CBSE Class 12 Mathematics Relations and Functions Class Test Set 19

This class test on Chapter 1 Relations and Functions for Class 12 CBSE Mathematics has 1, 2 and 3 mark questions on relations, equivalence classes and one-one and onto functions. Full solutions are given after every question.

Question 1. Show that the number of equivalence relations on the set {1, 2, 3} containing (1, 2) and (2, 1) is two. [1 Mark]
Answer:
Let A = {1, 2, 3}
Possible equivalence relations are
R₁ = {(1, 1), (2, 2), (3, 3)}
R₂ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R₃ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}
Relations R₂ and R₃ contain both (1, 2) and (2, 1)
∴ Two possible equivalence relations

Question 2. Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. Check whether R is reflexive, symmetric or transitive. [1 Mark]
Answer:
Since (1, 1) ∉ R ∴ R is not reflexive
Since (2, 3) ∈ R but (3, 2) ∉ R
∴ R is not symmetric
Since (1, 3) ∈ R and (3, 1) ∈ R but (1, 1) ∉ R
∴ R is not transitive

Question 3. 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. Check whether f is a bijection or not. [1 Mark]
Answer:
Yes, f is one-one, as different elements of A have different images.

But 7 ∈ B has no pre-image in A, so f is not onto.
Hence f is not a bijection.

Question 4. Let R be the equivalence relation in the set A = {0, 1, 2, 3, 4, 5} given by R = {(a, b) : 2 divides (a – b)}. Write the equivalence class [0]. [2 Marks]
Answer:
[0] = {0, 2, 4}

Question 5. Show that the function f : R → {x ∈ R : –1 < x < 1} defined by f(x) = \( \frac{x}{1 + |x|} \), x ∈ R is one one and onto function. [3 Marks]
Answer:
Given f : R → {x ∈ R : –1 < x < 1} defined by f(x) = \( \frac{x}{1 + |x|} \)
For one-one: Let f(x₁) = f(x₂)
⇒ \( \frac{x_1}{1 + |x_1|} = \frac{x_2}{1 + |x_2|} \)
Since f(x) has the same sign as x, the case of one positive & other negative is ruled out, and f(x) = 0 only for x = 0. So we have
Case I: When both positive
f(x₁) = f(x₂)
⇒ \( \frac{x_1}{1 + x_1} = \frac{x_2}{1 + x_2} \Rightarrow x_1 = x_2 \)
Case II: When both negative
⇒ \( \frac{x_1}{1 - x_1} = \frac{x_2}{1 - x_2} \Rightarrow x_1 = x_2 \)
∴ f is one-one
For onto: Let y ∈ R such that –1 < y < 1
When y is negative, then there exists x = \( \frac{y}{1 + y} \) (negative) such that
\( f(x) = f\left(\frac{y}{1+y}\right) = \frac{\left(\frac{y}{1+y}\right)}{1 + \left|\frac{y}{1+y}\right|} = y \)
When y is non-negative, then there exists x = \( \frac{y}{1 - y} \) (≥ 0) such that
\( f(x) = f\left(\frac{y}{1-y}\right) = \frac{\left(\frac{y}{1-y}\right)}{1 + \frac{y}{1-y}} = y \)
∴ f is onto
Hence f is one-one and onto.

Question 6. Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric. [3 Marks]
Answer:
As a ≤ a ∀ a ∈ R, therefore, (a, a) ∈ R for all a ∈ R.
⇒ R is reflexive.
Also (a, b) ∈ R and (b, c) ∈ R
⇒ a ≤ b and b ≤ c ⇒ a ≤ c
⇒ (a, c) ∈ R
⇒ R is transitive.
However, R is not symmetric as (a, b) ∈ R ⇏ (b, a) ∈ R; for example (3, 4) ∈ R but (4, 3) ∉ R.
(∵ 3 ≤ 4 but 4 > 3)

Question 7. Let N denote the set of all natural numbers and R be the relation on N × N defined by (a, b) R (c, d) ⇔ ad(b + c) = bc(a + d). Check whether R is an equivalence relation on N × N. [3 Marks]
Answer:
Reflexive:
Let (a, b) ∈ N × N i.e. a, b ∈ N such that
if (a, b) R (a, b)
⇒ ab(b + a) = ba(a + b) [by commu. of add. and mult. on N]
Which is true ∀ (a, b) ∈ N × N
So R is reflexive on N × N.
Symmetric:
Let (a, b), (c, d) ∈ N × N be such that
if (a, b) R (c, d)
⇒ ad(b + c) = bc(a + d)
⇒ cb(d + a) = da(c + b)
⇒ (c, d) R (a, b)
Thus, (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N
So, R is symmetric on N × N.
Transitive:
Let (a, b), (c, d), (e, f) ∈ N × N such that
if (a, b) R (c, d) and (c, d) R (e, f).
(a, b) R (c, d) ⇒ ad(b + c) = bc(a + d) ⇒ \( \frac{b+c}{bc} = \frac{a+d}{ad} \Rightarrow \frac{1}{c} + \frac{1}{b} = \frac{1}{d} + \frac{1}{a} \) ....(i)
and, (c, d) R (e, f) ⇒ cf(d + e) = de(c + f) ⇒ \( \frac{d+e}{de} = \frac{c+f}{cf} \Rightarrow \frac{1}{e} + \frac{1}{d} = \frac{1}{f} + \frac{1}{c} \) ....(ii)
Adding (i) and (ii), we get
\( \left(\frac{1}{c} + \frac{1}{b}\right) + \left(\frac{1}{e} + \frac{1}{d}\right) = \left(\frac{1}{d} + \frac{1}{a}\right) + \left(\frac{1}{f} + \frac{1}{c}\right) \)
⇒ \( \frac{1}{b} + \frac{1}{e} = \frac{1}{a} + \frac{1}{f} \Rightarrow \frac{b+e}{be} = \frac{a+f}{af} \)
⇒ af(b + e) = be(a + f) ⇒ (a, b) R (e, f)
So, R is transitive on N × N.
Since R is reflexive, symmetric and transitive on set N × N.
Hence R is an equivalence relation on N × N.

Question 8. Show that the relation R on the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a – b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1 i.e. equivalence class [1]. [3 Marks]
Answer:
(i) Reflexive:
Let a ∈ A
|a – a| = 0 which is divisible by 4.
So, (a, a) ∈ R ∀ a ∈ A
Hence R is reflexive.
(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.
(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.
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
Required set is {1, 5, 9}.

Question 9. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = \( \left(\frac{x-2}{x-3}\right) \). Is f one-one and onto? Justify your answer. [3 Marks]
Answer:
Let x₁, x₂ ∈ A be such that
f(x₁) = f(x₂)
⇒ \( \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \)
⇒ (x₁ – 2)(x₂ – 3) = (x₂ – 2)(x₁ – 3)
⇒ x₁x₂ – 2x₂ – 3x₁ + 6 = x₁x₂ – 2x₁ – 3x₂ + 6
⇒ x₂ = x₁
∴ f is one-one.
Let y ∈ B = R – {1}, then f(x) = y
if \( \frac{x-2}{x-3} = y \), x ≠ 3
i.e., if x – 2 = yx – 3y
i.e., if x – xy = 2 – 3y
i.e., if x(1 – y) = 2 – 3y
i.e., if \( x = \frac{2 - 3y}{1 - y} \) ∈ A (where y ≠ 1)
Hence, f is onto.

Question 10. Show that the function f : R → R given by f(x) = ax + b, where a, b ∈ R, a ≠ 0 is a bijection. [3 Marks]
Answer:
For one-one:
Let x₁, x₂ ∈ R such that
f(x₁) = f(x₂)
⇒ ax₁ + b = ax₂ + b
⇒ x₁ = x₂
Hence f is one-one
For onto:
Let y ∈ R, then y = f(x) if y = ax + b
i.e. \( x = \frac{y - b}{a} \)
\( f(x) = f\left(\frac{y-b}{a}\right) = a\left(\frac{y-b}{a}\right) + b = y \)
∴ f is onto function
Since f is one-one and onto
∴ f is bijective function

Question 11. Consider the function f : [0, π/2] → R given by f(x) = sin x and g : [0, π/2] → R given by g(x) = cos x. Show that f and g are one-one but f + g is not one-one. [3 Marks]
Answer:
Given \( f : \left[0, \frac{\pi}{2}\right] \to R,\ f(x) = \sin x \)
Checking one-one for f
let f(x₁) = f(x₂)
⇒ sin x₁ = sin x₂
So x₁ = x₂ (∵ sin x is increasing on [0, π/2])
∴ f is one-one
Also \( g : \left[0, \frac{\pi}{2}\right] \to R,\ g(x) = \cos x \)
Checking one-one for g
let g(x₁) = g(x₂)
⇒ cos x₁ = cos x₂
So x₁ = x₂ (∵ cos x is decreasing on [0, π/2])
∴ g is one-one
Now \( f + g : \left[0, \frac{\pi}{2}\right] \to R \), where (f + g)(x) = sin x + cos x
(f + g)(0) = sin 0 + cos 0 = 1
\( (f + g)\left(\frac{\pi}{2}\right) = \sin\frac{\pi}{2} + \cos\frac{\pi}{2} = 1 \)
Different elements have same image.
∴ f + g is not one-one.