CBSE Class 12 Mathematics Relations and Functions Previous Year Questions Set 01

This page has the CBSE previous year board questions of Chapter 1 Relations and Functions for Class 12 Mathematics with complete step-wise solutions. The questions are grouped by marks, from 1-mark very short answers to 6-mark long answers.

Very Short Answer Questions (1 Mark)

Question 1. If R = {(x, y) : x + 2y = 8} is a relation on N, write the range of R. [CBSE 2014, 1M]
Answer:
Given x + 2y = 8; x, y ∈ N
⇒ \( y = \frac{8 - x}{2} \)
R = {(2, 3), (4, 2), (6, 1)}
Range of R = {3, 2, 1} [1]

Question 2. The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1), (1, 1)} is [CBSE 2020, 1M]
(A) symmetric and transitive, but not reflexive
(B) reflexive and symmetric, but not transitive
(C) symmetric, but neither reflexive nor transitive
(D) an equivalence relation
Answer:
Ans. (C)
Let A = {1, 2, 3} and R = {(1, 2), (2, 1), (1, 1)}
R is not reflexive as (2, 2) & (3, 3) ∉ R.
R is symmetric relation as (a, b) ∈ R & (b, a) ∈ R ∀ a, b ∈ A.
R is not transitive relation because (2, 1) ∈ R & (1, 2) ∈ R but (2, 2) ∉ R. [1]

Question 3. Let A = {1, 3, 5}. Then the number of equivalence relations in A containing (1, 3) is [CBSE 2020, 1M]
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
Ans. (B)
R₁ = {(1, 1), (3, 3), (5, 5), (1, 3), (3, 1)}
and R₂ = {(1, 1), (3, 3), (5, 5), (1, 3), (3, 1), (1, 5), (5, 1), (3, 5), (5, 3)}
Hence, the number of equivalence relations in A containing (1, 3) are 2. [1]

Long Answer-I Questions (4 Marks)

Question 4. Show that the relation R defined by (a, b) R (c, d) ⇔ a + d = b + c on the set N × N is an equivalence relation. [CBSE 2008, 2010 4M]
Answer:
(i) Reflexive:
Let (a, b) ∈ N × N such that
if (a, b) R (a, b) ⇒ a + b = b + a
which is true ∀ (a, b) ∈ N × N
∴ R is reflexive [½]
(ii) Symmetric:
Let (a, b), (c, d) ∈ N × N such that
if (a, b) R (c, d) ⇒ a + d = b + c
⇒ b + c = a + d ⇒ c + b = d + a
⇒ (c, d) R (a, b)
∴ R is symmetric [1½]
(iii) Transitive:
Let (a, b), (c, d), (e, f) ∈ N × N such that
if (a, b) R (c, d) ⇒ a + d = b + c .......(i)
& (c, d) R (e, f) ⇒ c + f = d + e .......(ii)
from equation (i) + (ii)
⇒ a + d + c + f = b + c + d + e
⇒ a + f = b + e ⇒ (a, b) R (e, f)
∴ R is transitive. [1½]
Since R is reflexive, symmetric and transitive on N × N.
Hence R is equivalence relation. [½]

Question 5. Let f : N → N be defined by f(n) = \( \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases} \) for all n ∈ N. Find whether the function f is bijective. [CBSE 2009, 4M]
Answer:
f(n) = \( \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases} \)
for f to be bijective it should be one-one and onto [1]
Let x₁ = 3 & x₂ = 4
here \( f(3) = \frac{3+1}{2} = 2 \) & \( f(4) = \frac{4}{2} = 2 \) [1]
∵ f(3) = f(4) but 3 ≠ 4
f(x₁) = f(x₂) ⇒ x₁ ≠ x₂ [1]
i.e. f is not one-one
Hence f is not bijective. [1]

Question 6. Show that f : N → N, given by f(x) = \( \begin{cases} x + 1, & \text{if } x \text{ is odd} \\ x - 1, & \text{if } x \text{ is even} \end{cases} \) is both one-one and onto. [CBSE 2012, 4M]
Answer:
f : N → N, such that
f(x) = \( \begin{cases} x + 1, & \text{if } x \text{ is odd} \\ x - 1, & \text{if } x \text{ is even} \end{cases} \)
For one-one:
Case-I: Let x₁ and x₂ both are even, x₁, x₂ ∈ N.
if f(x₁) = f(x₂)
x₁ – 1 = x₂ – 1 ⇒ x₁ = x₂
∴ f is one-one. [½]
Case-II: Let x₁ and x₂ both are odd, x₁, x₂ ∈ N
if f(x₁) = f(x₂)
x₁ + 1 = x₂ + 1 ⇒ x₁ = x₂
∴ f is one-one [½]
Case-III: Let x₁ = odd and x₂ = even, x₁, x₂ ∈ N
if f(x₁) = f(x₂)
x₁ + 1 = x₂ – 1
x₂ – x₁ = 2
Which is not possible, since the difference between an odd and an even number can never be 2.
Thus in this case
x₁ ≠ x₂ ⇒ f(x₁) ≠ f(x₂)
∴ f is one-one. [1]
For onto:
Case-I: When x is odd, in this case x + 1 is even.
f(x + 1) = x + 1 – 1 = x [½]
Case-II: When x is even, in this case x – 1 is odd.
f(x – 1) = x – 1 + 1 = x [½]
Thus every x ∈ N has its pre-image in N.
So range = co-domain [1]
Hence f is onto.

Question 7. Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive. [CBSE 2019, 4M]
Answer:
Given, R = {(a, b) : b = a + 1}
is defined on the set A = {1, 2, 3, 4, 5, 6}
⇒ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
Now, 1 ∈ A but (1, 1) ∉ R.
Therefore, R is not reflexive. [1]
For 1, 2 ∈ A, it can be observed that (1, 2) ∈ R but (2, 1) ∉ R. Therefore, R is not symmetric. [1½]
For 1, 2, 3 ∈ A, (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R.
Therefore, R is not transitive.
Hence, R is neither reflexive nor symmetric nor transitive. [1½]

Question 8. Check whether the relation R in the set N of natural numbers given by R = {(a, b) : a is divisor of b} is reflexive, symmetric or transitive. Also determine whether R is an equivalence relation. [CBSE 2020, 4M]
Answer:
Given relation R in the set N given by
R = {(a, b) : a is divisor of b}
Reflexive:
a is divisor of a, ∀ a ∈ N
so (a, a) ∈ R.
So R is reflexive relation [1]
Symmetric:
If a is divisor of b then (a, b) ∈ R
But it is not necessary that b is also divisor of a.
so (b, a) ∉ R
So R is not symmetric relation. [1]
Transitive:
If a is divisor of b and b is divisor of c
then, (a, b) ∈ R & (b, c) ∈ R
then we can say a is also divisor of c
i.e. (a, c) ∈ R
so R is transitive relation [1]
Here R is reflexive and transitive but not symmetric.
So R is not equivalence relation on N. [1]

Long Answer-II Questions (6 Marks)

Question 9. Let A = {x ∈ Z : 0 ≤ x ≤ 12}. Show that R = {(a, b) : a, b ∈ A, |a – b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2]. [CBSE 2010, 2018, 6M]
Answer:
(i) Reflexive:
Let a ∈ A
|a – a| = 0 which is divisible by 4.
So, (a, a) ∈ R ∀ a ∈ A
Hence R is reflexive. [1]
(ii) Symmetric:
Let a, b ∈ A such that (a, b) ∈ R
i.e. |a – b| is divisible by 4.
⇒ |–(b – a)| = |b – a| is divisible by 4.
Hence (b, a) ∈ R.
So, R is symmetric. [1]
(iii) Transitive:
Let a, b, c ∈ A such that (a, b), (b, c) ∈ R
i.e. |a – b| & |b – c| is divisible by 4.
Let |a – b| = 4k₁
& |b – c| = 4k₂
⇒ (a – b) = ±4k₁ ...........(i)
& (b – c) = ±4k₂ ...........(ii)
Adding equations (i) & (ii)
⇒ (a – b) + (b – c) = ±4k₁ ± 4k₂
⇒ a – c = 4k, where k = ±k₁ ± k₂ is an integer
⇒ a – c is divisible by 4.
⇒ |a – c| is divisible by 4.
Hence, (a, c) ∈ R
So, R is transitive.
Since R is reflexive, symmetric and transitive on A. [2]
Hence R is an equivalence relation.
Further, let (x, 1) ∈ R, x ∈ A
⇒ |x – 1| is divisible by 4
⇒ x – 1 = 0, 4, 8, .........
⇒ x = 1, 5, 9
Equivalence class of [1] = {1, 5, 9}. [1]
Now equivalence class of [2]
Let (x, 2) ∈ R, x ∈ A
⇒ |x – 2| = 0, 4, 8, .........
x = 2, 6, 10
Equivalence class of [2] = {2, 6, 10}. [1]