MATHS Assertion (A) & Reason (B) MCQ

4. A is false; R is true.

Solution:

Assertion: We have,

f(x) = 2 - 3x, $\text{x}\in\text{R},$ x > 0

Let f(x) = y, then y = 2 - 3x

⇒ 3x = 2 - y 

$\Rightarrow\text{x}=\frac{2-\text{y}}{3}$

$\because\text{x}>0$

$\Rightarrow\frac{2-\text{y}}{3}>0$

$\Rightarrow2-\text{y}>0$

$\Rightarrow2>\text{y}$

$\therefore\text{y}<2$

Hence, range of $\text{f}=(-\infty,2)$

Reason: Now, f(x) = x² + 2

Let y = f(x), then

y = x² + 5

$\Rightarrow\text{x}=\sqrt{\text{y}-2}$

x assumes real values, if $\text{y}-2\geq0$

$\Rightarrow\text{y}\geq2$

$\Rightarrow\text{y}\in[2,\infty)$

$\therefore$ Range of $\text{f}=[2,\infty)$

2. A is true, R is true; R is not a correct explanation of A.

Solution:

Assertion: A = Set of first elements = {x, y, z}

B = Set of second elements = {1, 2}

$\therefore$ A is correct.

Reason: n( A) = 3, n{(B) = 2 n{(A . B) = n(A) . n(B) = 3 . 2 = 6

2. A is true, R is true; R is not a correct explanation of A.

Solution:

Assertion: Given, R = {$(\text{x},\text{y}):\mid\text{x}-\text{y}\mid$ is odd, $\text{x}\in\text{A},\text{y}\in\text{B}$}

The relation R in Roster form is R = {(1, 4), (1, 6), (2 9), (3, 4), (3, 6), (5, 4), (5, 6)}

$\therefore$ Domain of R = {1, 2, 3, 5}

So, A is true. 

Reason: It is also true $\mid\text{x}\mid$is always positive. 

4. A is false; R is true.

Solution:

Assertion: We have,

f (x) = 2x + 3, g(x) = x² + 7 

g [f(x)] = 8

⇒ g(2x +  3) = 8

⇒ (2x + 3)² + 7 = 8

⇒​​​​​​(2x + 3)² = 1

$\Rightarrow2\text{x}+3=\pm1,$

2x + 3 = -1

or 2x + 3 = 1, then

⇒ x = -1, x = -2

Reason: Now, $\text{f}(\text{x})=\frac{4^{\text{x}}}{4^{\text{x}}+2}$

$\text{f}(1-\text{x})=\frac{4^{1-\text{x}}}{4^{1-\text{x}}+2}$

$\therefore\text{f}(\text{x})+\text{f}(1-\text{x})=\frac{4^{\text{x}}}{4^{\text{x}}+2}+\frac{4^{1-\text{x}}}{4^{1-\text{x}}+2}$

$=\frac{4^{\text{x}}}{4^{\text{x}}+2}+\frac{\frac{4}{4^{\text{x}}}}{4+2\cdot4^{\text{x}}}$

$=\frac{4^{\text{x}}}{4^{\text{x}}+2}+\frac{2}{4^{\text{x}}+2}$

$=\frac{4^{\text{x}}+2}{4^{\text{x}}+2}=1.$

4. A is false; R is true.

Solution:

Assertion: We have, $\text{f}(\text{x})=\sqrt{\text{x}-1}$

f(x) is defined, if $\text{x}-1\geq0$ i. e. $\text{x}\geq0$

$\therefore$ Domain of $\text{f}=[1,\infty)$

Hence, A is incorrect.

Reason: Let f(x) = y

Then, $\text{y}=\sqrt{\text{x}-1}$

$\because\text{x}\geq1$

$\therefore$ Range of $\text{f}=[0,\infty).$

3. A is true; R is false.

Solution:

Assertion Given, (4x + 3, y) = (3x + 5, -2)

Two ordered pairs are equal when their corresponding elements are equal.

4x + 3 = 3x + 5 and y = -2

4x - 3x = 5 - 3

x = 2

Reason: Now, A = {-1, 3, 4}

$\therefore$ A . A = {(-1, -1), (-1, 3), (-1, 4), (3, -1), (3, 3), (3, 4), (4, -1), (4, 3), (4, 4)}

$\therefore$ Assertion is true and Reason is false.

4. A is false; R is true.

Solution:

Assertion: In Set - builder form,

R = {(x, y) : y = x², x, $\text{y}\in\text{N}$ and $-2,\leq\text{x}\leq2$} $[\therefore0\in\text{N}]$

Reason: The relation shown in figure is represented in Roster form as

R = {(-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4)}

We observe that, second element of each ordered pair is the square of first element.

1. A is true, R is true; R is a correct explanation of A.

Solution:

Assertion: The total number of relation that can be defined from a set A to a set B is the number of possible subset of A . B.

If n( A) = p and n(B) = q, then n(A . B) = pq and the total number of relation is 2^pq.

Given, A = {1, 2} and B = {3, 4}

$\therefore$ A . B = ((1, 3), (1, 4), (2 3), (2, 4)}

Since, n{(A . B) = 4, the number of subsets of A . B is 2⁴.

Therefore, the number of relation from A to B will be 2⁴ = 16.

3. A is true; R is false.

Solution:

Assertion: In arrow diagram, every element of P has its unique image in Q.

So, it represent a function.

Reason: Domain of f = R - {2}.

Domain of g = R

$\therefore\text{D}_{\text{f}}\neq\text{D}_{\text{g}}$

We know that, two functions are equal when their domain and range are equal and same element in their domain produce same image.

$\therefore\text{f}\neq\text{g}$

1. A is true, R is true; R is a correct explanation of A.

Solution:

Assertion: P and Q are two non - empty sets.

The cartesian product P . Q is the set of all ordered pairs of elements from P and Q, i.e.

$\text{P}\cdot\text{Q}=\{(\text{p},\text{q}):\text{p}\in\text{P}$ and $\text{q}\in\text{Q}\}.$

Reason: Now, A = {red, blue}, B = {b, c, s} A . B = set of all ordered pairs 

= {(red, 5), (red, c), (red, s), (blue, 3), (blue, c), (blue, s)}

3. A is true; R is false.

Solution:

Assertion: Given,

$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$

$\text{f}(\text{x}^{3})=\text{x}^{3}+\frac{1}{\text{x}^{3}}$

$[\text{f}(\text{x})]^{3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^{3}$

$=\text{x}^{3}+\frac{1}{\text{x}^{3}}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$

$=\text{f}(\text{x}^{3})+3\text{f}(\text{x})$

$=\text{f}(\text{x}^{3})+3\text{f}\big(\frac{1}{\text{x}}\big)$$\big[\because\text{f}\big(\frac{1}{\text{x}}\big)=\frac{1}{\text{x}}+\text{x}=\text{f}(\text{x})\big]$

Reason: Now, we have, 

f(x) = (x - a)² (x - b)²

f(a + b) = (a + b - a)² (a + b - b)² = b²a²