Class 12 Mathematics

Chapter 1 Solutions — Relations and Functions

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Solutions

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Step-by-step NCERT solutions for Relations and Functions (Chapter 1, CBSE Class 12 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Relations and Functions textbook chapter.

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All 35 questions in Relations and Functions are solved in the PDF. Here's what's inside, exercise by exercise:

Exercise 1.1

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…
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.
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.
Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.
Check whether the relation R in R defined by R = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive.
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.
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.
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}.
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.
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.
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.
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?
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?
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.
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.
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

Exercise 1.2

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

Miscellaneous Exercise

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.
Show that the function f : R → R given by f(x) = x³ is injective.
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.
Find the number of all onto functions from the set {1, 2, 3, ..., n} to itself.
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 – 1/2| – 1, x ∈ A. Are f and g equal? Justify your answer.
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
Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is (A) 1 (B) 2 (C) 3 (D) 4

Sample

Worked examples

3 questions from Relations and Functions, solved step by step.

Exercise 1.1

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.

Not reflexive: We need a ≤ a² for all real a. Take a = 1/2: 1/2 ≤ (1/2)² = 1/4 is false. So (1/2, 1/2) ∉ R, and R is not reflexive.

Not symmetric: Take a = 1, b = 4. Then a ≤ b² since 1 ≤ 16, so (1, 4) ∈ R. But b ≤ a² requires 4 ≤ 1, which is false. So (4, 1) ∉ R, and R is not symmetric.

Not transitive: Take a = 25, b = −5, c = 2. Then a ≤ b²: 25 ≤ (−5)² = 25, true, so (25, −5) ∈ R. Also b ≤ c²: −5 ≤ 4, true, so (−5, 2) ∈ R. But a ≤ c²: 25 ≤ 4 is false. So (25, 2) ∉ R, and R is not transitive.

Answer: R is neither reflexive (1/2 ≤ 1/4 fails), nor symmetric ((1,4) ∈ R but (4,1) ∉ R), nor transitive ((25,−5) ∈ R, (−5,2) ∈ R but (25,2) ∉ R).

Exercise 1.2

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.

Not one-one: f(1.2) = [1.2] = 1 and f(1.7) = [1.7] = 1, but 1.2 ≠ 1.7. So f is not one-one.

Not onto: For any x ∈ R, [x] is an integer. So the range of f is Z ⊂ R. An element like 1/2 ∈ R is not in the range of f. So f is not onto.

Answer: f is neither one-one (f(1.2) = f(1.7) = 1) nor onto (1/2 ∈ R has no pre-image).

Miscellaneous Exercise

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

Suppose f(x1) = f(x2), i.e., x1³ = x2³.

Then x1³ − x2³ = 0 ⇒ (x1 − x2)(x1² + x1*x2 + x2²) = 0.

The factor x1² + x1*x2 + x2² = (x1 + x2/2)² + 3x2²/4 ≥ 0, and it equals 0 only if x1 = x2 = 0.

So either x1 − x2 = 0 (giving x1 = x2) or x1 = x2 = 0. In both cases x1 = x2.

Therefore f is injective.

Answer: f(x) = x³ is injective: x1³ = x2³ implies x1 = x2 for real numbers.

All 35 questions in Relations and Functions are solved step by step in the Solutions PDF above.

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Chapter 1 Solutions — Relations and Functions

Overview

What these solutions cover

Exercise 1.1

Exercise 1.2

Miscellaneous Exercise

Worked examples

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.

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.

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

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