Which is the best order to practise Class 11 Maths NCERT solutions?

Follow the NCERT chapter order: Sets, Relations and Functions, Trigonometric Functions, Complex Numbers, Linear Inequalities, Permutations and Combinations, Binomial Theorem, Sequences and Series, Straight Lines, Conic Sections, Three Dimensional Geometry, Limits and Derivatives, Statistics and Probability — the same order used on this page.

How do I score full marks in Class 11 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 — algebra, trigonometry, coordinate geometry and limits — that JEE Main tests heavily. Use these solved problems to master the standard methods, then practise JEE-level problems for speed.

Which Class 11 maths chapters carry the most board marks?

Algebra and calculus dominate: sets, relations and functions, trigonometric functions, sequences and series, straight lines, conic sections, and limits and derivatives carry the largest share of the Class 11 board and JEE weightage.

Free PDF: Class 11 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 11 Maths Notes hub. These solutions complement that hub — same NCERT order, worked problems instead of theory.

02

Chapter 1 — Sets

A set is a well-defined collection of distinct objects, written in either roster form (listing elements) or set-builder form (stating a rule). This chapter covers types of sets, subsets, the empty set, power sets, union, intersection, difference and complement, Venn diagrams and De Morgan's laws, and the formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Every question below is from the NCERT Class 11 textbook (rationalised edition), worked out line by line in the board pattern.

Board pattern

Set questions are almost always short/1-mark or 4-mark problems. For 1-mark questions only the final set in roster form is expected; for proof questions you must state each side's element condition before concluding equality. Always finish with {} roster form and never repeat an element inside a set.

$Ex 1.1 — Sets and their representations, roster and set-builder form (6 Q)$
$Ex 1.2 — Empty, finite and infinite sets (6 Q)$
$Ex 1.3 — Subsets, equal sets, power set, universal set (9 Q)$
$Ex 1.4 — Union, intersection and difference of sets (12 Q)$
$Ex 1.5 — Complement of a set, De Morgan’s laws (7 Q)$
$Ex 1.6 — Cardinal numbers with Venn diagrams (10 Q)$
$Misc — Mixed problems on sets (16 Q)$

03

Exercise 1.1 — Sets and Their Representations

6Exercise questions

Step-by-step solution

1A set must be well defined: one must be able to decide unambiguously whether an object belongs to it.
2(i) It is a set: the months are January, June and July — a definite collection.
3(ii) Not a set: 'ten most talented writers' is subjective and cannot be decided precisely.
4(iii) Not a set: 'eleven best batsmen' is subjective, not well defined.
5(iv) It is a set: all 15-year-old boys in your class are clearly identifiable.
6(v) It is a set: {1, 2, 3, …, 99} — all natural numbers less than 100.
7(vi) It is a set: the novels of Munshi Prem Chand form a definite collection.

Final answer

Sets: (i), (iv), (v), (vi). Not sets: (ii), (iii).

Step-by-step solution

1The elements of A are exactly 1, 2, 3, 4, 5, 6.
2(i) 5 ∈ A. (ii) 8 is not in A, so 8 ∉ A. (iii) 0 ∉ A. (iv) 4 ∈ A. (v) 2 ∈ A. (vi) 10 ∉ A.

Final answer

(i) ∈ (ii) ∉ (iii) ∉ (iv) ∈ (v) ∈ (vi) ∉

Step-by-step solution

1(i) Integers strictly between −3 and 7: A = {−2, −1, 0, 1, 2, 3, 4, 5, 6}.
2(ii) Natural numbers less than 6: B = {1, 2, 3, 4, 5}.
3(iii) Two-digit numbers with digit sum 8: C = {17, 26, 35, 44, 53, 62, 71, 80}.
4(iv) Prime divisors of 60 = 2²·3·5: D = {2, 3, 5}.
5(v) Distinct letters of TRIGONOMETRY: E = {T, R, I, G, O, N, M, E, Y}.

Final answer

A = {−2, −1, 0, 1, 2, 3, 4, 5, 6}, B = {1, 2, 3, 4, 5}, C = {17, 26, 35, 44, 53, 62, 71, 80}, D = {2, 3, 5}, E = {T, R, I, G, O, N, M, E, Y}.

Step-by-step solution

1(i) Elements are the multiples of 3 from 3 to 12: {x : x = 3n, n ∈ N, 1 ≤ n ≤ 4}.
2(ii) Powers of 2 from 2 to 32: {x : x = 2ⁿ, n ∈ N, 1 ≤ n ≤ 5}.
3(iii) Powers of 5: {x : x = 5ⁿ, n ∈ N, 1 ≤ n ≤ 4}.
4(iv) Even natural numbers: {x : x = 2n, n ∈ N}.
5(v) Perfect squares up to 100: {x : x = n², n ∈ N, 1 ≤ n ≤ 10}.

Final answer

(i) {x : x = 3n, 1 ≤ n ≤ 4, n ∈ N}; (ii) {x : x = 2ⁿ, 1 ≤ n ≤ 5}; (iii) {x : x = 5ⁿ, 1 ≤ n ≤ 4}; (iv) {x : x = 2n, n ∈ N}; (v) {x : x = n², 1 ≤ n ≤ 10}.

Step-by-step solution

1(i) A = {1, 3, 5, 7, …} — all odd natural numbers.
2(ii) Integers strictly between −0.5 and 4.5: B = {0, 1, 2, 3, 4}.
3(iii) x² ≤ 4 gives −2 ≤ x ≤ 2; integers: C = {−2, −1, 0, 1, 2}.
4(iv) Distinct letters of LOYAL: D = {L, O, Y, A}.
5(v) Months without 31 days: E = {April, June, September, November}.
6(vi) Consonants before k: F = {b, c, d, f, g, h, j}.

Final answer

A = {1, 3, 5, 7, …}, B = {0, 1, 2, 3, 4}, C = {−2, −1, 0, 1, 2}, D = {L, O, Y, A}, E = {April, June, September, November}, F = {b, c, d, f, g, h, j}.

Step-by-step solution

1(i) Primes less than 8 are 2, 3, 5, 7 → (c).
2(ii) Factors of 8: 1, 2, 4, 8 → (d).
3(iii) Odd natural numbers less than 7: {1, 3, 5} → (a).
4(iv) Squares of integers: {0, 1, 4, 9, 16, …} → (e).
5(v) Even numbers between 6 and 10: {8} but options → (b) {4, 8}, the pair given in the book.

Final answer

(i)→(c), (ii)→(d), (iii)→(a), (iv)→(e), (v)→(b).

04

Exercise 1.2 — Empty, Finite and Infinite Sets

6Exercise questions

Step-by-step solution

1(i) No odd number is divisible by 2, so this set has no elements → it is the null set.
2(ii) 2 is an even prime number, so the set {2} is not empty.
3(iii) No natural number is both < 5 and > 7 → null set.
4(iv) Parallel lines never meet → null set.

Final answer

(i), (iii) and (iv) are null sets; (ii) is not.

Step-by-step solution

1(i) {Jan … Dec} has 12 elements → finite.
2(ii) Countably endless → infinite.
3(iii) Exactly 100 elements → finite.
4(iv) {101, 102, 103, …} → infinite.
5(v) Only finitely many primes below 99 → finite.

Final answer

Finite: (i), (iii), (v). Infinite: (ii), (iv).

Step-by-step solution

1(i) y = c for any real c gives a parallel line → infinite.
2(ii) 26 letters → finite.
3(iii) {5, 10, 15, …} → infinite.
4(iv) The current animal population is finite.
5(v) A circle through the origin has centre anywhere on the plane → infinite many.

Final answer

Infinite: (i), (iii), (v). Finite: (ii), (iv).

Step-by-step solution

1(i) Same elements (order does not matter) → A = B.
2(ii) 12 ∈ A but 12 ∉ B; also 18 ∈ B but 18 ∉ A → A ≠ B.
3(iii) B = {2, 4, 6, 8, 10} = A → equal.
4(iv) A = {10, 20, 30, …} but B contains 15, 25, … → A ≠ B.

Final answer

A = B for (i) and (iii); A ≠ B for (ii) and (iv).

Step-by-step solution

1(i) x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = −2, −3. So B = {−2, −3} ≠ {2, 3}.
2(ii) A = {F, O, L, W}, B = {W, O, L, F} — same letters → A = B.

Final answer

(i) Not equal: B = {−2, −3}. (ii) Equal.

Step-by-step solution

1A, B, C all list a definite number of elements → finite.
2D = {1, 2, 3, 4, 5} → finite.
3All four sets are finite.

Final answer

All of A, B, C, D are finite sets.

05

Exercise 1.3 — Subsets, Equal Sets and the Power Set

9Exercise questions

Step-by-step solution

1(i) Every element 2, 3, 4 is in the second set → {2, 3, 4} ⊂ {1, 2, 3, 4, 5}.
2(ii) i, o, u are not in {a, b, c, d, e} → ⊄.
3(iii) Every Class 11 student of the school is a student of the school → ⊂.
4(iv) The set of all circles is bigger than just radius-1 circles → ⊄ (with the roles as given: the second is a subset of the first).

Final answer

(i) ⊂ (ii) ⊄ (iii) ⊂ (iv) ⊄

Step-by-step solution

1(i) a and b are both in {b, c, a} → it is a subset, so the statement ⊄ is False.
2(ii) Vowels are a, e, i, o, u; {a, e} ⊂ vowels → True.
3(iii) 2 ∉ {1, 3, 5} → False.
4(iv) a ∈ {a, b, c} → True.
5(v) {a} is a set, not an element of {a, b, c} → False.
6(vi) {2, 4} and the divisors of 36 include 2, 4 → True.

Final answer

True: (ii), (iv), (vi). False: (i), (iii), (v).

Step-by-step solution

1A = {1, 2, {3, 4}, 5} — note {3, 4} is a single element of A.
2(i) Incorrect: the set {3, 4} is an element, not a subset. (ii) Correct: {3, 4} ∈ A.
3(iii) Correct: {{3, 4}} ⊂ A. (iv) Correct: 1 ∈ A. (v) Incorrect: 1 is an element, ⊂ is used between sets.
4(vi) Correct. (vii) Incorrect: {1, 2, 5} ∈ A would need it to be an element. (viii) Incorrect: 3 ∉ A (only {3, 4} is).
5(ix) Incorrect: ∅ is not an element of A. (x) Correct: every set contains ∅. (xi) Incorrect: {∅} is not a subset of A.

Final answer

Incorrect: (i), (v), (vii), (viii), (ix), (xi). Correct: (ii), (iii), (iv), (vi), (x).

Step-by-step solution

1A set with n elements has 2ⁿ subsets.
2(i) {a} has 2 subsets: ∅, {a}.
3(ii) {a, b} has 4: ∅, {a}, {b}, {a, b}.
4(iii) {1, 2, 3} has 8: ∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.
5(iv) ∅ has 1 subset: ∅ itself.

Final answer

(i) ∅, {a}. (ii) ∅, {a}, {b}, {a, b}. (iii) ∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}. (iv) ∅.

Step-by-step solution

1P(A) is the set of all subsets of A.
2A = ∅ has exactly one subset, namely ∅.
3So P(A) = {∅} has 1 element.

Final answer

P(A) has 1 element.

Step-by-step solution

1(i) Open at −4, closed at 6 → (−4, 6].
2(ii) Both open → (−12, −10).
3(iii) Closed at 0, open at 7 → [0, 7).
4(iv) Both closed → [3, 4].

Final answer

(i) (−4, 6] (ii) (−12, −10) (iii) [0, 7) (iv) [3, 4]

Step-by-step solution

1(i) {x : x ∈ R, −3 < x < 0}.
2(ii) {x : x ∈ R, 6 ≤ x ≤ 12}.
3(iii) {x : x ∈ R, 6 < x ≤ 12}.
4(iv) {x : x ∈ R, −23 ≤ x < 5}.

Final answer

(i) {x ∈ R : −3 < x < 0}; (ii) {x ∈ R : 6 ≤ x ≤ 12}; (iii) {x ∈ R : 6 < x ≤ 12}; (iv) {x ∈ R : −23 ≤ x < 5}.

Step-by-step solution

1A universal set must contain all elements under discussion.
2(i) The set of all triangles.
3(ii) The set of all triangles.

Final answer

In both cases the set of all triangles can be the universal set.

Step-by-step solution

1The exercise asks you to write down the elements of A or B that belong to the indicated subset.
2A ∪ B = {2, 3, 5, 6, 8, 9}.
3A ∩ B = {5}.

Final answer

A ∪ B = {2, 3, 5, 6, 8, 9}, A ∩ B = {5}.

06

Exercise 1.4 — Union, Intersection and Difference of Sets

12Exercise questions

Step-by-step solution

1Union A ∪ B = set of all elements in A or B.
2(i) X ∪ Y = {1, 2, 3, 5}.
3(ii) A ∪ B = {a, b, c, e, i, o, u}.
4(iii) A = {3, 6, 9, …}, B = {1, 2, 3, 4, 5}, so A ∪ B = {1, 2, 3, 4, 5, 6, 9, …}.
5(iv) A = {2, 3, 4, 5, 6}, B = {7, 8, 9}, so A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}.
6(v) A ∪ ∅ = {1, 2, 3}.

Final answer

(i) {1,2,3,5} (ii) {a,b,c,e,i,o,u} (iii) {1,2,3,4,5,6,9,…} (iv) {2,3,4,5,6,7,8,9} (v) {1,2,3}

Step-by-step solution

1Every element of A is in B → A ⊂ B is true.
2A ∪ B = {a, b, c} = B.

Final answer

Yes, A ⊂ B and A ∪ B = {a, b, c}.

Step-by-step solution

1If A ⊂ B then every element of A is already in B.
2So A ∪ B = B.

Final answer

A ∪ B = B.

Step-by-step solution

1(i) A ∪ B = {1, 2, 3, 4, 5, 6}.
2(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}.
3(iii) B ∪ C = {3, 4, 5, 6, 7, 8}.
4(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}.
5(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}.
6(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
7(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}.

Final answer

Listed in the steps above.

Step-by-step solution

1(i) X ∩ Y = {1, 3}.
2(ii) A ∩ B = {a}.
3(iii) A = {3, 6, 9, …}, B = {2, 4, 6, 8, …}, common multiples of 2 and 3 → {6, 12, 18, …}.
4(iv) A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, so A ∩ B = {3, 4}.

Final answer

(i) {1,3} (ii) {a} (iii) {6,12,18,…} (iv) {3,4}

Step-by-step solution

1(i) A ∩ B = {7, 9, 11}.
2(ii) B ∩ C = {11, 13}.
3(iii) A ∩ C ∩ D = ∅ (D has only 15, 17).
4(iv) A ∩ C = {11}.
5(v) B ∩ D = ∅.
6(vi) B ∪ C = {7, 9, 11, 13, 15}, so A ∩ (B ∪ C) = {7, 9, 11}.
7(vii) A ∩ D = ∅.
8(viii) B ∪ D = {7, 9, 11, 13, 15, 17}, so A ∩ (B ∪ D) = {7, 9, 11}.
9(ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7,9,11,13,15} = {7, 9, 11}.
10(x) (A ∪ D) ∩ (B ∪ C) = {3,5,7,9,11,15,17} ∩ {7,9,11,13,15} = {7, 9, 11, 15}.

Final answer

Listed in the steps.

Step-by-step solution

1(i) A ∩ B = B = even naturals.
2(ii) A ∩ C = C = odd naturals.
3(iii) A ∩ D = D = prime numbers.
4(iv) B ∩ C = ∅ (no number is both even and odd).
5(v) B ∩ D = {2} (only even prime).
6(vi) C ∩ D = odd primes = {3, 5, 7, 11, …}.

Final answer

(i) B (ii) C (iii) D (iv) ∅ (v) {2} (vi) {3, 5, 7, 11, …}

Step-by-step solution

1(i) The second set is {4, 5, 6} which shares 4 with the first → not disjoint.
2(ii) Share the element e → not disjoint.
3(iii) No integer is both even and odd → disjoint.

Final answer

Only (iii) is a disjoint pair.

Step-by-step solution

1A − B = {3, 6, 9, 15, 18, 21} (drop 12).
2A − C = {3, 9, 15, 18, 21} (drop 6, 12).
3A − D = {3, 6, 9, 12, 18, 21} (drop 15).
4B − A = {4, 8, 16, 20}. C − A = {2, 4, 8, 10, 14, 16}.
5D − A = {5, 10, 20}.
6B − C = {20} (drop 4, 8, 12, 16). B − D = {4, 8, 12, 16}.
7C − B = {2, 6, 10, 14}.
8D − B = {5, 10, 15}.
9C − D = {2, 4, 6, 8, 12, 14, 16}.
10D − C = {5, 15, 20}.

Final answer

Listed in the steps.

Step-by-step solution

1(i) X − Y = {a, c}.
2(ii) Y − X = {f, g}.
3(iii) X ∩ Y = {b, d}.

Final answer

(i) {a, c} (ii) {f, g} (iii) {b, d}

Step-by-step solution

1R − Q contains all real numbers that are not rational.
2Those are precisely the irrational numbers.

Final answer

R − Q = the set of irrational numbers.

Step-by-step solution

1Sets are equal when they contain exactly the same elements; order is irrelevant.
2All four pairs contain the same elements in different order → all equal.

Final answer

All four pairs are equal sets.

07

Exercise 1.5 — Complement of a Set

7Exercise questions

Step-by-step solution

1Complement = elements of U not in the set.
2(i) A′ = {5, 6, 7, 8, 9}.
3(ii) B′ = {1, 3, 5, 7, 9}.
4(iii) A ∪ C = {1, 2, 3, 4, 5, 6}, so (A ∪ C)′ = {7, 8, 9}.
5(iv) A ∪ B = {1, 2, 3, 4, 6, 8}, so (A ∪ B)′ = {5, 7, 9}.
6(v) (A′)′ = A = {1, 2, 3, 4}.
7(vi) B − C = {2, 8}, so (B − C)′ = {1, 3, 4, 5, 6, 7, 9}.

Final answer

Listed in the steps.

Step-by-step solution

1(i) A′ = {d, e, f, g, h}.
2(ii) B′ = {a, b, c, h}.
3(iii) C′ = {b, d, f, h}.
4(iv) D′ = {b, c, d, e}.

Final answer

(i) {d,e,f,g,h} (ii) {a,b,c,h} (iii) {b,d,f,h} (iv) {b,c,d,e}

Step-by-step solution

1(i) Odd naturals.
2(ii) Even naturals.
3(iii) Naturals that are not positive multiples of 3: {1, 2, 4, 5, 7, …}.
4(iv) Non-prime naturals: {1, 4, 6, 8, 9, …}.
5(v) Naturals not divisible by both 3 and 5 (i.e. not divisible by 15).
6(vi) Non-perfect-square naturals: {2, 3, 5, 6, 7, …}.
7(vii) Non-perfect-cube naturals: {2, 3, 4, 5, 6, 7, 9, …}.
8(viii) {3} has complement N − {3}.
9(ix) 2x + 5 = 9 gives x = 2, so the complement is N − {2}.
10(x) Complement: {x : x ∈ N and x < 7} = {1, 2, 3, 4, 5, 6}.
11(xi) 2x + 1 > 10 → x > 4.5 → x ≥ 5; complement = {1, 2, 3, 4}.

Final answer

Listed in the steps.

Step-by-step solution

1(i) A ∪ B = {2, 3, 4, 5, 6, 7, 8}, so (A ∪ B)′ = {1, 9}.
2A′ = {1, 3, 5, 7, 9} and B′ = {1, 4, 6, 8, 9}, so A′ ∩ B′ = {1, 9} = (A ∪ B)′. Verified.
3(ii) A ∩ B = {2}, so (A ∩ B)′ = {1, 3, 4, 5, 6, 7, 8, 9}.
4A′ ∪ B′ = {1, 3, 4, 5, 6, 7, 8, 9} = (A ∩ B)′. Verified.

Final answer

Both De Morgan's laws verified.

Step-by-step solution

1Sketch two overlapping circles A and B inside rectangle U.
2(i) (A ∪ B)′ shades only the outside region (neither circle).
3(ii) A′ ∩ B′ is the same outside region — consistent with (A ∪ B)′.
4(iii) (A ∩ B)′ shades everything except the overlap of A and B.
5(iv) A′ ∪ B′ is everything except the overlap — same as (iii).

Final answer

Venn diagrams as described; (i)=(ii) and (iii)=(iv).

Step-by-step solution

1A′ = triangles not having any angle different from 60°.
2That means every angle equals 60° → equilateral triangles.

Final answer

A′ is the set of all equilateral triangles.

Step-by-step solution

1(i) A ∪ A′ = U.
2(ii) ∅′ = U, so ∅′ ∩ A = A.
3(iii) A ∩ A′ = ∅.
4(iv) U′ = ∅, so U′ ∩ A = ∅.

Final answer

(i) U (ii) A (iii) ∅ (iv) ∅

08

Exercise 1.6 — Cardinal Numbers and Venn Diagrams

10Exercise questions

Step-by-step solution

1Use n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y).
238 = 17 + 23 − n(X ∩ Y).
3n(X ∩ Y) = 40 − 38 = 2.

Final answer

n(X ∩ Y) = 2.

Step-by-step solution

1n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y).
218 = 8 + 15 − n(X ∩ Y) → n(X ∩ Y) = 23 − 18 = 5.

Final answer

n(X ∩ Y) = 5.

Step-by-step solution

1Let H = Hindi speakers, E = English speakers. n(H) = 250, n(E) = 200, n(H ∪ E) = 400.
2n(H ∩ E) = n(H) + n(E) − n(H ∪ E) = 250 + 200 − 400 = 50.

Final answer

50 people speak both languages.

Step-by-step solution

1n(S ∪ T) = n(S) + n(T) − n(S ∩ T) = 21 + 32 − 11 = 42.

Final answer

n(S ∪ T) = 42.

Step-by-step solution

1n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y).
260 = 40 + n(Y) − 10 → n(Y) = 30.

Final answer

n(Y) = 30.

Step-by-step solution

1n(C ∪ T) = 70 (everyone likes at least one), n(C) = 37, n(T) = 52.
2n(C ∩ T) = 37 + 52 − 70 = 19.

Final answer

19 people like both.

Step-by-step solution

1Let C = cricket, T = tennis. n(C) = 40, n(C ∩ T) = 10.
2The group total is n(C ∪ T) = 65 (assuming everyone likes at least one).
3Number liking tennis = n(T) = n(C ∪ T) − n(C) + n(C ∩ T) = 65 − 40 + 10 = 35.
4Tennis but not cricket = n(T) − n(C ∩ T) = 35 − 10 = 25.

Final answer

25 like tennis but not cricket; 35 like tennis.

Step-by-step solution

1n(F) = 50, n(S) = 20, n(F ∩ S) = 10.
2At least one = n(F ∪ S) = 50 + 20 − 10 = 60.

Final answer

60 people speak at least one of the languages.

Step-by-step solution

1n(M ∪ P) = 20, n(M) = 12, n(M ∩ P) = 4.
2n(P) = 20 − 12 + 4 = 12.
3Teach only Physics = n(P) − n(M ∩ P) = 12 − 4 = 8.

Final answer

8 teach only Physics; 12 teach Physics.

Step-by-step solution

1n(T ∪ C) = 150 + 225 − 100 = 275.
2Total 600, so neither = 600 − 275 = 325.

Final answer

325 students take neither.

09

Miscellaneous Exercise — Mixed Problems on Sets

16Exercise questions

Step-by-step solution

1x² − 8x + 12 = 0 → (x − 2)(x − 6) = 0 → x = 2, 6. So A = {2, 6}.
2B = {2, 4, 6}, C = {2, 4, 6, 8, …}, D = {6}.
3D ⊂ A, A ⊂ B, B ⊂ C.

Final answer

D ⊂ A ⊂ B ⊂ C.

Step-by-step solution

1(i) False: example A = {{1}}, x = 1, B = {{1}}. Then x ∈ A and A ∈ B but x ∉ B.
2(ii) False: e.g. A = {1}, B = {{1}}, C = {{1}, 2}: A ⊂ B and B ∈ C but A ∉ C.
3(iii) True: subset relation is transitive — every element of A is in B and every element of B is in C.
4(iv) False: A = {1}, B = {1, 2}, C = {2} gives A ⊄ B (false premise) — pick A = {3}, B = {1, 2}, C = {2, 3}: A ⊄ B, B ⊄ C yet A ⊄ C is true too, so the statement fails.
5(v) False: x ∈ A and A not a subset of B does not force x ∈ B.
6(vi) True: contrapositive of A ⊂ B — if x ∈ A then x ∈ B, so x ∉ B forces x ∉ A.

Final answer

True: (iii), (vi). False: (i), (ii), (iv), (v).

Step-by-step solution

1Take x ∈ B. Then x ∈ A ∪ B = A ∪ C, so x ∈ A or x ∈ C.
2If x ∈ A, then x ∈ A ∩ B = A ∩ C, hence x ∈ C.
3So in every case x ∈ C; thus B ⊂ C.
4By symmetry x ∈ C ⇒ x ∈ B, so C ⊂ B. Hence B = C.

Final answer

B = C.

Step-by-step solution

1(i) ⇒ (ii): if A ⊂ B every element of A is in B, so A − B has no elements.
2(ii) ⇒ (iii): A − B = ∅ means A ⊂ B, so A ∪ B = B.
3(iii) ⇒ (iv): A ∪ B = B implies A ⊂ B, so A ∩ B = A.
4(iv) ⇒ (i): A ∩ B = A means every element of A is in B, i.e. A ⊂ B.
5All four conditions chain, hence are equivalent.

Final answer

All four are equivalent.

Step-by-step solution

1Take x ∈ C − B, so x ∈ C and x ∉ B.
2Since A ⊂ B, x ∉ B implies x ∉ A (contrapositive).
3Hence x ∈ C and x ∉ A, i.e. x ∈ C − A.
4Therefore C − B ⊂ C − A.

Final answer

C − B ⊂ C − A, shown element-wise.

Step-by-step solution

1P(A) = P(B) means A and B have identical collections of subsets.
2Since A ∈ P(A), A ∈ P(B), so A ⊂ B.
3Similarly B ∈ P(B) = P(A), so B ⊂ A.
4Hence A = B.

Final answer

A = B.

Step-by-step solution

1P(A) ∪ P(B) ⊆ P(A ∪ B) is always true since each subset of A or B is a subset of A ∪ B.
2But equality fails: take A = {1}, B = {2}. Then {1, 2} ∈ P(A ∪ B) yet {1, 2} is neither a subset of A nor of B.
3So the statement is not true in general.

Final answer

Not true in general; counterexample A = {1}, B = {2}.

Step-by-step solution

1A ∩ B collects elements of A that are in B; A − B collects elements of A not in B. Together they give every element of A exactly once.
2Hence A = (A ∩ B) ∪ (A − B).
3B − A ⊆ B, and any element of B not in A lies in B − A. So A ∪ (B − A) = A ∪ B.

Final answer

Both identities shown.

Step-by-step solution

1(i) A ∩ B ⊂ A, so A ∪ (A ∩ B) ⊂ A; also A ⊂ A ∪ (A ∩ B) trivially. Hence equality.
2(ii) A ⊂ A ∪ B, so A ⊂ A ∩ (A ∪ B); and A ∩ (A ∪ B) ⊂ A. Hence equality.

Final answer

Both absorption laws hold.

Step-by-step solution

1Take A = {1}, B = {1, 2}, C = {1, 3}.
2A ∩ B = {1} = A ∩ C, but B ≠ C.

Final answer

Counterexample: A = {1}, B = {1, 2}, C = {1, 3}.

Step-by-step solution

1Take x ∈ A. Then x ∈ A ∪ X = B ∪ X, so x ∈ B or x ∈ X.
2If x ∈ X then x ∈ A ∩ X = ∅, impossible. So x ∈ B; hence A ⊂ B.
3Symmetrically B ⊂ A. Therefore A = B.

Final answer

A = B.

Step-by-step solution

1Let A = {1, 2}, B = {2, 3}, C = {1, 3}.
2A ∩ B = {2}, B ∩ C = {3}, A ∩ C = {1} — all non-empty.
3A ∩ B ∩ C = ∅ since no element is in all three.

Final answer

A = {1, 2}, B = {2, 3}, C = {1, 3}.

Step-by-step solution

1n(T ∪ C) = 150 + 225 − 100 = 275.
2Neither = 600 − 275 = 325.

Final answer

325 students like neither.

Step-by-step solution

1n(H ∪ E) = n(H) + n(E) − n(H ∩ E) = 100 + 50 − 25 = 125.

Final answer

125 students.

Step-by-step solution

1(i) n(H ∪ T ∪ I) = 25 + 26 + 26 − 9 − 11 − 8 + 3 = 52.
2Exactly one = n(H) − n(H∩T) − n(H∩I) + n(H∩T∩I) = 25 − 11 − 9 + 3 = 8.
3For T: 26 − 11 − 8 + 3 = 10. For I: 26 − 9 − 8 + 3 = 12.
4(ii) Total exactly one = 8 + 10 + 12 = 30.

Final answer

(i) 52 read at least one. (ii) 30 read exactly one.

Step-by-step solution

1Only C = n(C) − n(C ∩ A) − n(C ∩ B) + n(A ∩ B ∩ C).
2= 29 − 12 − 14 + 8 = 11.

Final answer

11 people liked product C only.

10

Chapter 2 — Relations and Functions

This chapter introduces ordered pairs, Cartesian products, relations between two sets, and functions as special relations where each input has exactly one output. You will practise writing relations in roster and set-builder form, finding domains and ranges, and classifying functions. Every question below is from the NCERT Class 11 textbook (rationalised edition), worked line by line.

Board pattern

A relation from A to B is any subset of A × B; a function is a relation in which no first coordinate repeats. Board questions often ask for domain (all first coordinates) and range (all second coordinates) — right the roster form before attempting these. When A has m elements and B has n elements, the number of relations from A to B is 2ᵐⁿ.

$Ex 2.1 — Ordered pairs and Cartesian products (10 Q)$
$Ex 2.2 — Relations, domain, codomain and range (9 Q)$
$Ex 2.3 — Functions and their domains and ranges (5 Q)$
$Misc — Mixed problems on relations and functions (12 Q)$

11

Exercise 2.1 — Ordered Pairs and Cartesian Products

10Exercise questions

Step-by-step solution

1Two ordered pairs are equal iff their first coordinates match and their second coordinates match.
2x/3 + 1 = 5/3 → x/3 = 2/3 → x = 2.
3y − 2/3 = 1/3 → y = 1.

Final answer

x = 2, y = 1.

Step-by-step solution

1A has 3 elements; B has 3 elements.
2n(A × B) = n(A) × n(B) = 3 × 3 = 9.

Final answer

A × B has 9 elements.

Step-by-step solution

1G × H: first coordinate from G, second from H → {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}.
2H × G: first from H, second from G → {(5,7), (5,8), (4,7), (4,8), (2,7), (2,8)}.

Final answer

G × H = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}; H × G = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}.

Step-by-step solution

1(i) False. P × Q has 2 × 2 = 4 elements: P × Q = {(m,n), (m,m), (n,n), (n,m)}.
2(ii) True — this is exactly the definition of the Cartesian product.
3(iii) True — B ∩ ∅ = ∅, so A × ∅ = ∅.

Final answer

(i) False → P × Q = {(m,n),(m,m),(n,n),(n,m)}. (ii) True. (iii) True.

Step-by-step solution

1A × A × A = ordered triples with each coordinate in {−1, 1}.
2{(-1,-1,-1), (-1,-1,1), (-1,1,-1), (-1,1,1), (1,-1,-1), (1,-1,1), (1,1,-1), (1,1,1)}.

Final answer

A × A × A = {(-1,-1,-1), (-1,-1,1), (-1,1,-1), (-1,1,1), (1,-1,-1), (1,-1,1), (1,1,-1), (1,1,1)}.

Step-by-step solution

1The first coordinates are a and b → A = {a, b}.
2The second coordinates are x and y → B = {x, y}.

Final answer

A = {a, b}, B = {x, y}.

Step-by-step solution

1B ∩ C = ∅ (no common elements).
2So A × (B ∩ C) = A × ∅ = ∅.
3A × B = 8 pairs, A × C = 4 pairs, with no ordered pair in common since seconds are disjoint → (A × B) ∩ (A × C) = ∅.
4Both sides equal ∅, hence verified.

Final answer

Verified: both sides equal ∅.

Step-by-step solution

1A × B = {(1,3), (1,4), (2,3), (2,4)} — 4 elements.
2Number of subsets = 2⁴ = 16.
3They are: ∅; singletons {(1,3)}, {(1,4)}, {(2,3)}, {(2,4)}; three pairs sharing one element; the 6 two-element subsets; 4 three-element subsets; and A × B itself.

Final answer

A × B = {(1,3),(1,4),(2,3),(2,4)}; it has 2⁴ = 16 subsets.

Step-by-step solution

1First coordinates give A; they are x, y, z (distinct) → A = {x, y, z}.
2Second coordinates are 1 and 2 → B = {1, 2}.

Final answer

A = {x, y, z}, B = {1, 2}.

Step-by-step solution

1n(A × A) = n(A)² = 9 → n(A) = 3.
2−1, 0 appear as first coordinates, and 0, 1 appear as second coordinates → A = {−1, 0, 1}.
3A × A = {(-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)}.
4Remaining elements besides (−1, 0) and (0, 1): (-1,-1), (-1,1), (0,-1), (0,0), (1,-1), (1,0), (1,1).

Final answer

A = {−1, 0, 1}; remaining: (-1,-1), (-1,1), (0,-1), (0,0), (1,-1), (1,0), (1,1).

12

Exercise 2.2 — Relations, Domain, Codomain and Range

9Exercise questions

Step-by-step solution

13x − y = 0 → y = 3x. For x = 1, 2, 3, 4: y = 3, 6, 9, 12, all in A.
2x = 5 gives y = 15 ∉ A, so stop there. R = {(1,3), (2,6), (3,9), (4,12)}.
3Domain = {1, 2, 3, 4}.
4Codomain = A = {1, 2, 3, …, 14}.
5Range = {3, 6, 9, 12}.

Final answer

R = {(1,3),(2,6),(3,9),(4,12)}; domain {1,2,3,4}, codomain {1,…,14}, range {3,6,9,12}.

Step-by-step solution

1x < 4, x ∈ N → x = 1, 2, 3. Then y = x + 5 → 6, 7, 8.
2R = {(1,6), (2,7), (3,8)}.
3Domain = {1, 2, 3}; range = {6, 7, 8}.

Final answer

R = {(1,6),(2,7),(3,8)}; domain {1,2,3}, range {6,7,8}.

Step-by-step solution

1Check every pair: x − y odd when x and y have opposite parity.
2x = 1: y = 4, 6 give odd differences → (1,4), (1,6).
3x = 2: y = 9 gives odd difference → (2,9).
4x = 3: y = 4, 6 → (3,4), (3,6).
5x = 5: y = 4, 6 → (5,4), (5,6).
6R = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}.

Final answer

R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}.

Step-by-step solution

1(i) Set-builder form: R = {(x, y) : y = x − 2, x ∈ P, y ∈ Q}.
2(ii) Roster form: R = {(5,3), (6,4), (7,5)}.
3Domain = {5, 6, 7}; range = {3, 4, 5}.

Final answer

R = {(x,y) : y = x − 2, x ∈ P, y ∈ Q} = {(5,3),(6,4),(7,5)}; domain {5,6,7}, range {3,4,5}.

Step-by-step solution

1b divisible by a means a | b, so b ≥ a.
2a = 1: all b → (1,1),(1,2),(1,3),(1,4),(1,6).
3a = 2: (2,2),(2,4),(2,6). a = 3: (3,3),(3,6). a = 4: (4,4). a = 6: (6,6).
4R = {(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)}.
5Domain = {1, 2, 3, 4, 6}; range = {1, 2, 3, 4, 6}.

Final answer

R as listed above; domain = {1,2,3,4,6}, range = {1,2,3,4,6}.

Step-by-step solution

1Roster form: R = {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}.
2Domain = {0, 1, 2, 3, 4, 5}.
3Range = {5, 6, 7, 8, 9, 10}.

Final answer

Domain = {0,1,2,3,4,5}, range = {5,6,7,8,9,10}.

Step-by-step solution

1Primes less than 10: 2, 3, 5, 7.
2x³ for each: 8, 27, 125, 343.
3R = {(2,8), (3,27), (5,125), (7,343)}.

Final answer

R = {(2,8),(3,27),(5,125),(7,343)}.

Step-by-step solution

1n(A) = 3, n(B) = 2 → n(A × B) = 6.
2Every subset of A × B is a relation from A to B.
3Number of relations = 2⁶ = 64.

Final answer

2⁶ = 64 relations.

Step-by-step solution

1a and b are any integers, and the difference of two integers is always an integer.
2So R contains every (a, b) with a, b ∈ Z.
3Domain = Z; range = Z.

Final answer

Domain = Z, range = Z.

13

Exercise 2.3 — Functions

5Exercise questions

Step-by-step solution

1(i) Each first coordinate appears once → function. Domain = {2,5,8,11,14,17}, range = {1}.
2(ii) Each first coordinate distinct → function. Domain = {2,4,6,8,10,12,14}, range = {1,2,3,4,5,6,7}.
3(iii) First coordinate 1 repeats with different images → NOT a function.

Final answer

(i) and (ii) are functions; (iii) is not a function.

Step-by-step solution

1(i) |x| is defined for all real x → domain = R.
2|x| ≥ 0 so −|x| ≤ 0, and every non-positive value is attained → range = (−∞, 0].
3(ii) 9 − x² ≥ 0 → x² ≤ 9 → −3 ≤ x ≤ 3 → domain = [−3, 3].
4x² ∈ [0, 9] → 9 − x² ∈ [0, 9] → √(9 − x²) ∈ [0, 3] → range = [0, 3].

Final answer

(i) Domain R, range (−∞, 0]. (ii) Domain [−3, 3], range [0, 3].

Step-by-step solution

1(i) f(0) = 2(0) − 5 = −5.
2(ii) f(7) = 2(7) − 5 = 9.
3(iii) f(−3) = 2(−3) − 5 = −11.

Final answer

f(0) = −5, f(7) = 9, f(−3) = −11.

Step-by-step solution

1(i) t(0) = 0 + 32 = 32.
2(ii) t(28) = 9(28)/5 + 32 = 252/5 + 32 = 50.4 + 32 = 82.4.
3(iii) t(−10) = 9(−10)/5 + 32 = −18 + 32 = 14.
4(iv) 212 = 9C/5 + 32 → 180 = 9C/5 → C = 100.

Final answer

t(0) = 32, t(28) = 82.4, t(−10) = 14, and C = 100 when t(C) = 212.

Step-by-step solution

1(i) For x > 0, −3x < 0 so 2 − 3x < 2; as x → 0+, f → 2 and as x → ∞, f → −∞. Range = (−∞, 2).
2(ii) x² ≥ 0 → x² + 2 ≥ 2, and every value ≥ 2 occurs. Range = [2, ∞).
3(iii) f(x) = x takes every real value → range = R.

Final answer

(i) (−∞, 2). (ii) [2, ∞). (iii) R.

14

Miscellaneous Exercise — Mixed Problems on Relations and Functions

12Exercise questions

Step-by-step solution

1f: the two rules overlap only at x = 3, where x² = 9 and 3x = 9 agree. So f assigns one value to each input → a function.
2g: at x = 2, rule one gives g(2) = 4 but rule two gives g(2) = 6 → two different values for one input.
3Hence g is not a function.

Final answer

f is a function; g is not (x = 2 maps to both 4 and 6).

Step-by-step solution

1f(1.1) = (1.1)² = 1.21 and f(1) = 1.
2(1.21 − 1) / (0.1) = 0.21 / 0.1 = 2.1.

Final answer

2.1.

Step-by-step solution

1Denominator must not be zero: x² − 8x + 12 ≠ 0.
2x² − 8x + 12 = (x − 2)(x − 6) → zero at x = 2 and x = 6.
3Domain = R − {2, 6}.

Final answer

Domain = R − {2, 6}.

Step-by-step solution

1x − 1 ≥ 0 → x ≥ 1 → domain = [1, ∞).
2√(x − 1) ≥ 0 and takes every non-negative value → range = [0, ∞).

Final answer

Domain = [1, ∞), range = [0, ∞).

Step-by-step solution

1The absolute value is defined for all reals → domain = R.
2|x − 1| ≥ 0 and attains every non-negative value → range = [0, ∞).

Final answer

Domain = R, range = [0, ∞).

Step-by-step solution

1Let y = x²/(1 + x²) with x ∈ R. Clearly y ≥ 0 and y < 1 because 1 + x² > x².
2Also every value in [0, 1) is attained: given y in [0,1), solve y = x²/(1+x²) → y + yx² = x² → x² = y/(1 − y) which is always ≥ 0.
3Range = [0, 1).

Final answer

Range = [0, 1).

Step-by-step solution

1(f + g)(x) = (x + 1) + (2x − 3) = 3x − 2, domain R.
2(f − g)(x) = (x + 1) − (2x − 3) = −x + 4, domain R.
3(f/g)(x) = (x + 1)/(2x − 3), domain R − {3/2}.

Final answer

f + g = 3x − 2 on R; f − g = −x + 4 on R; f/g = (x + 1)/(2x − 3) on R − {3/2}.

Step-by-step solution

1Use f(0) = −1: b = −1.
2Use f(1) = 1: a + b = 1 → a − 1 = 1 → a = 2.
3Check f(2) = 2(2) − 1 = 3 ✓ and f(−1) = −2 − 1 = −3 ✓.

Final answer

a = 2, b = −1.

Step-by-step solution

1(i) False: (a, a) ∈ R would need a = a², i.e. a = 1 only; for a = 2, 2 ≠ 4.
2(ii) False: (4, 2) ∈ R since 4 = 2², but (2, 4) ∉ R since 2 ≠ 16.
3(iii) False: (a, c) needs a = c²; e.g. (4, 2) and (2, c) demands 2 = c², impossible for c ∈ N. Try (9,3) and (3, √3) ∉ N. So false.

Final answer

All three statements are false.

Step-by-step solution

1(i) Every first coordinate lies in A and every second coordinate lies in B → f ⊂ A × B, so yes, f is a relation from A to B.
2(ii) The element 2 of A has two images: (2, 9) and (2, 11).
3A function cannot send one input to two outputs → f is not a function.

Final answer

(i) True — f is a relation. (ii) False — 2 has two images, so not a function.

Step-by-step solution

1Take the input 0: (0, a + 0) comes from a b = 0; choose b = 0, a = 2 → (0, 2); choose a = 0, b = 3 → (0, 3).
2So the first coordinate ab = 0 has two different images 2 and 3.
3Hence f is not a function.

Final answer

No — ab = 0 has multiple images, so f is not a function.

Step-by-step solution

19 = 3² → h.p.f = 3. 10 = 2·5 → 5. 11 prime → 11. 12 = 2²·3 → 3. 13 prime → 13.
2Images: {3, 5, 11, 13}.
3Range = {3, 5, 11, 13}.

Final answer

Range = {3, 5, 11, 13}.

15

Chapter 3 — Trigonometric Functions

This chapter converts between degree and radian measure, defines the six trigonometric functions on the unit circle, and develops the compound-angle, double-angle and sum-to-product identities together with general solutions of trigonometric equations. These identities are the workhorses of Class 11 and reappear in Class 12 calculus. Every question below is from the NCERT Class 11 textbook (rationalised edition), worked line by line.

Board pattern

Proof questions earn method marks: state the identity you use, apply it to both sides (or LHS only), and stop when both sides match. For general solutions always include '+ nπ' or '+ 2nπ' with the correct period — forgetting the period loses the mark. Remember π radians = 180° and the sign of each function in the four quadrants.

$Ex 3.1 — Radian and degree measure, arc length (7 Q)$
$Ex 3.2 — Values of trigonometric functions (10 Q)$
$Ex 3.3 — Identities: compound, double, sum-to-product (25 Q)$
$Ex 3.4 — Trigonometric equations, general solutions (9 Q)$
$Misc — Mixed problems on trigonometric functions (10 Q)$

16

Exercise 3.1 — Angles and their Measures

7Exercise questions

Step-by-step solution

1Use 180° = π radians, so 1° = π/180.
2(i) 25° = 25π/180 = 5π/36 rad.
3(ii) −47°30′ = −(47.5)° = −47.5 × π/180 = −19π/72 rad.
4(iii) 240° = 240π/180 = 4π/3 rad.
5(iv) 520° = 520π/180 = 26π/9 rad.

Final answer

(i) 5π/36 (ii) −19π/72 (iii) 4π/3 (iv) 26π/9 radians.

Step-by-step solution

1(i) (11/16) × 180/π = (11/16) × 180 × 7/22 = 39.375° = 39°22′30″.
2(ii) −4 rad = −4 × 180 × 7/22 = −2520/11 = −229 1/11° = −229°5′27″.
3(iii) (5π/3)(180/π) = 300°.
4(iv) (7π/6)(180/π) = 210°.

Final answer

(i) 39°22′30″ (ii) −229°5′27″ (iii) 300° (iv) 210°.

Step-by-step solution

11 revolution = 2π radians.
2360 revolutions per minute = 360/60 = 6 revolutions per second.
3Angle per second = 6 × 2π = 12π radians.

Final answer

12π radians per second.

Step-by-step solution

1Arc length l = rθ → θ = l/r = 22/100 rad.
2θ = 0.22 rad = 0.22 × 180/π = 0.22 × 180 × 7/22 = 12.6°.
312.6° = 12°36′.

Final answer

12°36′.

Step-by-step solution

1Radius r = 20 cm. Chord length 20 = 2r sin(θ/2) → 20 = 40 sin(θ/2).
2sin(θ/2) = 1/2 → θ/2 = π/6 → θ = π/3.
3Minor arc = rθ = 20 × π/3 = 20π/3 cm.

Final answer

20π/3 cm.

Step-by-step solution

1l = r₁θ₁ = r₂θ₂ with the same arc length l.
2θ₁ = 60° = π/3, θ₂ = 75° = 5π/12.
3r₁/r₂ = θ₂/θ₁ = (5π/12)/(π/3) = 5/4.

Final answer

r₁ : r₂ = 5 : 4.

Step-by-step solution

1θ = l/r with r = 75 cm.
2(i) θ = 10/75 = 2/15 rad.
3(ii) θ = 15/75 = 1/5 rad.
4(iii) θ = 21/75 = 7/25 rad.

Final answer

(i) 2/15 (ii) 1/5 (iii) 7/25 radians.

17

Exercise 3.2 — Trigonometric Functions of an Angle

10Exercise questions

Step-by-step solution

1sin²x = 1 − cos²x = 1 − 1/4 = 3/4.
2Third quadrant → sin x < 0 → sin x = −√3/2.
3tan x = sin x/cos x = (−√3/2)/(−1/2) = √3.
4cosec x = −2/√3, sec x = −2, cot x = 1/√3.

Final answer

sin x = −√3/2, tan x = √3, cosec x = −2/√3, sec x = −2, cot x = 1/√3.

Step-by-step solution

1cos²x = 1 − 9/25 = 16/25 → cos x = ±4/5.
2Second quadrant → cos x = −4/5.
3tan x = (3/5)/(−4/5) = −3/4.
4cosec x = 5/3, sec x = −5/4, cot x = −4/3.

Final answer

cos x = −4/5, tan x = −3/4, cosec x = 5/3, sec x = −5/4, cot x = −4/3.

Step-by-step solution

1tan x = 1/cot x = 4/3 (cot is positive in the third quadrant).
2Draw a right triangle with opposite 4, adjacent 3 → hypotenuse 5.
3Third quadrant → sin x = −4/5, cos x = −3/5.
4cosec x = −5/4, sec x = −5/3.

Final answer

tan x = 4/3, sin x = −4/5, cos x = −3/5, cosec x = −5/4, sec x = −5/3.

Step-by-step solution

1cos x = 1/sec x = 5/13.
2sin²x = 1 − 25/169 = 144/169 → sin x = ±12/13.
3Fourth quadrant → sin x = −12/13.
4cosec x = −13/12, tan x = −12/5, cot x = −5/12.

Final answer

cos x = 5/13, sin x = −12/13, cosec x = −13/12, tan x = −12/5, cot x = −5/12.

Step-by-step solution

1Triangle: opposite 5, adjacent 12 → hypotenuse 13.
2Second quadrant → sin x = 5/13, cos x = −12/13.
3cosec x = 13/5, sec x = −13/12, cot x = −12/5.

Final answer

cot x = −12/5, sin x = 5/13, cos x = −12/13, cosec x = 13/5, sec x = −13/12.

Step-by-step solution

1Subtract full revolutions: 765° − 720° = 45°.
2sin 765° = sin 45° = √2/2.

Final answer

√2/2.

Step-by-step solution

1cos(−1710°) = cos(1710°). 1710° = 4 × 360° + 270°.
2cos 1710° = cos 270° = 0.

Final answer

0.

Step-by-step solution

1−1410° + 4 × 360° = −1410° + 1440° = 30°.
2cosec 30° = 2.

Final answer

2.

Step-by-step solution

119π/3 = 18π/3 + π/3 = 6π + π/3.
2tan has period π: tan(6π + π/3) = tan(π/3) = √3.

Final answer

√3.

Step-by-step solution

1−15π/4 = −12π/4 − 3π/4 → cot(−15π/4) = cot(−3π/4) since cot has period π: −12π/4 = −3π is an integer multiple of π.
2cot(−3π/4) = −cot(3π/4) = −(−1) = 1.

Final answer

1.

18

Exercise 3.3 — Trigonometric Identities

25Exercise questions

Step-by-step solution

1sin(π/6) = 1/2 → sin²(π/6) = 1/4.
2cos(π/3) = 1/2 → cos²(π/3) = 1/4. tan(π/4) = 1 → tan²(π/4) = 1.
3LHS = 1/4 + 1/4 − 1 = −1/2 = RHS.

Final answer

Proved: LHS = −1/2.

Step-by-step solution

1sin²(π/6) = 1/4 so 2 sin²(π/6) = 1/2.
2cosec(7π/6) = 1/sin(7π/6) = 1/(−1/2) = −2 → cosec²(7π/6) = 4.
3cos²(π/3) = 1/4, so 4 × 1/4 = 1.
4LHS = 1/2 + 1 = 3/2 = RHS.

Final answer

Proved: LHS = 3/2.

Step-by-step solution

1cot(π/6) = √3 → cot² = 3. tan(π/6) = 1/√3 → 3 tan² = 3 × 1/3 = 1.
2cosec(5π/6) = 1/sin(5π/6) = 1/(1/2) = 2.
3LHS = 3 + 2 + 1 = 6 = RHS.

Final answer

Proved: LHS = 6.

Step-by-step solution

1sin(3π/4) = √2/2 → sin² = 1/2. cos(π/4) = √2/2 → cos² = 1/2.
2sec(π/3) = 2 → sec² = 4.
3LHS = 2(1/2) + 2(1/2) + 2(4) = 1 + 1 + 8 = 10 = RHS.

Final answer

Proved: LHS = 10.

Step-by-step solution

1sin 75° = sin(45° + 30°).
2= sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2).
3= (√6 + √2)/4.

Final answer

(√6 + √2)/4.

Step-by-step solution

1Use cos A cos B − sin A sin B = cos(A + B) with A = π/4 − x, B = π/4 − y.
2LHS = cos((π/4 − x) + (π/4 − y)) = cos(π/2 − (x + y)) = sin(x + y) = RHS.

Final answer

Proved.

Step-by-step solution

1Use tan(A + B) = (tan A + tan B)/(1 − tan A tan B) with tan(π/4) = 1.
2tan(π/4 + x) = (1 + tan x)/(1 − tan x) and tan(π/4 − x) = (1 − tan x)/(1 + tan x).
3Ratio = [(1 + tan x)/(1 − tan x)] / [(1 − tan x)/(1 + tan x)] = ((1 + tan x)/(1 − tan x))² = RHS.

Final answer

Proved.

Step-by-step solution

1cos(π + x) = −cos x; cos(−x) = cos x; sin(π − x) = sin x; cos(π/2 + x) = −sin x.
2Numerator = (−cos x)(cos x) = −cos²x. Denominator = (sin x)(−sin x) = −sin²x.
3LHS = cos²x/sin²x = cot²x = RHS.

Final answer

Proved.

Step-by-step solution

1cos(3π/2 + x) = sin x; cos(2π + x) = cos x.
2cot(3π/2 − x) = tan x; cot(2π + x) = cot x.
3LHS = sin x cos x (tan x + cot x) = sin x cos x (sin x/cos x + cos x/sin x).
4= sin x cos x × (sin²x + cos²x)/(sin x cos x) = 1 = RHS.

Final answer

Proved.

Step-by-step solution

1This is cos A cos B + sin A sin B = cos(A − B) with A = (n + 2)x, B = (n + 1)x.
2LHS = cos((n + 2)x − (n + 1)x) = cos x = RHS.

Final answer

Proved.

Step-by-step solution

1Use cos(A + x) − cos(A − x) = −2 sin A sin x.
2Here A = 3π/4: LHS = −2 sin(3π/4) sin x = −2(√2/2) sin x = −√2 sin x = RHS.

Final answer

Proved.

Step-by-step solution

1sin²A − sin²B = sin(A + B) sin(A − B).
2LHS = sin(6x + 4x) sin(6x − 4x) = sin 10x sin 2x = RHS.

Final answer

Proved.

Step-by-step solution

1Use cos²A − cos²B = −sin(A + B) sin(A − B) (derived from sum-to-product or identities).
2LHS = −sin(2x + 6x) sin(2x − 6x) = −sin 8x sin(−4x) = sin 8x sin 4x = RHS.

Final answer

Proved.

Step-by-step solution

1Group sin 2x + sin 6x = 2 sin 4x cos 2x (sum-to-product).
2LHS = 2 sin 4x cos 2x + 2 sin 4x = 2 sin 4x (cos 2x + 1).
3cos 2x + 1 = 2 cos²x → LHS = 2 sin 4x × 2 cos²x = 4 cos²x sin 4x = RHS.

Final answer

Proved.

Step-by-step solution

1sin 5x + sin 3x = 2 sin 4x cos x (sum-to-product).
2sin 5x − sin 3x = 2 cos 4x sin x.
3LHS = cot 4x · 2 sin 4x cos x = 2 cos 4x cos x.
4RHS = cot x · 2 cos 4x sin x = 2 cos x cos 4x × (cos x/sin x × sin x)... = 2 cos 4x cos x. Both sides equal, hence proved (correcting: RHS = (cos x/sin x)·2 cos 4x sin x = 2 cos 4x cos x).

Final answer

Proved — both sides equal 2 cos 4x cos x.

Step-by-step solution

1cos 9x − cos 5x = −2 sin 7x sin 2x.
2sin 17x − sin 3x = 2 cos 10x sin 7x.
3Ratio = (−2 sin 7x sin 2x)/(2 cos 10x sin 7x) = −sin 2x / cos 10x = RHS.

Final answer

Proved.

Step-by-step solution

1sin 5x + sin 3x = 2 sin 4x cos x.
2cos 5x + cos 3x = 2 cos 4x cos x.
3Ratio = (2 sin 4x cos x)/(2 cos 4x cos x) = tan 4x = RHS.

Final answer

Proved.

Step-by-step solution

1sin x − sin y = 2 cos((x+y)/2) sin((x−y)/2).
2cos x + cos y = 2 cos((x+y)/2) cos((x−y)/2).
3Ratio = sin((x−y)/2)/cos((x−y)/2) = tan((x−y)/2) = RHS.

Final answer

Proved.

Step-by-step solution

1sin x + sin 3x = 2 sin 2x cos x.
2cos x + cos 3x = 2 cos 2x cos x.
3Ratio = sin 2x/cos 2x = tan 2x = RHS.

Final answer

Proved.

Step-by-step solution

1sin x − sin 3x = −2 cos 2x sin x (using sin 3x − sin x = 2 cos 2x sin x).
2sin²x − cos²x = −cos 2x.
3Ratio = (−2 cos 2x sin x)/(−cos 2x) = 2 sin x = RHS.

Final answer

Proved.

Step-by-step solution

1cos 4x + cos 2x = 2 cos 3x cos x, and sin 4x + sin 2x = 2 sin 3x cos x.
2Numerator = 2 cos 3x cos x + cos 3x = cos 3x(2 cos x + 1).
3Denominator = 2 sin 3x cos x + sin 3x = sin 3x(2 cos x + 1).
4Ratio = cos 3x/sin 3x = cot 3x = RHS.

Final answer

Proved.

Step-by-step solution

1Write 3x = x + 2x: cot 3x = cot(x + 2x) = (cot x cot 2x − 1)/(cot x + cot 2x).
2Cross-multiply: cot 3x (cot x + cot 2x) = cot x cot 2x − 1.
3cot x cot 3x + cot 2x cot 3x = cot x cot 2x − 1.
4Rearrange: cot x cot 2x − cot 2x cot 3x − cot 3x cot x = 1 = RHS.

Final answer

Proved.

Step-by-step solution

1tan 2x = 2t/(1 − t²) where t = tan x.
2tan 4x = tan(2x + 2x) = 2 tan 2x/(1 − tan² 2x) = [4t/(1 − t²)] / [1 − 4t²/(1 − t²)²].
3= 4t(1 − t²)/[(1 − t²)² − 4t²] = 4t(1 − t²)/(1 − 2t² + t⁴ − 4t²).
4= 4 tan x (1 − tan²x)/(1 − 6 tan²x + tan⁴x) = RHS.

Final answer

Proved.

Step-by-step solution

1cos 4x = cos(2 · 2x) = 1 − 2 sin²2x.
2sin 2x = 2 sin x cos x → sin²2x = 4 sin²x cos²x.
3LHS = 1 − 2(4 sin²x cos²x) = 1 − 8 sin²x cos²x = RHS.

Final answer

Proved.

Step-by-step solution

1cos 6x = cos(2 · 3x) = 2 cos²3x − 1.
2cos 3x = 4 cos³x − 3 cos x, so cos²3x = 16 cos⁶x − 24 cos⁴x + 9 cos²x.
32 cos²3x − 1 = 32 cos⁶x − 48 cos⁴x + 18 cos²x − 1 = RHS.

Final answer

Proved.

19

Exercise 3.4 — Trigonometric Equations

9Exercise questions

Step-by-step solution

1tan(π/3) = √3 and tan(4π/3) = √3; principal solutions are the ones in [0, 2π): π/3 and 4π/3.
2General solution: tan x = tan α with α = π/3 gives x = nπ + π/3, n ∈ Z.

Final answer

Principal: x = π/3, 4π/3. General: x = nπ + π/3, n ∈ Z.

Step-by-step solution

1sec x = 2 → cos x = 1/2.
2cos(π/3) = 1/2 and cos(5π/3) = 1/2 → principal solutions π/3 and 5π/3.
3General: x = 2nπ ± π/3, n ∈ Z.

Final answer

Principal: π/3, 5π/3. General: x = 2nπ ± π/3.

Step-by-step solution

1cot x = −√3 → tan x = −1/√3 = tan(−π/6).
2In [0, 2π): tan x = −1/√3 at x = 5π/6 and 11π/6 → principal solutions.
3General: x = nπ − π/6, n ∈ Z.

Final answer

Principal: 5π/6, 11π/6. General: x = nπ − π/6.

Step-by-step solution

1cosec x = −2 → sin x = −1/2.
2sin(7π/6) = −1/2 and sin(11π/6) = −1/2 → principal solutions 7π/6, 11π/6.
3General: x = nπ + (−1)ⁿ(7π/6), n ∈ Z.

Final answer

Principal: 7π/6, 11π/6. General: x = nπ + (−1)ⁿ(7π/6).

Step-by-step solution

1cos 4x − cos 2x = 0 → −2 sin 3x sin x = 0.
2sin 3x = 0 → 3x = nπ → x = nπ/3.
3sin x = 0 → x = nπ, already included in nπ/3 for n multiple of 3.
4General solution: x = nπ/3, n ∈ Z.

Final answer

x = nπ/3, n ∈ Z.

Step-by-step solution

1cos 3x + cos x = 2 cos 2x cos x.
2Equation: 2 cos 2x cos x − cos 2x = 0 → cos 2x (2 cos x − 1) = 0.
3cos 2x = 0 → 2x = (2n + 1)π/2 → x = (2n + 1)π/4.
42 cos x − 1 = 0 → cos x = 1/2 → x = 2nπ ± π/3.

Final answer

x = (2n+1)π/4 or x = 2nπ ± π/3, n ∈ Z.

Step-by-step solution

1sin 2x = 2 sin x cos x → cos x (2 sin x + 1) = 0.
2cos x = 0 → x = (2n + 1)π/2.
3sin x = −1/2 → x = nπ + (−1)ⁿ(7π/6).

Final answer

x = (2n+1)π/2 or x = nπ + (−1)ⁿ(7π/6), n ∈ Z.

Step-by-step solution

1sec²2x = 1 + tan²2x, so 1 + tan²2x = 1 − tan 2x.
2tan²2x + tan 2x = 0 → tan 2x (tan 2x + 1) = 0.
3tan 2x = 0 → 2x = nπ → x = nπ/2.
4tan 2x = −1 → 2x = nπ − π/4 → x = nπ/2 − π/8.

Final answer

x = nπ/2 or x = nπ/2 − π/8, n ∈ Z.

Step-by-step solution

1sin x + sin 5x = 2 sin 3x cos 2x.
22 sin 3x cos 2x + sin 3x = 0 → sin 3x (2 cos 2x + 1) = 0.
3sin 3x = 0 → 3x = nπ → x = nπ/3.
4cos 2x = −1/2 → 2x = 2nπ ± 2π/3 → x = nπ ± π/3.

Final answer

x = nπ/3 or x = nπ ± π/3, n ∈ Z.

20

Miscellaneous Exercise — Mixed Problems on Trigonometric Functions

10Exercise questions

Step-by-step solution

12 cos(π/13) cos(9π/13) = cos(10π/13) + cos(8π/13) (using 2 cos A cos B = cos(A+B) + cos(A−B)).
2LHS = cos(10π/13) + cos(8π/13) + cos(3π/13) + cos(5π/13).
3Group: cos(10π/13) + cos(3π/13) = 2 cos(13π/26) cos(7π/26) = 2 cos(π/2) cos(7π/26) = 0.
4Similarly cos(8π/13) + cos(5π/13) = 2 cos(π/2) cos(3π/26) = 0.
5Hence LHS = 0.

Final answer

Proved: LHS = 0.

Step-by-step solution

1sin 3x + sin x = 2 sin 2x cos x and cos 3x − cos x = −2 sin 2x sin x.
2LHS = 2 sin 2x cos x sin x + (−2 sin 2x sin x) cos x = 2 sin 2x sin x cos x − 2 sin 2x sin x cos x = 0 = RHS.

Final answer

Proved.

Step-by-step solution

1Expand: cos²x + 2 cos x cos y + cos²y + sin²x − 2 sin x sin y + sin²y.
2= (cos²x + sin²x) + (cos²y + sin²y) + 2(cos x cos y − sin x sin y) = 2 + 2 cos(x + y).
32 + 2 cos(x + y) = 2(1 + cos(x + y)) = 4 cos²((x + y)/2) = RHS.

Final answer

Proved.

Step-by-step solution

1Expand: cos²x − 2 cos x cos y + cos²y + sin²x − 2 sin x sin y + sin²y.
2= 2 − 2(cos x cos y + sin x sin y) = 2 − 2 cos(x − y) = 2(1 − cos(x − y)).
3= 4 sin²((x − y)/2) = RHS.

Final answer

Proved.

Step-by-step solution

1(sin x + sin 7x) + (sin 3x + sin 5x) = 2 sin 4x cos 3x + 2 sin 4x cos x.
2= 2 sin 4x (cos 3x + cos x) = 2 sin 4x · 2 cos 2x cos x.
3= 4 cos x cos 2x sin 4x = RHS.

Final answer

Proved.

Step-by-step solution

1sin 7x + sin 5x = 2 sin 6x cos x; sin 9x + sin 3x = 2 sin 6x cos 3x.
2Numerator = 2 sin 6x (cos x + cos 3x).
3cos 7x + cos 5x = 2 cos 6x cos x; cos 9x + cos 3x = 2 cos 6x cos 3x.
4Denominator = 2 cos 6x (cos x + cos 3x). Ratio = tan 6x = RHS.

Final answer

Proved.

Step-by-step solution

1LHS = sin 3x − sin x + sin 2x = 2 cos 2x sin x + sin 2x.
2= 2 cos 2x sin x + 2 sin x cos x = 2 sin x (cos 2x + cos x).
3cos 2x + cos x = 2 cos(3x/2) cos(x/2).
4LHS = 4 sin x cos(x/2) cos(3x/2) = RHS.

Final answer

Proved.

Step-by-step solution

1cos x = 1/√(1 + tan²x) with sign negative in QII → cos x = −3/5.
2sin x = tan x · cos x = (−4/3)(−3/5) = 4/5.
3x/2 in quadrant I (since 90° < x < 180° → 45° < x/2 < 90°), so all values positive.
4sin(x/2) = √((1 − cos x)/2) = √((1 + 3/5)/2) = √(4/5) = 2/√5.
5cos(x/2) = √((1 + cos x)/2) = √((1 − 3/5)/2) = √(1/5) = 1/√5.
6tan(x/2) = sin(x/2)/cos(x/2) = 2.

Final answer

sin(x/2) = 2/√5, cos(x/2) = 1/√5, tan(x/2) = 2.

Step-by-step solution

1Third quadrant: π < x < 3π/2 → π/2 < x/2 < 3π/4 → sin(x/2) > 0, cos(x/2) < 0.
2sin(x/2) = √((1 − cos x)/2) = √((1 + 1/3)/2) = √(2/3).
3cos(x/2) = −√((1 + cos x)/2) = −√((1 − 1/3)/2) = −√(1/3).
4tan(x/2) = sin(x/2)/cos(x/2) = √(2/3)/(−√(1/3)) = −√2.

Final answer

sin(x/2) = √(2/3), cos(x/2) = −√(1/3), tan(x/2) = −√2.

Step-by-step solution

1QII: π/2 < x < π → π/4 < x/2 < π/2 → all of sin½, cos½ positive.
2cos x = −√(1 − 1/16) = −√15/4.
3sin(x/2) = √((1 − cos x)/2) = √((1 + √15/4)/2) = √((4 + √15)/8).
4cos(x/2) = √((1 + cos x)/2) = √((1 − √15/4)/2) = √((4 − √15)/8).
5tan(x/2) = √((1 − cos x)/(1 + cos x)) = √((4 + √15)/(4 − √15)).

Final answer

sin(x/2) = √((4+√15)/8), cos(x/2) = √((4−√15)/8), tan(x/2) = √((4+√15)/(4−√15)).

21

Chapter 4 — Complex Numbers and Quadratic Equations

Complex numbers extend the real number line to the complex plane: a complex number is z = a + ib with i² = −1. This chapter practises the four operations, powers of i, the multiplicative inverse, modulus, conjugate and polar form, and then uses complex numbers to solve quadratic equations with negative discriminants.

Board pattern

The shortcut i⁴ = 1 collapses every power of i to a remainder mod 4. The multiplicative inverse of a + ib is (a − ib)/(a² + b²) — always rationalise the denominator. For quadratic equations, compute D = b² − 4ac first; if D < 0 write √−D as (√|D|)i and finish with the quadratic formula.

$Ex 4.1 — Operations and multiplicative inverse (14 Q)$
$Misc — Modulus, conjugate, polar form, quadratic equations (14 Q)$

22

Exercise 4.1 — Operations on Complex Numbers

14Exercise questions

Step-by-step solution

1(5i)(−3/5 i) = 5 × (−3/5) × i × i = −3i².
2i² = −1, so −3i² = 3.
3In a + ib form: 3 + 0i.

Final answer

3 + 0i.

Step-by-step solution

1Use i⁴ = 1: i⁹ = i⁴·² · i = i, and i¹⁹ = i⁴·⁴ · i³ = i³ = −i.
2i⁹ + i¹⁹ = i + (−i) = 0.
3In a + ib form: 0 + 0i.

Final answer

0 + 0i.

Step-by-step solution

139 = 4·9 + 3, so i³⁹ = i³ = −i.
2i⁻³⁹ = 1/(−i). Multiply numerator and denominator by i: 1/(−i) × i/i = i/(−i²) = i.
3In a + ib form: 0 + i.

Final answer

0 + i.

Step-by-step solution

1Expand: 3(7 + 7i) = 21 + 21i, and i(7 + 7i) = 7i + 7i².
2Sum = 21 + 21i + 7i + 7i² = 21 + 28i + 7(−1).
3= 14 + 28i.

Final answer

14 + 28i.

Step-by-step solution

1Subtract term by term: (1 − i) + (1 − 6i).
2= 1 + 1 − i − 6i = 2 − 7i.

Final answer

2 − 7i.

Step-by-step solution

1Separate real and imaginary parts: (1/5 − 4) + i(2/5 − 5/2).
2Real: 1/5 − 4 = (1 − 20)/5 = −19/5.
3Imaginary: 2/5 − 5/2 = (4 − 25)/10 = −21/10.
4Form: −19/5 + (−21/10)i.

Final answer

−19/5 − 21/10 i.

Step-by-step solution

1Add the first two brackets: (1/3 + 4) + i(7/3 + 1/3) = 13/3 + 8/3 i.
2Subtract (−4/3 + i): 13/3 + 8/3 i + 4/3 − i.
3= (13/3 + 4/3) + i(8/3 − 1) = 17/3 + 5/3 i.

Final answer

17/3 + 5/3 i.

Step-by-step solution

1(1 − i)² = 1 − 2i + i² = −2i.
2(1 − i)⁴ = (−2i)² = 4i² = −4.
3Form: −4 + 0i.

Final answer

−4 + 0i.

Step-by-step solution

1Expand (a + b)³ with a = 1/3, b = 3i.
2a³ = 1/27; 3a²b = 3(1/9)(3i) = i; 3ab² = 3(1/3)(9i²) = 9i² = −9; b³ = 27i³ = −27i.
3Sum: 1/27 − 9 + i − 27i = (1 − 243)/27 − 26i.
4= −242/27 − 26i.

Final answer

−242/27 − 26i.

Step-by-step solution

1Expand (a + b)³ with a = −2, b = −i/3.
2a³ = −8; 3a²b = 3(4)(−i/3) = −4i; 3ab² = 3(−2)(i²/9) = −6/9 (−1) = 2/3; b³ = −i³/27 = i/27.
3Sum: −8 + 2/3 − 4i + i/27 = −22/3 + i(−4 + 1/27) = −22/3 − 107/27 i.

Final answer

−22/3 − 107/27 i.

Step-by-step solution

1Inverse = 1/(4 − 3i). Rationalise: multiply by (4 + 3i)/(4 + 3i).
21/(4 − 3i) × (4 + 3i)/(4 + 3i) = (4 + 3i)/(16 + 9).
3= (4 + 3i)/25 = 4/25 + 3/25 i.
4Powers-of-i check: modulus² of 4−3i is 25 ⟶ inverse modulus 1/5. Correct.

Final answer

4/25 + 3/25 i.

Step-by-step solution

1Inverse = 1/(√5 + 3i). Rationalise with (√5 − 3i).
2= (√5 − 3i)/(5 + 9) = (√5 − 3i)/14.
3= √5/14 − 3/14 i.

Final answer

√5/14 − 3/14 i.

Step-by-step solution

1Inverse = 1/(−i). Multiply numerator and denominator by i: 1/(−i) × i/i = i/(−i²) = i/1 = i.
2Check: (−i)(i) = −i² = 1. ✓

Final answer

i.

Step-by-step solution

1Denominator: (√3 + √2 i) − (√3 − √2 i) = 2√2 i.
2Numerator: (3 + i√5)(3 − i√5) = 3² − (i√5)² = 9 + 5 = 14.
3Expression = 14/(2√2 i) = 7/(√2 i).
4Rationalise: 7/(√2 i) × i/i = 7i/(√2 i²) = −7i/√2.
5Form: 0 + (−7/√2)i.

Final answer

0 − (7/√2)i.

23

Miscellaneous Exercise — Mixed Problems on Complex Numbers

14Exercise questions

Step-by-step solution

1i¹⁸ = i⁴·⁴·i² = i² = −1.
2i²⁵ = i⁴·⁶·i = i, so (1/i)²⁵ = 1/i = −i.
3Bracket = −1 − i. Cube: (−1 − i)³ = −(1 + i)³.
4(1 + i)³ = 1 + 3i + 3i² + i³ = 1 + 3i − 3 − i = −2 + 2i.
5Answer = −(−2 + 2i) = 2 − 2i.

Final answer

2 − 2i.

Step-by-step solution

1Write z₁ = a + ib, z₂ = c + id with a = Re z₁, b = Im z₁, c = Re z₂, d = Im z₂.
2z₁z₂ = (a + ib)(c + id) = ac + iad + ibc + i²bd = (ac − bd) + i(ad + bc).
3Re(z₁z₂) = ac − bd = Re(z₁)Re(z₂) − Im(z₁)Im(z₂).

Final answer

Proved.

Step-by-step solution

11/(1 − 4i) = (1 + 4i)/17, and 2/(1 + i) = 2(1 − i)/2 = 1 − i.
2First bracket = (1 + 4i)/17 − (1 − i) = (1 + 4i − 17 + 17i)/17 = (−16 + 21i)/17.
3(3 − 4i)/(5 + i) = (3 − 4i)(5 − i)/26 = (15 − 3i − 20i + 4i²)/26 = (11 − 23i)/26.
4Product = (−16 + 21i)(11 − 23i)/(17 · 26).
5(−16 + 21i)(11 − 23i) = −176 + 368i + 231i − 483i² = (−176 + 483) + 599i = 307 + 599i.
6So expression = 307/442 + 599/442 i.

Final answer

307/442 + 599/442 i.

Step-by-step solution

1Take modulus of both sides: |x − iy| = |√((a − ib)/(c − id))|.
2|x − iy|² = |(a − ib)/(c − id)| = |a − ib| / |c − id|.
3|a − ib| = √(a² + b²), |c − id| = √(c² + d²), and |x − iy|² = x² + y².
4Hence x² + y² = √((a² + b²)/(c² + d²)).
5Square both sides: (x² + y²)² = (a² + b²)/(c² + d²).

Final answer

Proved.

Step-by-step solution

1(i) (2 − i)² = 4 − 4i + i² = 3 − 4i. Ratio = (1 + 7i)(3 + 4i)/25 = (3 + 4i + 21i + 28i²)/25 = (−25 + 25i)/25 = −1 + i.
2r = |−1 + i| = √2. cos θ = −1/√2, sin θ = 1/√2 → θ = 3π/4 (II quadrant).
3(i) Polar form: √2(cos(3π/4) + i sin(3π/4)).
4(ii) (1 + 3i)/(1 − 2i) = (1 + 3i)(1 + 2i)/5 = (1 + 2i + 3i + 6i²)/5 = (−5 + 5i)/5 = −1 + i.
5Same as before: r = √2, θ = 3π/4.
6(ii) Polar form: √2(cos(3π/4) + i sin(3π/4)).

Final answer

Both equal √2(cos(3π/4) + i sin(3π/4)).

Step-by-step solution

1Multiply by 3: 9x² − 12x + 20 = 0.
2a = 9, b = −12, c = 20. D = b² − 4ac = 144 − 720 = −576.
3√D = √−576 = 24i.
4x = (12 ± 24i)/18 = (2 ± 4i)/3.

Final answer

x = (2 ± 4i)/3.

Step-by-step solution

1Multiply by 2: 2x² − 4x + 3 = 0.
2D = (−4)² − 4(2)(3) = 16 − 24 = −8, so √D = 2√2 i.
3x = (4 ± 2√2 i)/4 = 1 ± (√2/2)i.

Final answer

x = 1 ± (√2/2)i.

Step-by-step solution

1D = (−10)² − 4(27)(1) = 100 − 108 = −8, so √D = 2√2 i.
2x = (10 ± 2√2 i)/(2 · 27) = (5 ± √2 i)/27.

Final answer

x = (5 ± √2 i)/27.

Step-by-step solution

1D = (−28)² − 4(21)(10) = 784 − 840 = −56, so √D = 2√14 i.
2x = (28 ± 2√14 i)/(2 · 21) = (14 ± √14 i)/21.

Final answer

x = (14 ± √14 i)/21.

Step-by-step solution

1z₁ + z₂ + 1 = (2 − i) + (1 + i) + 1 = 4.
2z₁ − z₂ + 1 = (2 − i) − (1 + i) + 1 = 2 − 2i.
3Ratio = 4/(2 − 2i) = 4(1 + i)/(2(1 − i)(1 + i))·... Simplest: 4/(2 − 2i) = 2/(1 − i) = 2(1 + i)/2 = 1 + i.
4|1 + i| = √2.

Final answer

√2.

Step-by-step solution

1(x + i)² = x² + 2xi + i² = (x² − 1) + 2xi.
2So a = (x² − 1)/(2x² + 1) and b = 2x/(2x² + 1).
3a² + b² = [(x² − 1)² + 4x²]/(2x² + 1)² = (x⁴ − 2x² + 1 + 4x²)/(2x² + 1)².
4= (x⁴ + 2x² + 1)/(2x² + 1)² = (x² + 1)²/(2x² + 1)².

Final answer

Proved.

Step-by-step solution

1Since |β| = 1, write 1 = ββ̄.
21 − ᾱβ = β(β̄ − ᾱ).
3|1 − ᾱβ| = |β|·|β̄ − ᾱ| = 1 · |(β − α)̄| = |β − α|.
4Hence |(β − α)/(1 − ᾱβ)| = |β − α|/|β − α| = 1.

Final answer

1.

Step-by-step solution

1Take modulus of both sides of the given product.
2|(a + ib)(c + id)(e + if)(g + ih)| = |a + ib||c + id||e + if||g + ih|.
3|a + ib| = √(a² + b²), etc., and |A + iB| = √(A² + B²).
4Squaring: (a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B².

Final answer

Proved.

Step-by-step solution

1(1 + i)/(1 − i) = (1 + i)²/2 = (1 + 2i + i²)/2 = 2i/2 = i.
2Equation becomes iᵐ = 1.
3i⁴ = 1, so the least positive m = 4.

Final answer

m = 4.

24

Chapter 5 — Linear Inequalities

A linear inequality keeps the usual algebra of equations but tracks the sign of the inequality. Multiplying or dividing both sides by a NEGATIVE number reverses the sign — the one rule every student forgets. This chapter practises one-variable inequalities, number-line graphs, and compound inequalities, then applies them to real-world range problems.

Board pattern

Always state the solution set in interval or set-builder form at the end. When you divide by a negative number, flip the inequality sign — lose that and you lose the whole question. For 'natural number' / 'integer' parts, list the finite solution set explicitly; for real numbers give the interval.

$Ex 5.1 — Solving linear inequalities, graphs, word problems (26 Q)$
$Misc — Compound inequalities, temperature, mixtures, IQ (14 Q)$

25

Exercise 5.1 — Solving Linear Inequalities

26Exercise questions

Step-by-step solution

1Divide both sides by 24: x < 100/24 = 25/6 ≈ 4.17.
2(i) Natural numbers {1, 2, 3, ...} less than 4.17: x ∈ {1, 2, 3, 4}.
3(ii) Integers less than 4.17: {..., −2, −1, 0, 1, 2, 3, 4}.

Final answer

(i) {1, 2, 3, 4}, (ii) {…, −1, 0, 1, 2, 3, 4}.

Step-by-step solution

1Divide by −12 and reverse: x < 30/(−12) = −5/2.
2(i) No natural number is < −2.5 → empty set.
3(ii) Integers less than −2.5: {..., −5, −4, −3}.

Final answer

(i) No solution, (ii) {…, −5, −4, −3}.

Step-by-step solution

15x < 10 → x < 2.
2(i) Integers less than 2: {..., −1, 0, 1}.
3(ii) Real numbers: x ∈ (−∞, 2).

Final answer

(i) {…, −1, 0, 1}, (ii) (−∞, 2).

Step-by-step solution

13x > 2 − 8 = −6 → x > −2.
2(i) Integers greater than −2: {−1, 0, 1, 2, ...}.
3(ii) Real numbers: x ∈ (−2, ∞).

Final answer

(i) {−1, 0, 1, 2, …}, (ii) (−2, ∞).

Step-by-step solution

1Bring x terms together: 4x − 5x < 7 − 3.
2−x < 4 → multiply by −1 and reverse sign: x > −4.

Final answer

x ∈ (−4, ∞).

Step-by-step solution

13x − 5x > −1 + 7 → −2x > 6.
2Divide by −2 and reverse: x < −3.

Final answer

x ∈ (−∞, −3).

Step-by-step solution

1Expand: 3x − 3 ≤ 2x − 6.
23x − 2x ≤ −6 + 3 → x ≤ −3.

Final answer

x ∈ (−∞, −3].

Step-by-step solution

1Expand: 6 − 3x ≥ 2 − 2x.
26 − 2 ≥ 3x − 2x → 4 ≥ x → x ≤ 4.

Final answer

x ∈ (−∞, 4].

Step-by-step solution

1LHS = (3x + 2x)/6 = 5x/6.
25x/6 < 11 → 5x < 66 → x < 66/5 = 13.2.

Final answer

x ∈ (−∞, 66/5).

Step-by-step solution

1x/3 − x/2 > 1 → (2x − 3x)/6 > 1 → −x/6 > 1.
2−x > 6 → x < −6.

Final answer

x ∈ (−∞, −6).

Step-by-step solution

1Cross-multiply (LCM 15): 9(x − 2) ≤ 25(2 − x).
29x − 18 ≤ 50 − 25x → 34x ≤ 68 → x ≤ 2.

Final answer

x ∈ (−∞, 2].

Step-by-step solution

1Multiply by 6: 3(3x/5 + 4) ≥ 2(x − 6).
29x/5 + 12 ≥ 2x − 12 → 24 ≥ 2x − 9x/5 = (10x − 9x)/5 = x/5.
324 ≥ x/5 → x ≤ 120.

Final answer

x ∈ (−∞, 120].

Step-by-step solution

1Expand: 4x + 6 − 10 < 6x − 12 → 4x − 4 < 6x − 12.
2−4 + 12 < 6x − 4x → 8 < 2x → x > 4.

Final answer

x ∈ (4, ∞).

Step-by-step solution

1LHS = 37 − 3x − 5 = 32 − 3x. RHS = 9x − 8x + 24 = x + 24.
232 − 3x ≥ x + 24 → 8 ≥ 4x → x ≤ 2.

Final answer

x ∈ (−∞, 2].

Step-by-step solution

1RHS = [5(5x−2) − 3(7x−3)]/15 = (25x − 10 − 21x + 9)/15 = (4x − 1)/15.
2x/4 < (4x − 1)/15 → 15x < 16x − 4.
3−x < −4 → x > 4.

Final answer

x ∈ (4, ∞).

Step-by-step solution

1RHS = [5(3x−2) − 4(2−x)]/20 = (15x − 10 − 8 + 4x)/20 = (19x − 18)/20.
2(2x − 1)/3 ≥ (19x − 18)/20 → 20(2x − 1) ≥ 3(19x − 18).
340x − 20 ≥ 57x − 54 → 34 ≥ 17x → x ≤ 2.

Final answer

x ∈ (−∞, 2].

Step-by-step solution

13x − 2x < 1 + 2 → x < 3.
2Graph: open circle at 3, ray extending left to −∞.

Final answer

x ∈ (−∞, 3). Graph: open circle at 3, arrow to the left.

Step-by-step solution

15x − 3x ≥ −5 + 3 → 2x ≥ −2 → x ≥ −1.
2Graph: closed circle at −1, ray extending right.

Final answer

x ∈ [−1, ∞). Graph: filled circle at −1, arrow to the right.

Step-by-step solution

13 − 3x < 2x + 8 → −5 < 5x → x > −1.
2Graph: open circle at −1, ray extending right.

Final answer

x ∈ (−1, ∞). Graph: open circle at −1, arrow right.

Step-by-step solution

1RHS = (4x − 1)/15 (as computed in Q15).
2x/2 ≥ (4x − 1)/15 → 15x ≥ 8x − 2 → 7x ≥ −2 → x ≥ −2/7.
3Graph: closed circle at −2/7, ray extending right.

Final answer

x ∈ [−2/7, ∞). Graph: filled circle at −2/7, arrow right.

Step-by-step solution

1Let the third test marks be x.
2(70 + 75 + x)/3 ≥ 60 → 145 + x ≥ 180.
3x ≥ 35.

Final answer

At least 35 marks.

Step-by-step solution

1Let the fifth examination mark be x.
2(87 + 92 + 94 + 95 + x)/5 ≥ 90 → 368 + x ≥ 450.
3x ≥ 82.

Final answer

At least 82 marks.

Step-by-step solution

1Consecutive odd positive pairs under 10: (1,3), (3,5), (5,7), (7,9).
2Sums: 4, 8, 12, 16.
3Sum > 11 for (5,7) and (7,9).

Final answer

(5, 7) and (7, 9).

Step-by-step solution

1Let the smaller be 2n; pairs (6,8), (8,10), (10,12), ...
2Sums: 14, 18, 22, 26, ...
3Sum < 23 for (6,8), (8,10), (10,12).

Final answer

(6, 8), (8, 10) and (10, 12).

Step-by-step solution

1Let shortest = x, longest = 3x, third = 3x − 2.
2Perimeter ≥ 61: x + 3x + (3x − 2) ≥ 61 → 7x − 2 ≥ 61.
37x ≥ 63 → x ≥ 9.

Final answer

Minimum shortest side = 9 cm.

Step-by-step solution

1Let shortest = x, second = x + 3, third = 2x.
2Board limit: x + (x + 3) + 2x ≤ 91 → 4x + 3 ≤ 91 → x ≤ 22.
3Third at least 5 cm longer than second: 2x ≥ (x + 3) + 5 → x ≥ 8.
4Combine: 8 ≤ x ≤ 22.

Final answer

Shortest board between 8 cm and 22 cm.

26

Miscellaneous Exercise — Compound Inequalities and Applications

14Exercise questions

Step-by-step solution

1Add 4 to all three parts: 6 ≤ 3x ≤ 9.
2Divide by 3: 2 ≤ x ≤ 3.

Final answer

x ∈ [2, 3].

Step-by-step solution

1Divide all parts by −3 and reverse the signs: −2 ≥ 2x − 4 > −4.
2Add 4: 2 ≥ 2x > 0 → 1 ≥ x > 0.

Final answer

x ∈ (0, 1].

Step-by-step solution

1Subtract 4: −7 ≤ −7x/2 ≤ 14.
2Multiply by 2: −14 ≤ −7x ≤ 28.
3Divide by −7, reversing signs: 2 ≥ x ≥ −4 → −4 ≤ x ≤ 2.

Final answer

x ∈ [−4, 2].

Step-by-step solution

1Multiply by 5: −75 < 3(x − 2) ≤ 0.
2Divide by 3: −25 < x − 2 ≤ 0.
3Add 2: −23 < x ≤ 2.

Final answer

x ∈ (−23, 2].

Step-by-step solution

1Subtract 4: −16 < −3x/(−5) ≤ −2, i.e. −16 < 3x/5 ≤ −2.
2Multiply by 5: −80 < 3x ≤ −10.
3Divide by 3: −80/3 < x ≤ −10/3.

Final answer

x ∈ (−80/3, −10/3].

Step-by-step solution

1Multiply by 2: 14 ≤ 3x + 11 ≤ 22.
2Subtract 11: 3 ≤ 3x ≤ 11.
3Divide by 3: 1 ≤ x ≤ 11/3.

Final answer

x ∈ [1, 11/3].

Step-by-step solution

15x + 1 > −24 → 5x > −25 → x > −5.
25x − 1 < 24 → 5x < 25 → x < 5.
3Together: −5 < x < 5. Graph: open circles at −5 and 5, segment between them.

Final answer

x ∈ (−5, 5). Open segment from −5 to 5.

Step-by-step solution

12(x − 1) < x + 5 → 2x − 2 < x + 5 → x < 7.
23(x + 2) > 2 − x → 3x + 6 > 2 − x → 4x > −4 → x > −1.
3Together: −1 < x < 7. Graph: open segment from −1 to 7.

Final answer

x ∈ (−1, 7).

Step-by-step solution

13x − 7 > 2(x − 6) → 3x − 7 > 2x − 12 → x > −5.
26 − x > 11 − 2x → 2x − x > 11 − 6 → x > 5.
3Both must hold → x > 5. Graph: open circle at 5, ray right.

Final answer

x ∈ (5, ∞).

Step-by-step solution

1First: 10x − 35 − 6x − 9 ≤ 0 → 4x ≤ 44 → x ≤ 11.
2Second: 2x + 19 ≤ 6x + 47 → −28 ≤ 4x → x ≥ −7.
3Together: −7 ≤ x ≤ 11. Graph: closed segment from −7 to 11.

Final answer

x ∈ [−7, 11].

Step-by-step solution

168 ≤ (9/5)C + 32 ≤ 77.
2Subtract 32: 36 ≤ (9/5)C ≤ 45.
3Multiply by 5/9: 20 ≤ C ≤ 25.

Final answer

Between 20 °C and 25 °C.

Step-by-step solution

1Let x litres of 2% solution be added. Total mixture = 640 + x.
2Acid content: 8% of 640 + 2% of x = 1024/20 + x/50 (working in percent·litres).
3Condition: 4% of (640 + x) < acid < 6% of (640 + x).
44%: (4/100)(640 + x) < (2x/100) + (8×640/100) → 2560 + 4x < 2x + 5120 → x < 1280.
56%: (2x/100) + 5120/10 … rewrite: (2x/100) + (8)(640)/100 < (6/100)(640+x) → 2x + 5120 < 3840 + 6x → 1280 < 4x → x > 320.
6So 320 < x < 1280.

Final answer

More than 320 but less than 1280 litres.

Step-by-step solution

1Let x litres of water be added. Total = 1125 + x; acid = 45% of 1125 = 506.25.
2More than 25%: 506.25 > (25/100)(1125 + x) → 50625 > 28125 + 25x → x < 900.
3Less than 30%: 506.25 < (30/100)(1125 + x) → 50625 < 33750 + 30x → x > 562.5.
4So 562.5 < x < 900.

Final answer

Between 562.5 and 900 litres.

Step-by-step solution

1CA = 12, so IQ = (MA/12) × 100.
280 ≤ (100 MA)/12 ≤ 140 → multiply by 12: 960 ≤ 100 MA ≤ 1680.
3Divide by 100: 9.6 ≤ MA ≤ 16.8.

Final answer

Mental age between 9.6 and 16.8 years.

27

Chapter 6 — Permutations and Combinations

This chapter is the counting toolbox of combinatorics. The fundamental principle of multiplication (and addition) lets you count compound events; factorials and the nPr / nCr formulas count arrangements and selections. The two big traps: telling a permutation (order matters) from a combination (order does not) and handling repetitions inside a word like MISSISSIPPI.

Board pattern

For word-arrangement questions, always divide by the factorial of each repeated letter. For selection questions, ask 'does the order matter?' — ordered → nPr, unordered → nCr. Write the formula out explicitly before evaluating so the examiner can award method marks.

$Ex 6.1 — Fundamental principle of counting (6 Q)$
$Ex 6.2 — Factorials, n! / (n - r)! (5 Q)$
$Ex 6.3 — Permutations: digits, words, n P_{r} (11 Q)$
$Ex 6.4 — Combinations: n C_{r}, selections (9 Q)$
$Misc — Mixed counting problems (11 Q)$

28

Exercise 6.1 — Fundamental Principle of Counting

6Exercise questions

Step-by-step solution

1Each of the three places has 5 choices when repetition is allowed.
2(i) 5 × 5 × 5 = 125.
3(ii) First place 5 choices, second 4, third 3 → 5 × 4 × 3 = 60.

Final answer

(i) 125, (ii) 60.

Step-by-step solution

1The unit's digit must be even: 2, 4 or 6 → 3 choices.
2Hundred's and ten's places: 6 choices each (repetition allowed).
3Total = 6 × 6 × 3 = 108.

Final answer

108.

Step-by-step solution

1First letter: 10 choices, then 9, 8, 7 for the next places.
2Total = 10 × 9 × 8 × 7 = 5040.

Final answer

5040.

Step-by-step solution

1The first two digits are fixed as 6 and 7.
2Remaining digits available: 8 (0, 1, 2, 3, 4, 5, 8, 9).
3Last three places: 8 × 7 × 6 = 336.

Final answer

336.

Step-by-step solution

1Each toss has 2 outcomes (Head or Tail).
2Total = 2 × 2 × 2 = 8.

Final answer

8.

Step-by-step solution

1Top flag: 5 choices, bottom flag: 4 choices (order matters).
2Total = 5 × 4 = 20.

Final answer

20.

29

Exercise 6.2 — Factorials and n! / (n − r)!

5Exercise questions

Step-by-step solution

1(i) 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320.
2(ii) 4! − 3! = 24 − 6 = 18.

Final answer

(i) 40320, (ii) 18.

Step-by-step solution

13! + 4! = 6 + 24 = 30.
27! = 5040.
330 ≠ 5040, so the statement is false.

Final answer

No (30 ≠ 5040).

Step-by-step solution

18! / (6! × 2!) = (8 × 7 × 6!) / (6! × 2).
2= 8 × 7 / 2 = 28.

Final answer

28.

Step-by-step solution

11/6! + 1/7! = 1/6! + 1/(7 × 6!) = (1 + 1/7)/6! = (8/7)/6! = 8/7!.
2Rewrite right side: x/8! = x/(8 × 7!).
3(8 × 7!)/8! × 8/7! → x = 8 × 8 = 64.

Final answer

x = 64.

Step-by-step solution

1(i) 6!/4! = 6 × 5 × 4!/4! = 6 × 5 = 30.
2(ii) 9!/4! = 9 × 8 × 7 × 6 × 5 × 4!/4! = 15120.

Final answer

(i) 30, (ii) 15120.

30

Exercise 6.3 — Permutations

11Exercise questions

Step-by-step solution

1This is the number of permutations of 9 digits taken 3 at a time: ⁹P₃.
2= 9 × 8 × 7 = 504.

Final answer

504.

Step-by-step solution

1Thousands place: 9 choices (1 to 9, since 0 cannot lead).
2Hundreds place: 9 choices (0 plus 8 remaining digits).
3Tens: 8 choices, Units: 7 choices.
4Total = 9 × 9 × 8 × 7 = 4536.

Final answer

4536.

Step-by-step solution

1For an even number, unit's digit is 2, 4 or 6 → 3 choices.
2Hundred's digit: 5 choices (remaining digits).
3Ten's digit: 4 choices.
4Total = 3 × 5 × 4 = 60.

Final answer

60.

Step-by-step solution

1Total 4-digit numbers: ⁵P₄ = 5 × 4 × 3 × 2 = 120.
2Even numbers: unit's digit is 2 or 4 → 2 choices.
3Remaining 3 places from 4 digits: 4 × 3 × 2 = 24 ways each case.
4Even numbers = 2 × 24 = 48.

Final answer

120 total, 48 even.

Step-by-step solution

1Chairman: 8 choices.
2Vice chairman: 7 choices (excluding the chairman).
3Total = 8 × 7 = 56.

Final answer

56.

Step-by-step solution

1ⁿ⁻¹P₃ = (n−1)(n−2)(n−3), ⁿP₄ = n(n−1)(n−2)(n−3).
2Ratio = 1/n = 1/9.
3n = 9.

Final answer

n = 9.

Step-by-step solution

1(i) 5!/(5−r)! = 2 · 6!/(7−r)! → (7−r)(6−r) = 12.
2r² − 13r + 30 = 0 → (r − 3)(r − 10) = 0 → r = 3 (r = 10 is impossible).
3(ii) 5!/(5−r)! = 6!/(7−r)! → (7−r)(6−r) = 6.
4r² − 13r + 36 = 0 → (r − 4)(r − 9) = 0 → r = 4.

Final answer

(i) r = 3, (ii) r = 4.

Step-by-step solution

1EQUATION has 8 distinct letters.
2Number of arrangements = 8! = 40320.

Final answer

40320.

Step-by-step solution

1MONDAY has 6 distinct letters.
2(i) ⁶P₄ = 6 × 5 × 4 × 3 = 360.
3(ii) 6! = 720.
4(iii) First letter is O or A → 2 choices, remaining 5 letters in 5! = 120 ways.
5Total = 2 × 120 = 240.

Final answer

(i) 360, (ii) 720, (iii) 240.

Step-by-step solution

1MISSISSIPPI has 11 letters: I×4, S×4, P×2, M×1.
2Total distinct arrangements = 11!/(4!·4!·2!) = 34650.
3All four I's together: treat IIII as one block → 8 objects: 8!/(4!·2!) = 840.
4Not together = 34650 − 840 = 33810.

Final answer

33810.

Step-by-step solution

1PERMUTATIONS has 12 letters with T repeated twice.
2(i) Fix P first, S last; middle 10 letters with T twice: 10!/2! = 1814400.
3(ii) Vowels E, U, A, I, O form one block g with 7 consonants → 8 objects, T twice.
48!/2! × 5! = 20160 × 120 = 2419200.
5(iii) Pairs of positions for P, S with exactly 4 letters between: (1,6), (2,7), …, (7,12) → 7 pairs, each swappable.
67 × 2 × 10!/2! = 14 × 1814400 = 25401600.

Final answer

(i) 1814400, (ii) 2419200, (iii) 25401600.

31

Exercise 6.4 — Combinations

9Exercise questions

Step-by-step solution

1ⁿC₈ = ⁿC₂ ⇒ n = 8 + 2 = 10.
2¹⁰C₂ = 10 × 9 / 2 = 45.

Final answer

45.

Step-by-step solution

1²ⁿC₃/ⁿC₃ = [2n(2n−1)(2n−2)/6] / [n(n−1)(n−2)/6] = 4(2n−1)/(n−2).
2(i) 4(2n−1)/(n−2) = 12 → 8n − 4 = 12n − 24 → n = 5.
3(ii) 4(2n−1)/(n−2) = 11 → 8n − 4 = 11n − 22 → n = 6.

Final answer

(i) n = 5, (ii) n = 6.

Step-by-step solution

1A chord joins 2 points (order does not matter).
2Number = ²¹C₂ = 21 × 20 / 2 = 210.

Final answer

210.

Step-by-step solution

1Choose 3 boys from 5: ⁵C₃ = 10.
2Choose 3 girls from 4: ⁴C₃ = 4.
3Total = 10 × 4 = 40.

Final answer

40.

Step-by-step solution

1Select 3 red from 6: ⁶C₃ = 20.
2Select 3 white from 5: ⁵C₃ = 10.
3Select 3 blue from 5: ⁵C₃ = 10.
4Total = 20 × 10 × 10 = 2000.

Final answer

2000.

Step-by-step solution

1Choose 1 ace from 4: ⁴C₁ = 4.
2Choose remaining 4 cards from the 48 non-aces: ⁴⁸C₄ = 194580.
3Total = 4 × 194580 = 778320.

Final answer

778320.

Step-by-step solution

1Choose 4 bowlers from 5: ⁵C₄ = 5.
2Choose remaining 7 players from the other 12: ¹²C₇ = 792.
3Total = 5 × 792 = 3960.

Final answer

3960.

Step-by-step solution

1Choose 2 black from 5: ⁵C₂ = 10.
2Choose 3 red from 6: ⁶C₃ = 20.
3Total = 10 × 20 = 200.

Final answer

200.

Step-by-step solution

1The 2 compulsory courses are fixed.
2Choose the remaining 3 courses from the other 7: ⁷C₃ = 35.

Final answer

35.

32

Miscellaneous Exercise on Chapter 6

11Exercise questions

Step-by-step solution

1DAUGHTER: vowels A, U, E (3); consonants D, G, H, T, R (5).
2Choose 2 vowels: ³C₂ = 3; choose 3 consonants: ⁵C₃ = 10.
3Arrange the 5 chosen letters: 5! = 120.
4Total = 3 × 10 × 120 = 3600.

Final answer

3600.

Step-by-step solution

1EQUATION: 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N).
2Vowel-block and consonant-block: 2 ways to order the blocks.
3Inside blocks: 5! × 3! = 120 × 6 = 720.
4Total = 2 × 720 = 1440.

Final answer

1440.

Step-by-step solution

1(i) Exact 3 girls: ⁴C₃ × ⁹C₄ = 4 × 126 = 504.
2(ii) At least 3 girls: (3 g, 4 b) + (4 g, 3 b) = 504 + ⁴C₄·⁹C₃ = 504 + 84 = 588.
3(iii) At most 3 girls: 0g, 1g, 2g, 3g cases.
4⁴C₀·⁹C₇ + ⁴C₁·⁹C₆ + ⁴C₂·⁹C₅ + 504 = 36 + 336 + 756 + 504 = 1632.

Final answer

(i) 504, (ii) 588, (iii) 1632.

Step-by-step solution

1EXAMINATION: 11 letters with A, I, N each repeated twice.
2Treat NN as one block → 10 objects with A, I repeated twice.
3Number = 10!/(2!·2!) = 3628800/4 = 907200.

Final answer

907200.

Step-by-step solution

1Divisible by 10 ⇒ must end in 0.
2Remaining 5 digits arranged in the first 5 places: 5! = 120.

Final answer

120.

Step-by-step solution

1Choose 2 vowels: ⁵C₂ = 10; choose 2 consonants: ²¹C₂ = 210.
2Arrange the 4 selected letters: 4! = 24.
3Total = 10 × 210 × 24 = 50400.

Final answer

50400.

Step-by-step solution

1Cases by (from Part I, from Part II): (3, 5), (4, 4), (5, 3).
2(3, 5): ⁵C₃ × ⁷C₅ = 10 × 21 = 210.
3(4, 4): ⁵C₄ × ⁷C₄ = 5 × 35 = 175.
4(5, 3): ⁵C₅ × ⁷C₃ = 1 × 35 = 35.
5Total = 210 + 175 + 35 = 420.

Final answer

420.

Step-by-step solution

1Choose 1 king from 4: ⁴C₁ = 4.
2Choose 4 more cards from the 48 non-kings: ⁴⁸C₄ = 194580.
3Total = 4 × 194580 = 778320.

Final answer

778320.

Step-by-step solution

1Even places in a row of 9 seats: positions 2, 4, 6, 8 → 4 places.
2Arrange 4 women in even places: 4! = 24.
3Arrange 5 men in odd places: 5! = 120.
4Total = 24 × 120 = 2880.

Final answer

2880.

Step-by-step solution

1Case 1: all 3 join → choose remaining 7 from 22: ²²C₇ = 170544.
2Case 2: none joins → choose 10 from 22: ²²C₁₀ = 646646.
3Total = 170544 + 646646 = 817190.

Final answer

817190.

Step-by-step solution

1ASSASSINATION: 13 letters — A×3, S×4, I×2, N×2, T, O.
2All 4 S's together → treat as one block → 10 objects.
3Number = 10!/(3!·2!·2!) = 3628800/24 = 151200.

Final answer

151200.

33

Chapter 7 — Binomial Theorem

The binomial theorem expands (x + y)ⁿ as a sum of n+1 terms whose coefficients are the binomial numbers ⁿCᵣ. The expansion is symmetric — the coefficients read the same forwards and backwards (⁰C₀, ⁰C₁, …, ⁰Cₙ = Pascal's row) — and the (r+1)th term is ⁿCᵣ xⁿ⁻ʳ yʳ. That term formula is the workhorse for evaluating numbers like (99)⁵ and for divisibility proofs.

Board pattern

In a written expansion, always pair each term with its coefficient explicitly (write ⁵C₂(1)³(−2x)² before simplifying). For 'indicate which is larger' and approximation questions, the first one or two terms alone often decide the answer — quote the inequality formed by dropping the positive tail.

$Ex 7.1 — Expansions, evaluations, proofs (14 Q)$
$Misc — Factor theorem, approximations, expansions (6 Q)$

34

Exercise 7.1 — Using the Binomial Theorem

14Exercise questions

Step-by-step solution

1(1 − 2x)⁵ = Σ ⁵Cᵣ (1)⁵⁻ʳ (−2x)ʳ.
2= 1 − 5(2x) + 10(2x)² − 10(2x)³ + 5(2x)⁴ − (2x)⁵.
3= 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵.

Final answer

1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵.

Step-by-step solution

1(2/x − x/2)⁵ = Σ ⁵Cᵣ (2/x)⁵⁻ʳ (−x/2)ʳ.
2= 32/x⁵ − 5·16/x⁴·x/2 + 10·8/x³·x²/4 − 10·4/x²·x³/8 + 5·2/x·x⁴/16 − x⁵/32.
3= 32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32.

Final answer

32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32.

Step-by-step solution

1(2x − 3)⁶ = Σ ⁶Cᵣ (2x)⁶⁻ʳ (−3)ʳ.
2= 64x⁶ − 6·32x⁵·3 + 15·16x⁴·9 − 20·8x³·27 + 15·4x²·81 − 6·2x·243 + 729.
3= 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729.

Final answer

64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729.

Step-by-step solution

1(x/3 + 1/x)⁵ = Σ ⁵Cᵣ (x/3)⁵⁻ʳ (1/x)ʳ.
2= x⁵/243 + 5x⁴/81·1/x + 10x³/27·1/x² + 10x²/9·1/x³ + 5x/3·1/x⁴ + 1/x⁵.
3= x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵.

Final answer

x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵.

Step-by-step solution

1(x + 1/x)⁶ = Σ ⁶Cᵣ x⁶⁻ʳ (1/x)ʳ.
2= x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶.

Final answer

x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶.

Step-by-step solution

196 = 100 − 4, so (96)³ = (100 − 4)³.
2= 100³ − 3·100²·4 + 3·100·4² − 4³.
3= 1000000 − 120000 + 4800 − 64 = 884736.

Final answer

884736.

Step-by-step solution

1102 = 100 + 2, so (102)⁵ = (100 + 2)⁵.
2= 100⁵ + 5·100⁴·2 + 10·100³·4 + 10·100²·8 + 5·100·16 + 32.
3= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 = 11040808032.

Final answer

11040808032.

Step-by-step solution

1101 = 100 + 1, so (101)⁴ = (100 + 1)⁴.
2= 100⁴ + 4·100³ + 6·100² + 4·100 + 1.
3= 100000000 + 4000000 + 60000 + 400 + 1 = 104060401.

Final answer

104060401.

Step-by-step solution

199 = 100 − 1, so (99)⁵ = (100 − 1)⁵.
2= 100⁵ − 5·100⁴ + 10·100³ − 10·100² + 5·100 − 1.
3= 10000000000 − 5000000 + 100000 − 10000 + 500 − 1 = 9509900499.

Final answer

9509900499.

Step-by-step solution

1(1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰ = 1 + ¹⁰⁰⁰⁰C₁(0.1) + positive terms.
2= 1 + 10000 × 0.1 + (positive) = 1001 + (positive).
31001 > 1000, hence (1.1)¹⁰⁰⁰⁰ is larger.

Final answer

(1.1)¹⁰⁰⁰⁰.

Step-by-step solution

1(a+b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
2(a−b)⁴ = a⁴ − 4a³b + 6a²b² − 4ab³ + b⁴.
3Difference = 8a³b + 8ab³ = 8ab(a² + b²).
4With a = √3, b = √2: 8√3√2(3 + 2) = 8√6 × 5 = 40√6.

Final answer

8ab(a² + b²); hence 40√6.

Step-by-step solution

1Even powers survive the addition: (x+1)⁶ + (x−1)⁶ = 2(x⁶ + 15x⁴ + 15x² + 1).
2With x = √2: 2[(√2)⁶ + 15(√2)⁴ + 15(√2)² + 1].
3= 2[8 + 60 + 30 + 1] = 2 × 99 = 198.

Final answer

2(x⁶ + 15x⁴ + 15x² + 1); hence 198.

Step-by-step solution

19ⁿ⁺¹ = (1 + 8)ⁿ⁺¹ = 1 + (n+1)·8 + ⁰ⁿ⁺¹C₂·8² + … + 8ⁿ⁺¹.
2= (1 + 8n + 8) + 64·(⁰ⁿ⁺¹C₂ + ⁰ⁿ⁺¹C₃·8 + …) = 9 + 8n + 64k.
3Then 9ⁿ⁺¹ − 8n − 9 = 64k, which is divisible by 64.

Final answer

9ⁿ⁺¹ − 8n − 9 = 64k (shown).

Step-by-step solution

1By the binomial theorem, (1 + x)ⁿ = Σᵣ₌₀ⁿ ⁿCᵣ xʳ.
2Put x = 3: (1 + 3)ⁿ = Σᵣ₌₀ⁿ ⁿCᵣ 3ʳ.
3Hence Σᵣ₌₀ⁿ 3ʳ ⁿCᵣ = 4ⁿ.

Final answer

Proved: LHS = 4ⁿ.

35

Miscellaneous Exercise on Chapter 7

6Exercise questions

Step-by-step solution

1Write aⁿ = (a − b + b)ⁿ = [(a − b) + b]ⁿ.
2Expand: aⁿ = (a−b)ⁿ + n(a−b)ⁿ⁻¹b + … + n(a−b)bⁿ⁻¹ + bⁿ.
3Take bⁿ to the left: aⁿ − bⁿ = (a−b)·[(a−b)ⁿ⁻¹ + n(a−b)ⁿ⁻²b + … + nbⁿ⁻¹].
4Every term in the bracket is an integer, so (a − b) divides aⁿ − bⁿ.

Final answer

Proved: a − b is a factor of aⁿ − bⁿ.

Step-by-step solution

1Odd terms survive the subtraction: (x+y)⁶ − (x−y)⁶ = 2(6x⁵y + 20x³y³ + 6xy⁵).
2= 12x⁵y + 40x³y³ + 12xy⁵ = xy(12x⁴ + 40x²y² + 12y⁴).
3With x = √3, y = √2: xy = √6, x⁴ = 9, x²y² = 6, y⁴ = 4.
4= √6(108 + 240 + 48) = 396√6.

Final answer

396√6.

Step-by-step solution

1Let x = a², y = √(a² − 1), so y² = a² − 1.
2(x+y)⁴ + (x−y)⁴ = 2x⁴ + 12x²y² + 2y⁴.
3= 2a⁸ + 12a⁴(a²−1) + 2(a²−1)².
4= 2a⁸ + 12a⁶ − 12a⁴ + 2a⁴ − 4a² + 2 = 2(a⁸ + 6a⁶ − 5a⁴ − 2a² + 1).

Final answer

2(a⁸ + 6a⁶ − 5a⁴ − 2a² + 1).

Step-by-step solution

1(0.99)⁵ = (1 − 0.01)⁵.
2First three terms: 1 − 5(0.01) + 10(0.01)².
3= 1 − 0.05 + 0.001 = 0.951.

Final answer

0.951 (approx).

Step-by-step solution

1Let u = x + 2/x, so the expression is (u + 1)⁴ = u⁴ + 4u³ + 6u² + 4u + 1.
2u² = x² + 4 + 4/x²; u³ = x³ + 6x + 12/x + 8/x³; u⁴ = x⁴ + 8x² + 24 + 32/x² + 16/x⁴.
3Substitute: x⁴ + 4x³ + (8+6)x² + (24+4)x + (24+1) + (48+8)/x + (32+24)/x² + 32/x³ + 16/x⁴.
4= x⁴ + 4x³ + 14x² + 28x + 49 + 56/x + 56/x² + 32/x³ + 16/x⁴.

Final answer

x⁴ + 4x³ + 14x² + 28x + 49 + 56/x + 56/x² + 32/x³ + 16/x⁴.

Step-by-step solution

1Treat −2ax + 3a² as B: (3x² + B)³ = 27x⁶ + 27x⁴B + 9x²B² + B³.
227x⁴B = −54ax⁵ + 81a²x⁴.
3B² = 4a²x² − 12a³x + 9a⁴ → 9x²B² = 36a²x⁴ − 108a³x³ + 81a⁴x².
4B³ = (4a²x² − 12a³x + 9a⁴)(−2ax + 3a²) = −8a³x³ + 36a⁴x² − 54a⁵x + 27a⁶.
5Sum: 27x⁶ − 54ax⁵ + 117a²x⁴ − 116a³x³ + 117a⁴x² − 54a⁵x + 27a⁶.

Final answer

27x⁶ − 54ax⁵ + 117a²x⁴ − 116a³x³ + 117a⁴x² − 54a⁵x + 27a⁶.

36

Chapter 8 — Sequences and Series

A sequence lists numbers in a fixed order; its series is the running sum. Two progressions dominate the chapter: the arithmetic progression (AP), where each term grows by a constant difference d, and the geometric progression (GP), where each term multiplies by a constant ratio r. Their nth terms and sums, plus the arithmetic and geometric means (A.M. and G.M.), power the word problems at the end.

Board pattern

Quote the formula in symbols before substituting: aₙ = a + (n−1)d, Sₙ = n/2 (2a + (n−1)d), aₙ = arⁿ⁻¹, Sₙ = a(rⁿ−1)/(r−1). For instalment and depreciation questions, identify the AP or GP hidden inside the story and state the first term and common difference/ratio explicitly. For a GP that forms an AP after modification, write both progressions before solving.

$Ex 8.1 — Sequences and nth terms (14 Q)$
$Ex 8.2 — Geometric progressions, G.M., applications (32 Q)$
$Misc — Mixed series and word problems (18 Q)$

37

Exercise 8.1 — Sequences

14Exercise questions

Step-by-step solution

1a₁ = 1(3) = 3, a₂ = 2(4) = 8, a₃ = 3(5) = 15.
2a₄ = 4(6) = 24, a₅ = 5(7) = 35.

Final answer

3, 8, 15, 24, 35.

Step-by-step solution

1a₁ = 1/2, a₂ = 2/3, a₃ = 3/4.
2a₄ = 4/5, a₅ = 5/6.

Final answer

1/2, 2/3, 3/4, 4/5, 5/6.

Step-by-step solution

1a₁ = 2, a₂ = 4, a₃ = 8.
2a₄ = 16, a₅ = 32.

Final answer

2, 4, 8, 16, 32.

Step-by-step solution

1a₁ = −1/6, a₂ = 1/6, a₃ = 3/6 = 1/2.
2a₄ = 5/6, a₅ = 7/6.

Final answer

−1/6, 1/6, 1/2, 5/6, 7/6.

Step-by-step solution

1a₁ = (−1)⁰·5² = 25, a₂ = (−1)¹·5³ = −125.
2a₃ = 625, a₄ = −3125, a₅ = 15625.

Final answer

25, −125, 625, −3125, 15625.

Step-by-step solution

1a₁ = 1(6)/4 = 3/2, a₂ = 2(9)/4 = 9/2, a₃ = 3(14)/4 = 21/2.
2a₄ = 4(21)/4 = 21, a₅ = 5(30)/4 = 75/2.

Final answer

3/2, 9/2, 21/2, 21, 75/2.

Step-by-step solution

1a₁₇ = 4(17) − 3 = 68 − 3 = 65.
2a₂₄ = 4(24) − 3 = 96 − 3 = 93.

Final answer

65, 93.

Step-by-step solution

1a₇ = 7²/2⁷ = 49/128.

Final answer

49/128.

Step-by-step solution

1a₉ = (−1)⁸·9³ = 729.

Final answer

729.

Step-by-step solution

1a₂₀ = 20(20 − 2)/(20 + 3) = 20 × 18/23 = 360/23.

Final answer

360/23.

Step-by-step solution

1a₁ = 3, a₂ = 3(3) + 2 = 11, a₃ = 3(11) + 2 = 35.
2a₄ = 3(35) + 2 = 107, a₅ = 3(107) + 2 = 323.
3Series: 3 + 11 + 35 + 107 + 323 + …

Final answer

3, 11, 35, 107, 323; series 3 + 11 + 35 + 107 + 323 + ….

Step-by-step solution

1a₁ = −1, a₂ = −1/2, a₃ = (−1/2)/3 = −1/6.
2a₄ = (−1/6)/4 = −1/24, a₅ = (−1/24)/5 = −1/120.
3Series: −1 − 1/2 − 1/6 − 1/24 − 1/120.

Final answer

−1, −1/2, −1/6, −1/24, −1/120; series −1 − 1/2 − 1/6 − 1/24 − 1/120.

Step-by-step solution

1a₃ = a₂ − 1 = 1, a₄ = a₃ − 1 = 0, a₅ = a₄ − 1 = −1.
2Series: 2 + 2 + 1 + 0 + (−1).

Final answer

2, 2, 1, 0, −1; series 2 + 2 + 1 + 0 − 1.

Step-by-step solution

1a₃ = 1 + 1 = 2, a₄ = 2 + 1 = 3, a₅ = 3 + 2 = 5, a₆ = 5 + 3 = 8.
2n = 1: a₂/a₁ = 1; n = 2: a₃/a₂ = 2.
3n = 3: a₄/a₃ = 3/2; n = 4: a₅/a₄ = 5/3; n = 5: a₆/a₅ = 8/5.

Final answer

1, 2, 3/2, 5/3, 8/5.

38

Exercise 8.2 — Geometric Progressions

32Exercise questions

Step-by-step solution

1a = 5/2, r = (5/4)/(5/2) = 1/2.
2a₂₀ = (5/2)(1/2)¹⁹ = 5/2²⁰.
3aₙ = (5/2)(1/2)ⁿ⁻¹ = 5/2ⁿ.

Final answer

5/2²⁰ and 5/2ⁿ.

Step-by-step solution

1a₈ = ar⁷ = a·2⁷ = 128a = 192 → a = 3/2.
2a₁₂ = (3/2)·2¹¹ = (3/2)(2048) = 3072.

Final answer

3072.

Step-by-step solution

1p = ar⁴, q = ar⁷, s = ar¹⁰.
2q² = a²r¹⁴ and ps = (ar⁴)(ar¹⁰) = a²r¹⁴.
3Hence q² = ps.

Final answer

q² = ps (shown).

Step-by-step solution

1Let terms be a, ar, ar², ar³ with a = −3.
24th = (2nd)² ⇒ ar³ = (ar)² ⇒ r = a = −3.
37th term = ar⁶ = (−3)(−3)⁶ = (−3)⁷ = −2187.

Final answer

−2187.

Step-by-step solution

1(i) a = 2, r = √2; 2·(√2)ⁿ⁻¹ = 128 = 2⁷ → (√2)ⁿ⁻¹ = 2⁶ → (n−1)/2 = 6 → n = 13.
2(ii) a = √3, r = √3; (√3)ⁿ = 729 = 3⁶ → 3ⁿᐟ² = 3⁶ → n = 12.
3(iii) a = r = 1/3; (1/3)ⁿ = 1/19683 = (1/3)⁹ → n = 9.

Final answer

(i) 13th, (ii) 12th, (iii) 9th.

Step-by-step solution

1x² = (−2/7)(−7/2) = 1.
2x = ±1.

Final answer

x = ±1.

Step-by-step solution

1a = 0.15, r = 0.1.
2S₂₀ = 0.15(1 − (0.1)²⁰)/(1 − 0.1) = (1/6)(1 − 10⁻²⁰).

Final answer

(1/6)(1 − 10⁻²⁰).

Step-by-step solution

1a = √7, r = √21/√7 = √3.
2Sₙ = √7[(√3)ⁿ − 1]/(√3 − 1).

Final answer

√7(3ⁿᐟ² − 1)/(√3 − 1).

Step-by-step solution

1a₁ = 1, r = −a.
2Sₙ = 1(1 − (−a)ⁿ)/(1 − (−a)) = (1 − (−a)ⁿ)/(1 + a).

Final answer

(1 − (−a)ⁿ)/(1 + a).

Step-by-step solution

1a = x³, r = x².
2Sₙ = x³(x²ⁿ − 1)/(x² − 1).

Final answer

x³(x²ⁿ − 1)/(x² − 1).

Step-by-step solution

1Σ(2 + 3ᵏ) = 2·11 + (3 + 3² + … + 3¹¹).
2= 22 + 3(3¹¹ − 1)/(3 − 1) = 22 + (3¹² − 3)/2.
3= 22 + (531441 − 3)/2 = 22 + 265719 = 265741.

Final answer

265741.

Step-by-step solution

1Let the terms be a/r, a, ar; product a³ = 1 → a = 1.
2Sum: 1/r + 1 + r = 39/10 → 1/r + r = 29/10 → 10r² − 29r + 10 = 0.
3r = (29 ± 21)/20 → r = 5/2 or r = 2/5.
4Terms: (2/5, 1, 5/2) or (5/2, 1, 2/5).

Final answer

r = 5/2 or 2/5; terms (2/5, 1, 5/2) or (5/2, 1, 2/5).

Step-by-step solution

1a = 3, r = 3, Sₙ = 3(3ⁿ − 1)/(3 − 1) = 3(3ⁿ − 1)/2.
23(3ⁿ − 1)/2 = 120 → 3ⁿ − 1 = 80 → 3ⁿ = 81 = 3⁴.
3n = 4.

Final answer

4 terms.

Step-by-step solution

1First three sum: a(1 + r + r²) = 16.
2Next three: ar³(1 + r + r²) = 128 ⇒ 16r³ = 128 → r³ = 8 → r = 2.
3a(1 + 2 + 4) = 16 → 7a = 16 → a = 16/7.
4Sₙ = (16/7)(2ⁿ − 1)/(2 − 1) = (16/7)(2ⁿ − 1).

Final answer

a = 16/7, r = 2, Sₙ = (16/7)(2ⁿ − 1).

Step-by-step solution

1ar⁶ = 64 → 729r⁶ = 64 → r⁶ = 64/729 = (2/3)⁶ → r = 2/3.
2S₇ = 729(1 − (2/3)⁷)/(1 − 2/3) = 729·3(1 − 128/2187).
3= 2187 − 128 = 2059.

Final answer

2059.

Step-by-step solution

1a + ar = −4 and ar⁴ = 4ar² → r² = 4 → r = ±2.
2r = 2: a(1 + 2) = −4 → a = −4/3.
3r = −2: a(1 − 2) = −4 → a = 4.
4G.P.s: (−4/3, −8/3, …) or (4, −8, …).

Final answer

a = −4/3 with r = 2, or a = 4 with r = −2.

Step-by-step solution

1x = ar³, y = ar⁹, z = ar¹⁵.
2y² = a²r¹⁸ and xz = a²r¹⁸.
3y² = xz, so x, y, z are in G.P.

Final answer

Proved (y² = xz).

Step-by-step solution

1kth term = 8(10ᵏ − 1)/9.
2Sₙ = (8/9)Σ(10ᵏ − 1) = (8/9)[10(10ⁿ − 1)/9 − n].
3= 80(10ⁿ − 1)/81 − 8n/9.

Final answer

80(10ⁿ − 1)/81 − 8n/9.

Step-by-step solution

1Products: 2·128 = 256, 4·32 = 128, 8·8 = 64, 16·2 = 32, 32·(1/2) = 16.
2Sum = 256 + 128 + 64 + 32 + 16 = 496.

Final answer

496.

Step-by-step solution

1Products are aA, aArR, aAr²R², …, aA(rR)ⁿ⁻¹.
2Each term is the previous × (rR).
3They form a G.P. with first term aA and common ratio rR.

Final answer

G.P. with common ratio rR (shown).

Step-by-step solution

1Let the numbers be a, ar, ar², ar³: ar² − a = 9 and ar − ar³ = 18.
2a(r² − 1) = 9 and ar(1 − r²) = 18 → −r·a(r² − 1) = 18 → −r·9 = 18 → r = −2.
3a(4 − 1) = 9 → a = 3.
4Numbers: 3, −6, 12, −24.

Final answer

3, −6, 12, −24.

Step-by-step solution

1a = A r^(p−1), b = A r^(q−1), c = A r^(r−1).
2Exponent of A: (q−r) + (r−p) + (p−q) = 0.
3Exponent of r: (p−1)(q−r) + (q−1)(r−p) + (r−1)(p−q) = 0.
4Result: A⁰r⁰ = 1.

Final answer

Proved: the product equals 1.

Step-by-step solution

1b = arⁿ⁻¹ and P = aⁿr^(n(n−1)/2).
2P² = a²ⁿr^(n(n−1)) and (ab)ⁿ = (a·arⁿ⁻¹)ⁿ = a²ⁿr^(n(n−1)).
3P² = (ab)ⁿ.

Final answer

P² = (ab)ⁿ (shown).

Step-by-step solution

1Sum of first n terms: S = a(1 − rⁿ)/(1 − r).
2Sum of (n+1)th to (2n)th: arⁿ(1 − rⁿ)/(1 − r).
3Ratio = [a(1−rⁿ)/(1−r)] / [arⁿ(1−rⁿ)/(1−r)] = 1/rⁿ.

Final answer

Ratio = 1/rⁿ (shown).

Step-by-step solution

1Let b = ar, c = ar², d = ar³.
2LHS = a²(1 + r² + r⁴)·a²r²(1 + r² + r⁴) = a⁴r²(1 + r² + r⁴)².
3RHS = (a²r + a²r³ + a²r⁵)² = a⁴r²(1 + r² + r⁴)².
4Both sides are equal.

Final answer

Identity proved.

Step-by-step solution

1Let 3, x, y, 81 be a G.P.: 3r³ = 81 → r³ = 27 → r = 3.
2x = 3·3 = 9 and y = 9·3 = 27.

Final answer

9 and 27.

Step-by-step solution

1Set (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) = √(ab).
2Cross-multiply and divide both sides by (ab)ⁿᐟ².
3This gives n = −1/2 (with the simplification aⁿ⁺¹ = a·aⁿ).

Final answer

n = −1/2.

Step-by-step solution

1Let the numbers be x and y: x + y = 6√(xy).
2Divide by y: t + 1 = 6√t where t = x/y.
3Let u = √t: u² − 6u + 1 = 0 → u = 3 ± 2√2.
4x/y = (3 + 2√2)² but (3+2√2)/(3−2√2) = (3+2√2)², so the ratio is (3 + 2√2) : (3 − 2√2).

Final answer

Ratio is (3 + 2√2) : (3 − 2√2).

Step-by-step solution

1Let the numbers be a and b: A = (a+b)/2, G = √(ab).
2a, b are the roots of x² − 2Ax + G² = 0.
3x = A ± √(A² − G²) = A ± √((A − G)(A + G)).
4Hence the numbers are A ± √((A+G)(A−G)).

Final answer

Proved: numbers are A ± √((A+G)(A−G)).

Step-by-step solution

1After n hours: 30·2ⁿ (G.P. with a = 30, r = 2).
22nd hour: 30·2² = 120; 4th hour: 30·2⁴ = 480.
3nth hour: 30·2ⁿ.

Final answer

120, 480, and 30·2ⁿ.

Step-by-step solution

1Amount = P(1 + r/100)ⁿ with P = 500, r = 10, n = 10.
2= 500(1.1)¹⁰.
3= 500 × 2.5937 ≈ Rs 1296.87.

Final answer

500(1.1)¹⁰ ≈ Rs 1296.87.

Step-by-step solution

1Sum of roots = 2 × A.M. = 16.
2Product of roots = G.M.² = 25.
3Equation: x² − 16x + 25 = 0.

Final answer

x² − 16x + 25 = 0.

39

Miscellaneous Exercise on Chapter 8

18Exercise questions

Step-by-step solution

1f(n) = 3ⁿ (multiplying the property repeatedly from f(1) = 3).
2Σ f(x) = 3 + 9 + … + 3ⁿ = 3(3ⁿ − 1)/2 = 120.
33ⁿ − 1 = 80 → 3ⁿ = 81 = 3⁴ → n = 4.

Final answer

n = 4.

Step-by-step solution

1Sₙ = 5(2ⁿ − 1)/(2 − 1) = 5(2ⁿ − 1) = 315 → 2ⁿ − 1 = 63 → 2ⁿ = 64 → n = 6.
2Last term = arⁿ⁻¹ = 5·2⁵ = 160.

Final answer

6 terms, last term 160.

Step-by-step solution

1Terms: 1, r, r², …, so r² + r⁴ = 90.
2r⁴ + r² − 90 = 0 → (r² + 10)(r² − 9) = 0 → r² = 9.
3r = ±3.

Final answer

r = ±3.

Step-by-step solution

1Let the numbers be a, ar, ar²: a(1 + r + r²) = 56.
2(a − 1), (ar − 7), (ar² − 21) in AP → 2(ar − 7) = (a − 1) + (ar² − 21) → a(r−1)² = 8.
3Dividing: (1 + r + r²)/(r − 1)² = 7 → 2r² − 5r + 2 = 0 → r = 2 or r = 1/2.
4r = 2 gives a = 8 (numbers 8, 16, 32); r = 1/2 gives a = 32 (numbers 32, 16, 8).

Final answer

8, 16, 32 (or in reverse order).

Step-by-step solution

1Let the terms be a, ar, …, ar²ⁿ⁻¹.
2Sum of all = a(1 − r²ⁿ)/(1 − r); odd places = a(1 − r²ⁿ)/(1 − r²).
3a(1−r²ⁿ)/(1−r) = 5a(1−r²ⁿ)/(1−r²) → 1/(1−r) = 5/(1−r²).
41 − r² = 5 − 5r → r² − 5r + 4 = 0 → r = 4 (r = 1 rejected).

Final answer

r = 4.

Step-by-step solution

1From (a + bx)(b − cx) = (a − bx)(b + cx): 2b²x = 2acx → b² = ac.
2Similarly the second equality gives c² = bd.
3b² = ac and c² = bd imply a, b, c, d are in G.P.

Final answer

Proved: a, b, c, d in G.P.

Step-by-step solution

1S = a(rⁿ − 1)/(r − 1), P = aⁿr^(n(n−1)/2), R = (1/a)(1 + 1/r + … + 1/rⁿ⁻¹).
2R = (rⁿ − 1)/(a rⁿ⁻¹(r − 1)).
3P²Rⁿ = a²ⁿ r^(n(n−1)) · (rⁿ−1)ⁿ/(aⁿ r^(n(n−1)) (r−1)ⁿ) = [a(rⁿ − 1)/(r − 1)]ⁿ = Sⁿ.

Final answer

P²Rⁿ = Sⁿ (proved).

Step-by-step solution

1b = ar, c = ar², d = ar³.
2aⁿ + bⁿ = aⁿ(1 + rⁿ), bⁿ + cⁿ = aⁿrⁿ(1 + rⁿ), cⁿ + dⁿ = aⁿr²ⁿ(1 + rⁿ).
3Each term is the previous multiplied by rⁿ, so they form a G.P.

Final answer

Proved: the three terms are in G.P.

Step-by-step solution

1a + b = 3, ab = p; c + d = 12, cd = q.
2Let a, b, c, d = a, ar, ar², ar³: a(1 + r) = 3 and ar²(1 + r) = 12.
3Divide: r² = 4 → r = 2; then a = 1. Terms are 1, 2, 4, 8.
4p = 2, q = 32 → (q + p) : (q − p) = 34 : 30 = 17 : 15.

Final answer

Ratio = 17 : 15 (proved).

Step-by-step solution

1A/G = m/n → [(a+b)/2]/√(ab) = m/n.
2Let a = kr, b = k/r: (k(r + 1/r)/2)/k = m/n → (r + 1/r)/2 = m/n.
3r + 1/r = 2m/n → n r² − 2m r + n = 0 → r = (m ± √(m² − n²))/n.
4a : b = r² : 1 = (m + √(m²−n²)) : (m − √(m²−n²)).

Final answer

a : b = (m + √(m² − n²)) : (m − √(m² − n²)).

Step-by-step solution

1(i) 5 + 55 + … = (5/9)(9 + 99 + 999 + …) = (5/9)[(10 + 10² + … + 10ⁿ) − n].
2= (5/9)[10(10ⁿ − 1)/9 − n] = 50(10ⁿ − 1)/81 − 5n/9.
3(ii) 0.6 + 0.66 + … = (6/9)Σ(1 − 10ᵏ⁻¹/10ᵏ⁻¹·…) → Sₙ = (2/3)[n − (1/9)(1 − 10⁻ⁿ)].
4= 2n/3 − 2(1 − 10⁻ⁿ)/27.

Final answer

(i) 50(10ⁿ − 1)/81 − 5n/9; (ii) 2n/3 − 2(1 − 10⁻ⁿ)/27.

Step-by-step solution

1kth term = (2k)(2k + 2) = 4k(k + 1).
220th term = 4·20·21 = 1680.

Final answer

1680.

Step-by-step solution

1Balance = Rs 6000; number of instalments = 6000/500 = 12.
2Interest paid: 12% of 6000, 5500, …, 500 per year.
3= 720 + 660 + … + 60 (AP with 12 terms, d = −60).
4Sum = 12/2(720 + 60) = 4680.
5Total cost = 12000 + 4680 = Rs 16680.

Final answer

Rs 16680.

Step-by-step solution

1Balance = Rs 18000; number of instalments = 18.
2Interest: 10% of 18000, 17000, …, 1000 → 1800, 1700, …, 100.
3Sum = 18/2(1800 + 100) = 9 × 1900 = 17100.
4Total cost = 22000 + 17100 = Rs 39100.

Final answer

Rs 39100.

Step-by-step solution

1Letters per set form a G.P.: 4, 16, 64, …, 4⁸.
2Total letters up to set 8 = 4(4⁸ − 1)/(4 − 1) = 4(65536 − 1)/3 = 87380.
3Cost = 87380 × 0.50 = Rs 43690.

Final answer

Rs 43690.

Step-by-step solution

1Yearly interest = 5% of 10000 = Rs 500.
2Amount in the 15th year = 10000 + 14 × 500 = Rs 17000.
3Amount after 20 years = 10000 + 20 × 500 = Rs 20000.

Final answer

Rs 17000 (15th year) and Rs 20000 (after 20 years).

Step-by-step solution

1Value after n years = 15625(0.8)ⁿ.
2After 5 years = 15625(4/5)⁵ = 15625 × 1024/3125.
3= 5 × 1024 = Rs 5120.

Final answer

Rs 5120.

Step-by-step solution

1Work = 150d, where d = planned days; actual days = d + 8.
2Workers on day k: 150 − 4(k − 1), so total work = (d+8)·150 − 4(d+8)(d+7)/2.
3= (d + 8)(150 − 2(d + 7)) = (d + 8)(136 − 2d).
4(d + 8)(136 − 2d) = 150d → d² − 49d + 544 = 0 → (d − 17)(d − 32) = 0.
5d = 17 (32 would give a negative tail); total days = 17 + 8 = 25.

Final answer

25 days.

40

Chapter 9 — Straight Lines

Coordinate geometry turns lines into equations. The slope m of a line, the angle between two lines via their slopes, and the relationship between perpendicular and parallel slopes unlock the whole chapter. From there, lines are written in point-slope, two-point, slope-intercept, and intercept forms, and closed with the perpendicular distance of a point from a line.

Board pattern

Two lines with slopes m₁ and m₂ are parallel when m₁ = m₂ and perpendicular when m₁m₂ = −1. Distance from (x₁, y₁) to Ax + By + C = 0 is |Ax₁ + By₁ + C|/√(A² + B²). For intercept sums/products, form the quadratic in a and b directly. In proof questions, substitute the point into the equation of the line rather than arguing verbally.

$Ex 9.1 — Slope, parallel, perpendicular (11 Q)$
$Ex 9.2 — Equations of a line (19 Q)$
$Ex 9.3 — Point-line distance (17 Q)$
$Misc — Concurrency, images, applications (23 Q)$

41

Exercise 9.1 — Slope of a Line

11Exercise questions

Step-by-step solution

1Label A(−4,5), B(0,7), C(5,−5), D(−4,−2).
2Shoelace sum₁ = xᵢyᵢ₊₁ = (−4)(7) + (0)(−5) + (5)(−2) + (−4)(5) = −58.
3sum₂ = yᵢxᵢ₊₁ = 5(0) + 7(5) + (−5)(−4) + (−2)(−4) = 63.
4Area = ½|sum₁ − sum₂| = ½|−58 − 63| = 121/2 = 60.5 sq units.

Final answer

Area = 121/2 = 60.5 sq units.

Step-by-step solution

1Base on y-axis with midpoint (0,0) → endpoints (0, a) and (0, −a).
2Third vertex on x-axis; height of the triangle = √(4a² − a²) = √3a.
3Vertices: (0, a), (0, −a), (√3a, 0) — also (−√3a, 0) on the other side.

Final answer

(0, a), (0, −a), (±√3a, 0).

Step-by-step solution

1(i) PQ ∥ y-axis ⇒ x₁ = x₂ → distance = |y₂ − y₁|.
2(ii) PQ ∥ x-axis ⇒ y₁ = y₂ → distance = |x₂ − x₁|.

Final answer

(i) |y₂ − y₁|, (ii) |x₂ − x₁|.

Step-by-step solution

1Let the point be (x, 0).
2(x − 7)² + 36 = (x − 3)² + 16 → −14x + 85 = −6x + 25.
38x = 60 → x = 15/2. Point: (15/2, 0).

Final answer

(15/2, 0).

Step-by-step solution

1Mid-point of PB = (4, −2).
2Slope through (0,0) and (4,−2) = (−2 − 0)/(4 − 0) = −1/2.

Final answer

−1/2.

Step-by-step solution

1Slope of (4,4)→(3,5) = (5−4)/(3−4) = −1.
2Slope of (4,4)→(−1,−1) = (−1−4)/(−1−4) = 1.
3Product = −1 → the two sides are perpendicular → right angled at (4, 4).

Final answer

Right angled at (4, 4) — shown.

Step-by-step solution

130° from the y-axis ⇒ 90° − 30° = 60° from the positive x-axis.
2Slope = tan 60° = √3.

Final answer

√3.

Step-by-step solution

1Slope of (−2,−1)→(4,0) = 1/6; slope of (3,3)→(−3,2) = (2−3)/(−3−3) = 1/6.
2Slope of (4,0)→(3,3) = −3; slope of (−2,−1)→(−3,2) = 3/(−1) = −3.
3Opposite sides are parallel → parallelogram.

Final answer

Opposite sides parallel — shown.

Step-by-step solution

1Slope = (−2 + 1)/(4 − 3) = −1.
2tan θ = −1 ⇒ θ = 135° (measured anticlockwise from the x-axis).

Final answer

135°.

Step-by-step solution

1Let the slopes be m and 2m: tan θ = |(2m − m)/(1 + 2m²)| = |m/(1 + 2m²)| = 1/3.
2m/(1 + 2m²) = 1/3 → 2m² − 3m + 1 = 0 → (2m − 1)(m − 1) = 0.
3m = 1/2 or m = 1 → slopes are (1/2, 1) or (1, 2).

Final answer

(1/2, 1) or (1, 2).

Step-by-step solution

1By definition, m = (k − y₁)/(h − x₁).
2Cross-multiplying: k − y₁ = m(h − x₁).

Final answer

k − y₁ = m(h − x₁) — shown.

42

Exercise 9.2 — Various Forms of the Equation of a Line

19Exercise questions

Step-by-step solution

1x-axis: all points have y = 0 → y = 0.
2y-axis: all points have x = 0 → x = 0.

Final answer

x-axis: y = 0; y-axis: x = 0.

Step-by-step solution

1Point-slope form: y − 3 = (1/2)(x + 4).
22y − 6 = x + 4 → x − 2y + 10 = 0.

Final answer

x − 2y + 10 = 0.

Step-by-step solution

1Point-slope form: y − 0 = m(x − 0).
2y = mx.

Final answer

y = mx.

Step-by-step solution

1m = tan 75° = 2 + √3.
2y − 2√3 = (2 + √3)(x − 2).
3y = (2 + √3)x − 4 → (2 + √3)x − y − 4 = 0.

Final answer

(2 + √3)x − y − 4 = 0.

Step-by-step solution

1Passes through (−3, 0) with slope −2.
2y − 0 = −2(x + 3) → y = −2x − 6 → 2x + y + 6 = 0.

Final answer

2x + y + 6 = 0.

Step-by-step solution

1Passes through (0, 2) with slope tan 30° = 1/√3.
2y = (1/√3)x + 2 → x − √3y + 2√3 = 0.

Final answer

x − √3 y + 2√3 = 0.

Step-by-step solution

1Slope = (−4 − 1)/(2 + 1) = −5/3.
2y − 1 = (−5/3)(x + 1) → 3y − 3 = −5x − 5.
35x + 3y + 2 = 0.

Final answer

5x + 3y + 2 = 0.

Step-by-step solution

1Mid-point of PQ = (0, 2).
2Median is the line through R(4, 5) and (0, 2): slope = 3/4.
3y − 2 = (3/4)(x − 0) → 3x − 4y + 8 = 0.

Final answer

3x − 4y + 8 = 0.

Step-by-step solution

1Slope of the given line = (6 − 5)/(−3 − 2) = −1/5.
2Perpendicular slope = 5.
3y − 5 = 5(x + 3) → 5x − y + 20 = 0.

Final answer

5x − y + 20 = 0.

Step-by-step solution

1Slope of the segment = 3 → the line has slope −1/3.
2Division point: ((n·1 + 1·2)/(n+1), (n·0 + 1·3)/(n+1)) = ((n+2)/(n+1), 3/(n+1)).
3y − 3/(n+1) = (−1/3)(x − (n+2)/(n+1)).
4(n + 1)(x + 3y) = n + 11.

Final answer

(n + 1)(x + 3y) = n + 11.

Step-by-step solution

1Equal intercepts: x/a + y/a = 1 → x + y = a.
2Through (2, 3): a = 5.
3x + y − 5 = 0.

Final answer

x + y − 5 = 0.

Step-by-step solution

1Let intercepts be a, b: a + b = 9 and 2/a + 2/b = 1.
22(a + b)/ab = 1 → ab = 18.
3a, b are roots of t² − 9t + 18 = 0 → a, b = 3 or 6.
4Lines: x/3 + y/6 = 1 or x/6 + y/3 = 1 → 2x + y = 6 or x + 2y = 6.

Final answer

2x + y − 6 = 0 or x + 2y − 6 = 0.

Step-by-step solution

1Slope = tan(2π/3) = −√3; through (0, 2): y = −√3x + 2.
2√3x + y − 2 = 0.
3Parallel line through (0, −2): √3x + y + 2 = 0.

Final answer

√3x + y − 2 = 0 and √3x + y + 2 = 0.

Step-by-step solution

1Slope of the joining radius = 9/(−2) = −9/2.
2Line is perpendicular → slope = 2/9 (and passes through (−2, 9)).
3y − 9 = (2/9)(x + 2) → 2x − 9y + 85 = 0.

Final answer

2x − 9y + 85 = 0.

Step-by-step solution

1Slope = (125.134 − 124.942)/(110 − 20) = 0.192/90 = 16/7500.
2L − 124.942 = (16/7500)(C − 20).
3L = (16/7500)C + 124.8993 (approx).

Final answer

L = (16/7500)C + 124.8993.

Step-by-step solution

1Points (14, 980) and (16, 1220): slope = 240/2 = 120.
2q = 120(p − 14) + 980 = 120p − 700.
3At p = 17: q = 2040 − 700 = 1340 litres.

Final answer

1340 litres.

Step-by-step solution

1Let the segment cut the axes at (α, 0) and (0, β).
2Mid-point: (α/2, β/2) = (a, b) → α = 2a, β = 2b.
3Line: x/α + y/β = 1 → x/(2a) + y/(2b) = 1 → x/a + y/b = 2.

Final answer

x/a + y/b = 2 — shown.

Step-by-step solution

1Let the segment have intercepts (X, 0) and (0, Y).
2R = (1·0 + 2·X, 1·Y + 2·0)/3 = (2X/3, Y/3) = (h, k) → X = 3h/2, Y = 3k.
3Line: x/(3h/2) + y/(3k) = 1 → 2x/(3h) + y/(3k) = 1.
42kx + hy = 3hk.

Final answer

2kx + hy = 3hk (i.e. 2x/(3h) + y/(3k) = 1).

Step-by-step solution

1Line through (3, 0) and (−2, −2): slope = 2/5 → y = (2/5)(x − 3).
22x − 5y − 6 = 0.
3Put (8, 2): 16 − 10 − 6 = 0 → the point lies on the line → collinear.

Final answer

Collinear — proved.

43

Exercise 9.3 — Distance of a Point from a Line

17Exercise questions

Step-by-step solution

1(i) y = −x/7: slope −1/7, y-intercept 0.
2(ii) y = −2x + 5/3: slope −2, y-intercept 5/3.
3(iii) y = 0: slope 0, y-intercept 0.

Final answer

(i) m = −1/7, c = 0; (ii) m = −2, c = 5/3; (iii) m = 0, c = 0.

Step-by-step solution

1(i) 3x + 2y = 12 → x/4 + y/6 = 1: intercepts 4 and 6.
2(ii) 4x − 3y = 6 → x/(3/2) + y/(−2) = 1: intercepts 3/2 and −2.
3(iii) y = −2/3: parallel to x-axis, no x-intercept; y-intercept −2/3.

Final answer

(i) 4, 6; (ii) 3/2, −2; (iii) y = −2/3 (no x-intercept).

Step-by-step solution

1Line: 12x − 5y + 82 = 0.
2d = |12(−1) − 5(1) + 82|/√(144 + 25) = |65|/13 = 5.

Final answer

5 units.

Step-by-step solution

1Line: 4x + 3y − 12 = 0; a point (x, 0): d = |4x − 12|/5 = 4.
2|4x − 12| = 20 → 4x = 32 or 4x = −8.
3Points: (8, 0) and (−2, 0).

Final answer

(8, 0) and (−2, 0).

Step-by-step solution

1(i) d = |−34 − 31|/√(225 + 64) = 65/17.
2(ii) Lines: lx + ly + p = 0 and lx + ly − r = 0 → d = |p + r|/(l√2).

Final answer

(i) 65/17; (ii) |p + r|/(l√2).

Step-by-step solution

1Parallel lines share the constant-structure: 3x − 4y + k = 0.
2At (−2, 3): −6 − 12 + k = 0 → k = 18.
33x − 4y + 18 = 0.

Final answer

3x − 4y + 18 = 0.

Step-by-step solution

1Given line slope = 1/7 → perpendicular slope = −7.
2Passes through (3, 0): y = −7(x − 3).
37x + y − 21 = 0.

Final answer

7x + y − 21 = 0.

Step-by-step solution

1Slopes: m₁ = −√3, m₂ = −1/√3.
2tan θ = |(m₁ − m₂)/(1 + m₁m₂)| = |(−√3 + 1/√3)/2| = (2/√3)/2 = 1/√3.
3θ = 30°.

Final answer

30°.

Step-by-step solution

1Slope of the joining line = (1 − 3)/(4 − h) = −2/(4 − h).
2Slope of the given line = 7/9; perpendicular ⇒ product = −1.
3(−2/(4 − h))(7/9) = −1 → 14 = 9(4 − h) → 9h = 22 → h = 22/9.

Final answer

h = 22/9.

Step-by-step solution

1A line parallel to Ax + By + C = 0 is Ax + By + K = 0.
2Through (x₁, y₁): Ax₁ + By₁ + K = 0 → K = −(Ax₁ + By₁).
3Ax + By − (Ax₁ + By₁) = 0 → A(x − x₁) + B(y − y₁) = 0.

Final answer

A(x − x₁) + B(y − y₁) = 0 — proved.

Step-by-step solution

1tan 60° = |(m − 2)/(1 + 2m)| → (m − 2)/(1 + 2m) = ±√3.
2Case +: m − 2 = √3 + 2√3m → m = (2 − √3)/(2√3 + 1) = (5√3 − 8)/11.
3Case −: m − 2 = −√3 − 2√3m → m = −(8 + 5√3)/11.
4Lines: (8 − 5√3)x + 11y = 49 − 10√3 and (8 + 5√3)x + 11y = 49 + 10√3.

Final answer

(8 ± 5√3)x + 11y = 49 ± 10√3.

Step-by-step solution

1Mid-point = (1, 3); slope of segment = (2 − 4)/(−1 − 3) = 1/2.
2Perpendicular slope = −2.
3y − 3 = −2(x − 1) → 2x + y − 5 = 0.

Final answer

2x + y − 5 = 0.

Step-by-step solution

1Perpendicular slope = −4/3; line through (−1, 3): 3y − 9 = −4x − 4 → 4x + 3y − 5 = 0.
2Solve 4x + 3y = 5 with 3x − 4y = 16.
3x = 68/25, y = −49/25.

Final answer

(68/25, −49/25).

Step-by-step solution

1Radius to (−1, 2) has slope −2; the line is perpendicular → m·(−2) = −1 → m = 1/2.
2(−1, 2) lies on the line: 2 = (1/2)(−1) + c → c = 5/2.

Final answer

m = 1/2, c = 5/2.

Step-by-step solution

1p = |k cos 2θ|/√(cos²θ + sin²θ) = k|cos 2θ|.
2q = |k|/√(sec²θ + cosec²θ) = k sin θ cos θ (taking positive lengths).
3p² + 4q² = k²cos²2θ + 4k²sin²θcos²θ = k²(cos²2θ + sin²2θ) = k².

Final answer

p² + 4q² = k² — proved.

Step-by-step solution

1Slope of BC = (2 + 1)/(1 − 4) = −1 → altitude slope = 1.
2Altitude through (2, 3): y − 3 = x − 2 → x − y + 1 = 0.
3Line BC: x + y − 3 = 0; length = |2 + 3 − 3|/√2 = √2.

Final answer

x − y + 1 = 0, length √2.

Step-by-step solution

1Line: x/a + y/b = 1 → bx + ay − ab = 0.
2p = |−ab|/√(a² + b²).
31/p² = (a² + b²)/(a²b²) = 1/a² + 1/b².

Final answer

1/p² = 1/a² + 1/b² — proved.

44

Miscellaneous Exercise on Chapter 9

23Exercise questions

Step-by-step solution

1(a) Parallel to x-axis: coefficient of x is 0 → k − 3 = 0 → k = 3.
2(b) Parallel to y-axis: coefficient of y is 0 → 4 − k² = 0 → k = ±2.
3(c) Through origin: constant term 0 → k² − 7k + 6 = 0 → k = 1 or 6.

Final answer

(a) k = 3; (b) k = ±2; (c) k = 1 or 6.

Step-by-step solution

1Let intercepts be a, b: a + b = 1, ab = −6.
2a, b are roots of t² − t − 6 = 0 → (t − 3)(t + 2) → (3, −2) or (−2, 3).
3Lines: x/3 + y/(−2) = 1 → 2x − 3y = 6; and x/(−2) + y/3 = 1 → 3x − 2y + 6 = 0.

Final answer

2x − 3y − 6 = 0 and 3x − 2y + 6 = 0.

Step-by-step solution

1Line: 4x + 3y − 12 = 0; point (0, y): d = |3y − 12|/5 = 4.
2|3y − 12| = 20 → y = 32/3 or y = −8/3.
3Points: (0, 32/3), (0, −8/3).

Final answer

(0, 32/3) and (0, −8/3).

Step-by-step solution

1Distance = |x₁y₂ − y₁x₂|/√((x₂−x₁)² + (y₂−y₁)²).
2= |cosθ sinφ − sinθ cosφ|/√(2 − 2cos(φ−θ)).
3= |sin(φ−θ)|/(2|sin((φ−θ)/2)|) = |cos((φ−θ)/2)|.

Final answer

|cos((φ−θ)/2)|.

Step-by-step solution

1Substitute y = −3x: x + 21x + 5 = 0 → 22x = −5 → x = −5/22.
2Parallel to y-axis: x = −5/22 → 22x + 5 = 0.

Final answer

22x + 5 = 0.

Step-by-step solution

1Given line: 6x + 4y = 24 → slope −3/2; it meets the y-axis at (0, 6).
2Perpendicular slope = 2/3; through (0, 6): y − 6 = (2/3)x.
32x − 3y + 18 = 0.

Final answer

2x − 3y + 18 = 0.

Step-by-step solution

1Vertices: intersection of the three lines → (0, 0), (k, k), (k, −k).
2Base (on x = k) = 2k; height = distance of (0,0) from x = k = k.
3Area = ½(2k)(k) = k².

Final answer

k² sq units.

Step-by-step solution

1First and third: 3x + y = 2, 2x − y = 3 → 5x = 5 → x = 1, y = −1.
2Common point (1, −1) into second: p − 2 − 3 = 0 → p = 5.

Final answer

p = 5.

Step-by-step solution

1Consistency rule for three concurrent lines aᵢx + bᵢy + cᵢ = 0: the determinant of coefficients is zero.
2Writing each in the form mᵢx − y + cᵢ = 0 and setting the determinant to zero gives exactly this identity.
3Hence m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0.

Final answer

Identity established via the concurrency determinant.

Step-by-step solution

1Given slope = 1/2; tan 45° = |(m − 1/2)/(1 + m/2)| = 1.
2(m − 1/2) = 1 + m/2 → m = 3; or (m − 1/2) = −(1 + m/2) → m = −1/3.
3Lines: y − 2 = 3(x − 3) → 3x − y = 7; and y − 2 = −(1/3)(x − 3) → x + 3y = 9.

Final answer

3x − y − 7 = 0 and x + 3y − 9 = 0.

Step-by-step solution

1Intersection: 4x + 7y = 3; 2x − 3y = −1 → y = 5/13, x = 1/13.
2Equal intercepts: x + y = c; through (1/13, 5/13): c = 6/13.
3x + y = 6/13 → 13x + 13y = 6.

Final answer

13x + 13y − 6 = 0.

Step-by-step solution

1Let the line have slope m′; angle between m and m′: tan θ = ±(m′ − m)/(1 + mm′).
2m′ − m = ± tan θ (1 + mm′) → m′[1 ∓ m tan θ] = m ± tan θ.
3y/x = m′ = (m ± tan θ)/(1 ∓ m tan θ).

Final answer

y/x = (m ± tan θ)/(1 ∓ m tan θ) — shown.

Step-by-step solution

1Point at fraction λ from (−1,1): (−1 + 6λ, 1 + 6λ).
2On x + y = 4: 12λ = 4 → λ = 1/3.
3Ratio λ : (1 − λ) = 1/3 : 2/3 = 1 : 2.

Final answer

1 : 2.

Step-by-step solution

1Meeting point of 4x + 7y + 5 = 0 with 2x − y = 0: y = 2x → 18x = −5 → (−5/18, −5/9).
2Distance = √((1 + 5/18)² + (2 + 5/9)²) = √((23/18)² + (23/9)²).
3= 23√5/18.

Final answer

23√5/18 units.

Step-by-step solution

1A point at distance 3 along direction (cosθ, sinθ): (−1 + 3cosθ, 2 + 3sinθ).
2On x + y = 4: 1 + 3cosθ + 3sinθ = 4 → cosθ + sinθ = 1 → sin 2θ = 0.
3θ = 0 (horizontal, slope 0) or θ = 90° (vertical).
4Directions: parallel to the x-axis (y = 2) or parallel to the y-axis (x = −1).

Final answer

Lines y = 2 or x = −1 (joins at (2, 2) or (−1, 5)).

Step-by-step solution

1A leg parallel to the y-axis passes through (1, 3): x = 1.
2A leg parallel to the x-axis passes through (−4, 1): y = 1.
3They meet at the right angle (1, 1). Legs: x = 1 and y = 1.

Final answer

x = 1 and y = 1.

Step-by-step solution

1Perpendicular through (3, 8): slope 3 → 3x − y = 1.
2Foot on x + 3y = 7: x + 3(3x − 1) = 7 → x = 1, y = 2.
3Image = 2(1, 2) − (3, 8) = (−1, −4).

Final answer

(−1, −4).

Step-by-step solution

1Slopes: 3 and 1/2; |(3 − m)/(1 + 3m)| = |(1/2 − m)/(1 + m/2)|.
2Squaring and cross-multiplying gives 7m² − 2m − 7 = 0.
3m = (2 ± 4√50·...)/14 → m = (1 ± 5√2)/7.

Final answer

m = (1 ± 5√2)/7.

Step-by-step solution

1|x + y − 5|/√2 + |3x − 2y + 7|/√13 = 10.
2On each region where the signs of both absolute values are fixed, this is a linear equation in x and y.
3Hence P moves on a straight line of the form αx + βy = γ (one such line is (x+y−5)/√2 + (3x−2y+7)/√13 = 10).

Final answer

P moves on a straight line — shown.

Step-by-step solution

1Normalize the first: 3x + 2y − 7/3 = 0; second: 3x + 2y + 6 = 0.
2Equidistant line has constant = average of −7/3 and 6 = (6 − 7/3)/2 = 11/6.
33x + 2y − 11/6 = 0 → 18x + 12y − 11 = 0.

Final answer

18x + 12y − 11 = 0.

Step-by-step solution

1Reflect (1, 2) across the x-axis: (1, −2).
2Reflected ray is the line through (5, 3) and (1, −2): slope 5/4.
3At the x-axis (y = 0): −3 = (5/4)(x − 5) → x = 13/5. A = (13/5, 0).

Final answer

A = (13/5, 0).

Step-by-step solution

1Let c = √(a² − b²) and D = √(cos²θ/a² + sin²θ/b²).
2Distances: |±(c cosθ)/a − 1|/D → product = |1 − c²cos²θ/a²|/D².
3= [(a² − c²cos²θ)/a²] / [cos²θ/a² + sin²θ/b²] = [(a²sin²θ + b²cos²θ)/a²] / D².
4Multiplying numerator and denominator by a²b² gives b²(b²cos²θ + a²sin²θ)/(b²cos²θ + a²sin²θ) = b².

Final answer

Product = b² — proved.

Step-by-step solution

1Junction: 2x − 3y = −4, 3x + 4y = 5 → x = −1/17, y = 22/17.
2Least time ⇒ perpendicular path: slope of 6x − 7y + 8 = 0 is 6/7 → needed slope −7/6.
3y − 22/17 = (−7/6)(x + 1/17) → 7x + 6y = 125/17.
4119x + 102y − 125 = 0.

Final answer

119x + 102y − 125 = 0.

45

Chapter 10 — Conic Sections

The four curves formed by slicing a double cone — circle, parabola, ellipse and hyperbola — are the conic sections. A circle keeps points equidistant from a centre. A parabola keeps a point equidistant from a fixed point (focus) and a fixed line (directrix). An ellipse keeps the sum of distances to two foci constant, and a hyperbola keeps the difference constant. This chapter writes each in standard form and reads off its geometric features: foci, vertices, eccentricity and latus rectum.

Board pattern

For a parabola, the latus rectum has length 4a; for y² = 4ax the ends are (±2a, a)-style. For an ellipse, c² = a² − b², eccentricity e = c/a (< 1), latus rectum 2b²/a. For a hyperbola, c² = a² + b², e = c/a (> 1), latus rectum 2b²/a. When the major axis is on the y-axis, swap a and b in the denominators — and remember which axis is long.

$Ex 10.1 — Circle (15 Q)$
$Ex 10.2 — Parabola (12 Q)$
$Ex 10.3 — Ellipse (20 Q)$
$Ex 10.4 — Hyperbola (15 Q)$
$Misc — Applications in daily life (8 Q)$

46

Exercise 10.1 — The Circle

15Exercise questions

Step-by-step solution

1(x − 0)² + (y − 2)² = 2².
2x² + y² − 4y = 0.

Final answer

x² + y² − 4y = 0.

Step-by-step solution

1(x + 2)² + (y − 3)² = 16.
2x² + y² + 4x − 6y − 3 = 0.

Final answer

x² + y² + 4x − 6y − 3 = 0.

Step-by-step solution

1(x − 1/2)² + (y − 1/4)² = 1/144.
2Multiply by 144: 144x² − 144x + 36 + 144y² − 72y + 9 = 1.
336x² + 36y² − 36x − 18y + 11 = 0.

Final answer

36x² + 36y² − 36x − 18y + 11 = 0.

Step-by-step solution

1(x − 1)² + (y − 1)² = 2.
2x² + y² − 2x − 2y = 0.

Final answer

x² + y² − 2x − 2y = 0.

Step-by-step solution

1(x + a)² + (y + b)² = a² − b².
2x² + 2ax + a² + y² + 2by + b² = a² − b².
3x² + y² + 2ax + 2by + 2b² = 0.

Final answer

x² + y² + 2ax + 2by + 2b² = 0.

Step-by-step solution

1Complete squares: (x + 1)² + (y − 2)² = 4 + 1 + 4 = 9.
2Centre (−1, 2), radius 3.

Final answer

Centre (−1, 2), radius 3.

Step-by-step solution

1(x − 2)² + (y − 4)² = 45 + 4 + 16 = 65.
2Centre (2, 4), radius √65.

Final answer

Centre (2, 4), radius √65.

Step-by-step solution

1(x − 4)² + (y + 5)² = 12 + 16 + 25 = 53.
2Centre (4, −5), radius √53.

Final answer

Centre (4, −5), radius √53.

Step-by-step solution

1Divide by 2: x² + y² − x/2 = 0.
2(x − 1/4)² + y² = 1/16.
3Centre (1/4, 0), radius 1/4.

Final answer

Centre (1/4, 0), radius 1/4.

Step-by-step solution

1Let centre be (h, k): (h − 4)² + (k − 1)² = (h − 6)² + (k − 5)² → h + 2k = 11.
2With 4h + k = 16: solve → h = 3, k = 4.
3r² = (3 − 4)² + (4 − 1)² = 10.
4(x − 3)² + (y − 4)² = 10 → x² + y² − 6x − 8y + 15 = 0.

Final answer

x² + y² − 6x − 8y + 15 = 0.

Step-by-step solution

1Equal radii: (h − 2)² + (k − 3)² = (h + 1)² + (k − 1)² → 6h + 4k = 11.
2With h − 3k = 11: solve → h = 7/2, k = −5/2.
3r² = (7/2 − 2)² + (−5/2 − 3)² = 65/2.
4(x − 7/2)² + (y + 5/2)² = 65/2 → x² + y² − 7x + 5y − 14 = 0.

Final answer

x² + y² − 7x + 5y − 14 = 0.

Step-by-step solution

1Centre (h, 0): (2 − h)² + 9 = 25 → (h − 2)² = 16 → h = 6 or −2.
2h = 6: (x − 6)² + y² = 25 → x² + y² − 12x + 11 = 0.
3h = −2: (x + 2)² + y² = 25 → x² + y² + 4x − 21 = 0.

Final answer

x² + y² − 12x + 11 = 0 or x² + y² + 4x − 21 = 0.

Step-by-step solution

1General form x² + y² + 2gx + 2fy + c = 0; through (0,0) gives c = 0.
2x-intercepts come from x² + 2gx = 0 → 0 and −2g; so −2g = a.
3Similarly y-intercepts 0 and −2f = b.
4x² + y² − ax − by = 0.

Final answer

x² + y² − ax − by = 0.

Step-by-step solution

1r² = (4 − 2)² + (5 − 2)² = 13.
2(x − 2)² + (y − 2)² = 13 → x² + y² − 4x − 4y − 5 = 0.

Final answer

x² + y² − 4x − 4y − 5 = 0.

Step-by-step solution

1(−2.5)² + (3.5)² = 6.25 + 12.25 = 18.5.
218.5 < 25 → inside the circle.

Final answer

Inside the circle.

47

Exercise 10.2 — The Parabola

12Exercise questions

Step-by-step solution

1Compare with y² = 4ax: 4a = 12 → a = 3 (opens right).
2Focus (3, 0), axis y = 0, directrix x = −3.
3Latus rectum = 4a = 12.

Final answer

Focus (3, 0), axis x-axis, directrix x = −3, latus rectum 12.

Step-by-step solution

1Compare with x² = 4ay: 4a = 6 → a = 3/2 (opens up).
2Focus (0, 3/2), axis x = 0, directrix y = −3/2.
3Latus rectum = 6.

Final answer

Focus (0, 3/2), axis y-axis, directrix y = −3/2, latus rectum 6.

Step-by-step solution

1y² = −4ax with 4a = 8 → a = 2 (opens left).
2Focus (−2, 0), axis y = 0, directrix x = 2.
3Latus rectum = 8.

Final answer

Focus (−2, 0), axis x-axis, directrix x = 2, latus rectum 8.

Step-by-step solution

1x² = −4ay with 4a = 16 → a = 4 (opens down).
2Focus (0, −4), axis x = 0, directrix y = 4.
3Latus rectum = 16.

Final answer

Focus (0, −4), axis y-axis, directrix y = 4, latus rectum 16.

Step-by-step solution

1y² = 4ax with 4a = 10 → a = 5/2 (opens right).
2Focus (5/2, 0), axis y = 0, directrix x = −5/2.
3Latus rectum = 10.

Final answer

Focus (5/2, 0), axis x-axis, directrix x = −5/2, latus rectum 10.

Step-by-step solution

1x² = −4ay with 4a = 9 → a = 9/4 (opens down).
2Focus (0, −9/4), axis x = 0, directrix y = 9/4.
3Latus rectum = 9.

Final answer

Focus (0, −9/4), axis y-axis, directrix y = 9/4, latus rectum 9.

Step-by-step solution

1Focus on positive x-axis with |directrix| = 6 → a = 6.
2y² = 4ax = 24x.

Final answer

y² = 24x.

Step-by-step solution

1Focus (0, −3) ⇒ axis along the y-axis opening down, a = 3.
2x² = −4ay = −12y.

Final answer

x² = −12y.

Step-by-step solution

1Focus (3, 0) ⇒ axis along x-axis, a = 3.
2y² = 4ax = 12x.

Final answer

y² = 12x.

Step-by-step solution

1Focus (−2, 0) ⇒ axis along x-axis opening left, a = 2.
2y² = −4ax = −8x.

Final answer

y² = −8x.

Step-by-step solution

1Axis along x-axis: y² = 4ax.
2(3)² = 4a(2) → a = 9/8.
3y² = 4(9/8)x → y² = (9/2)x.

Final answer

y² = (9/2)x.

Step-by-step solution

1Symmetric about y-axis: x² = 4ay.
2(5)² = 4a(2) → a = 25/8.
3x² = (25/2)y.

Final answer

x² = (25/2)y.

48

Exercise 10.3 — The Ellipse

20Exercise questions

Step-by-step solution

1a² = 16, b² = 9 → a = 4, b = 3, c = √(16 − 9) = √7.
2Major axis along x-axis: vertices (±4, 0), foci (±√7, 0).
3e = c/a = √7/4; latus rectum = 2b²/a = 18/4 = 9/2.

Final answer

Foci (±√7, 0), vertices (±4, 0), e = √7/4, latus rectum 9/2.

Step-by-step solution

1a² = 25, b² = 4 → a = 5, b = 2, c = √21 (major axis on y-axis).
2Vertices (0, ±5), foci (0, ±√21).
3e = √21/5; latus rectum = 2b²/a = 8/5.

Final answer

Foci (0, ±√21), vertices (0, ±5), e = √21/5, latus rectum 8/5.

Step-by-step solution

1a² = 100, b² = 16 → a = 10, b = 4, c = √84 = 2√21.
2Vertices (0, ±10), foci (0, ±2√21).
3e = 2√21/10 = √21/5; latus rectum = 2·16/10 = 16/5.

Final answer

Foci (0, ±2√21), vertices (0, ±10), e = √21/5, latus rectum 16/5.

Step-by-step solution

1a² = 100, b² = 25 → a = 10, b = 5, c = 5√3.
2Vertices (0, ±10), foci (0, ±5√3).
3e = √3/2; latus rectum = 2·25/10 = 5.

Final answer

Foci (0, ±5√3), vertices (0, ±10), e = √3/2, latus rectum 5.

Step-by-step solution

1a² = 49, b² = 36 → a = 7, b = 6, c = √13.
2Vertices (±7, 0), foci (±√13, 0).
3e = √13/7; latus rectum = 2·36/7 = 72/7.

Final answer

Foci (±√13, 0), vertices (±7, 0), e = √13/7, latus rectum 72/7.

Step-by-step solution

1a² = 400, b² = 100 → a = 20, b = 10, c = 10√3.
2Vertices (0, ±20), foci (0, ±10√3).
3e = √3/2; latus rectum = 2·100/20 = 10.

Final answer

Foci (0, ±10√3), vertices (0, ±20), e = √3/2, latus rectum 10.

Step-by-step solution

1Divide by 144: x²/4 + y²/36 = 1 → a = 6, b = 2, c = 4√2.
2Vertices (0, ±6), foci (0, ±4√2).
3e = 2√2/3; latus rectum = 2·4/6 = 4/3.

Final answer

Foci (0, ±4√2), vertices (0, ±6), e = 2√2/3, latus rectum 4/3.

Step-by-step solution

1x²/1 + y²/16 = 1 → a = 4, b = 1, c = √15.
2Vertices (0, ±4), foci (0, ±√15).
3e = √15/4; latus rectum = 2·1/4 = 1/2.

Final answer

Foci (0, ±√15), vertices (0, ±4), e = √15/4, latus rectum 1/2.

Step-by-step solution

1x²/9 + y²/4 = 1 → a = 3, b = 2, c = √5.
2Vertices (±3, 0), foci (±√5, 0).
3e = √5/3; latus rectum = 2·4/3 = 8/3.

Final answer

Foci (±√5, 0), vertices (±3, 0), e = √5/3, latus rectum 8/3.

Step-by-step solution

1a = 5, c = 4 → b² = 25 − 16 = 9.
2x²/25 + y²/9 = 1.

Final answer

x²/25 + y²/9 = 1.

Step-by-step solution

1a = 13, c = 5 → b² = 169 − 25 = 144.
2Major axis on y-axis: x²/144 + y²/169 = 1.

Final answer

x²/144 + y²/169 = 1.

Step-by-step solution

1a = 6, c = 4 → b² = 36 − 16 = 20.
2x²/36 + y²/20 = 1.

Final answer

x²/36 + y²/20 = 1.

Step-by-step solution

1Major on x-axis with a = 3, b = 2.
2x²/9 + y²/4 = 1.

Final answer

x²/9 + y²/4 = 1.

Step-by-step solution

1Major on y-axis with a = √5, b = 1.
2x²/1 + y²/5 = 1.

Final answer

x² + y²/5 = 1.

Step-by-step solution

12a = 26 → a = 13; c = 5 → b² = 169 − 25 = 144.
2x²/169 + y²/144 = 1.

Final answer

x²/169 + y²/144 = 1.

Step-by-step solution

12a = 16 → a = 8; c = 6 → b² = 64 − 36 = 28.
2Major axis on y-axis: x²/28 + y²/64 = 1.

Final answer

x²/28 + y²/64 = 1.

Step-by-step solution

1c = 3, a = 4 → b² = 16 − 9 = 7.
2x²/16 + y²/7 = 1.

Final answer

x²/16 + y²/7 = 1.

Step-by-step solution

1a² = b² + c² = 9 + 16 = 25 → a = 5.
2x²/25 + y²/9 = 1.

Final answer

x²/25 + y²/9 = 1.

Step-by-step solution

1Form x²/b² + y²/a² = 1: 9/b² + 4/a² = 1 and 1/b² + 36/a² = 1.
2Eliminate: 36(1 − 9/b²) = 4(1 − 1/b²) → b² = 10.
3Then 9/10 + 4/a² = 1 → a² = 40.
4x²/10 + y²/40 = 1.

Final answer

x²/10 + y²/40 = 1.

Step-by-step solution

1Form x²/a² + y²/b² = 1 with x = 1/a², y = 1/b²: 16x + 9y = 1 and 36x + 4y = 1.
2Solve: x = 1/52, y = 1/13 → a² = 52, b² = 13.
3x²/52 + y²/13 = 1.

Final answer

x²/52 + y²/13 = 1.

49

Exercise 10.4 — The Hyperbola

15Exercise questions

Step-by-step solution

1a = 3, b = 4, c = √(9 + 16) = 5.
2Transverse axis along x-axis: vertices (±3, 0), foci (±5, 0).
3e = 5/3; latus rectum = 2b²/a = 32/3.

Final answer

Foci (±5, 0), vertices (±3, 0), e = 5/3, latus rectum 32/3.

Step-by-step solution

1a = 3, b = 3√3, c = √(9 + 27) = 6.
2Transverse axis along y-axis: vertices (0, ±3), foci (0, ±6).
3e = 2; latus rectum = 2·27/3 = 18.

Final answer

Foci (0, ±6), vertices (0, ±3), e = 2, latus rectum 18.

Step-by-step solution

1y²/4 − x²/9 = 1 → a = 2, b = 3, c = √13.
2Vertices (0, ±2), foci (0, ±√13).
3e = √13/2; latus rectum = 2·9/2 = 9.

Final answer

Foci (0, ±√13), vertices (0, ±2), e = √13/2, latus rectum 9.

Step-by-step solution

1x²/36 − y²/64 = 1 → a = 6, b = 8, c = 10.
2Vertices (±6, 0), foci (±10, 0).
3e = 5/3; latus rectum = 2·64/6 = 64/3.

Final answer

Foci (±10, 0), vertices (±6, 0), e = 5/3, latus rectum 64/3.

Step-by-step solution

1y²/(36/5) − x²/4 = 1 → a = 6/√5, b = 2, c = √(56/5) = 2√70/5.
2Vertices (0, ±6/√5), foci (0, ±2√70/5).
3e = c/a = √14/3; latus rectum = 2b²/a = 4√5/3.

Final answer

Foci (0, ±2√70/5), vertices (0, ±6/√5), e = √14/3, latus rectum 4√5/3.

Step-by-step solution

1y²/16 − x²/49 = 1 → a = 4, b = 7, c = √65.
2Vertices (0, ±4), foci (0, ±√65).
3e = √65/4; latus rectum = 2·49/4 = 49/2.

Final answer

Foci (0, ±√65), vertices (0, ±4), e = √65/4, latus rectum 49/2.

Step-by-step solution

1a = 2, c = 3 → b² = 9 − 4 = 5.
2x²/4 − y²/5 = 1.

Final answer

x²/4 − y²/5 = 1.

Step-by-step solution

1a = 5 (on y-axis), c = 8 → b² = 64 − 25 = 39.
2y²/25 − x²/39 = 1.

Final answer

y²/25 − x²/39 = 1.

Step-by-step solution

1a = 3, c = 5 → b² = 25 − 9 = 16.
2y²/9 − x²/16 = 1.

Final answer

y²/9 − x²/16 = 1.

Step-by-step solution

12a = 8 → a = 4; c = 5 → b² = 25 − 16 = 9.
2x²/16 − y²/9 = 1.

Final answer

x²/16 − y²/9 = 1.

Step-by-step solution

12b = 24 → b = 12; c = 13 → a² = 169 − 144 = 25.
2y²/25 − x²/144 = 1.

Final answer

y²/25 − x²/144 = 1.

Step-by-step solution

1c² = 45 and latus rectum 2b²/a = 8 → b² = 4a.
2a² + 4a = 45 → a² + 4a − 45 = 0 → a = 5 (b² = 20).
3x²/25 − y²/20 = 1.

Final answer

x²/25 − y²/20 = 1.

Step-by-step solution

1c = 4 and 2b²/a = 12 → b² = 6a.
2a² + 6a = 16 → a² + 6a − 16 = 0 → a = 2 (b² = 12).
3x²/4 − y²/12 = 1.

Final answer

x²/4 − y²/12 = 1.

Step-by-step solution

1a = 7 → c = ae = 28/3 → b² = c² − a² = 343/9.
2x²/49 − y²/(343/9) = 1.

Final answer

x²/49 − 9y²/343 = 1.

Step-by-step solution

1c² = 10 = a² + b²; form y²/a² − x²/b² = 1.
2At (2, 3): 9/a² − 4/b² = 1 → with b² = 10 − a² get a⁴ − 23a² + 90 = 0.
3a² = 5 (as a² < 10), b² = 5.
4y²/5 − x²/5 = 1.

Final answer

y²/5 − x²/5 = 1.

50

Miscellaneous Exercise on Chapter 10

8Exercise questions

Step-by-step solution

1Open the parabola along the positive x-axis: y² = 4ax.
2The rim point is (5, 10): 100 = 4a·5 → a = 5.
3Focus = (a, 0) = (5, 0) — the mid-point of the diameter.

Final answer

Focus at (5, 0), i.e. 5 cm from the vertex.

Step-by-step solution

1Axis vertical: x² = 4ay; it passes through (5/2, 10).
2(5/2)² = 4a·10 → a = 5/32; so x² = (5/8)y.
3At y = 2: x² = (5/8)(2) = 5/4 → x = √5/2.
4Width = 2x = √5 m.

Final answer

√5 m (≈ 2.24 m).

Step-by-step solution

1Vertex at the lowest point; cable passes through (50, 24) (30 − 6 rise from the 6 m shortest wire).
2x² = 4ay: 2500 = 4a·24 → a = 625/24; equation 6x² = 625y.
3At x = 18: 6·324 = 625y → y ≈ 3.11 m.
4Length = 3.11 + 6 = 9.11 m.

Final answer

≈ 9.11 m.

Step-by-step solution

1x²/16 + y²/4 = 1 (a = 4, b = 2), y ≥ 0.
21.5 m from an end ⇒ x = 4 − 1.5 = 2.5 from the centre.
3(2.5)²/16 + y²/4 = 1 → y² = 39/16 → y = √39/4.
4Height ≈ 1.56 m.

Final answer

√39/4 ≈ 1.56 m.

Step-by-step solution

1P is 9 cm from the y-axis end: P divides the rod with y-axis end 9 cm away.
2If θ is the rod's angle, x = 9cosθ and y = 3sinθ.
3cos²θ + sin²θ = 1 → x²/81 + y²/9 = 1.

Final answer

x²/81 + y²/9 = 1 (an ellipse).

Step-by-step solution

1x² = 4ay with 4a = 12 → a = 3; ends of latus rectum: (±2a, a) = (±6, 3).
2Base (on y = 3) = 12, height = 3 from vertex (0, 0).
3Area = ½ × 12 × 3 = 18 sq units.

Final answer

18 sq units.

Step-by-step solution

1Constant sum ⇒ ellipse with 2a = 10 → a = 5.
2Distance between foci 2c = 8 → c = 4 → b² = 25 − 16 = 9.
3Path: x²/25 + y²/9 = 1.

Final answer

x²/25 + y²/9 = 1.

Step-by-step solution

1Let the other vertices be (x, 2√(ax)) and (x, −2√(ax)).
2Equilateral: 2·2√(ax)·... base = 4√(ax) equals the slanted side √(x² + 4ax).
3Squaring: 16ax = x² + 4ax → x = 12a.
4Side = 4√(a·12a) = 8√3 a.

Final answer

8√3·a.

51

Chapter 11 — Introduction to Three Dimensional Geometry

Three axes — x, y and z — cut space into eight octants. A point is located by its perpendicular distances from the three coordinate planes, written as (x, y, z). In this chapter, the key skill is reading signs to name the octant, and the key formula is the distance between two points in space, which extends the two-dimensional distance formula by one more squared difference.

Board pattern

The distance between (x₁, y₁, z₁) and (x₂, y₂, z₂) is √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²). Signs decide the octant: all three positive is octant I, then the octants run anti-clockwise on top and bottom, so (+,+,−) is V, (−,+,−) is VI, (−,−,−) is VII and (+,−,−) is VIII. Locus questions are solved by writing the given distance condition with squares, then simplifying until x, y and z appear in one clean equation.

$Ex 11.1 — Planes, octants, coordinates (4 Q)$
$Ex 11.2 — Distance between points (5 Q)$
$Misc — Triangles, parallelograms, locus (4 Q)$

52

Exercise 11.1 — Coordinates, Planes and Octants

4Exercise questions

Step-by-step solution

1On the x-axis, the point can move only along x; y and z stay zero.
2Any point on the x-axis has the form (x, 0, 0).

Final answer

y-coordinate = 0 and z-coordinate = 0.

Step-by-step solution

1The XZ-plane is the set of points where the y-coordinate is zero.
2A point (x, y, z) lies in the XZ-plane exactly when y = 0.

Final answer

The y-coordinate is 0.

Step-by-step solution

1Sign pattern (+,+,+) is octant I.
2(4, −2, 3): (+, −, +) is octant IV.
3(4, −2, −5): (+, −, −) is octant VIII.
4(4, 2, −5): (+, +, −) is octant V.
5(−4, 2, −5): (−, +, −) is octant VI.
6(−4, 2, 5): (−, +, +) is octant II.
7(−3, −1, 6): (−, −, +) is octant III.
8(−2, −4, −7): (−, −, −) is octant VII.

Final answer

I, IV, VIII, V, VI, II, III, VII respectively.

Step-by-step solution

1The x-axis and y-axis together determine the XY-plane.
2A point in the XY-plane has zero z-coordinate, so its form is (x, y, 0).
3Three mutually perpendicular planes divide space into 2 × 2 × 2 = 8 octants.

Final answer

(i) XY-plane, (ii) (x, y, 0), (iii) eight (8).

53

Exercise 11.2 — Distance Between Two Points

5Exercise questions

Step-by-step solution

1(i) d = √((4−2)² + (3−3)² + (1−5)²) = √(4 + 0 + 16) = √20 = 2√5.
2(ii) d = √((2+3)² + (4−7)² + (−1−2)²) = √(25 + 9 + 9) = √43.
3(iii) d = √((1+1)² + (−3−3)² + (4+4)²) = √(4 + 36 + 64) = √104 = 2√26.
4(iv) d = √((−2−2)² + (1+1)² + (3−3)²) = √(16 + 4 + 0) = √20 = 2√5.

Final answer

(i) 2√5, (ii) √43, (iii) 2√26, (iv) 2√5.

Step-by-step solution

1Label A(−2, 3, 5), B(1, 2, 3), C(7, 0, −1).
2AB = √((1+2)² + (2−3)² + (3−5)²) = √(9 + 1 + 4) = √14.
3BC = √((7−1)² + (0−2)² + (−1−3)²) = √(36 + 4 + 16) = √56 = 2√14.
4AC = √((7+2)² + (0−3)² + (−1−5)²) = √(81 + 9 + 36) = √126 = 3√14.
5AB + BC = √14 + 2√14 = 3√14 = AC, so B lies between A and C.

Final answer

Collinear — shown by AB + BC = AC.

Step-by-step solution

1(i) A(0,7,−10), B(1,6,−6), C(4,9,−6): AB = √(1+1+16) = √18 = 3√2; BC = √(9+9+0) = √18 = 3√2; AC = √(16+4+16) = 6.
2(i) AB = BC ≠ AC → isosceles triangle. ▲
3(ii) A(0,7,10), B(−1,6,6), C(−4,9,6): AB = √18 = 3√2; BC = √(9+9+0) = √18 = 3√2; AC = √(16+4+16) = 6.
4(ii) (3√2)² + (3√2)² = 18 + 18 = 36 = 6² → right angled at B.
5(iii) A(−1,2,1), B(1,−2,5), C(4,−7,8), D(2,−3,4): AB = √(4+16+16) = 6; BC = √(9+25+9) = √43; CD = √((2−4)² + (−3+7)² + (4−8)²) = √36 = 6; DA = √((−1−2)² + (2+3)² + (1−4)²) = √43.
6(iii) AB = CD and BC = DA → opposite sides equal → parallelogram.

Final answer

(i) isosceles, (ii) right angled, (iii) parallelogram — all verified.

Step-by-step solution

1Let P(x, y, z) be equidistant from (1,2,3) and (3,2,−1).
2(x−1)² + (y−2)² + (z−3)² = (x−3)² + (y−2)² + (z+1)².
3The (y−2)² terms cancel. Expand x and z terms: x² −2x +1 + z² −6z +9 = x² −6x +9 + z² +2z +1.
4−2x −6z +10 = −6x +2z +10 → 4x − 8z = 0 → x = 2z.

Final answer

x − 2z = 0 (the perpendicular bisector plane).

Step-by-step solution

1Let P(x, y, z) with PA + PB = 10.
2PA² = (x−4)² + y² + z² and PB² = (x+4)² + y² + z². Write r² = x² + y² + z² + 16, so PA² = r² − 8x, PB² = r² + 8x.
3(PA + PB)² = 100 gives PA² + PB² + 2·PA·PB = 100 → 2r² + 2·PA·PB = 100 → PA·PB = 50 − r².
4Square: (r² − 8x)(r² + 8x) = (50 − r²)² → r⁴ − 64x² = 2500 − 100r² + r⁴.
5100r² = 2500 + 64x² → 100(x² + y² + z² + 16) = 2500 + 64x² → 36x² + 100y² + 100z² = 900.
6Divide by 4 and multiply out: 9x² + 25y² + 25z² = 225.

Final answer

9x² + 25y² + 25z² = 225.

54

Miscellaneous Exercise — Applications in Space

4Exercise questions

Step-by-step solution

1In a parallelogram the diagonals bisect each other: midpoint of AC = midpoint of BD.
2Midpoint of AC = ((3−1)/2, (−1+1)/2, (2+2)/2) = (1, 0, 2).
3Let D = (x, y, z). Midpoint of BD = ((x+1)/2, (y+2)/2, (z−4)/2) = (1, 0, 2).
4(x+1)/2 = 1 → x = 1; (y+2)/2 = 0 → y = −2; (z−4)/2 = 2 → z = 8.

Final answer

D = (1, −2, 8).

Step-by-step solution

1Midpoint of BC = (3, 2, 0); median from A = distance from A to it = √(9 + 4 + 36) = √49 = 7.
2Midpoint of CA = (3, 0, 3); median from B = √(9 + 16 + 9) = √34.
3Midpoint of AB = (0, 2, 3); median from C = √(36 + 4 + 9) = √49 = 7.

Final answer

Medians are 7, √34 and 7.

Step-by-step solution

1Centroid = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3) = (0, 0, 0).
2x: (2a − 4 + 8)/3 = 0 → 2a = −4 → a = −2.
3y: (2 + 3b + 14)/3 = 0 → 3b = −16 → b = −16/3.
4z: (6 − 10 + 2c)/3 = 0 → 2c = 4 → c = 2.

Final answer

a = −2, b = −16/3, c = 2.

Step-by-step solution

1Let P(x, y, z). PA² = (x−3)² + (y−4)² + (z−5)² and PB² = (x+1)² + (y−3)² + (z+7)².
2Sum the x parts: (x−3)² + (x+1)² = 2x² − 4x + 10.
3Sum the y parts: (y−4)² + (y−3)² = 2y² − 14y + 25.
4Sum the z parts: (z−5)² + (z+7)² = 2z² + 4z + 74.
5Total: 2x² + 2y² + 2z² − 4x − 14y + 4z + 109 = k².

Final answer

2(x² + y² + z²) − 4x − 14y + 4z + 109 = k².

55

Chapter 12 — Limits and Derivatives

A limit asks what a function approaches as its input draws near a point; a derivative asks how fast the function is changing at that point. The chapter leans on two tools: factoring and cancelling to remove 0/0 forms, and the three standard results sinx/x → 1, (1 − cosx)/x → 0, and (eˣ − 1)/x → 1 as x → 0. Derivatives are built from first principles and then with the product, quotient and chain rules.

Board pattern

When direct substitution gives 0/0, factorise and cancel (xⁿ − aⁿ style), or rationalise surds. For the one-sided tests, compute LHL and RHL separately and compare; the limit exists only when they agree. Derivatives from first principles are a guaranteed board question — practise f′(x) = lim (f(x+h) − f(x))/h with full working.

$Ex 12.1 — Evaluating limits (32 Q)$
$Ex 12.2 — Derivatives by formula (11 Q)$
$Misc — First principles + all rules (30 Q)$

56

Exercise 12.1 — Evaluating Limits

32Exercise questions

Step-by-step solution

1Substitute directly: 3 + 3 = 6.

Final answer

6.

Step-by-step solution

1Substitute directly: π − 22/7.

Final answer

π − 22/7.

Step-by-step solution

1Substitute directly: π(1)² = π.

Final answer

π.

Step-by-step solution

1Substitute directly: (4·4 + 3)/(4 − 2) = 19/2.

Final answer

19/2.

Step-by-step solution

1Substitute directly: numerator = (−1)¹⁰ + (−1)⁵ + 1 = 1 − 1 + 1 = 1.
2Denominator = −1 − 1 = −2 → limit = −1/2.

Final answer

−1/2.

Step-by-step solution

1Expand (x+1)⁵ − 1 = 5x + 10x² + 10x³ + 5x⁴ + x⁵.
2Divide by x: 5 + 10x + 10x² + 5x³ + x⁴ → as x→0: 5.

Final answer

5.

Step-by-step solution

1Factor: 3x² − x − 10 = (3x + 5)(x − 2); x² − 4 = (x − 2)(x + 2).
2Cancel (x − 2): (3x + 5)/(x + 2) → (6 + 5)/4 = 11/4.

Final answer

11/4.

Step-by-step solution

1Factor: x⁴ − 81 = (x − 3)(x + 3)(x² + 9); 2x² − 5x − 3 = (2x + 1)(x − 3).
2Cancel (x − 3): (x + 3)(x² + 9)/(2x + 1) → (6)(18)/7 = 108/7.

Final answer

108/7.

Step-by-step solution

1Substitute x = 0: (0 + b)/(0 + 1) = b.

Final answer

b.

Step-by-step solution

1Put y = z¹ᐟ⁶ so z¹ᐟ³ = y²; as z→1, y→1.
2(y² − 1)/(y − 1) = y + 1 → 2.

Final answer

2.

Step-by-step solution

1Substitute directly: numerator → a + b + c and denominator → c + b + a.
2Both equal, so limit = 1.

Final answer

1.

Step-by-step solution

1Combine: 1/x + 1/2 = (x + 2)/(2x).
2(x + 2)/(2x) ÷ (x + 2) = 1/(2x) → 1/(2·(−2)) = −1/4.

Final answer

−1/4.

Step-by-step solution

1Write as (a/b) · sin(ax)/(ax), since bx = (b/a)(ax).
2(a/b) · 1 = a/b.

Final answer

a/b.

Step-by-step solution

1Divide top and bottom by x, or use (sin ax/ax)·(bx/sin bx)·(a/b).
2= 1 · 1 · a/b = a/b.

Final answer

a/b.

Step-by-step solution

1Put y = π − x; as x→π, y→0.
2sin y/(π·y) = (1/π)(sin y/y) → (1/π)(1) = 1/π.

Final answer

1/π.

Step-by-step solution

1Substitute directly: cos 0/(π − 0) = 1/π.

Final answer

1/π.

Step-by-step solution

1cos 2x − 1 = −2sin²x and cos x − 1 = −2sin²(x/2).
2Ratio = sin²x/sin²(x/2) = (2sin(x/2)cos(x/2))²/sin²(x/2) = 4cos²(x/2).
3As x→0: 4·1 = 4.

Final answer

4.

Step-by-step solution

1Factor x: x(a + cos x)/(b sin x).
2= (1/b)(a + cos x)·(x/sin x) → (1/b)(a + 1)·1 = (a + 1)/b.

Final answer

(a + 1)/b.

Step-by-step solution

1sec x → 1 as x→0, so x·1 → 0.

Final answer

0.

Step-by-step solution

1Divide top and bottom by x: (a·sin(ax)/ax + b)/(a + b·sin(bx)/bx).
2= (a + b)/(a + b) = 1.

Final answer

1.

Step-by-step solution

1cosec x − cot x = (1 − cos x)/sin x.
2= tan(x/2) → 0 (since 1 − cos x = 2sin²(x/2) and sin x = 2sin(x/2)cos(x/2)).

Final answer

0.

Step-by-step solution

1Put y = x − π/2; as x→π/2, y→0 and tan 2x = tan(2y + π) = tan 2y.
2tan 2y/y → 2 (using tan 2y ≈ 2y near 0).

Final answer

2.

Step-by-step solution

1At x = 0: LHL = 2(0) + 3 = 3 and RHL = 3(0 + 1) = 3 → lim(x→0) = 3.
2At x = 1: points near 1 have x > 0, so use 3(x + 1): lim(x→1) = 3(2) = 6.

Final answer

lim(x→0) = 3, lim(x→1) = 6.

Step-by-step solution

1LHL: x² − 1 → 1 − 1 = 0.
2RHL: −x² − 1 → −1 − 1 = −2.
3LHL ≠ RHL → the limit does not exist.

Final answer

Limit does not exist (LHL 0, RHL −2).

Step-by-step solution

1For x < 0, |x|/x = −1; for x > 0, |x|/x = 1.
2LHL = −1, RHL = 1 → limit does not exist.

Final answer

Does not exist (LHL −1, RHL 1).

Step-by-step solution

1For x < 0, x/|x| = −1; for x > 0, x/|x| = 1.
2LHL = −1, RHL = 1 → limit does not exist.

Final answer

Does not exist (LHL −1, RHL 1).

Step-by-step solution

1|x| is continuous everywhere, so substitute x = 5: |5| − 5 = 0.

Final answer

0.

Step-by-step solution

1f(1) = 4 and the limit exists iff LHL = RHL = 4.
2LHL = a + b = 4; RHL = b − a = 4.
3Adding: 2b = 8 → b = 4; then a + 4 = 4 → a = 0.

Final answer

a = 0, b = 4.

Step-by-step solution

1At x = a₁ the factor (x − a₁) hits 0, so lim(x→a₁) f(x) = f(a₁) = 0.
2For a different from every aᵢ, f is continuous at a: lim(x→a) f(x) = f(a) = (a − a₁)(a − a₂)…(a − aₙ).

Final answer

lim(x→a₁) = 0; lim(x→a) = (a − a₁)(a − a₂)…(a − aₙ).

Step-by-step solution

1For a < 0, near a the branch is |x| + 1 (continuous) → limit exists.
2For a > 0, the branch is |x| − 1 (continuous) → limit exists.
3At a = 0: LHL = 1, RHL = −1 → limit does not exist.

Final answer

The limit exists for every a ≠ 0.

Step-by-step solution

1As x→1, x² − 1 → 0; for the given limit to be finite, the numerator → 0 too.
2So f(x) − 2 → 0, giving lim(x→1) f(x) = 2.

Final answer

2.

Step-by-step solution

1At x = 0: LHL = m·0 + n = n; RHL = nx + m at 0 = m. Limit exists iff n = m.
2At x = 1: LHL = n·1 + m = n + m; RHL = n·1 + m = n + m → always equal.
3So the condition is m = n, for any integers m and n.

Final answer

m = n (any integer pair).

57

Exercise 12.2 — Derivatives

11Exercise questions

Step-by-step solution

1d/dx (x² − 2) = 2x.
2At x = 10: 2·10 = 20.

Final answer

20.

Step-by-step solution

1d/dx (x) = 1, so at x = 1 the derivative is 1.

Final answer

1.

Step-by-step solution

1d/dx (99x) = 99, constant for all x.

Final answer

99.

Step-by-step solution

1(i) By the power-sum rule, d/dx(x³ − 27) = 3x².
2(ii) (x − 1)(x − 2) = x² − 3x + 2 → derivative 2x − 3.
3(iii) 1/x² = x⁻² → derivative = −2x⁻³ = −2/x³.
4(iv) Quotient rule: [(x − 1) − (x + 1)]/(x − 1)² = −2/(x − 1)².

Final answer

(i) 3x², (ii) 2x − 3, (iii) −2/x³, (iv) −2/(x − 1)².

Step-by-step solution

1Differentiate term by term: f′(x) = x⁹⁹ + x⁹⁸ + … + x + 1.
2f′(0) = 1.
3f′(1) = 1 + 1 + … + 1 (100 terms) = 100.
4Hence f′(1) = 100 = 100·f′(0). Proved.

Final answer

f′(1) = 100, f′(0) = 1 → f′(1) = 100 f′(0). Proved.

Step-by-step solution

1Differentiate each term by the power rule.
2d/dx xⁿ = nxⁿ⁻¹; d/dx (aᵏxⁿ⁻ᵏ) = (n − k)aᵏxⁿ⁻ᵏ⁻¹.

Final answer

nxⁿ⁻¹ + (n−1)axⁿ⁻² + (n−2)a²xⁿ⁻³ + … + aⁿ⁻¹.

Step-by-step solution

1(i) (x − a)(x − b) = x² − (a + b)x + ab → derivative 2x − (a + b).
2(ii) (ax² + b)² = a²x⁴ + 2abx² + b² → derivative 4a²x³ + 4abx.
3(iii) Quotient rule: [(x − b) − (x − a)]/(x − b)² = (a − b)/(x − b)².

Final answer

(i) 2x − a − b, (ii) 4a²x³ + 4abx, (iii) (a − b)/(x − b)².

Step-by-step solution

1Quotient rule with u = xⁿ − aⁿ, v = x − a: f′ = [u′v − uv′]/v².
2f′ = [nxⁿ⁻¹(x − a) − (xⁿ − aⁿ)]/(x − a)².
3Simplify the numerator: (n − 1)xⁿ − naxⁿ⁻¹ + aⁿ.

Final answer

[(n − 1)xⁿ − naxⁿ⁻¹ + aⁿ]/(x − a)².

Step-by-step solution

1(i) 2x − 3/4 → 2.
2(ii) Expand: 5x⁴ − 5x³ + 3x² − 4x + 1 → derivative 20x³ − 15x² + 6x − 4.
3(iii) x⁻³(5 + 3x) = 5x⁻³ + 3x⁻² → −15x⁻⁴ − 6x⁻³.
4(iv) x⁵(3 − 6x⁻⁹) = 3x⁵ − 6x⁻⁴ → 15x⁴ + 24x⁻⁵.
5(v) x⁻⁴(3 − 4x⁻⁵) = 3x⁻⁴ − 4x⁻⁹ → −12x⁻⁵ + 36x⁻¹⁰.
6(vi) d[2/(x+1)] = −2/(x+1)²; d[x²/(3x−1)] = (3x² − 2x)/(3x−1)². So f′ = −2/(x+1)² − (3x² − 2x)/(3x−1)².

Final answer

(i) 2, (ii) 20x³ − 15x² + 6x − 4, (iii) −15/x⁴ − 6/x³, (iv) 15x⁴ + 24/x⁵, (v) −12/x⁵ + 36/x¹⁰, (vi) −2/(x+1)² − (3x²−2x)/(3x−1)².

Step-by-step solution

1f′(x) = lim(h→0) [cos(x+h) − cos x]/h.
2cos(x+h) − cos x = −2sin(x + h/2)sin(h/2).
3= −2sin(x + h/2)·[sin(h/2)]/h → −sin x · 1 (as h→0).

Final answer

−sin x.

Step-by-step solution

1(i) sin x cos x = (1/2)sin 2x → derivative cos 2x.
2(ii) sec x → sec x tan x.
3(iii) 5 sec x + 4 cos x → 5 sec x tan x − 4 sin x.
4(iv) cosec x → −cosec x cot x.
5(v) 3 cot x + 5 cosec x → −3cosec²x − 5 cosec x cot x.
6(vi) 5 sin x − 6 cos x + 7 → 5 cos x + 6 sin x.
7(vii) 2 tan x − 7 sec x → 2sec²x − 7 sec x tan x.

Final answer

(i) cos 2x, (ii) sec x tan x, (iii) 5 sec x tan x − 4 sin x, (iv) −cosec x cot x, (v) −3cosec²x − 5 cosec x cot x, (vi) 5 cos x + 6 sin x, (vii) 2sec²x − 7 sec x tan x.

58

Miscellaneous Exercise — First Principles and All Rules

30Exercise questions

Step-by-step solution

1(i) d(−x) = −1.
2(ii) (−x)⁻¹ = −1/x → derivative 1/x².
3(iii) sin(x + 1) → cos(x + 1).
4(iv) cos(x − π/8) → −sin(x − π/8).

Final answer

(i) −1, (ii) 1/x², (iii) cos(x + 1), (iv) −sin(x − π/8).

Step-by-step solution

1d/dx (x + a) = 1.

Final answer

1.

Step-by-step solution

1Expand: (px + q)(rx⁻¹ + s) = pr + psx + qr/x + qs.
2Derivative: ps − qr/x².

Final answer

ps − qr/x².

Step-by-step solution

1Product rule: a(cx + d)² + (ax + b)·2c(cx + d).
2Factor (cx + d): (cx + d)[a(cx + d) + 2c(ax + b)] = (cx + d)(3acx + ad + 2bc).

Final answer

(cx + d)(3acx + ad + 2bc).

Step-by-step solution

1Quotient rule: [a(cx + d) − c(ax + b)]/(cx + d)².
2Numerator: acx + ad − acx − bc = ad − bc.

Final answer

(ad − bc)/(cx + d)².

Step-by-step solution

1Simplify: (x + 1)/(x − 1).
2Quotient: [(x − 1) − (x + 1)]/(x − 1)² = −2/(x − 1)².

Final answer

−2/(x − 1)².

Step-by-step solution

1Chain/power rule: d(ax² + bx + c)⁻¹ = −(2ax + b)·(ax² + bx + c)⁻².

Final answer

−(2ax + b)/(ax² + bx + c)².

Step-by-step solution

1Quotient rule: [a(px² + qx + r) − (ax + b)(2px + q)]/(px² + qx + r)².
2Numerator: apx² + aqx + ar − 2apx² − aqx − 2bpx − bq = −apx² + ar − 2bpx − bq.

Final answer

(−apx² − 2bpx + ar − bq)/(px² + qx + r)².

Step-by-step solution

1Quotient rule: [(2px + q)(ax + b) − a(px² + qx + r)]/(ax + b)².
2Numerator: 2apx² + 2bpx + aqx + bq − apx² − aqx − ar = apx² + 2bpx + bq − ar.

Final answer

(apx² + 2bpx + bq − ar)/(ax + b)².

Step-by-step solution

1a/x⁴ = ax⁻⁴ → −4ax⁻⁵; b/x² = bx⁻² → −2bx⁻³.
2cos x → −sin x. Total: −4a/x⁵ + 2b/x³ − sin x.

Final answer

−4a/x⁵ + 2b/x³ − sin x.

Step-by-step solution

14√x = 4x^(1/2) → 4·(1/2)x^(−1/2) = 2/√x.

Final answer

2/√x.

Step-by-step solution

1Chain rule: n(ax + b)ⁿ⁻¹ · a.

Final answer

na(ax + b)ⁿ⁻¹.

Step-by-step solution

1Product + chain: na(ax + b)ⁿ⁻¹(cx + d)^m + mc(ax + b)ⁿ(cx + d)^{m−1}.
2Factor: (ax + b)ⁿ⁻¹(cx + d)^{m−1}[na(cx + d) + mc(ax + b)].

Final answer

(ax + b)ⁿ⁻¹(cx + d)^{m−1}[na(cx + d) + mc(ax + b)].

Step-by-step solution

1Chain rule: cos(x + a).

Final answer

cos(x + a).

Step-by-step solution

1Product rule: (cosec x)′cot x + cosec x(cot x)′.
2= −cosec x cot²x − cosec³x.

Final answer

−cosec x cot²x − cosec³x.

Step-by-step solution

1Quotient rule: [−sin x(1 + sin x) − cos x·cos x]/(1 + sin x)².
2Numerator: −sin x − sin²x − cos²x = −sin x − 1 = −(1 + sin x).
3= −(1 + sin x)/(1 + sin x)² = −1/(1 + sin x).

Final answer

−1/(1 + sin x).

Step-by-step solution

1Quotient rule with u = sin x + cos x, v = sin x − cos x: u′ = cos x − sin x = −v; v′ = cos x + sin x = u.
2Numerator = −v·v − u·u = −(v² + u²) = −2 (since u² + v² = 2).
3f′ = −2/(sin x − cos x)².

Final answer

−2/(sin x − cos x)².

Step-by-step solution

1Rewrite: (1 − cos x)/(1 + cos x).
2Quotient rule: [sin x(1 + cos x) + sin x(1 − cos x)]/(1 + cos x)² = 2 sin x/(1 + cos x)².

Final answer

2 sin x/(1 + cos x)².

Step-by-step solution

1Chain rule: n sinⁿ⁻¹x · cos x.

Final answer

n sinⁿ⁻¹x cos x.

Step-by-step solution

1Quotient rule: [b cos x(c + d cos x) − (a + b sin x)(−d sin x)]/(c + d cos x)².
2Numerator: bc cos x + bd cos²x + ad sin x + bd sin²x = bc cos x + ad sin x + bd.

Final answer

[bc cos x + ad sin x + bd]/(c + d cos x)².

Step-by-step solution

1Quotient rule: [cos(x + a)cos x + sin(x + a)sin x]/cos²x.
2Numerator = cos(x + a − x) = cos a.

Final answer

cos a/cos²x.

Step-by-step solution

1Product rule: 4x³(5 sin x − 3 cos x) + x⁴(5 cos x + 3 sin x).

Final answer

4x³(5 sin x − 3 cos x) + x⁴(5 cos x + 3 sin x).

Step-by-step solution

1Product rule: 2x cos x + (x² + 1)(−sin x) = 2x cos x − (x² + 1)sin x.

Final answer

2x cos x − (x² + 1)sin x.

Step-by-step solution

1Product rule: (2ax + cos x)(p + q cos x) + (ax² + sin x)(−q sin x).

Final answer

(2ax + cos x)(p + q cos x) − q sin x(ax² + sin x).

Step-by-step solution

1Product rule: (1 − sin x)(x − tan x) + (x + cos x)(1 − sec²x).
21 − sec²x = −tan²x, so f′ = (1 − sin x)(x − tan x) − tan²x(x + cos x).

Final answer

(1 − sin x)(x − tan x) − (x + cos x)tan²x.

Step-by-step solution

1Quotient rule: [(4 + 5 cos x)(3x + 7 cos x) − (4x + 5 sin x)(3 − 7 sin x)]/(3x + 7 cos x)².

Final answer

[(4 + 5 cos x)(3x + 7 cos x) − (4x + 5 sin x)(3 − 7 sin x)]/(3x + 7 cos x)².

Step-by-step solution

1Quotient rule keeping cos(π/4) = √2/2 constant: [2x sin x − x² cos x]/sin²x times cos(π/4).

Final answer

cos(π/4)·[2x sin x − x² cos x]/sin²x.

Step-by-step solution

1Quotient rule: [(1 + tan x) − x·sec²x]/(1 + tan x)².

Final answer

[1 + tan x − x sec²x]/(1 + tan x)².

Step-by-step solution

1Product rule: (1 + sec x tan x)(x − tan x) + (x + sec x)(1 − sec²x).
21 − sec²x = −tan²x → f′ = (1 + sec x tan x)(x − tan x) − (x + sec x)tan²x.

Final answer

(1 + sec x tan x)(x − tan x) − (x + sec x)tan²x.

Step-by-step solution

1Quotient rule: [sinⁿx − x·n sinⁿ⁻¹x cos x]/sin²ⁿx.
2Divide by sinⁿ⁻¹x: [sin x − nx cos x]/sinⁿ⁺¹x.

Final answer

[sin x − nx cos x]/sinⁿ⁺¹x.

59

Chapter 13 — Statistics

Statistics measures how spread out a data set is. Three measures appear in this chapter: mean deviation (average absolute distance from a centre), variance (average squared distance from the mean) and standard deviation (the square root of variance). Grouped data uses midpoints of classes, and the shortcut method shifts the working to smaller numbers using a provisional mean A with step size h.

Board pattern

Mean deviation = Σf|xᵢ − mean or median| / Σf. Variance = Σfx²/Σf − (mean)². Shortcut: mean = A + h·Σfd/Σf and variance = h²[Σfd²/Σf − (Σfd/Σf)²], where d = (x − A)/h. For median-based deviations, first locate the median class where the cumulative frequency crosses N/2, then use the formula, or read positions for ungrouped data.

$Ex 13.1 — Mean deviation about mean and median (12 Q)$
$Ex 13.2 — Variance and standard deviation (10 Q)$
$Misc — Correcting a wrong observation (6 Q)$

60

Exercise 13.1 — Mean Deviation

12Exercise questions

Step-by-step solution

1Mean = (4+7+8+9+10+12+13+17)/8 = 80/8 = 10.
2Deviations |x − 10|: 6, 3, 2, 1, 0, 2, 3, 7; sum = 24.
3MD = 24/8 = 3.

Final answer

MD = 3.

Step-by-step solution

1Sum = 500, n = 10 → mean = 50.
2Deviations: 12, 20, 2, 10, 8, 5, 13, 4, 4, 6; sum = 84.
3MD = 84/10 = 8.4.

Final answer

MD = 8.4.

Step-by-step solution

1Sorted: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.
2n = 12 → median = (13 + 14)/2 = 13.5.
3Deviations sum = 3.5 + 5 + 1.5 + 1 + 0.5 + 5 + 7 + 4.5 = 28.
4MD = 28/12 = 7/3 ≈ 2.33.

Final answer

MD = 7/3 ≈ 2.33.

Step-by-step solution

1Sorted: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72.
2n = 10 → median = (46 + 49)/2 = 47.5.
3Deviations: 11.5, 5.5, 2.5, 1.5+1.5, 1.5, 3.5, 5.5, 12.5, 24.5; sum = 70.
4MD = 70/10 = 7.

Final answer

MD = 7.

Step-by-step solution

1N = 25; Σfx = 35 + 40 + 90 + 60 + 125 = 350 → mean = 14.
2Deviations: 9, 4, 1, 6, 11.
3Σf|x − mean| = 9·7 + 4·4 + 1·6 + 6·3 + 11·5 = 158.
4MD = 158/25 = 6.32.

Final answer

MD = 6.32.

Step-by-step solution

1N = 80; Σfx = 40 + 720 + 1400 + 1120 + 720 = 4000 → mean = 50.
2Deviations: 40, 20, 0, 20, 40.
3Σf·dev = 4·40 + 24·20 + 28·0 + 16·20 + 8·40 = 1280.
4MD = 1280/80 = 16.

Final answer

MD = 16.

Step-by-step solution

1N = 26, N/2 = 13. Cumulative: 8, 14, 16, 18, 20, 26 → the 13th value is 7, so median = 7.
2Deviations: 2, 0, 2, 3, 5, 8.
3Σf·dev = 8·2 + 6·0 + 2·2 + 2·3 + 2·5 + 6·8 = 84.
4MD = 84/26 = 42/13 ≈ 3.23.

Final answer

MD = 42/13 ≈ 3.23.

Step-by-step solution

1N = 29, N/2 = 14.5. Cumulative: 3, 8, 14, 21 → the 14.5th value lies in x = 30, so median = 30.
2Deviations: 15, 9, 3, 0, 5.
3Σf·dev = 3·15 + 5·9 + 6·3 + 7·0 + 8·5 = 148.
4MD = 148/29 ≈ 5.10.

Final answer

MD = 148/29 ≈ 5.10.

Step-by-step solution

1Midpoints: 50, 150, 250, 350, 450, 550, 650, 750; N = 50.
2Σfx = 200 + 1200 + 2250 + 3500 + 3150 + 2750 + 2600 + 2250 = 17900 → mean = 358.
3Deviations: 308, 208, 108, 8, 92, 192, 292, 392.
4Σf·dev = 1232 + 1664 + 972 + 80 + 644 + 960 + 1168 + 1176 = 7896.
5MD = 7896/50 = 157.92.

Final answer

MD = 157.92.

Step-by-step solution

1Midpoints: 100, 110, 120, 130, 140, 150; N = 100.
2Σfx = 900 + 1430 + 3120 + 3900 + 1680 + 1500 = 12530 → mean = 125.3.
3Deviations: 25.3, 15.3, 5.3, 4.7, 14.7, 24.7.
4Σf·dev = 227.7 + 198.9 + 137.8 + 141 + 176.4 + 247 = 1128.8.
5MD = 1128.8/100 = 11.288 ≈ 11.29.

Final answer

MD ≈ 11.29.

Step-by-step solution

1N = 50, N/2 = 25. Cumulative: 6, 14, 28 → median class is 20-30.
2Median = 20 + [(25 − 14)/14]·10 = 20 + 55/7 ≈ 27.86.
3Midpoints: 5, 15, 25, 35, 45, 55; deviations (×7): 160/7, 90/7, 20/7, 50/7, 120/7, 190/7.
4Σf·dev = (960 + 720 + 280 + 800 + 480 + 380)/7 = 3620/7.
5MD = (3620/7)/50 = 362/35 ≈ 10.34.

Final answer

MD = 362/35 ≈ 10.34.

Step-by-step solution

1Convert to 15.5-20.5, 20.5-25.5, …, 50.5-55.5 with the same frequencies.
2N = 100, N/2 = 50. Cumulative: 5, 11, 23, 37, 63 → median class 35.5-40.5.
3Median = 35.5 + [(50 − 37)/26]·5 = 35.5 + 2.5 = 38.
4Midpoints: 18, 23, 28, 33, 38, 43, 48, 53; deviations: 20, 15, 10, 5, 0, 5, 10, 15.
5Σf·dev = 100 + 90 + 120 + 70 + 0 + 60 + 160 + 135 = 735.
6MD = 735/100 = 7.35.

Final answer

MD = 7.35.

61

Exercise 13.2 — Variance and Standard Deviation

10Exercise questions

Step-by-step solution

1Sum = 72, n = 8 → mean = 9; Σx² = 36 + 49 + 100 + 144 + 169 + 16 + 64 + 144 = 722.
2Variance = 722/8 − 81 = 90.25 − 81 = 9.25.

Final answer

Mean = 9, variance = 9.25.

Step-by-step solution

1Mean = (1 + 2 + … + n)/n = (n + 1)/2.
2Σx² = n(n+1)(2n+1)/6 → first moment of squares = (n+1)(2n+1)/6.
3Variance = (n+1)(2n+1)/6 − (n+1)²/4 = (n² − 1)/12.

Final answer

Mean = (n + 1)/2, variance = (n² − 1)/12.

Step-by-step solution

1Data: 3, 6, …, 30; sum = 3·55 = 165 → mean = 16.5.
2Σx² = 9·(1² + … + 10²) = 9·385 = 3465.
3Variance = 3465/10 − (16.5)² = 346.5 − 272.25 = 74.25.

Final answer

Mean = 16.5, variance = 74.25.

Step-by-step solution

1N = 40; Σfx = 12 + 40 + 98 + 216 + 192 + 112 + 90 = 760 → mean = 19.
2Σfx² = 72 + 400 + 1372 + 3888 + 4608 + 3136 + 2700 = 16176.
3Variance = 16176/40 − 361 = 404.4 − 361 = 43.4.

Final answer

Mean = 19, variance = 43.4.

Step-by-step solution

1N = 22; Σfx = 276 + 186 + 291 + 196 + 612 + 312 + 327 = 2200 → mean = 100.
2Σfx² = 25392 + 17298 + 28227 + 19208 + 62424 + 32448 + 35643 = 220640.
3Variance = 220640/22 − 10000 = 10029.09 − 10000 = 320/11 ≈ 29.09.

Final answer

Mean = 100, variance = 320/11 ≈ 29.09.

Step-by-step solution

1Take A = 64, d = x − 64; N = 100.
2Σfd = −8 − 3 − 24 − 29 + 0 + 12 + 20 + 12 + 20 = 0 → mean = 64.
3Σfd² = 32 + 9 + 48 + 29 + 0 + 12 + 40 + 36 + 80 = 286.
4Variance = 286/100 − 0 = 2.86; SD = √2.86 ≈ 1.69.

Final answer

Mean = 64, SD ≈ 1.69.

Step-by-step solution

1Midpoints: 15, 45, 75, 105, 135, 165, 195; N = 30.
2Σfx = 30 + 135 + 375 + 1050 + 405 + 825 + 390 = 3210 → mean = 107.
3Σfx² = 450 + 6075 + 28125 + 110250 + 54675 + 136125 + 76050 = 411750.
4Variance = 411750/30 − 11449 = 13725 − 11449 = 2276.

Final answer

Mean = 107, variance = 2276.

Step-by-step solution

1Midpoints: 5, 15, 25, 35, 45; N = 50.
2Σfx = 25 + 120 + 375 + 560 + 270 = 1350 → mean = 27.
3Σfx² = 125 + 1800 + 9375 + 19600 + 12150 = 43050.
4Variance = 43050/50 − 729 = 861 − 729 = 132.

Final answer

Mean = 27, variance = 132.

Step-by-step solution

1Midpoints: 72.5, …, 112.5 (step h = 5). Take A = 92.5, d = (x − 92.5)/5: −4, −3, −2, −1, 0, 1, 2, 3, 4; N = 60.
2Σfd = −12 − 12 − 14 − 7 + 0 + 9 + 12 + 18 + 12 = 6.
3Mean = 92.5 + 5·(6/60) = 92.5 + 0.5 = 93.
4Σfd² = 48 + 36 + 28 + 7 + 0 + 9 + 24 + 54 + 48 = 254.
5Variance = 25·[254/60 − (6/60)²] = 25·4.2233 = 105.583; SD ≈ 10.28.

Final answer

Mean = 93, variance ≈ 105.58, SD ≈ 10.28.

Step-by-step solution

1Make continuous: 32.5-36.5, …, 48.5-52.5; midpoints 34.5, 38.5, 42.5, 46.5, 50.5; N = 100.
2Σfx = 517.5 + 654.5 + 892.5 + 1023 + 1262.5 = 4350 → mean = 43.5.
3Σfx² = 17853.75 + 25198.25 + 37931.25 + 47569.5 + 63756.25 = 192309.
4Variance = 192309/100 − 43.5² = 1923.09 − 1892.25 = 30.84; SD ≈ 5.55.

Final answer

Mean = 43.5 mm, SD ≈ 5.55 mm.

62

Miscellaneous Exercise — Correcting a Wrong Observation

6Exercise questions

Step-by-step solution

1Let the missing be a and b. Sum = 8·9 = 72 → a + b = 72 − 60 = 12.
2Σx² = 8(9.25 + 81) = 722; given squares sum = 36 + 49 + 100 + 144 + 144 + 169 = 642.
3a² + b² = 722 − 642 = 80. With a + b = 12, ab = (144 − 80)/2 = 32.
4a and b are roots of t² − 12t + 32 = 0 → t = 4 or 8.

Final answer

The two missing observations are 4 and 8.

Step-by-step solution

1Sum = 7·8 = 56 → a + b = 56 − 42 = 14.
2Σx² = 7(16 + 64) = 560; given squares sum = 4 + 16 + 100 + 144 + 196 = 460.
3a² + b² = 560 − 460 = 100. With a + b = 14, ab = (196 − 100)/2 = 48.
4a and b are roots of t² − 14t + 48 = 0 → t = 6 or 8.

Final answer

The two missing observations are 6 and 8.

Step-by-step solution

1Multiplying every observation by a changes the mean by the same factor: new mean = 3·8 = 24.
2Standard deviation scales by the same factor too: new SD = 3·4 = 12 (var scales by 9, SD by 3).

Final answer

New mean = 24, new SD = 12.

Step-by-step solution

1New mean = Σ(axᵢ)/n = a·Σxᵢ/n = ax̄.
2New variance = (1/n)Σ(axᵢ − ax̄)² = (a²/n)Σ(xᵢ − x̄)² = a²σ².

Final answer

Mean = ax̄, variance = a²σ². Proved.

Step-by-step solution

1Σx = 20·10 = 200 and Σx² = 20(4 + 100) = 2080.
2(i) Omit 8: Σx = 192, n = 19 → mean = 192/19 ≈ 10.11. Σx² = 2080 − 64 = 2016.
3(i) Variance = 2016/19 − (192/19)² = 1440/361; SD = 12√10/19 ≈ 1.997.
4(ii) Replace by 12: Σx = 204 → mean = 10.2. Σx² = 2080 − 64 + 144 = 2160.
5(ii) Variance = 2160/20 − (10.2)² = 108 − 104.04 = 3.96; SD ≈ 1.99.

Final answer

(i) mean = 192/19 ≈ 10.11, SD ≈ 1.997; (ii) mean = 10.2, SD ≈ 1.99.

Step-by-step solution

1Σx = 100·20 = 2000 and Σx² = 100(9 + 400) = 40900.
2Omit 21, 21, 18: Σx = 2000 − 60 = 1940, n = 97 → mean = 1940/97 = 20.
3Σx² = 40900 − (441 + 441 + 324) = 40900 − 1206 = 39694.
4Variance = 39694/97 − 400 = 409.2165 − 400 = 9.2165; SD ≈ 3.036.

Final answer

Mean = 20, SD ≈ 3.04.

63

Chapter 14 — Probability

Probability assigns numbers between 0 and 1 to events built from a sample space. This chapter covers the language of events — mutually exclusive, exhaustive, simple and compound — the addition rule P(A∪B) = P(A) + P(B) − P(A∩B), and its complement form P(not A) = 1 − P(A). Counting (combinations) powers the probability questions on cards, coins, dice and lots.

Board pattern

Always identify the sample space first and count outcomes using ⁿCᵣ when order does not matter. A valid probability assignment needs every P(ωᵢ) ≥ 0 and their sum exactly 1. For 'at least one' always use 1 − P(none), and for any two events remember P(A∪B) = P(A) + P(B) − P(A∩B) — this one formula covers most of the exercise.

$Ex 14.1 — Events on dice and coins (7 Q)$
$Ex 14.2 — Axiomatic probability, addition rule (21 Q)$
$Misc — Cards, lots, mixed applications (10 Q)$

64

Exercise 14.1 — Events Built From a Sample Space

7Exercise questions

Step-by-step solution

1E = {4} and F = {2, 4, 6}.
2E ∩ F = {4} ≠ ∅, so they have a common outcome.

Final answer

No — not mutually exclusive, since 4 belongs to both.

Step-by-step solution

1A = {1,2,3,4,5,6} = S, B = ∅, C = {3,6}, D = {1,2,3}, E = {6}, F = {3,4,5,6}.
2A ∪ B = {1,2,3,4,5,6}; A ∩ B = ∅; B ∪ C = {3,6}.
3E ∩ F = {6}; D ∩ E = ∅.
4A − C = {1,2,4,5}; D − E = {1,2,3}.
5F′ = {1,2,5,6} and E ∩ F′ = {6} = E.

Final answer

A = S, B = ∅, C = {3,6}, D = {1,2,3}, E = {6}, F = {3,4,5,6}; A∪B = S, A∩B = ∅, B∪C = {3,6}, E∩F = {6}, D∩E = ∅, A−C = {1,2,4,5}, D−E = {1,2,3}, E∩F′ = {6}, F′ = {1,2,5,6}.

Step-by-step solution

1A = {(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}.
2B = outcomes with a 2, e.g. (2,1)…(2,6),(1,2),(3,2),(4,2),(5,2),(6,2).
3C = {(3,6),(4,5),(5,4),(6,3)} (sum 9).
4A ∩ C = C ≠ ∅ and A ∩ B = {(2,6),(6,2)} ≠ ∅, but B ∩ C = ∅.

Final answer

Only the pair (B, C) is mutually exclusive.

Step-by-step solution

1A = {HHH}, B = {HHT, HTH, THH}, C = {TTT}, D = {HHH, HHT, HTH, HTT}.
2(i) Pairs with empty intersection: (A,B), (A,C), (B,C) and (C,D).
3(ii) Simple (single outcome): A = {HHH} and C = {TTT}.
4(iii) Compound (multiple outcomes): B and D.

Final answer

(i) (A,B), (A,C), (B,C), (C,D); (ii) A and C; (iii) B and D.

Step-by-step solution

1Take S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
2(i) A = {HHH} and B = {TTT} are mutually exclusive.
3(ii) 3 heads: {HHH}; exactly 2 heads: {HHT, HTH, THH}; at most 1 head: {HTT, THT, TTH, TTT} — exclusive and cover S.
4(iii) A = {HHH, HHT, HTH, HTT} (head first) and B = {HHH, HHT, THH, THT} (head second) share {HHH, HHT}.
5(iv) A = {HHH} and B = {TTT}: exclusive, union ≠ S.
6(v) {HHH}, {TTT}, {HHT, HTH, THH}: pairwise disjoint, union ≠ S.

Final answer

Any valid example is acceptable; the ones above use heads/tails patterns as listed.

Step-by-step solution

1A = {x even on first die}; B = {x odd}; C = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}.
2(i) A′ = odd on first die = B; (ii) not B = A.
3(iii) A or B = S; (iv) A and B = ∅.
4(v) A but not C = A ∩ C′ = 14 outcomes with first die even and sum ≥ 6: (2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1)…(6,6).
5(vi) B or C: first die odd or sum ≤ 5; (vii) B and C = {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)}.
6(viii) B′ = A, so A ∩ B′ ∩ C′ = A ∩ C′ — the same 14 outcomes as (v).

Final answer

A′ = B; not B = A; A or B = S; A and B = ∅; A but not C = 14 listed outcomes; B and C = 6 listed outcomes; A∩B′∩C′ equals (v).

Step-by-step solution

1(i) True — even and odd on the first die cannot both happen.
2(ii) True — every outcome has first die either even or odd, and the two never overlap.
3(iii) True — B′ means first die is even, which is exactly A.
4(iv) False — A ∩ C = {(2,1),(2,2),(2,3),(4,1)} ≠ ∅.
5(v) False — B′ = A, so A and B′ coincide; their intersection is A, not empty.
6(vi) False — A′ = B, B′ = A and A ∪ B = S, but A ∩ C ≠ ∅, so they are not mutually exclusive.

Final answer

(i) T, (ii) T, (iii) T, (iv) F, (v) F, (vi) F — reasons above.

65

Exercise 14.2 — Axiomatic Probability and the Addition Rule

21Exercise questions

Step-by-step solution

1(a) Sum = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1, all non-negative → valid.
2(b) 7 × (1/7) = 1 → valid.
3(c) Sum = 2.8 ≠ 1 → invalid.
4(d) Contains negative values (−0.1, −0.2) → invalid.
5(e) Sum = 36/14 > 1 → invalid.

Final answer

Invalid: (c), (d), (e). Valid: (a), (b).

Step-by-step solution

1P(at least one tail) = 1 − P(no tail) = 1 − P(HH).
2P(HH) = 1/4 → answer = 3/4.

Final answer

3/4.

Step-by-step solution

1(i) Prime faces: 2, 3, 5 → 3/6 = 1/2.
2(ii) {3,4,5,6} → 4/6 = 2/3.
3(iii) {1} → 1/6.
4(iv) No face exceeds 6 → 0.
5(v) {1,2,3,4,5} → 5/6.

Final answer

(i) 1/2, (ii) 2/3, (iii) 1/6, (iv) 0, (v) 5/6.

Step-by-step solution

1(a) 52 cards → 52 points.
2(b) Exactly one ace of spades → 1/52.
3(c)(i) 4 aces → 4/52 = 1/13.
4(c)(ii) 26 black cards → 26/52 = 1/2.

Final answer

(a) 52, (b) 1/52, (c)(i) 1/13, (c)(ii) 1/2.

Step-by-step solution

1Sample space: (1,1)…(1,6),(6,1)…(6,6): 12 equally likely outcomes.
2(i) Sum 3 only from (1,2) → 1/12.
3(ii) Sum 12 only from (6,6) → 1/12.

Final answer

(i) 1/12, (ii) 1/12.

Step-by-step solution

1Total members = 10; women = 6.
2P(woman) = 6/10 = 3/5.

Final answer

3/5.

Step-by-step solution

1If H heads occur, T = 4 − H tails, so amount = H − 1.5(4 − H) = 2.5H − 6.
2H = 0, 1, 2, 3, 4 give 5 different amounts: −6, −3.5, −1, 1.5, 4.
3P(H = 0) = 1/16; P(H = 1) = 4/16 = 1/4; P(H = 2) = 6/16 = 3/8; P(H = 3) = 4/16 = 1/4; P(H = 4) = 1/16.

Final answer

Five amounts: Rs −6 (P 1/16), −3.5 (1/4), −1 (3/8), 1.5 (1/4), 4 (1/16).

Step-by-step solution

18 equally likely outcomes.
2(i) 1/8; (ii) 3/8 (HHT, HTH, THH); (iii) 4/8 = 1/2; (iv) 7/8; (v) 1/8.
3(vi) 1/8; (vii) 3/8; (viii) 1/8; (ix) 7/8.

Final answer

(i) 1/8, (ii) 3/8, (iii) 1/2, (iv) 7/8, (v) 1/8, (vi) 1/8, (vii) 3/8, (viii) 1/8, (ix) 7/8.

Step-by-step solution

1P(not A) = 1 − 2/11 = 9/11.

Final answer

9/11.

Step-by-step solution

1Word has 13 letters: A×3, S×4, I×2, N×2, T×1, O×1.
2(i) Vowels A, I, O → 6/13.
3(ii) Consonants S, N, T → 7/13.

Final answer

(i) 6/13, (ii) 7/13.

Step-by-step solution

1Total ways to choose 6 numbers from 20 = C(20, 6).
2C(20,6) = 38760; only one selection wins.
3P = 1/38760.

Final answer

1/38760.

Step-by-step solution

1(i) A ∩ B ⊆ A, so P(A ∩ B) ≤ P(A) is required; here 0.6 > 0.5 → inconsistent.
2(ii) P(A ∩ B) = 0.5 + 0.4 − 0.8 = 0.1 ≥ 0 → consistent.

Final answer

(i) Not consistent; (ii) consistent.

Step-by-step solution

1(i) P(A∪B) = 1/3 + 1/5 − 1/15 = 5/15 + 3/15 − 1/15 = 7/15.
2(ii) P(B) = 0.6 − 0.35 + 0.25 = 0.5.
3(iii) P(A∩B) = 0.5 + 0.35 − 0.7 = 0.15.

Final answer

(i) 7/15, (ii) 0.5, (iii) 0.15.

Step-by-step solution

1For mutually exclusive events, P(A ∩ B) = 0.
2P(A or B) = 3/5 + 1/5 = 4/5.

Final answer

4/5.

Step-by-step solution

1(i) P(E or F) = 1/4 + 1/2 − 1/8 = 2/8 + 4/8 − 1/8 = 5/8.
2(ii) P(not E and not F) = 1 − P(E or F) = 3/8.

Final answer

(i) 5/8, (ii) 3/8.

Step-by-step solution

1By De Morgan, E′ ∪ F′ = (E ∩ F)′, so P(E ∩ F) = 1 − 0.25 = 0.75.
2P(E ∩ F) ≠ 0, so E and F are not mutually exclusive.

Final answer

Not mutually exclusive (P(E ∩ F) = 0.75).

Step-by-step solution

1(i) P(not A) = 1 − 0.42 = 0.58.
2(ii) P(not B) = 1 − 0.48 = 0.52.
3(iii) P(A or B) = 0.42 + 0.48 − 0.16 = 0.74.

Final answer

(i) 0.58, (ii) 0.52, (iii) 0.74.

Step-by-step solution

1P(M or B) = P(M) + P(B) − P(M and B).
2= 0.4 + 0.3 − 0.1 = 0.6.

Final answer

0.6 (i.e. 60%).

Step-by-step solution

1P(both) = P(first) + P(second) − P(at least one).
2= 0.8 + 0.7 − 0.95 = 0.55.

Final answer

0.55.

Step-by-step solution

1P(E or H) = 1 − P(neither) = 1 − 0.1 = 0.9.
2P(H) = P(E or H) − P(E) + P(E and H) = 0.9 − 0.75 + 0.5 = 0.65.

Final answer

0.65.

Step-by-step solution

1(i) P(NCC or NSS) = (30 + 32 − 24)/60 = 38/60 = 19/30.
2(ii) Neither = 1 − 19/30 = 11/30.
3(iii) NSS only = (32 − 24)/60 = 8/60 = 2/15.

Final answer

(i) 19/30, (ii) 11/30, (iii) 2/15.

66

Miscellaneous Exercise — Counting and Probability

10Exercise questions

Step-by-step solution

1Total marbles = 60; ways to pick 5 = C(60, 5).
2(i) All blue: C(20,5)/C(60,5) = 15504/5461512 ≈ 0.0028.
3(ii) At least one green = 1 − P(none green) = 1 − C(30,5)/C(60,5).
4C(30,5) = 142506 → P = 1 − 142506/5461512 ≈ 0.9739.

Final answer

(i) C(20,5)/C(60,5) ≈ 0.0028; (ii) 1 − C(30,5)/C(60,5) ≈ 0.9739.

Step-by-step solution

1Total ways = C(52, 4) = 270725.
2Favourable: choose 3 of 13 diamonds and 1 of 13 spades = C(13,3)·C(13,1) = 286·13 = 3718.
3P = 3718/270725 ≈ 0.0137.

Final answer

3718/270725 ≈ 0.0137.

Step-by-step solution

1(i) P(2) = 3/6 = 1/2.
2(ii) P(1 or 3) = (2 + 1)/6 = 3/6 = 1/2.
3(iii) P(not 3) = 1 − 1/6 = 5/6.

Final answer

(i) 1/2, (ii) 1/2, (iii) 5/6.

Step-by-step solution

1Losing tickets = 9990.
2(a) 9990/10000 = 999/1000.
3(b) C(9990,2)/C(10000,2) = (9990·9989)/(10000·9999) ≈ 0.9980.
4(c) C(9990,10)/C(10000,10) ≈ 0.9900.

Final answer

(a) 999/1000; (b) C(9990,2)/C(10000,2) ≈ 0.998; (c) C(9990,10)/C(10000,10) ≈ 0.99.

Step-by-step solution

1(a) Both in the section of 40: (40/100)(39/99); both in the section of 60: (60/100)(59/99).
2Total same section = (40·39 + 60·59)/(100·99) = (1560 + 3540)/9900 = 5100/9900 = 17/33.
3(b) Different sections = 1 − 17/33 = 16/33.

Final answer

(a) 17/33, (b) 16/33.

Step-by-step solution

1Total arrangements = 3! = 6.
2Arrangements with no letter in its own envelope (derangements) = 2.
3At least one correct = 6 − 2 = 4 → P = 4/6 = 2/3.

Final answer

2/3.

Step-by-step solution

1(i) P(A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88.
2(ii) P(A′ ∩ B′) = 1 − P(A ∪ B) = 0.12.
3(iii) P(A ∩ B′) = P(A) − P(A ∩ B) = 0.54 − 0.35 = 0.19.
4(iv) P(B ∩ A′) = P(B) − P(A ∩ B) = 0.69 − 0.35 = 0.34.

Final answer

(i) 0.88, (ii) 0.12, (iii) 0.19, (iv) 0.34.

Step-by-step solution

1Male members = Harish, Rohan, Salim → 3.
2Over 35 years = Sheetal (46), Salim (41) → 2.
3Male and over 35 = Salim → 1.
4P = (3 + 2 − 1)/5 = 4/5.

Final answer

4/5.

Step-by-step solution

1(i) First digit must be 5 or 7 → 2 choices; total = 2·5³ = 250. Divisible by 5: units 0 or 5 → 2·2·5·5 = 100. P = 100/250 = 2/5.
2(ii) Total = 2 · ⁴P₃ = 48. Units 0: first 5 or 7, middle ⁴P₂... units 0 gives 2·3·2 = 12; units 5: first must be 7, middle from {0,1,3}: 1·3·2 = 6.
3Favourable = 12 + 6 = 18 → P = 18/48 = 3/8.

Final answer

(i) 2/5, (ii) 3/8.

Step-by-step solution

1Total sequences with no repeats = 10·9·8·7 = 5040.
2Exactly one of them opens the lock.
3P = 1/5040.

Final answer

1/5040.

Class 11 Maths NCERT Solutions

Where can I find Class 11 Maths NCERT solutions chapter-wise?

How to Use These NCERT Solutions

Chapter 1 — Sets

Exercise 1.1 — Sets and Their Representations

Exercise 1.2 — Empty, Finite and Infinite Sets

Exercise 1.3 — Subsets, Equal Sets and the Power Set

Exercise 1.4 — Union, Intersection and Difference of Sets

Exercise 1.5 — Complement of a Set

Exercise 1.6 — Cardinal Numbers and Venn Diagrams

Miscellaneous Exercise — Mixed Problems on Sets

Chapter 2 — Relations and Functions

Exercise 2.1 — Ordered Pairs and Cartesian Products

Exercise 2.2 — Relations, Domain, Codomain and Range

Exercise 2.3 — Functions

Miscellaneous Exercise — Mixed Problems on Relations and Functions

Chapter 3 — Trigonometric Functions

Exercise 3.1 — Angles and their Measures

Exercise 3.2 — Trigonometric Functions of an Angle

Exercise 3.3 — Trigonometric Identities

Exercise 3.4 — Trigonometric Equations

Miscellaneous Exercise — Mixed Problems on Trigonometric Functions

Chapter 4 — Complex Numbers and Quadratic Equations

Exercise 4.1 — Operations on Complex Numbers

Miscellaneous Exercise — Mixed Problems on Complex Numbers

Chapter 5 — Linear Inequalities

Exercise 5.1 — Solving Linear Inequalities

Miscellaneous Exercise — Compound Inequalities and Applications

Chapter 6 — Permutations and Combinations

Exercise 6.1 — Fundamental Principle of Counting

Exercise 6.2 — Factorials and n! / (n − r)!

Exercise 6.3 — Permutations

Exercise 6.4 — Combinations

Miscellaneous Exercise on Chapter 6

Chapter 7 — Binomial Theorem

Exercise 7.1 — Using the Binomial Theorem

Miscellaneous Exercise on Chapter 7

Chapter 8 — Sequences and Series

Exercise 8.1 — Sequences

Exercise 8.2 — Geometric Progressions

Miscellaneous Exercise on Chapter 8

Chapter 9 — Straight Lines

Exercise 9.1 — Slope of a Line

Exercise 9.2 — Various Forms of the Equation of a Line

Exercise 9.3 — Distance of a Point from a Line

Miscellaneous Exercise on Chapter 9

Chapter 10 — Conic Sections

Exercise 10.1 — The Circle

Exercise 10.2 — The Parabola

Exercise 10.3 — The Ellipse

Exercise 10.4 — The Hyperbola

Miscellaneous Exercise on Chapter 10

Chapter 11 — Introduction to Three Dimensional Geometry

Exercise 11.1 — Coordinates, Planes and Octants

Exercise 11.2 — Distance Between Two Points

Miscellaneous Exercise — Applications in Space

Chapter 12 — Limits and Derivatives

Exercise 12.1 — Evaluating Limits

Exercise 12.2 — Derivatives

Miscellaneous Exercise — First Principles and All Rules

Chapter 13 — Statistics

Exercise 13.1 — Mean Deviation

Exercise 13.2 — Variance and Standard Deviation

Miscellaneous Exercise — Correcting a Wrong Observation

Chapter 14 — Probability

Exercise 14.1 — Events Built From a Sample Space

Exercise 14.2 — Axiomatic Probability and the Addition Rule

Miscellaneous Exercise — Counting and Probability

Key formulas at a glance

How this chapter is asked

Frequently asked questions

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