(0.01)
2= 0.03 + 0.01 + 0.012 + 0.004 = 0.056.
3P(bus | L) = 0.03/0.056 = 30/56 = 15/28.

Final answer

\frac{15}{28} .

Step-by-step solution

1Binomial: n = 5, p = 0.1.
2P(X = 0) = (0.9)⁵ = 0.59049.
3P(X ≤ 2) = P(0) + P(1) + P(2) = (0.9)⁵ + 5(0.1)(0.9)⁴ + 10(0.01)(0.9)³
4= 0.59049 + 0.32805 + 0.07290 = 0.99144.

Final answer

P(none defective) = 0.59049; P(at most 2 defective) = 0.99144.

Final answer

Need exactly 3 reds in first 3 draws (all red drawn by 4th draw), so \frac{( 3 4 )}{( 3 10 )} = \frac{4}{120} = \frac{1}{30} .

Step-by-step solution

1P(both kings) = (4/52)(3/51) = 12/2652 = 1/221.

Final answer

\frac{1}{221} .

Step-by-step solution

1With replacement: p(red) = 5/8 each draw.
2P = (5/8)³ = 125/512.

Final answer

\frac{125}{512} .

Step-by-step solution

1P(no head) = (1/2)ⁿ = 1/64.
22ⁿ = 64 = 2⁶, so n = 6.

Final answer

n = 6.

Step-by-step solution

1P(A ∪ B) = P(A) + P(B) − P(A)P(B) (independent).
22/3 = 1/3 + P(B)(1 − 1/3) = 1/3 + (2/3)P(B).
31/3 = (2/3)P(B), so P(B) = 1/2.

Final answer

P (B) = \frac{1}{2} .

Step-by-step solution

1P(not solved) = (1 − 1/2)(1 − 1/3)(1 − 1/4) = (1/2)(2/3)(3/4) = 6/24 = 1/4.
2P(solved) = 1 − 1/4 = 3/4.

Final answer

\frac{3}{4} .

85

Exercise 13.3 — Bayes' Theorem

18Exercise questions

Step-by-step solution

1P(A) = 0.6, P(B) = 0.4, P(D|A) = 0.02, P(D|B) = 0.05.
2P(D) = 0.6(0.02) + 0.4(0.05) = 0.012 + 0.020 = 0.032.
3P(A|D) = 0.012/0.032 = 12/32 = 3/8.

Final answer

\frac{3}{8} .

Step-by-step solution

1P(TwoHeaded) = 1/3, P(H|TwoHeaded) = 1; P(Biased) = 1/3, P(H|Biased) = 3/4; P(Fair) = 1/3, P(H|Fair) = 1/2.
2P(H) = (1/3)(1) + (1/3)(3/4) + (1/3)(1/2) = 1/3 + 1/4 + 1/6 = 4/12 + 3/12 + 2/12 = 9/12 = 3/4.
3P(TwoHeaded|H) = (1/3)/(3/4) = 4/9.

Final answer

\frac{4}{9} .

Step-by-step solution

1P(Uᵢ) = 1/3 each. P(2W|I) = C(2,2)/C(5,2) = 1/10; P(2W|II) = C(4,2)/C(5,2) = 6/10; P(2W|III) = C(3,2)/C(7,2) = 3/21 = 1/7.
2P(2W) = (1/3)(1/10) + (1/3)(6/10) + (1/3)(1/7) = 1/30 + 6/30 + 1/21
3= 7/30 + 1/21 = 49/210 + 10/210 = 59/210.
4P(III|2W) = (1/21)/(59/210) = 10/59.

Final answer

\frac{10}{59} .

Step-by-step solution

1P(G) = 0.6, P(B) = 0.4, P(P|G) = 0.7, P(P|Bo) = 0.85.
2P(P) = 0.6(0.7) + 0.4(0.85) = 0.42 + 0.34 = 0.76.
3P(Bo|P) = 0.34/0.76 = 17/38.

Final answer

\frac{17}{38} .

Step-by-step solution

1P(all good) = C(7,3)/C(10,3) = 35/120 = 7/24.
2P(at least one good) = 1 − P(all defective) = 1 − C(3,3)/C(10,3) = 1 − 1/120 = 119/120.
3P(all good | at least one good) = (7/24)/(119/120) = (7/24)(120/119) = 35/119 = 5/17.

Final answer

\frac{5}{17} .

Step-by-step solution

1P(R) = (1/2)(1/2) + (1/2)(1/4) = 1/4 + 1/8 = 3/8.
2P(1st|R) = (1/4)/(3/8) = 2/3.

Final answer

\frac{2}{3} .

Step-by-step solution

1P(reported six) = P(reported six|actual 6)P(6) + P(reported six|not 6)P(not 6)
2= (3/4)(1/6) + (1/4)(5/6) = 3/24 + 5/24 = 8/24 = 1/3.
3P(actual 6 | reported 6) = (3/4 × 1/6)/(1/3) = (3/24)/(1/3) = (1/8)/(1/3) = 3/8.

Final answer

\frac{3}{8} .

Step-by-step solution

1P(D) = 0.005, P(Dᶜ) = 0.995. P(+|D) = 0.99, P(+|Dᶜ) = 0.01.
2P(+) = 0.99(0.005) + 0.01(0.995) = 0.00495 + 0.00995 = 0.01490.
3P(D|+) = 0.00495/0.01490 = 495/1490 ≈ 0.3322.

Final answer

\frac{99}{299} \approx 0.3322.

Step-by-step solution

1P(Below) = 0.25, P(Mid) = 0.55, P(Above) = 0.20.
2P(crime) = 0.25(0.02) + 0.55(0.01) + 0.20(0.03) = 0.005 + 0.0055 + 0.006 = 0.0165.
3P(Above|crime) = 0.006/0.0165 = 60/165 = 4/11.

Final answer

\frac{4}{11} .

Step-by-step solution

1P(both red) = (5/8)(3/7) = 15/56.
2P(at least one red) = 1 − P(both blue) = 1 − (3/8)(2/7) = 1 − 6/56 = 50/56.
3P(both red | at least one red) = (15/56)/(50/56) = 15/50 = 3/10.

Final answer

\frac{3}{10} .

Step-by-step solution

1P(at least one H) = 1 − 1/8 = 7/8.
2P(exactly two H) = 3/8.
3P(exactly two H | at least one H) = (3/8)/(7/8) = 3/7.

Final answer

\frac{3}{7} .

Step-by-step solution

1P(not late | bus) = 0.90, P(not late | taxi) = 0.95, P(not late | bike) = 0.88, P(not late | car) = 0.99.
2P(not late) = 0.3(0.90) + 0.2(0.95) + 0.1(0.88) + 0.4(0.99) = 0.27 + 0.19 + 0.088 + 0.396 = 0.944.
3P(car | not late) = 0.396/0.944 = 396/944 = 99/236.

Final answer

\frac{99}{236} .

Step-by-step solution

1P(win) = (1/3)(1/5) + (1/3)(1/4) + (1/3)(1/6) = 1/15 + 1/12 + 1/18
2= 12/180 + 15/180 + 10/180 = 37/180.
3P(B|win) = (1/12)/(37/180) = (15/180)/(37/180) = 15/37.

Final answer

\frac{15}{37} .

Step-by-step solution

1P(D) = 0.6(0.02) + 0.4(0.03) = 0.012 + 0.012 = 0.024.
2P(2nd|D) = 0.012/0.024 = 1/2.

Final answer

\frac{1}{2} .

Step-by-step solution

1P(R) = (1/2)(3/7) + (1/2)(5/11) = 3/14 + 5/22 = 33/154 + 35/154 = 68/154 = 34/77.
2P(I|R) = (3/14)/(34/77) = (3/14)(77/34) = 231/476 = 33/68.

Final answer

\frac{33}{68} .

Step-by-step solution

1P(D) = 0.30(0.02) + 0.45(0.03) + 0.25(0.05) = 0.006 + 0.0135 + 0.0125 = 0.032.
2P(C|D) = 0.0125/0.032 = 125/320 = 25/64.

Final answer

\frac{25}{64} .

Step-by-step solution

1P(exactly one hits) = P(A only) + P(B only) + P(C only)
2= (4/5)(1/4)(1/3) + (1/5)(3/4)(1/3) + (1/5)(1/4)(2/3)
3= 4/60 + 3/60 + 2/60 = 9/60 = 3/20.
4P(B|exactly one) = (3/60)/(9/60) = 3/9 = 1/3.

Final answer

\frac{1}{3} .

Step-by-step solution

1P(D₂) = P(D₁)P(D₂|D₁) + P(not D₁)P(D₂|not D₁)
2= (3/10)(2/9) + (7/10)(3/9) = 6/90 + 21/90 = 27/90 = 3/10.
3P(D₁|D₂) = (6/90)/(27/90) = 6/27 = 2/9.

Final answer

\frac{2}{9} .

86

Exercise 13.4 — Random Variables and Probability Distributions

20Exercise questions

Step-by-step solution

1X = 0, 1, 2. P(X = 0) = 1/4, P(X = 1) = 2/4 = 1/2, P(X = 2) = 1/4.

Final answer

Distribution: 0 → 1/4, 1 → 1/2, 2 → 1/4.

Step-by-step solution

1X = 0,1,2,3. P(0) = 1/8, P(1) = 3/8, P(2) = 3/8, P(3) = 1/8.

Final answer

Distribution: 0→1/8, 1→3/8, 2→3/8, 3→1/8.

Step-by-step solution

1Sum of probabilities = 1 to find k. For NCERT standard: ΣP(X = xᵢ) = 1.
2Mean E(X) = Σ xᵢ P(X = xᵢ).

Final answer

k found by normalizing; μ = E (X) = \sum x_{i} P (x_{i}) .

Step-by-step solution

1P(success) = 2/6 = 1/3, P(failure) = 2/3.
2X ~ Binomial(2, 1/3). P(X = 0) = 4/9, P(X = 1) = 4/9, P(X = 2) = 1/9.

Final answer

Distribution: 0→4/9, 1→4/9, 2→1/9.

Step-by-step solution

1E(X) = 0(0.4) + 1(0.3) + 2(0.2) + 3(0.1) = 0 + 0.3 + 0.4 + 0.3 = 1.0.
2E(X²) = 0(0.4) + 1(0.3) + 4(0.2) + 9(0.1) = 0 + 0.3 + 0.8 + 0.9 = 2.0.

Final answer

(i) E(X) = 1.0; (ii) E(X²) = 2.0.

Step-by-step solution

1(i) k(1+2+3+4+5+6) = 21k = 1 ⇒ k = 1/21.
2(ii) P(X > 4) = P(5) + P(6) = 5/21 + 6/21 = 11/21.
3(iii) P(X ≤ 4) = 1 − 11/21 = 10/21.

Final answer

(i) 1/21; (ii) 11/21; (iii) 10/21.

Step-by-step solution

1X ranges from 2 to 12. P(X = k) = (number of ways to sum to k)/36.
2P(2)=1/36, P(3)=2/36, P(4)=3/36, P(5)=4/36, P(6)=5/36, P(7)=6/36, P(8)=5/36, P(9)=4/36, P(10)=3/36, P(11)=2/36, P(12)=1/36.

Final answer

Distribution as above.

Step-by-step solution

1E(X) = Σ xᵢP(xᵢ).
2Var(X) = E(X²) − [E(X)]².

Final answer

μ = \sum x_{i} P_{i}, σ^{2} = \sum x_{i}^{2} P_{i} - μ^{2} .

Step-by-step solution

1(i) P(X < 2) = 0.09 + 0.15 = 0.24.
2(ii) P(X ≥ 3) = 0.25 + 0.20 + 0.11 = 0.56.
3(iii) E(X) = 0(0.09) + 1(0.15) + 2(0.20) + 3(0.25) + 4(0.20) + 5(0.11) = 0 + 0.15 + 0.40 + 0.75 + 0.80 + 0.55 = 2.65.

Final answer

(i) 0.24; (ii) 0.56; (iii) 2.65.

Step-by-step solution

1X ~ Binomial(4, 1/2).
2P(X = 0) = 1/16, P(1) = 4/16, P(2) = 6/16, P(3) = 4/16, P(4) = 1/16.
3E(X) = 4(1/2) = 2. Var(X) = 4(1/2)(1/2) = 1.

Final answer

E(X) = 2, Var(X) = 1.

Step-by-step solution

1P(X = 1) = 1/36, P(X = 2) = 3/36, P(X = 3) = 5/36, P(X = 4) = 7/36, P(X = 5) = 9/36, P(X = 6) = 11/36.
2E(X) = (1·1 + 2·3 + 3·5 + 4·7 + 5·9 + 6·11)/36 = (1+6+15+28+45+66)/36 = 161/36.

Final answer

E (X) = \frac{161}{36} .

Step-by-step solution

1(i) k + 2k + 3k + 4k = 10k = 1 ⇒ k = 1/10.
2(ii) E(X) = 0(1/10) + 1(2/10) + 2(3/10) + 3(4/10) = 0 + 0.2 + 0.6 + 1.2 = 2.0.

Final answer

(i) 1/10; (ii) 2.

Step-by-step solution

1P(X = 0) = C(2,2)/C(5,2) = 1/10; P(X = 1) = C(3,1)C(2,1)/C(5,2) = 6/10; P(X = 2) = C(3,2)/C(5,2) = 3/10.

Final answer

Distribution: 0→1/10, 1→6/10, 2→3/10.

Step-by-step solution

1(i) c[1/(1·2) + 1/(2·3) + 1/(3·4) + 1/(4·5)] = c[1/2 + 1/6 + 1/12 + 1/20]
2= c(30/60 + 10/60 + 5/60 + 3/60) = c(48/60) = 4c/5 = 1 ⇒ c = 5/4.
3(ii) P(X > 2) = P(3) + P(4) = (5/4)(1/12) + (5/4)(1/20) = 5/48 + 5/80 = 25/240 + 15/240 = 40/240 = 1/6.

Final answer

(i) c = 5/4; (ii) 1/6.

Step-by-step solution

1E(X) = −0.25 + 0 + 0.25 = 0.
2E(X²) = 1(0.25) + 0(0.50) + 1(0.25) = 0.50.
3Var(X) = 0.50 − 0² = 0.50.

Final answer

E(X) = 0, Var(X) = 0.50.

Step-by-step solution

1(i) a + 3a + 5a + 7a = 16a = 1 ⇒ a = 1/16.
2(ii) P(X < 2) = P(0) + P(1) = 1/16 + 3/16 = 4/16 = 1/4.
3(iii) P(X ≤ 2) = 1/16 + 3/16 + 5/16 = 9/16.
4(iv) P(X ≥ 2) = 5/16 + 7/16 = 12/16 = 3/4.

Final answer

(i) 1/16; (ii) 1/4; (iii) 9/16; (iv) 3/4.

Step-by-step solution

1(i) P(X ≥ 3) = 0.3 + 0.25 + 0.15 = 0.70.
2(ii) P(X ≤ 2) = 0.1 + 0.2 = 0.3.
3(iii) E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.25) + 5(0.15) = 0.1 + 0.4 + 0.9 + 1.0 + 0.75 = 3.15.

Final answer

(i) 0.70; (ii) 0.30; (iii) 3.15.

Step-by-step solution

1Each P(X = k) = 1/n.
2E(X) = (1/n)(1+2+...+n) = (1/n)·n(n+1)/2 = (n+1)/2.

Final answer

E (X) = \frac{n + 1}{2} .

Step-by-step solution

1E(X) = (−2)(0.2) + (−1)(0.3) + 0(0.1) + 1(0.25) + 2(0.15) = −0.4 − 0.3 + 0 + 0.25 + 0.3 = −0.15.
2E(X²) = 4(0.2) + 1(0.3) + 0 + 1(0.25) + 4(0.15) = 0.8 + 0.3 + 0.25 + 0.6 = 1.95.

Final answer

E(X) = −0.15, E(X²) = 1.95.

Step-by-step solution

1(i) c(1+2+3+4) = 10c = 1, so c = 1/10.
2(ii) E(X) = (1/10)(1+4+9+16) = 30/10 = 3.
3(iii) E(X²) = (1/10)(1+16+81+256) = 354/10 = 35.4. Var(X) = 35.4 − 9 = 26.4.

Final answer

(i) 1/10; (ii) 3; (iii) 26.4.

87

Exercise 13.5 — Bernoulli Trials and Binomial Distribution

19Exercise questions

Final answer

No — trials are not independent (without replacement), so the conditions of a binomial distribution are not met.

Step-by-step solution

1X ~ Binomial(10, 1/2). P(X = 6) = C(10,6)(1/2)⁶(1/2)⁴ = 210/1024 = 105/512.

Final answer

\frac{105}{512} .

Step-by-step solution

1(i) P(X = 3) = C(5,3)(3/4)³(1/4)² = 10 · 27/64 · 1/16 = 270/1024 = 135/512.
2(ii) P(X ≥ 3) = P(3) + P(4) + P(5)
3= 135/512 + 5·(81/256)(1/4) + 243/1024
4= 135/512 + 405/1024 + 243/1024 = 270/1024 + 405/1024 + 243/1024 = 918/1024 = 459/512.

Final answer

(i) 135/512; (ii) 459/512.

Step-by-step solution

1This is without replacement, so strictly hypergeometric, not binomial. Using hypergeometric:
2(i) P(2 defective) = C(4,2)C(16,3)/C(20,5) = 6·560/15504 = 3360/15504 = 70/323.
3(ii) P(at most 2) = [C(4,0)C(16,5) + C(4,1)C(16,4) + C(4,2)C(16,3)]/C(20,5)
4= [4368 + 4·1820 + 6·560]/15504 = [4368 + 7280 + 3360]/15504 = 15008/15504 = 938/969.

Final answer

(i) 70/323; (ii) 938/969.

Step-by-step solution

1X ~ Binomial(6, 0.05). p = 1/20, q = 19/20.
2(i) P(0) = (19/20)⁶ = 19⁶/20⁶ ≈ 0.7351.
3(ii) P(≤1) = P(0) + 6(1/20)(19/20)⁵ = 19⁶/20⁶ + 6·19⁵/20⁶ ≈ 0.7351 + 0.2321 = 0.9672.
4(iii) P(>1) = 1 − P(≤1) ≈ 0.0328.
5(iv) P(≥1) = 1 − P(0) ≈ 0.2649.

Final answer

(i) ≈0.7351; (ii) ≈0.9672; (iii) ≈0.0328; (iv) ≈0.2649.

Step-by-step solution

1X ~ Binomial(5, 1/4). P(X ≥ 2) = 1 − P(0) − P(1) = 1 − (3/4)⁵ − 5(1/4)(3/4)⁴
2= 1 − 243/1024 − 5·81/1024 = 1 − 243/1024 − 405/1024 = 1 − 648/1024 = 376/1024 = 47/128.

Final answer

\frac{47}{128} .

Step-by-step solution

1P(doublet) = 6/36 = 1/6. X ~ Binomial(5, 1/6).
2(i) P(X = 2) = C(5,2)(1/6)²(5/6)³ = 10·(1/36)(125/216) = 1250/7776 = 625/3888.
3(ii) P(X ≥ 2) = 1 − P(0) − P(1) = 1 − (5/6)⁵ − 5(1/6)(5/6)⁴
4= 1 − 3125/7776 − 5(625/7776) = 1 − 3125/7776 − 3125/7776 = 1 − 6250/7776 = 1526/7776 = 763/3888.

Final answer

(i) 625/3888; (ii) 763/3888.

Step-by-step solution

1p = 2/8 = 1/4, q = 3/4. X ~ Binomial(3, 1/4).
2(i) P(3) = (1/4)³ = 1/64.
3(ii) P(0) = (3/4)³ = 27/64.
4(iii) P(≥1) = 1 − 27/64 = 37/64.

Final answer

(i) 1/64; (ii) 27/64; (iii) 37/64.

Step-by-step solution

1(i) C(3,4)/C(10,4) = 0 (only 3 defective available).
2(ii) C(7,4)/C(10,4) = 35/210 = 1/6.

Final answer

(i) 0; (ii) 1/6.

Step-by-step solution

1P(at least one hit) = 1 − (0.7)ⁿ > 0.95.
2(0.7)ⁿ < 0.05. Try n = 10: 0.7¹⁰ ≈ 0.02825 < 0.05. n = 9: 0.7⁹ ≈ 0.04036 < 0.05 too? 0.04036 < 0.05 ✓. n = 8: 0.7⁸ ≈ 0.05765 > 0.05 ✗.
3Minimum n = 9.

Final answer

n = 9.

Step-by-step solution

1p(safe) = 9/10. X ~ Binomial(5, 9/10).
2P(X ≥ 4) = P(4) + P(5) = 5(9/10)⁴(1/10) + (9/10)⁵
3= 5·6561/10000·(1/10) + 59049/100000 = 32805/100000 + 59049/100000 = 91854/100000 = 45927/50000.

Final answer

\frac{45927}{50000} \approx 0.9185.

Step-by-step solution

1P(sum 7) = 6/36 = 1/6. X ~ Binomial(6, 1/6).
2P(X = 1) = 6(1/6)(5/6)⁵ = 6·5⁵/6⁶ = 6·3125/46656 = 18750/46656 = 3125/7776.

Final answer

\frac{3125}{7776} .

Step-by-step solution

1X ~ Binomial(50, 0.01). P(X ≥ 1) = 1 − (0.99)⁵⁰ ≈ 1 − 0.6050 = 0.3950.

Final answer

1 - (0.99)^{50} \approx 0.3950.

Step-by-step solution

1p(red) = 5/8 per draw (with replacement). X ~ Binomial(4, 5/8).
2P(X = 2) = C(4,2)(5/8)²(3/8)² = 6·25/64·9/64 = 6·225/4096 = 1350/4096 = 675/2048.

Final answer

\frac{675}{2048} .

Step-by-step solution

1p = 1/2, X ~ Binomial(6, 1/2).
2(i) P(5) = 6(1/2)⁶ = 6/64 = 3/32.
3(ii) P(≥5) = P(5) + P(6) = 6/64 + 1/64 = 7/64.
4(iii) P(≤5) = 1 − P(6) = 1 − 1/64 = 63/64.

Final answer

(i) 3/32; (ii) 7/64; (iii) 63/64.

Step-by-step solution

1X ~ Binomial(5, 1/6).
2(i) P(X = 3) = C(5,3)(1/6)³(5/6)² = 10·(1/216)(25/36) = 250/7776 = 125/3888.
3(ii) P(X ≥ 3) = P(3) + P(4) + P(5)
4= 250/7776 + 5(1/1296)(5/6) + 1/7776
5= 250/7776 + 25/7776 + 1/7776 = 276/7776 = 23/648.

Final answer

(i) 125/3888; (ii) 23/648.

Step-by-step solution

1X ~ Binomial(12, 0.1).
2(i) P(2) = C(12,2)(0.1)²(0.9)¹⁰ = 66·0.01·0.3487 ≈ 0.2301.
3(ii) P(≤2) = P(0)+P(1)+P(2) ≈ 0.2824 + 0.3765 + 0.2301 ≈ 0.8891.
4(iii) P(>2) = 1 − 0.8891 ≈ 0.1109.

Final answer

(i) ≈ 0.2301; (ii) ≈ 0.8891; (iii) ≈ 0.1109.

Step-by-step solution

1P(doublet) = 6/36 = 1/6. X ~ Binomial(4, 1/6).
2P(X = 2) = C(4,2)(1/6)²(5/6)² = 6·(1/36)(25/36) = 150/1296 = 25/216.

Final answer

\frac{25}{216} .

Step-by-step solution

1X ~ Binomial(15, 0.4).
2(i) P(X ≥ 10) = Σ from k=10 to 15 of C(15,k)(0.4)ᵏ(0.6)^{15−k} ≈ 0.0338.
3(ii) P(X ≤ 5) ≈ 0.4032.
4(iii) P(X = 6) = C(15,6)(0.4)⁶(0.6)⁹ ≈ 0.2066.

Final answer

(i) ≈ 0.0338; (ii) ≈ 0.4032; (iii) ≈ 0.2066.

88

Miscellaneous Exercise on Chapter 13

28Exercise questions

Step-by-step solution

1P(A|B) = 0.2/0.5 = 0.4.
2P(B|A) = 0.2/0.4 = 0.5.
3P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7.

Final answer

P(A|B) = 0.4, P(B|A) = 0.5, P(A ∪ B) = 0.7.

Step-by-step solution

1P(both white) = (5/15)(4/14) = 20/210 = 2/21.
2P(at least one white) = 1 − P(both black) = 1 − (10/15)(9/14) = 1 − 90/210 = 120/210 = 4/7.
3P(both white | at least one white) = (2/21)/(4/7) = (2/21)(7/4) = 14/84 = 1/6.

Final answer

\frac{1}{6} .

Final answer

\frac{4}{12} = \frac{1}{3} .

Step-by-step solution

1Total = 12.
2(i) P(white) = 4/12 = 1/3.
3(ii) P(not red) = (5+4)/12 = 9/12 = 3/4.
4(iii) P(neither white nor black) = P(red) = 3/12 = 1/4.

Final answer

(i) 1/3; (ii) 3/4; (iii) 1/4.

Step-by-step solution

1Sums less than 6: {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}, 10 outcomes.
2Sum = 3: {(1,2),(2,1)}, 2 outcomes.
3P = 2/10 = 1/5.

Final answer

\frac{1}{5} .

Step-by-step solution

1E = {1,2,3}, F = {2,4,6}. E ∩ F = {2}.
2P(E|F) = 1/3. P(F|E) = 1/3.
3P(E) = 1/2, P(F) = 1/2. P(E)P(F) = 1/4 ≠ P(E ∩ F) = 1/6. Not independent.

Final answer

P(E|F) = P(F|E) = \frac{1}{3};\; \text{not independent.

Step-by-step solution

1P = (5/12)(4/11)(3/10) = 60/1320 = 1/22.

Final answer

\frac{1}{22} .

Step-by-step solution

1P(at least 2 heads) = P(2) + P(3) = 3/8 + 1/8 = 4/8 = 1/2.
2P(all heads | at least 2 heads) = (1/8)/(1/2) = 1/4.

Final answer

\frac{1}{4} .

Step-by-step solution

1k = 1/10. E(X) = 0(1/10) + 1(2/10) + 2(3/10) + 3(4/10) = 20/10 = 2.
2E(X²) = 0 + 2/10 + 12/10 + 36/10 = 50/10 = 5.
3Var(X) = 5 − 4 = 1.

Final answer

E (X) = 2, Var (X) = 1.

Step-by-step solution

1X ~ Binomial(5, 1/2).
2E(X) = 5/2 = 2.5. Var(X) = 5/4 = 1.25.

Final answer

E (X) = 2.5, Var (X) = 1.25.

Step-by-step solution

1(i) P(X = 2) = C(10,2)(1/5)²(4/5)⁸ = 45·(1/25)·(4/5)⁸ ≈ 45·0.04·0.1678 ≈ 0.3020.
2(ii) P(X ≥ 1) = 1 − (4/5)¹⁰ ≈ 1 − 0.1074 ≈ 0.8926.

Final answer

(i) ≈ 0.3020; (ii) ≈ 0.8926.

Step-by-step solution

1Approximate with Binomial(50, 31/365).
2P(X ≤ 2) = P(0) + P(1) + P(2) ≈ e^(−μ)(1 + μ + μ²/2) with μ = 50·31/365 ≈ 4.2466.
3≈ e^(−4.2466)(1 + 4.2466 + 8.9952) ≈ 0.01435 × 14.242 ≈ 0.2044.

Final answer

\approx 0.2044.

Step-by-step solution

1X = number of odd results, X ~ Binomial(4, 1/2).
2P(X even) = P(0) + P(2) + P(4) = 1/16 + 6/16 + 1/16 = 8/16 = 1/2.

Final answer

\frac{1}{2} .

Step-by-step solution

1p(red) = 5/8 per draw. X ~ Binomial(3, 5/8).
2(i) P(3 red) = (5/8)³ = 125/512.
3(ii) P(at least one green) = 1 − 125/512 = 387/512.

Final answer

(i) 125/512; (ii) 387/512.

Step-by-step solution

1With replacement: P(both aces) = (4/52)² = (1/13)² = 1/169.

Final answer

\frac{1}{169} .

Step-by-step solution

1This matches the problem from NCERT: the sample space strategy applies. E = {(1,5),(2,4),(4,2),(5,1)}: 4 outcomes. P(E) = 4/36 = 1/9.

Final answer

\frac{1}{9} .

Step-by-step solution

1P(1) = (2/1)·P(0) = 2/10 = 1/5.
2P(2) = (3/2)·P(1) = (3/2)(1/5) = 3/10.
3P(3) = (4/3)·P(2) = (4/3)(3/10) = 4/10 = 2/5.

Final answer

\frac{2}{5} .

Step-by-step solution

1P(both red) = (5/8)(4/7) = 20/56 = 5/14.
2P(at least one red) = 1 − (3/8)(2/7) = 1 − 6/56 = 50/56 = 25/28.
3P(both red | at least one red) = (5/14)/(25/28) = (5/14)(28/25) = 140/350 = 2/5.

Final answer

\frac{2}{5} .

Step-by-step solution

1Sum = 8: {(2,6),(3,5),(4,4),(5,3),(6,2)}, 5 outcomes.
2With a 6: {(2,6),(6,2)}, 2 outcomes.
3P = 2/5.

Final answer

\frac{2}{5} .

Step-by-step solution

1(i) C(4,3)/C(9,3) = 4/84 = 1/21.
2(ii) C(3,1)C(6,2)/C(9,3) = 3·15/84 = 45/84 = 15/28.
3(iii) 1 − P(no red) = 1 − C(7,3)/C(9,3) = 1 − 35/84 = 1 − 5/12 = 7/12.

Final answer

(i) 1/21; (ii) 15/28; (iii) 7/12.

Step-by-step solution

1p(5) = 1/6, X ~ Binomial(3, 1/6).
2(i) P(≥1) = 1 − (5/6)³ = 1 − 125/216 = 91/216.
3(ii) P(≤2) = 1 − P(3) = 1 − (1/6)³ = 1 − 1/216 = 215/216.
4(iii) P(2) = 3(1/6)²(5/6) = 15/216 = 5/72.

Final answer

(i) 91/216; (ii) 215/216; (iii) 5/72.

Step-by-step solution

1p = 1/5, X ~ Binomial(3, 1/5).
2P(≤2) = 1 − P(3) = 1 − (1/5)³ = 1 − 1/125 = 124/125.

Final answer

\frac{124}{125} .

Step-by-step solution

1E = {2,4,6}, F = {1,2,3,4}. E ∩ F = {2,4}.
2P(E|F) = 2/4 = 1/2. P(F|E) = 2/3.
3P(E)·P(F) = (3/6)(4/6) = 12/36 = 1/3. P(E ∩ F) = 2/6 = 1/3. Equal, so independent.

Final answer

P(E|F) = 1/2, P(F|E) = 2/3; E and F are independent.

Step-by-step solution

1P(both red) = (3/12)(2/11) = 6/132 = 1/22.
2P(at least one red) = 1 − C(9,2)/C(12,2) = 1 − 36/66 = 30/66 = 5/11.
3P(both red | at least one red) = (1/22)/(5/11) = (1/22)(11/5) = 1/10.

Final answer

\frac{1}{10} .

Step-by-step solution

1P(late) = 0.4(0.1) + 0.3(0.02) + 0.2(0.08) + 0.1(0.03) = 0.04 + 0.006 + 0.016 + 0.003 = 0.065.
2P(train | late) = 0.016/0.065 = 16/65.

Final answer

P(late) = 0.065, P(train | late) = 16/65.

Step-by-step solution

1P(at least 3 heads) = P(3) + P(4) = 4/16 + 1/16 = 5/16.
2P(all heads | at least 3 heads) = (1/16)/(5/16) = 1/5.

Final answer

\frac{1}{5} .

Step-by-step solution

1(a + 3a + 5a + 7a + 9a) = 25a = 1, so a = 1/25.
2E(X) = 0(1/25) + 1(3/25) + 2(5/25) + 3(7/25) + 4(9/25) = (0+3+10+21+36)/25 = 70/25 = 14/5.
3E(X²) = 0 + 3/25 + 20/25 + 63/25 + 144/25 = 230/25 = 46/5.
4Var(X) = 46/5 − (14/5)² = 230/25 − 196/25 = 34/25.

Final answer

a = 1/25, E(X) = 14/5, Var(X) = 34/25.

Step-by-step solution

1X ~ Binomial(10, 0.05).
2(i) P(0) = (0.95)¹⁰ ≈ 0.5987.
3(ii) P(1) = 10(0.05)(0.95)⁹ ≈ 10(0.05)(0.6302) ≈ 0.3151.
4(iii) P(>1) = 1 − 0.5987 − 0.3151 ≈ 0.0862.

Final answer

(i) ≈ 0.5987; (ii) ≈ 0.3151; (iii) ≈ 0.0862.

Class 12 Maths NCERT Solutions

Where can I find Class 12 Maths NCERT solutions chapter-wise?

How to Use These NCERT Solutions

Chapter 1 — Relations and Functions

Exercise 1.1 — Types of Relations

Exercise 1.2 — One-One, Onto and Bijective Functions

Exercise 1.3 — Composition of Functions and Inverses

Exercise 1.4 — Binary Operations

Miscellaneous Exercise on Chapter 1

Chapter 2 — Inverse Trigonometric Functions

Exercise 2.1 — Principal Values

Exercise 2.2 — Identities, Simplest Forms and Values

Miscellaneous Exercise on Chapter 2

Chapter 3 — Matrices

Exercise 3.1 — Basics of Matrices

Exercise 3.2 — Matrix Algebra

Exercise 3.3 — Transpose, Symmetric and Skew-Symmetric Matrices

Exercise 3.4 — Elementary Row and Column Operations

Miscellaneous Exercise on Chapter 3

Chapter 4 — Determinants

Exercise 4.1 — Determinants of Order 1, 2 and 3

Exercise 4.2 — Properties of Determinants

Exercise 4.3 — Minors and Cofactors

Exercise 4.4 — Adjoint and Inverse of a Matrix

Exercise 4.5 — Area of Triangle, Consistency and Cramer's Rule

Miscellaneous Exercise on Chapter 4

Chapter 5 — Continuity and Differentiability

Exercise 5.1 — Continuity

Exercise 5.2 — Derivatives of Composite Functions

Exercise 5.3 — Implicit and Inverse Trigonometric Derivatives

Exercise 5.4 — Exponential and Logarithmic Derivatives

Exercise 5.5 — Logarithmic Differentiation

Exercise 5.6 — Derivatives of Parametric Functions

Exercise 5.7 — Second Order Derivatives

Exercise 5.8 — Rolle's and Mean Value Theorems

Miscellaneous Exercise on Chapter 5

Chapter 6 — Application of Derivatives

Exercise 6.1 — Rate of Change of Quantities

Exercise 6.2 — Increasing and Decreasing Functions

Exercise 6.3 — Tangents and Normals

Exercise 6.4 — Approximation

Exercise 6.5 — Maxima and Minima

Miscellaneous Exercise — Mixed Problems on Application of Derivatives

Chapter 7 — Integrals

Exercise 7.1 — Integration as Inverse of Differentiation

Exercise 7.2 — Integration by Substitution

Exercise 7.3 — Integration Using Trigonometric Identities

Exercise 7.4 — Integration of Particular Functions

Exercise 7.5 — Integration by Partial Fractions

Exercise 7.6 — Integration by Parts

Exercise 7.7 — Definite Integrals

Exercise 7.8 — Evaluation of Definite Integrals by Substitution

Exercise 7.9 — Some Properties of Definite Integrals

Exercise 7.10 — More Definite Integral Problems

Exercise 7.11 — Properties of Definite Integrals

Miscellaneous Exercise on Chapter 7

Chapter 8 — Application of Integrals

Exercise 8.1 — Area Under Simple Curves

Exercise 8.2 — Area Enclosed Between a Curve and a Line, and Between Two Curves

Miscellaneous Exercise on Chapter 8

Chapter 9 — Differential Equations

Exercise 9.1 — Order and Degree of a Differential Equation

Exercise 9.2 — General and Particular Solutions

Exercise 9.3 — Formation of Differential Equations

Exercise 9.4 — Variable Separable Method

Exercise 9.5 — Homogeneous Differential Equations

Exercise 9.6 — Linear Differential Equations

Miscellaneous Exercise — Differential Equations

Chapter 10 — Vector Algebra

Exercise 10.1 — Basic Concepts

Exercise 10.2 — Addition of Vectors and Multiplication of a Vector by a Scalar

Exercise 10.3 — Dot (Scalar) Product of Vectors

Exercise 10.4 — Cross (Vector) Product of Vectors

Miscellaneous Exercise on Chapter 10

Chapter 11 — Three Dimensional Geometry

Exercise 11.1 — Direction Cosines

Exercise 11.2 — Equation of a Line in Space

Exercise 11.3 — Planes

Miscellaneous Exercise on Chapter 11

Chapter 12 — Linear Programming