Relations and Functions — Assertion-Reason Type Questions

Assertion (A): Number of relations from set A to set B = 2^{n(A) · n(B)}

Reason (R): A relation R from set A to set B is a subset of A × B.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true, but Reason (R) is false.

4. Assertion (A) is false, but Reason (R) is true.

Answer

By definition, a relation R from a set A to a set B is a subset of A × B.

∴ Reason (R) is true.

As every relation from A to B is a subset of A × B, the number of relations from A to B is the number of subsets of A × B.

We know that a set containing n elements has 2ⁿ subsets and n(A × B) = n(A) · n(B).

∴ Number of relations from A to B = 2^{n(A) · n(B)}

∴ Assertion (A) is true.

The formula in Assertion (A) is obtained exactly by counting the subsets of A × B, which is what Reason (R) states.

Hence, Option 1 is the correct option.

Assertion (A): Let A and B be two finite sets such that n(A) − n(B) = 1. If number of relations from set A to set B is 64 then n(A) = 3, n(B) = 2.

Reason (R): Number of relations from set A to set B = 2^{n(A) + n(B) + 1}

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true, but Reason (R) is false.

4. Assertion (A) is false, but Reason (R) is true.

Answer

We know that,

Number of relations from set A to set B = 2^{n(A) · n(B)} and not 2^{n(A) + n(B) + 1}.

∴ Reason (R) is false.

Given,

Number of relations from A to B = 64

⇒ 2^{n(A) · n(B)} = 64 = 2⁶

⇒ n(A) · n(B) = 6

Also given, n(A) − n(B) = 1.

The pairs of positive integers whose product is 6 are 1 and 6, 2 and 3, 3 and 2, 6 and 1.

Out of these, only n(A) = 3 and n(B) = 2 satisfy n(A) − n(B) = 1.

∴ Assertion (A) is true.

Hence, Option 3 is the correct option.

Assertion (A): Range of the function $f(x) = \dfrac{x}{|x|}$ is ${-1, 1}$.

Reason (R): $|x| = \begin{cases} x, & \text{if } x \ge 0 \\[1em] -x, & \text{if } x \lt 0 \end{cases}$

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true, but Reason (R) is false.

4. Assertion (A) is false, but Reason (R) is true.

Answer

By the definition of the modulus function,

$|x| = \begin{cases} x, & \text{if } x \ge 0 \\[1em] -x, & \text{if } x \lt 0 \end{cases}$

∴ Reason (R) is true.

Given,

$f(x) = \dfrac{x}{|x|}$, which is defined for every real number $x$ except 0.

Using the above definition,

when $x \gt 0$, $|x| = x$, so $f(x) = \dfrac{x}{x} = 1$,

when $x \lt 0$, $|x| = -x$, so $f(x) = \dfrac{x}{-x} = -1$.

So $f(x)$ takes only the values 1 and $-1$.

∴ $R_f = {-1, 1}$

∴ Assertion (A) is true.

The range of $f$ is obtained exactly by splitting $|x|$ into its two cases, which is what Reason (R) states.

Hence, Option 1 is the correct option.

Assertion (A): Domain of the function $f(x) = \dfrac{x + 2}{x^2 - 4x + 3}$ is $\textbf{R} - {1, 3}$.

Reason (R): Range of the function $f(x) = |x - 1|$ is $[0, \infty)$.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true, but Reason (R) is false.

4. Assertion (A) is false, but Reason (R) is true.

Answer

Given,

$f(x) = \dfrac{x + 2}{x^2 - 4x + 3}$

$f(x)$ is not defined when the denominator is zero.

⇒ $x^2 - 4x + 3 = 0$

⇒ $x^2 - 3x - x + 3 = 0$

⇒ $x(x - 3) - 1(x - 3) = 0$

⇒ $(x - 3)(x - 1) = 0$

⇒ $x = 3$ or $x = 1$

∴ $D_f = \textbf{R} - {1, 3}$

∴ Assertion (A) is true.

For the function $f(x) = |x - 1|$, we know that $|x - 1| \ge 0$ for every real number $x$, and for every $y \ge 0$ the number $x = 1 + y$ gives $|x - 1| = y$.

∴ $R_f = [0, \infty)$

∴ Reason (R) is true.

Reason (R) speaks about the range of a different function and it is of no use in finding the domain stated in Assertion (A).

Hence, Option 2 is the correct option.

Assertion (A): Relation R from a finite set A to a finite set B given by R = {(1, 2), (2, 3), (3, 4), (4, 5)} is a function.

Given that A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6}.

Reason (R): All relations are functions.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true, but Reason (R) is false.

4. Assertion (A) is false, but Reason (R) is true.

Answer

Every function is a relation, but a relation need not be a function. For example, the relation {(1, 2), (1, 3)} is not a function, as the element 1 has two different images.

∴ Reason (R) is false.

Given,

A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6} and R = {(1, 2), (2, 3), (3, 4), (4, 5)}

Each of the four elements 1, 2, 3 and 4 of A occurs exactly once as a first component of the ordered pairs of R, so every element of A has one and only one image in B.

∴ R is a function.

∴ Assertion (A) is true.

Hence, Option 3 is the correct option.

Assertion (A): Range of the greatest integer function $f(x) = [x]$ is $\textbf{I}$ (set of integers).

Reason (R): Domain of the function $f(x) = \dfrac{1}{\sqrt{x + 2}}$ is $[-2, \infty)$.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true, but Reason (R) is false.

4. Assertion (A) is false, but Reason (R) is true.

Answer

For the greatest integer function $f(x) = [x]$, the value $[x]$ is by definition the greatest integer not exceeding $x$, so every value of $f(x)$ is an integer.

Also, for any integer $n$, $f(n) = [n] = n$, so every integer is attained.

∴ $R_f = \textbf{I}$

∴ Assertion (A) is true.

Now, for $f(x) = \dfrac{1}{\sqrt{x + 2}}$, the expression $\sqrt{x + 2}$ must be a real number and it must not be zero.

⇒ $x + 2 \gt 0$

⇒ $x \gt -2$

∴ $D_f = (-2, \infty)$ and not $[-2, \infty)$, as $x = -2$ makes the denominator zero.

∴ Reason (R) is false.

Hence, Option 3 is the correct option.