Maths Assertion (A) & Reason (B) MCQ

4. A is false and R is true.

Solution:

By definition, a relation in A is said to be reflexive if xRx, $\forall\ \text{x}\in\text{A}.$

So R is true.

The number of reflexive relations on a set containing n elements is 2^n2-n.]

Here n = 3.

The number of reflexive relations on a set A = 2^9-3 = 2⁶.

Hence A is false.

3. A is true but R is false.

Solution:

f(x) is a one - one function if 

f(x₁) = f(x₂) > x₁ = x₂,

Hence R is false.

Let f(x₁) = f(x₂) for some $\text{x}_{1},\text{x}_{2}\in\text{R}$

⇒ (x₁)³ = (x₂)³

⇒ x₁ = x₂

Hence f(x) is one - one.

Hence A is true.

2. Both A and R are true but R is not the correct explanation of A.

Solution:

For any word x and x have atleast one (all) letter in common

$\therefore(\text{x},\text{x})\in\text{R},\forall\ \text{x}\in\text{W}$ 

$\therefore$ R is reflexive

Let $(\text{x},\text{y})\in\text{R},\text{x},\text{y}\in\text{W}$

⇒ x and y have atleast one letter in common

⇒ y and x have atleast one letter in common

$\Rightarrow(\text{y},\text{x})\in\text{R}$

$\therefore$ R is symmetric

Hence A is true, R is true; R is not a correct explanation for A.

1. Both A and R are true and R is the correct explanation of A.

Solution:

By definition, a relation in Z is said to be reflexive if xRx, $\forall\ \text{x}\in\text{Z}.$

So R is true.

a - a = 0

⇒ 2 divides a - a

⇒ aRa.

Hence R is reflexive and A is true. R is the correct explanation for A.

3. A is true but R is false.

Solution:

Given, A= {1, 2, 3}, B = {4, 5, 6, 7}

and f : A  → B is defined as f = {(1, 4), (2, 5), (3, 6)}

i.e., f(1) = 4, f(2) = 5 and f(3) = 6.

It can be seen that the images of distinct elements of A under f are distinct.

So, f is one - one.

So, A is true.

Range of f = {4, 5, 6}.

Co - domain = {4, 5, 6, 7}.

Since co - domain $\neq$ range, f(x) is not an onto function.

Hence R is false.