MATHS M.C.Q (1 Marks)

2. $2^{\text{n}^2}$

Solution:

Given, A finite set with n elements

Its Cartesian product with itself will have n² elements.

$\therefore$ Number of relations on $\text{A}=2^{\text{n}^2}$

1. 2^mn

Solution:

Given, n(A) = m

n(B) = n

$\therefore$ n(A × B) = mn

Then, the number of relations from A to is 2^mn

3. $\text{R}\subseteq\text{A}\times\text{B}$

Solution:

If R is a relation from set A to set B, then R is always a subset of A × B.

3. {-2, -1, 0, 1, 2}

Solution:

$\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$

We know that,

$(-2)^2+0^2\leq4$

$\Rightarrow(2)^2+0^2\leq4$

$\Rightarrow(-1)^2+0^2\leq4$

$\Rightarrow(1)^2+0^2\leq4$

$\Rightarrow(-1)^2+(1)^2\leq4$

$\Rightarrow0^2+0^2\leq4$

$\Rightarrow(1)^2+(1)^2\leq4$

$\Rightarrow(-1)^2+(-1)^2\leq4$

Hence, domain(R) = {-2, -1, 0, 1, 2}

1. {(3, 3), (3, 1), (5, 2)}

Solution:

A = {1, 2, 3}, B = {1, 3, 5}

R = {(1, 3), (2, 5), (3, 3)}

$\therefore$ R^-1 = {(3, 3), (3, 1), (5, 2)}

4. $\text{i}\ \phi\ 1$

Solution:

We have,

$|\text{i}|=\sqrt{1^2+0^2}=1$

Thus, $\text{i }\phi\ 1$ satisfies $\text{x}\ \phi\text{ y}\Leftrightarrow|\text{x}|=\text{y}$

1. {(8, 11), (10, 13)}

Solution:

R is a relation from {11, 12, 13} to {8, 10, 12}, defined by y = x - 3

Now, we have,

11 - 3 = 8

13 - 3 = 10

So, R = {(13, 10), (11, 8)}

$\therefore$ R^-1 = {(10, 13), (8, 11)}

3. {1}

Solution:

A = {1, 2, 3} and B = {1, 4, 6, 9}

R is a relation from A to B defined by: x is greater than y.

Then R = {(2, 1), (3, 1)}

$\therefore$ Range (R) = {1}

3. {2, 4, 6}

Solution:

x + 2y = 8

⇒ x = 8 - 2y

For y = 1, x = 6

y = 2, x = 4

y = 3, x = 2

Then R = {(2, 3), (4, 2), (6, 1)}

$\therefore$ Domain of R = {2, 4, 6}