NCERT Solutions for Class 12 Maths Chapter 1: Relations and Functions (NCERT 2026–27)

These Class 12 Maths Chapter 1 solutions cover Relations and Functions from the NCERT textbook (Reprint 2026–27). Every question of Exercise 1.1, Exercise 1.2 and the Miscellaneous Exercise is solved step by step — checking relations for reflexive, symmetric and transitive properties, proving equivalence relations, and testing functions for one-one (injective), onto (surjective) and bijective behaviour — with every answer cross-checked against the NCERT answer key.

Class: 12 Subject: Mathematics Chapter: 1 Chapter Name: Relations and Functions Exercises: 1.1, 1.2 & Miscellaneous Session: 2026–27
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Chapter 1 Overview

Chapter 1 of Class 12 Maths, Relations and Functions, deepens the ideas first met in Class 11. A relation R on a set A is just a subset of A × A, and we classify it as reflexive, symmetric or transitive. A relation that is all three is an equivalence relation, which neatly partitions a set into disjoint equivalence classes. The chapter then revisits functions, sorting them into one-one (injective), onto (surjective) and bijective maps, and introduces the composition of functions and the idea of an invertible function. The Class 12 Maths Chapter 1 solutions below work through every Exercise 1.1, Exercise 1.2 and Miscellaneous question in the exact order and numbering of the textbook.

Key Concepts & Definitions

Empty relation: R = φ ⊂ A × A — no element of A is related to any element of A.

Universal relation: R = A × A — every element is related to every element.

Reflexive: (a, a) ∈ R for every a ∈ A.

Symmetric: (a1, a2) ∈ R ⇒ (a2, a1) ∈ R for all a1, a2 ∈ A.

Transitive: (a1, a2) ∈ R and (a2, a3) ∈ R ⇒ (a1, a3) ∈ R.

Equivalence relation: a relation that is reflexive, symmetric and transitive at the same time.

One-one (injective): f(x1) = f(x2) ⇒ x1 = x2 — distinct inputs give distinct outputs.

Onto (surjective): for every y in the co-domain there is an x with f(x) = y, i.e. Range of f = co-domain.

Bijective: a function that is both one-one and onto; such a function is invertible.

Composition: (g°f)(x) = g(f(x)), for f : A → B and g : B → C.

Important Formulas & Results (Chapter 1)

Equivalence class: [a] = {b ∈ X : b R a}; equivalence classes partition X into disjoint subsets whose union is X.

One-one test: assume f(x1) = f(x2) and deduce x1 = x2.

Onto test: take any y in the co-domain and exhibit an x in the domain with f(x) = y.

Finite-set property: for a finite set X, a function f : X → X is one-one if and only if it is onto.

Number of one-one (and onto) maps from a set of n elements to itself = n!

Invertible function: f : X → Y is invertible ⇔ f is one-one and onto; then there is g : Y → X with g°f = IX and f°g = IY, written g = f−1.

Exercise 1.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. Determine whether each of the following relations are reflexive, symmetric and transitive: (i) Relation R in the set A = {1, 2, 3, …, 13, 14} defined as R = {(x, y) : 3x − y = 0} (ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x, y) : x − y is an integer} (v) Relation R in the set A of human beings in a town at a particular time given by   (a) R = {(x, y) : x and y work at the same place}   (b) R = {(x, y) : x and y live in the same locality}   (c) R = {(x, y) : x is exactly 7 cm taller than y}   (d) R = {(x, y) : x is wife of y}   (e) R = {(x, y) : x is father of y}

SOLUTION (i) R = {(x, y) : 3x − y = 0}, i.e. y = 3x, so R = {(1, 3), (2, 6), (3, 9), (4, 12)}. Reflexive: (1, 1) ∉ R since 3(1) ≠ 1, so not reflexive. Symmetric: (1, 3) ∈ R but (3, 1) ∉ R, so not symmetric. Transitive: (1, 3) ∈ R and (3, 9) ∈ R but (1, 9) ∉ R, so not transitive. ∴ Neither reflexive nor symmetric nor transitive. (ii) y = x + 5, x < 4 gives R = {(1, 6), (2, 7), (3, 8)}. Reflexive: (1, 1) ∉ R → not reflexive. Symmetric: (1, 6) ∈ R, (6, 1) ∉ R → not symmetric. Transitive: no pair (a, b), (b, c) both lie in R, so the condition is vacuously satisfied → transitive. ∴ Neither reflexive nor symmetric but transitive. (iii) R = {(x, y) : y is divisible by x}. Reflexive: x divides x for all x, so (x, x) ∈ R → reflexive. Symmetric: 6 is divisible by 2 so (2, 6) ∈ R, but 2 is not divisible by 6, (6, 2) ∉ R → not symmetric. Transitive: if y is divisible by x and z is divisible by y, then z is divisible by x → transitive. ∴ Reflexive and transitive but not symmetric. (iv) R = {(x, y) : x − y is an integer}. For integers x, y the difference x − y is always an integer. Reflexive: x − x = 0 is an integer → reflexive. Symmetric: if x − y is an integer then y − x is an integer → symmetric. Transitive: (x − y) + (y − z) = x − z is an integer → transitive. ∴ Reflexive, symmetric and transitive. (v)(a) “work at the same place”: x works where x works (reflexive); if x and y share a place so do y and x (symmetric); if x, y share and y, z share, then x, z share (transitive) → reflexive, symmetric and transitive. (v)(b) “live in the same locality”: by the same reasoning it is reflexive, symmetric and transitive. (v)(c) “x is exactly 7 cm taller than y”: x is not 7 cm taller than himself → not reflexive. If x is 7 cm taller than y, y is not 7 cm taller than x → not symmetric. If x is 7 cm taller than y and y is 7 cm taller than z, then x is 14 cm taller than z, not 7 → not transitive. → Neither reflexive nor symmetric nor transitive. (v)(d) “x is wife of y”: x is not her own wife → not reflexive. If x is wife of y, y is not wife of x → not symmetric. There is no x, y, z with x wife of y and y wife of z, so it is vacuously transitive → neither reflexive nor symmetric but transitive. (v)(e) “x is father of y”: x is not his own father → not reflexive. If x is father of y, y is not father of x → not symmetric. If x is father of y and y is father of z, then x is grandfather (not father) of z → not transitive. → Neither reflexive nor symmetric nor transitive.

2. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b2} is neither reflexive nor symmetric nor transitive.

SOLUTION Not reflexive: take a = ½. Then a ≤ a2 means ½ ≤ ¼, which is false, so (½, ½) ∉ R. Not symmetric: (1, 2) ∈ R because 1 ≤ 22 = 4, but (2, 1) ∉ R because 2 ≤ 12 = 1 is false. Not transitive: (3, −2) ∈ R since 3 ≤ (−2)2 = 4, and (−2, 1) ∈ R since −2 ≤ 12 = 1, but (3, 1) ∉ R since 3 ≤ 12 = 1 is false. ∴ R is neither reflexive nor symmetric nor transitive.

3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

SOLUTION R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}. Reflexive: (1, 1) ∉ R (since 1 ≠ 1 + 1), so not reflexive. Symmetric: (1, 2) ∈ R but (2, 1) ∉ R, so not symmetric. Transitive: (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R, so not transitive. ∴ Neither reflexive nor symmetric nor transitive.

4. Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.

SOLUTION Reflexive: a ≤ a is true for every real a, so (a, a) ∈ R. Transitive: if a ≤ b and b ≤ c, then a ≤ c, so (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R. Not symmetric: (2, 3) ∈ R since 2 ≤ 3, but (3, 2) ∉ R since 3 ≤ 2 is false. ∴ R is reflexive and transitive but not symmetric.

5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive.

SOLUTION Not reflexive: take a = ½. a ≤ a3 means ½ ≤ &frac18;, which is false, so (½, ½) ∉ R. Not symmetric: (1, 2) ∈ R since 1 ≤ 23 = 8, but (2, 1) ∉ R since 2 ≤ 13 = 1 is false. Not transitive: (3, &frac32;) ∈ R since 3 ≤ (1.5)3 = 3.375, and (&frac32;, &frac65;) ∈ R since 1.5 ≤ (1.2)3 = 1.728, but (3, &frac65;) ∉ R since 3 ≤ (1.2)3 = 1.728 is false. ∴ Neither reflexive nor symmetric nor transitive.

6. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

SOLUTION Symmetric: (1, 2) ∈ R and its reverse (2, 1) ∈ R; there are no other pairs, so R is symmetric. Not reflexive: (1, 1), (2, 2), (3, 3) are not in R. Not transitive: (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R. ∴ R is symmetric but neither reflexive nor transitive.

7. Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.

SOLUTION Reflexive: every book x has the same number of pages as itself, so (x, x) ∈ R. Symmetric: if x and y have the same number of pages, then y and x have the same number of pages, so (x, y) ∈ R ⇒ (y, x) ∈ R. Transitive: if x, y have equal pages and y, z have equal pages, then x, z have equal pages, so (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R. ∴ R is reflexive, symmetric and transitive, hence an equivalence relation.

8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a − b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

SOLUTION Reflexive: |a − a| = 0 is even, so (a, a) ∈ R. Symmetric: |a − b| = |b − a|, so (a, b) ∈ R ⇒ (b, a) ∈ R. Transitive: if |a − b| and |b − c| are both even, then a, b have the same parity and b, c have the same parity, so a, c have the same parity, hence |a − c| is even ⇒ (a, c) ∈ R. So R is an equivalence relation. The elements of {1, 3, 5} are all odd, so the difference of any two is even — they are related to each other. The elements of {2, 4} are both even, so they are related to each other. An odd and an even number differ by an odd number, so no element of {1, 3, 5} is related to any element of {2, 4}.

9. Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by (i) R = {(a, b) : |a − b| is a multiple of 4} (ii) R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.

SOLUTION Here A = {0, 1, 2, …, 12}. (i) Reflexive: |a − a| = 0 = 4×0 is a multiple of 4. Symmetric: |a − b| = |b − a|. Transitive: if |a − b| = 4m and |b − c| = 4n, then a − c is a multiple of 4. So R is an equivalence relation. Elements related to 1: those b with |1 − b| a multiple of 4, i.e. b = 1, 5, 9 (since |1−1|=0, |1−5|=4, |1−9|=8). ∴ {1, 5, 9}. (ii) R = {(a, b) : a = b}. Reflexive: a = a. Symmetric: a = b ⇒ b = a. Transitive: a = b, b = c ⇒ a = c. So R is an equivalence relation. Elements related to 1: only b with b = 1. ∴ {1}.

10. Give an example of a relation. Which is (i) Symmetric but neither reflexive nor transitive. (ii) Transitive but neither reflexive nor symmetric. (iii) Reflexive and symmetric but not transitive. (iv) Reflexive and transitive but not symmetric. (v) Symmetric and transitive but not reflexive.

SOLUTION (i) On A = {1, 2, 3}, R = {(1, 2), (2, 1)}. Symmetric; not reflexive ((1,1)∉R); not transitive ((1,2),(2,1)∈R but (1,1)∉R). (ii) On R, R = {(a, b) : a < b}. Transitive; not reflexive (a < a false); not symmetric (1 < 2 but 2 < 1 false). (iii) On A = {1, 2, 3}, R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}. Reflexive and symmetric; not transitive ((1,2),(2,3)∈R but (1,3)∉R). (iv) On R, R = {(a, b) : a ≤ b}. Reflexive and transitive; not symmetric (2 ≤ 3 but 3 ≤ 2 false). (v) On A = {1, 2, 3}, R = {(1, 1), (2, 2), (1, 2), (2, 1)}. Symmetric and transitive; not reflexive ((3, 3) ∉ R).

11. Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

SOLUTION Let d(P) denote the distance of point P from the origin O. Reflexive: d(P) = d(P), so (P, P) ∈ R. Symmetric: d(P) = d(Q) ⇒ d(Q) = d(P), so (P, Q) ∈ R ⇒ (Q, P) ∈ R. Transitive: d(P) = d(Q) and d(Q) = d(S) ⇒ d(P) = d(S), so R is transitive. Hence R is an equivalence relation. The points related to a fixed P ≠ (0, 0) are all points Q with d(Q) = d(P) = r (say), where r > 0. These are exactly the points at distance r from O, i.e. the circle with centre O and radius OP, which passes through P.

12. Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 is similar to T2}, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are related?

SOLUTION Reflexive: every triangle is similar to itself, so (T, T) ∈ R. Symmetric: if T1 is similar to T2, then T2 is similar to T1. Transitive: if T1 ~ T2 and T2 ~ T3, then T1 ~ T3. Hence R is an equivalence relation. Two triangles are similar when corresponding sides are proportional. For T1 (3, 4, 5) and T3 (6, 8, 10): 6/3 = 8/4 = 10/5 = 2, so T1 is similar to T3. T2 (5, 12, 13) has sides not proportional to either, so it is not related to them. ∴ T1 is related to T3.

13. Show that the relation R defined in the set A of all polygons as R = {(P1, P2) : P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

SOLUTION Reflexive: a polygon has the same number of sides as itself. Symmetric: if P1 and P2 have the same number of sides, so do P2 and P1. Transitive: if P1, P2 have equal sides and P2, P3 have equal sides, then P1, P3 have equal sides. Hence R is an equivalence relation. The triangle T (3, 4, 5) has 3 sides, so the elements related to it are all polygons with 3 sides. ∴ The set of all elements related to T is the set of all triangles in A.

14. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

SOLUTION Reflexive: every line is parallel to itself, so (L, L) ∈ R. Symmetric: if L1 is parallel to L2, then L2 is parallel to L1. Transitive: if L1 ∥ L2 and L2 ∥ L3, then L1 ∥ L3. Hence R is an equivalence relation. Lines parallel to y = 2x + 4 have the same slope, 2. So they are the lines y = 2x + c, where c is any real constant. ∴ The set of all related lines is {y = 2x + c : c ∈ R}.

15. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer. (A) R is reflexive and symmetric but not transitive. (B) R is reflexive and transitive but not symmetric. (C) R is symmetric and transitive but not reflexive. (D) R is an equivalence relation.

SOLUTION Reflexive: (1,1), (2,2), (3,3), (4,4) are all in R, so R is reflexive. Symmetric: (1, 2) ∈ R but (2, 1) ∉ R, so R is not symmetric. Transitive: check all chains — (1,3),(3,2)→(1,2)∈R; (1,2),(2,2)→(1,2)∈R; (3,2),(2,2)→(3,2)∈R; etc. No chain fails, so R is transitive. ∴ correct option is (B).

16. Let R be the relation in the set N given by R = {(a, b) : a = b − 2, b > 6}. Choose the correct answer. (A) (2, 4) ∈ R   (B) (3, 8) ∈ R   (C) (6, 8) ∈ R   (D) (8, 7) ∈ R

SOLUTION A pair (a, b) is in R when a = b − 2 and b > 6. (A) (2, 4): b = 4 is not > 6 → no. (B) (3, 8): b = 8 > 6 but a = 8 − 2 = 6 ≠ 3 → no. (C) (6, 8): b = 8 > 6 and a = 8 − 2 = 6 &checkmark; → yes. (D) (8, 7): b = 7 > 6 but a = 7 − 2 = 5 ≠ 8 → no. ∴ correct option is (C).

Exercise 1.2 Solutions

1. Show that the function f : R∗ → R∗ defined by f(x) = 1/x is one-one and onto, where R∗ is the set of all non-zero real numbers. Is the result true, if the domain R∗ is replaced by N with co-domain being same as R∗?

SOLUTION One-one: f(x1) = f(x2) ⇒ 1/x1 = 1/x2 ⇒ x1 = x2. Onto: for any non-zero y, choose x = 1/y (which is non-zero); then f(1/y) = 1/(1/y) = y. So every y in R∗ is an image, hence onto. Thus f is a bijection on R∗. If domain is N: define g : N → R∗ by g(x) = 1/x. It is still one-one. But it is not onto, because, for example, y = 2 ∈ R∗ has no natural number x with 1/x = 2 (that would need x = ½ ∉ N). ∴ The result is not true when the domain is replaced by N (answer: No).

2. Check the injectivity and surjectivity of the following functions: (i) f : N → N given by f(x) = x2 (ii) f : Z → Z given by f(x) = x2 (iii) f : R → R given by f(x) = x2 (iv) f : N → N given by f(x) = x3 (v) f : Z → Z given by f(x) = x3

SOLUTION (i) f(x) = x2 on N. Injective: for x1, x2 ∈ N, x12 = x22 ⇒ x1 = x2 (both positive), so injective. Surjective: 2 ∈ N is not a perfect square, so it has no pre-image → not surjective. ∴ Injective but not surjective. (ii) f(x) = x2 on Z. f(−1) = 1 = f(1), so not injective. 2 ∈ Z has no integer square root, so not surjective. ∴ Neither injective nor surjective. (iii) f(x) = x2 on R. f(−1) = 1 = f(1), so not injective. A negative number like −2 has no real pre-image (squares are ≥ 0), so not surjective. ∴ Neither injective nor surjective. (iv) f(x) = x3 on N. x13 = x23 ⇒ x1 = x2, so injective. 2 ∈ N is not a perfect cube, so not surjective. ∴ Injective but not surjective. (v) f(x) = x3 on Z. x13 = x23 ⇒ x1 = x2, so injective. But 2 ∈ Z has no integer cube root, so not surjective. ∴ Injective but not surjective.

3. Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

SOLUTION Not one-one: f(1.2) = [1.2] = 1 and f(1.7) = [1.7] = 1, so two different inputs share the same output. Not onto: the values of [x] are always integers, so a non-integer in the co-domain such as 0.5 has no pre-image. ∴ The greatest integer function is neither one-one nor onto.

4. Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is −x, if x is negative.

SOLUTION Not one-one: f(−1) = |−1| = 1 and f(1) = |1| = 1, so different inputs give the same output. Not onto: |x| ≥ 0 for all x, so no negative number such as −2 in the co-domain has a pre-image. ∴ The modulus function is neither one-one nor onto.

5. Show that the Signum Function f : R → R, given by f(x) = 1 if x > 0, f(x) = 0 if x = 0, f(x) = −1 if x < 0, is neither one-one nor onto.

SOLUTION Not one-one: f(1) = 1 and f(2) = 1, so two different positive inputs give the same value 1. Not onto: the range of the signum function is just {−1, 0, 1}, so any other real number in the co-domain (e.g. 2) has no pre-image. ∴ The signum function is neither one-one nor onto.

6. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

SOLUTION f(1) = 4, f(2) = 5, f(3) = 6 — the three distinct elements 1, 2, 3 have three distinct images 4, 5, 6. No two different elements of A have the same image, so f is one-one (injective).

7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f(x) = 3 − 4x (ii) f : R → R defined by f(x) = 1 + x2

SOLUTION (i) f(x) = 3 − 4x. One-one: f(x1) = f(x2) ⇒ 3 − 4x1 = 3 − 4x2 ⇒ x1 = x2. Onto: for any y, solve y = 3 − 4x ⇒ x = (3 − y)/4 ∈ R, so every y has a pre-image. ∴ f is one-one and onto (bijective). (ii) f(x) = 1 + x2. Not one-one: f(−1) = 2 = f(1). Not onto: 1 + x2 ≥ 1, so a value like 0 in the co-domain has no pre-image. ∴ f is neither one-one nor onto.

8. Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.

SOLUTION One-one: f(a1, b1) = f(a2, b2) ⇒ (b1, a1) = (b2, a2) ⇒ b1 = b2 and a1 = a2 ⇒ (a1, b1) = (a2, b2). Onto: take any (b, a) ∈ B × A; then (a, b) ∈ A × B and f(a, b) = (b, a). So every element of B × A has a pre-image. ∴ f is one-one and onto, hence bijective.

9. Let f : N → N be defined by f(n) = (n + 1)/2 if n is odd, and f(n) = n/2 if n is even, for all n ∈ N. State whether the function f is bijective. Justify your answer.

SOLUTION Not one-one: f(1) = (1 + 1)/2 = 1 (n = 1 odd) and f(2) = 2/2 = 1 (n = 2 even). Two different inputs 1 and 2 give the same value 1. Since f is not injective, it cannot be bijective. (It is, however, onto: any m ∈ N is hit by the even number 2m.) ∴ f is not bijective (answer: No).

10. Let A = R − {3} and B = R − {1}. Consider the function f : A → B defined by f(x) = (x − 2)/(x − 3). Is f one-one and onto? Justify your answer.

SOLUTION One-one: suppose f(x1) = f(x2): (x1 − 2)/(x1 − 3) = (x2 − 2)/(x2 − 3). Cross-multiplying: (x1 − 2)(x2 − 3) = (x2 − 2)(x1 − 3). Expanding both sides and cancelling x1x2 gives −3x1 − 2x2 = −3x2 − 2x1, i.e. −x1 = −x2, so x1 = x2. Hence one-one. Onto: let y ∈ B (y ≠ 1). Solve y = (x − 2)/(x − 3): y(x − 3) = x − 2 ⇒ yx − 3y = x − 2 ⇒ x(y − 1) = 3y − 2 ⇒ x = (3y − 2)/(y − 1), which is defined for y ≠ 1 and lies in A (x ≠ 3). So every y has a pre-image. Hence onto. ∴ f is one-one and onto (answer: Yes).

11. Let f : R → R be defined as f(x) = x4. Choose the correct answer. (A) f is one-one onto   (B) f is many-one onto   (C) f is one-one but not onto   (D) f is neither one-one nor onto.

SOLUTION f(−1) = (−1)4 = 1 = f(1), so f is not one-one (many-one). x4 ≥ 0 for all x, so negative numbers in the co-domain (e.g. −2) have no pre-image → not onto. ∴ correct option is (D).

12. Let f : R → R be defined as f(x) = 3x. Choose the correct answer. (A) f is one-one onto   (B) f is many-one onto   (C) f is one-one but not onto   (D) f is neither one-one nor onto.

SOLUTION One-one: 3x1 = 3x2 ⇒ x1 = x2. Onto: for any y ∈ R, take x = y/3; then f(y/3) = 3(y/3) = y. So every y has a pre-image. ∴ f is one-one and onto; correct option is (A).

Miscellaneous Exercise on Chapter 1

1. Show that the function f : R → {x ∈ R : −1 < x < 1} defined by f(x) = x / (1 + |x|), x ∈ R is one one and onto function.

SOLUTION For x ≥ 0, f(x) = x/(1 + x), which is increasing and takes values in [0, 1); for x < 0, f(x) = x/(1 − x), which is increasing and takes values in (−1, 0). So the range is exactly (−1, 1) and f is strictly increasing throughout. One-one: a strictly increasing function is one-one. (Algebraically, if f(x1) = f(x2) then, taking the same sign branch, the relations x/(1 + x) or x/(1 − x) each give x1 = x2; opposite signs give opposite-sign outputs, so cannot be equal.) Onto: let y ∈ (−1, 1). If 0 ≤ y < 1, solve y = x/(1 + x) ⇒ x = y/(1 − y) ≥ 0. If −1 < y < 0, solve y = x/(1 − x) ⇒ x = y/(1 + y) < 0. In each case x ∈ R maps to y. Hence onto. ∴ f is one-one and onto.

2. Show that the function f : R → R given by f(x) = x3 is injective.

SOLUTION Suppose f(x1) = f(x2), i.e. x13 = x23. Then x13 − x23 = 0 ⇒ (x1 − x2)(x12 + x1x2 + x22) = 0. The quadratic factor x12 + x1x2 + x22 = (x1 + x2/2)2 + 3x22/4 is zero only when x1 = x2 = 0. So in all cases x1 − x2 = 0, i.e. x1 = x2. Hence f is injective.

3. Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.

SOLUTION Reflexive: every set satisfies A ⊆ A, so (A, A) ∈ R. Symmetric: take a non-empty set X and a proper subset A ⊂ X with A ≠ X. Then A ⊆ X so (A, X) ∈ R, but X ⊈ A, so (X, A) ∉ R. Hence R is not symmetric. Since symmetry fails, R is not an equivalence relation (answer: No). (It is reflexive and transitive, but not symmetric.)

4. Find the number of all onto functions from the set {1, 2, 3, ……, n} to itself.

SOLUTION For a finite set, an onto function from the set to itself must also be one-one (the finite-set property). An onto map of {1, 2, …, n} onto itself is therefore a permutation of the n elements. The number of permutations of n distinct objects is n!. ∴ The number of onto functions is n!.

5. Let A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and f, g : A → B be functions defined by f(x) = x2 − x, x ∈ A and g(x) = 2|x − ½| − 1, x ∈ A. Are f and g equal? Justify your answer. (Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g(a) ∀ a ∈ A, are called equal functions).

SOLUTION Compute f(x) = x2 − x at each element: f(−1) = 1 + 1 = 2; f(0) = 0; f(1) = 1 − 1 = 0; f(2) = 4 − 2 = 2. Compute g(x) = 2|x − ½| − 1: g(−1) = 2(1.5) − 1 = 2; g(0) = 2(0.5) − 1 = 0; g(1) = 2(0.5) − 1 = 0; g(2) = 2(1.5) − 1 = 2. For every a ∈ A, f(a) = g(a). Hence f and g are equal functions (answer: Yes).

6. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is (A) 1   (B) 2   (C) 3   (D) 4

SOLUTION A reflexive relation must contain (1,1), (2,2), (3,3). For symmetry, with (1, 2) and (1, 3) we must also include (2, 1) and (3, 1). So far R = {(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)}. This R is not transitive, since (2, 1) ∈ R and (1, 3) ∈ R but (2, 3) ∉ R. Adding (2, 3) (and for symmetry (3, 2)) would force transitivity, so we keep R as it is. Exactly one such relation exists. ∴ correct option is (A) 1.

7. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is (A) 1   (B) 2   (C) 3   (D) 4

SOLUTION An equivalence relation containing (1, 2) must (by symmetry) also contain (2, 1) and (by reflexivity) all of (1,1), (2,2), (3,3). First relation: R1 = {(1,1),(2,2),(3,3),(1,2),(2,1)} — reflexive, symmetric and transitive. Second relation: the universal relation A × A (all 9 pairs), obtained when we additionally relate 3 to 1 and 2; this is forced once we add any further pair such as (2, 3). No equivalence relation lies strictly between these two, so there are exactly two. ∴ correct option is (B) 2.

Common Mistakes to Avoid

Watch out for these

  • Declaring a relation reflexive without checking every element — missing one pair (a, a) breaks reflexivity.
  • Confusing “not symmetric” with “asymmetric” — you only need one counter-example pair (a, b) ∈ R with (b, a) ∉ R.
  • Treating a relation with no chains as “not transitive” — if there is no pair (a, b), (b, c) in R, the relation is vacuously transitive.
  • For one-one, picking specific numbers instead of proving f(x1) = f(x2) ⇒ x1 = x2 in general.
  • For onto, forgetting to check the co-domain: x2, |x| and x4 are never onto R because their values are restricted.
  • Assuming “A ⊂ B” is an equivalence relation — it is reflexive and transitive but not symmetric.

Practice MCQs & Assertion–Reason

1. A relation R on a set A that is reflexive, symmetric and transitive is called:

(a) empty relation    (b) universal relation    (c) equivalence relation    (d) identity relation

2. The relation R = {(a, b) : a ≤ b} on R is:

(a) reflexive and symmetric    (b) reflexive and transitive but not symmetric    (c) symmetric and transitive    (d) an equivalence relation

3. The function f : R → R, f(x) = x4 is:

(a) one-one onto    (b) many-one onto    (c) one-one but not onto    (d) neither one-one nor onto

4. The function f : R → R, f(x) = 3x is:

(a) one-one onto    (b) many-one onto    (c) one-one not onto    (d) neither one-one nor onto

5. The number of all onto functions from {1, 2, 3, …, n} to itself is:

(a) n2    (b) 2n    (c) n!    (d) nn

6. For A = {1, 2, 3}, the number of equivalence relations containing (1, 2) is:

(a) 1    (b) 2    (c) 3    (d) 4

7. The greatest integer function f(x) = [x] on R is:

(a) one-one    (b) onto    (c) bijective    (d) neither one-one nor onto

8. If R = {(a, b) : a = b − 2, b > 6} on N, which pair belongs to R?

(a) (2, 4)    (b) (3, 8)    (c) (6, 8)    (d) (8, 7)

9. A function f : X → Y is invertible if and only if it is:

(a) only one-one    (b) only onto    (c) one-one and onto    (d) constant

10. The set of all lines parallel to y = 2x + 4 is:

(a) {y = 2x + c : c ∈ R}    (b) {y = −½x + c}    (c) {y = 4x + c}    (d) only y = 2x + 4

Answer key: 1-(c), 2-(b), 3-(d), 4-(a), 5-(c), 6-(b), 7-(d), 8-(c), 9-(c), 10-(a).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The relation R = {(a, b) : a ≤ b2} on R is not reflexive.

Reason: For a = ½, a ≤ a2 becomes ½ ≤ ¼, which is false.

A-R 2. Assertion: The function f : R → R, f(x) = x2 is not one-one.

Reason: f(−1) = f(1) = 1, so two distinct inputs have the same image.

A-R 3. Assertion: The relation “A ⊂ B” on the power set P(X) is an equivalence relation.

Reason: The subset relation is reflexive and transitive.

A-R 4. Assertion: For a finite set X, a one-one function f : X → X is necessarily onto.

Reason: For a finite set, the number of distinct images of a one-one map equals the size of the set, which forces it to be onto.

A-R 5. Assertion: The number of onto functions from {1, 2, 3} to itself is 6.

Reason: An onto function from a finite set to itself is a permutation, and 3! = 6.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • A relation R on A is a subset of A × A; it is reflexive if (a, a) ∈ R for all a, symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R, and transitive if (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R.
  • An equivalence relation is reflexive, symmetric and transitive; it partitions the set into disjoint equivalence classes.
  • A function is one-one (injective) if f(x1) = f(x2) ⇒ x1 = x2, and onto (surjective) if its range equals the co-domain.
  • A bijective (one-one and onto) function is invertible, with g°f = IX and f°g = IY.
  • For a finite set X, f : X → X is one-one if and only if it is onto.
  • The number of one-one (equivalently onto) maps of an n-element set to itself is n!.
  • Watch the co-domain: x2, x4 and |x| are never onto R because their outputs are restricted.

How to score full marks in this chapter

For every relation, prove or disprove all three properties (reflexive, symmetric, transitive) separately, giving a clear counter-example whenever one fails. For functions, prove one-one by assuming f(x1) = f(x2) and deriving x1 = x2, and prove onto by solving y = f(x) for x and showing it lies in the domain. State the finite-set property explicitly when counting one-one or onto maps, and always confirm your numerical answers (such as {1, 5, 9} or n!) against the structure of the problem.

Frequently Asked Questions

What is Class 12 Maths Chapter 1 Relations and Functions about?

Chapter 1 studies different types of relations (reflexive, symmetric, transitive and equivalence relations), equivalence classes, and types of functions (one-one/injective, onto/surjective and bijective), along with the composition of functions and the idea of an invertible function.

How many exercises are there in Class 12 Maths Chapter 1?

There are two main exercises — Exercise 1.1 (relations, 16 questions) and Exercise 1.2 (functions, 12 questions) — plus a Miscellaneous Exercise of 7 questions, all solved step by step on this page.

What is the difference between one-one and onto functions?

A function is one-one (injective) when distinct inputs always give distinct outputs, i.e. f(x₁) = f(x₂) implies x₁ = x₂. It is onto (surjective) when every element of the co-domain is the image of at least one input, i.e. the range equals the co-domain. A function that is both is bijective.

Are these Class 12 Maths Chapter 1 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026-27 session, with every answer verified against the book’s answer key.

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