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Class 12 Maths NCERT Solutions

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Relations and Functions Class 12 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 1, Relations and Functions — 54 questions from Ex 1.1 to Ex 1.4, each worked through step by step in the CBSE marking pattern. Types of relations, composition, inverses, and binary operations with identity and inverse elements.

Class:12Subject:MathsChapter:1
3 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 12 Maths Chapter 1?

Chapter 1 carries 4 exercise questions, numbered Ex 1.1 to Ex 1.4. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

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

Board pattern

Relations questions award marks chiefly for stating each property (reflexive, symmetric, transitive) and disproving it with a concrete counterexample. Always write the test — e.g. "(1,1) ∉ R, so R is not reflexive". A bare "yes" or "no" scores zero.
02

Exercise 1.1 — Types of Relations

16Exercise questions

Step-by-step solution

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

Final answer

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

Step-by-step solution

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

Final answer

None of the three properties.

Final answer

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

Step-by-step solution

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

Final answer

Reflexive and transitive, not symmetric.

Final answer

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

Final answer

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

Step-by-step solution

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

Final answer

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

Step-by-step solution

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

Final answer

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

Step-by-step solution

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

Final answer

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

Final answer

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

Final answer

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

Final answer

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

Step-by-step solution

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

Final answer

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

Final answer

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

Step-by-step solution

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

Final answer

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

Step-by-step solution

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

Final answer

Option (C) — (6,8) ∈ R.

03

Exercise 1.2 — One-One, Onto and Bijective Functions

12Exercise questions

Step-by-step solution

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

Final answer

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

Final answer

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

Step-by-step solution

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

Final answer

Neither one-one nor onto.

Final answer

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

Final answer

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

Step-by-step solution

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

Final answer

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

Final answer

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

Step-by-step solution

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

Final answer

f is a bijection.

Step-by-step solution

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

Final answer

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

Step-by-step solution

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

Final answer

f is bijective (one-one and onto).

Step-by-step solution

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

Final answer

Option (D).

Final answer

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

04

Exercise 1.3 — Composition of Functions and Inverses

14Exercise questions

Step-by-step solution

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

Final answer

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

Step-by-step solution

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

Final answer

Both distributivity identities; verified pointwise.

Step-by-step solution

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

Final answer

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

Step-by-step solution

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

Final answer

f = f⁻¹ (f is its own inverse).

Final answer

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

Step-by-step solution

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

Final answer

f⁻¹(y) = 2y/(1 − y).

Step-by-step solution

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

Final answer

f⁻¹(x) = (x − 3)/4.

Step-by-step solution

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

Final answer

f⁻¹(y) = √(y − 4), verified.

Step-by-step solution

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

Final answer

f⁻¹(y) = (√(y+6) − 1)/3.

Step-by-step solution

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

Final answer

The inverse is unique.

Final answer

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

Final answer

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

Step-by-step solution

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

Final answer

Option (C) — x.

Step-by-step solution

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

Final answer

Option (B) — g(y) = 4y/(4 − 3y).

05

Exercise 1.4 — Binary Operations

12Exercise questions

Step-by-step solution

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

Final answer

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

Final answer

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

Final answer

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

Step-by-step solution

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

Final answer

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

Step-by-step solution

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

Final answer

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

Step-by-step solution

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

Final answer

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

Final answer

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

Step-by-step solution

  1. 12∗3 = 2 + 4·9 = 38; then (2∗3)∗4 = 38 + 4·16 = 102.
  2. 23∗4 = 3 + 4·16 = 67; then 2∗(3∗4) = 2 + 4·67² = 2 + 17956 = 17958.
  3. 3102 ≠ 17958, so ∗ is not associative — the differing results are exactly the failure of associativity.

Final answer

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

Step-by-step solution

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

Final answer

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

Step-by-step solution

  1. 13^{a+b} ∈ ℤ⁺ for all a,b → binary operation.
  2. 2Commutative: 3^{a+b} = 3^{b+a}. Yes.
  3. 3Associative: (a∗b)∗c = 3^{(3^{a+b})+c} while a∗(b∗c) = 3^{a + 3^{b+c}}; these differ (exponent shapes differ), so not associative.

Final answer

Binary operation, commutative, not associative.

Step-by-step solution

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

Final answer

∗ is associative but not commutative.

Step-by-step solution

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

Final answer

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

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Composition

Inverse relation

Commutative binary operation

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • A function from A to B is a relation whose every element of A maps to exactly one element of B, so a relation with two outputs for one input is not a function.
  • A function has an inverse only when it is one-one (and onto its codomain) — otherwise two inputs collapse to the same output and the inverse is ambiguous.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Maths Chapter 1 (Relations and Functions)?

There are 4 exercise questions in this chapter, numbered Ex 1.1 to Ex 1.4. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Relations and Functions Class 12 Maths?

The formulas this chapter's questions actually turn on are: Composition, Inverse relation, Commutative binary operation. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Relations and Functions important for JEE Main?

Moderate — mostly definitions and properties, worth easy marks in boards and a standard JEE Main topic.

Next Chapters

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Next: Inverse Trigonometric FunctionsRelations and Functions revision notesAll Class 12 Maths NCERT Solutions

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