NCERT Solutions for Class 12 Mathematics Chapter 1: Relations and Functions – Free PDF Download

Class 12 Mathematics Chapter 1 – Relations and Functions: NCERT Solutions

Chapter 1 of Class 12 Mathematics opens the book with relations and functions — the language used throughout the rest of the syllabus (Chapter 2 Inverse Trigonometric Functions builds directly on the invertible-function ideas here). In the current 2026-27 rationalised syllabus this chapter has Exercise 1.1 (16 questions, types of relations), Exercise 1.2 (12 questions, types of functions) and a Miscellaneous Exercise (19 questions covering both, plus binary operations). These Class 12 Mathematics Chapter 1 solutions are also useful as quick revision notes before exams.

Why this chapter matters for boards

Reflexive/symmetric/transitive proofs and one-one/onto checks are near-guaranteed 2-5 mark questions every year, and the equivalence-relation and invertible-function ideas from this chapter reappear directly in Chapter 2 and in JEE/competitive syllabi.

Exercise 1.1 Solutions

1. Q1. Determine whether each of the following relations are reflexive, symmetric and transitive: (i) Relation R in the set A = {1,2,…,14} defined as R = {(x,y): 3x−y=0}. (ii) Relation R in N defined as R={(x,y): y=x+5 and x<4}. (iii) Relation R in A={1,2,3,4,5,6} as R={(x,y): y is divisible by x}. (iv) Relation R in Z defined as R={(x,y): x−y is an integer}. (v) Relation R in the set A of human beings in a town: (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}.
   Answer: (i) R={(1,3),(2,6),(3,9),(4,12)} — not reflexive [(1,1)∉R], not symmetric [(1,3)∈R but (3,1)∉R], not transitive [no chain exists since second elements never recur as first elements]. (ii) R={(1,6),(2,7),(3,8)} — not reflexive, not symmetric, not transitive (vacuously transitive is not claimed since it fails reflexive/symmetric; check shows no violation of transitivity as no two pairs chain, so it is trivially transitive, but neither reflexive nor symmetric). (iii) Reflexive (every number divides itself) and transitive (if y divisible by x and z divisible by y, z divisible by x), but not symmetric (2 divides 4 but 4 does not divide 2). (iv) Reflexive (x−x=0 is an integer), symmetric (if x−y is an integer so is y−x), transitive (sum of two integers is an integer) — R is an equivalence relation. (v)(a) Reflexive, symmetric, transitive — equivalence relation. (b) Reflexive, symmetric, transitive — equivalence relation. (c) Not reflexive (x is not 7cm taller than itself), not symmetric (if x is 7cm taller than y, y is not 7cm taller than x), not transitive. (d) Not reflexive, not symmetric (if x is wife of y, y is not wife of x), not transitive. (e) Not reflexive, not symmetric, not transitive (x father of y, y father of z does not make x father of z — x is grandfather).
2. Q2. 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.
   Answer: Not reflexive: take a=1/2, then 1/2≤(1/2)²=1/4 is false. Not symmetric: 1≤4=2² is true (1,2∈R) but 2≤1²=1 is false. Not transitive: 3≤(−2)²=4 true, −2≤1²=1 true, but 3≤1²=1 is false.
3. Q3. 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.
   Answer: Not reflexive [(1,1)∉R since 1≠2]. Not symmetric: (1,2)∈R but (2,1)∉R since 1≠3. Not transitive: (1,2)∈R,(2,3)∈R but (1,3)∉R.
4. Q4. Show that the relation R in R defined as R={(a,b): a≤b} is reflexive and transitive but not symmetric.
   Answer: Reflexive: a≤a always true. Transitive: a≤b and b≤c ⇒ a≤c. Not symmetric: 2≤3 true but 3≤2 false.
5. Q5. Check whether the relation R in R defined by R={(a,b): a≤b³} is reflexive, symmetric or transitive.
   Answer: Not reflexive: a=1/2, 1/2≤(1/2)³=1/8 is false. Not symmetric: 1≤2³=8 true but 2≤1³=1 false. Not transitive: 100≤5³=125 true, 5≤2³=8 true, but 100≤2³=8 is false. So not transitive.
6. Q6. Show that relation R in the set {1,2,3} given by R={(1,2),(2,1)} is symmetric but neither reflexive nor transitive.
   Answer: Symmetric: (1,2)∈R and (2,1)∈R, condition holds. Not reflexive: (1,1)∉R. Not transitive: (1,2)∈R,(2,1)∈R but (1,1)∉R.
7. Q7. Show that relation R in the set A of all books in a library, given by R={(x,y): x and y have same number of pages}, is an equivalence relation.
   Answer: Reflexive: x has same pages as itself. Symmetric: if x,y have same pages then y,x do too. Transitive: if x,y same pages and y,z same pages, then x,z same pages. So R is reflexive, symmetric and transitive — equivalence relation.
8. Q8. Show that the relation R in A={1,2,3,4,5} given by R={(a,b): |a−b| is even} is an equivalence relation. Show all elements of {1,3,5} are related to each other and all of {2,4} are related to each other, but no element of {1,3,5} is related to any element of {2,4}.
   Answer: Reflexive: |a−a|=0, even. Symmetric: |a−b|=|b−a|. Transitive: if |a−b| and |b−c| are even, a and b have the same parity, b and c have the same parity, so a and c have the same parity, making |a−c| even. So R is an equivalence relation. Within {1,3,5} every pairwise difference is even (all odd numbers), similarly within {2,4}; but between the sets, e.g. |1−2|=1 is odd, so no cross pair is related.
9. Q9. 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.
   Answer: (i) Reflexive (|a−a|=0), symmetric (|a−b|=|b−a|), transitive (sum of two multiples of 4 is a multiple of 4, and if a−b, b−c are multiples of 4 then a−c is too) — equivalence relation. Elements related to 1: {1,5,9}. (ii) Trivially reflexive, symmetric, transitive — equivalence relation. Elements related to 1: {1}.
10. Q10. 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.
    Answer: On A={1,2,3}: (i) R={(1,2),(2,1)}. (ii) R={(1,2)} alone is vacuously transitive and not reflexive/symmetric. (iii) R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)} (fails transitivity: 1,2 and 2,3 related but 1,3 not). (iv) R={(1,1),(2,2),(3,3),(1,2)} (reflexive, transitive, but (2,1)∉R). (v) R={(1,2),(2,1),(1,1)} (symmetric, transitive, but (2,2)∉R so not reflexive).
11. Q11. Show that the relation R in the set A of points in a plane given by R={(P,Q): distance of P from origin is same as distance of Q from origin} is an equivalence relation. Further show that the set of points related to a point P≠(0,0) is the circle passing through P with centre at the origin.
    Answer: Reflexive/symmetric/transitive all follow immediately since “equal distance from a fixed point” is an equality of real numbers, which is itself an equivalence relation — so R is an equivalence relation. For fixed P at distance r=OP from the origin, every point Q with OQ=r lies on the circle of radius r centred at O, and every point of that circle is at distance r — so the related set is exactly that circle.
12. Q12. Show that the relation R defined in the set A of all triangles as R={(T1,T2): T1 is similar to T2} is an equivalence relation. Consider T1 (sides 3,4,5), T2 (sides 5,12,13), T3 (sides 6,8,10). Which triangles are related?
    Answer: Similarity is reflexive (every triangle is similar to itself), symmetric, and transitive, so R is an equivalence relation. T1 (3,4,5) and T3 (6,8,10) have proportional sides (ratio 1:2), so T1 is related to T3. T2 (5,12,13) has no common ratio with T1 or T3, so T2 is related to neither.
13. Q13. 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 elements related to the right triangle T with sides 3,4,5?
    Answer: Reflexive/symmetric/transitive all hold since “same number of sides” is an equality of a count — equivalence relation. The set related to T is the set of all triangles (every polygon with exactly 3 sides), since T itself has 3 sides.
14. Q14. Let L be the set of all lines in the XY-plane and 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.
    Answer: Reflexive (every line is parallel to itself), symmetric, transitive (lines parallel to a common line are parallel to each other) — equivalence relation. The set of lines related to y=2x+4 is every line of the form y=2x+c (c any real number, same slope 2).
15. Q15. 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) reflexive and symmetric but not transitive (B) reflexive and transitive but not symmetric (C) symmetric and transitive but not reflexive (D) an equivalence relation.
    Answer: (B). (1,1),(2,2),(3,3),(4,4) all present — reflexive. (1,3)∈R but (3,1)∉R — not symmetric. Every chain closes: (1,3),(3,2)⇒(1,2)∈R; (1,2),(2,2)⇒(1,2)∈R — transitive holds.
16. Q16. 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.
    Answer: (C). a=6,b=8: 6=8−2 and 8>6, both conditions satisfied.

Exercise 1.2 Solutions

1. Q1. 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*?
   Answer: One-one: f(x1)=f(x2)⇒1/x1=1/x2⇒x1=x2. Onto: for any y∈R*, x=1/y∈R* gives f(x)=y. So f is a bijection on R*. If domain is restricted to N: f is still one-one, but not onto — e.g. y=2/3 has no x∈N with 1/x=2/3. So f is one-one but not onto when domain is N.
2. Q2. Check the injectivity and surjectivity of: (i) f:N→N, f(x)=x² (ii) f:Z→Z, f(x)=x² (iii) f:R→R, f(x)=x² (iv) f:N→N, f(x)=x³ (v) f:Z→Z, f(x)=x³.
   Answer: (i) Injective, not surjective (2 has no square root in N). (ii) Not injective (f(−1)=f(1)=1), not surjective. (iii) Not injective, not surjective. (iv) Injective, not surjective (2 is not a perfect cube). (v) Injective, not surjective.
3. Q3. Prove that the Greatest Integer Function f:R→R, f(x)=[x], is neither one-one nor onto.
   Answer: Not one-one: f(1.5)=f(1.9)=1 but 1.5≠1.9. Not onto: there is no real x with f(x)=1.5 since [x] is always an integer.
4. Q4. Show that the Modulus Function f:R→R, f(x)=|x|, is neither one-one nor onto.
   Answer: Not one-one: f(−1)=f(1)=1. Not onto: no x gives f(x)=−1 since |x|≥0 always.
5. Q5. Show that the Signum Function f:R→R defined by f(x)=1 (x>0), 0 (x=0), −1 (x<0), is neither one-one nor onto.
   Answer: Not one-one: f(2)=f(5)=1. Not onto: range is only {−1,0,1}, so no x maps to 2.
6. Q6. Let A={1,2,3}, B={4,5,6,7} and f={(1,4),(2,5),(3,6)} be a function from A to B. Show that f is one-one.
   Answer: Every element of A maps to a distinct element of B (1→4, 2→5, 3→6, all different) — f is one-one.
7. Q7. State whether each function is one-one, onto or bijective, with justification: (i) f:R→R, f(x)=3−4x (ii) f:R→R, f(x)=1+x².
   Answer: (i) One-one and onto — bijective. (ii) Not one-one (f(1)=f(−1)=2), not onto (range is [1,∞)) — neither one-one nor onto.
8. Q8. Let A and B be sets. Show that f:A×B→B×A defined by f(a,b)=(b,a) is bijective.
   Answer: One-one: f(a1,b1)=f(a2,b2)⇒(b1,a1)=(b2,a2)⇒a1=a2,b1=b2. Onto: for any (b,a)∈B×A, the pair (a,b)∈A×B maps to it. So f is bijective.
9. Q9. Let f:N→N be defined by f(n)=(n+1)/2 if n is odd, n/2 if n is even. State whether f is bijective. Justify.
   Answer: Not one-one: f(1)=1 and f(2)=1, but 1≠2. Since f is not one-one, it is not bijective.
10. Q10. Let A=R−{3}, B=R−{1}. Consider f:A→B, f(x)=(x−2)/(x−3). Is f one-one and onto? Justify.
    Answer: One-one and onto (solving y=(x−2)/(x−3) for x gives x=(3y−2)/(y−1), defined and in A for every y∈B) — f is one-one and onto.
11. Q11. Let f:R→R, f(x)=x⁴. Choose the correct answer: (A) one-one onto (B) many-one onto (C) one-one but not onto (D) neither one-one nor onto.
    Answer: (D). f(−1)=f(1)=1 (not one-one); negative numbers have no fourth root in R (not onto).
12. Q12. Let f:R→R, f(x)=3x. Choose the correct answer: (A) one-one onto (B) many-one onto (C) one-one but not onto (D) neither one-one nor onto.
    Answer: (A). Linear with non-zero slope — one-one; for any y, x=y/3 works — onto.

Miscellaneous Exercise Solutions

1. Q1. Let f:R→R be defined as f(x)=10x+7. Find g:R→R such that gof=fog=1_R.
   Answer: g(x)=(x−7)/10. Check: g(f(x))=x, f(g(x))=x. Both compositions give the identity.
2. Q2. Let f:W→W, f(n)=n−1 if n is odd, n+1 if n is even (W=whole numbers). Show f is invertible and find its inverse.
   Answer: f swaps consecutive pairs (0↔1, 2↔3, …). Applying f twice returns the original number, so f is its own inverse: f⁻¹=f.
3. Q3. If f:R→R is defined by f(x)=x²−3x+2, find f(f(x)).
   Answer: f(f(x)) = (x²−3x+2)²−3(x²−3x+2)+2 = x⁴−6x³+10x²−3x.
4. Q4. Show that f:R→{x∈R: −1<x<1} defined by f(x)=x/(1+|x|) is one-one and onto.
   Answer: f is strictly increasing on all of R (piecewise increasing, continuous at 0), hence one-one, and its range covers all of (−1,1) — onto.
5. Q5. Show that f:R→R, f(x)=x³, is injective.
   Answer: x1³=x2³⇒(x1−x2)(x1²+x1x2+x2²)=0; the second factor is ≥0 and =0 only when x1=x2=0, so x1=x2 always.
6. Q6. Give examples of f:N→Z and g:Z→Z such that gof is injective but g is not injective.
   Answer: Let f(x)=x and g(x)=|x|. g is not injective (g(−1)=g(1)=1), but gof(x)=|x|=x for all x∈N, which is injective.
7. Q7. Give examples of f:N→N and g:N→N such that gof is onto but f is not onto.
   Answer: Let f(x)=x+1 (not onto) and g(x)=x−1 if x>1, g(1)=1. Then gof(x)=x for all x∈N, which is onto.
8. Q8. Given a non-empty set X, let P(X) be the power set of X. Define ARB iff A⊂B for A,B∈P(X). Is R an equivalence relation on P(X)?
   Answer: Reflexive but not symmetric (A⊂B does not imply B⊂A) — R is not an equivalence relation.
9. Q9. Given a non-empty set X, define *:P(X)×P(X)→P(X) by A*B=A∩B. Show that X is the identity element for * and X is the only invertible element.
   Answer: A∩X=A for every A⊂X, so X is the identity. A∩B⊂A⊂X, so A∩B=X forces A=X — X is the only invertible element.
10. Q10. Find the number of all onto functions from the set {1,2,…,n} to itself.
    Answer: Must also be one-one (a bijection), so the count is n!.
11. Q11. Let S={a,b,c}, T={1,2,3}. Find F⁻¹ for: (i) F={(a,3),(b,2),(c,1)} (ii) F={(a,2),(b,1),(c,1)}.
    Answer: (i) F⁻¹={(3,a),(2,b),(1,c)}. (ii) F is not one-one, so F⁻¹ does not exist.
12. Q12. Consider *:R×R→R and o:R×R→R defined as a*b=|a−b| and aob=a. Show * is commutative but not associative, o is associative but not commutative.
    Answer: * commutative (|a−b|=|b−a|), not associative ((1*2)*3=2 but 1*(2*3)=0). o associative ((aob)oc=a=ao(boc)), not commutative (aob=a but boa=b).
13. Q13. Given a non-empty set X, let *:P(X)×P(X)→P(X), A*B=(A−B)∪(B−A). Show φ is the identity and every A is invertible with A⁻¹=A.
    Answer: A*φ=A (identity); A*A=φ (self-inverse), so every A is its own inverse.
14. Q14. Define * on {0,1,2,3,4,5} as a*b=a+b if a+b<6, a+b−6 if a+b≥6. Show 0 is the identity and each a≠0 has inverse 6−a.
    Answer: a*0=a (identity); a*(6−a)=6→0 (identity), so 6−a is the inverse of a.
15. Q15. Let A={−1,0,1,2}, B={−4,−2,0,2}, f(x)=x²−x and g(x)=2|x−½|−1 for x∈A. Are f and g equal? Justify.
    Answer: f and g agree at all four points of A (2,0,0,2 respectively), so f=g as functions on A.
16. Q16. Let A={1,2,3}. The 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.
    Answer: (A). Only one such non-transitive relation exists once reflexive+symmetric closure is forced.
17. Q17. Let A={1,2,3}. The number of equivalence relations containing (1,2) is: (A) 1 (B) 2 (C) 3 (D) 4.
    Answer: (B). Either {1,2},{3} stay separate blocks, or all three merge into one block.
18. Q18. Let f be the Signum Function and g the Greatest Integer Function. Do fog and gof coincide in (0,1]?
    Answer: They agree only at x=1; for x∈(0,1), fog gives 0 while gof gives 1 — they do not coincide throughout (0,1].
19. Q19. The number of binary operations on the set {a,b} is: (A) 10 (B) 16 (C) 20 (D) 8.
    Answer: (B). 2⁴=16 possible operations.

Related

• Solutions
• Extra Questions (HOTS)
• Revision Notes
• Class 12 Mathematics Formulas Handbook
• Class 12 Mathematics Book

Written by Satish

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