Fill In The Blanks[1 Marks ]

Question 11 Mark

A relation R in a set A is called ____________, if $(a_1, B_2)$ $\in\text{R}$ implies $(a_2, a_1) \in\text{R}$ for all $a_1, a_2$ $\in\text{A}.$

Answer

A relation R in a set A is called Symmetric if $(a_1, B_2) \in\text{R}$ implies $(a_2, a_1) \in\text{R}$ for all $a_1, a_2 \in\text{A}.$

View full question & answer→

Question 21 Mark

Fill in the blank.
Let f = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}. Then gof = ______ and fog = ______.

Answer

Let f = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}. Then gof = {(1, 3), (3, 1), (4, 3)} and fog = {(2, 5), (5, 2), (1, 5)}.
Solution:
Given that, f = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}.
$\therefore$ gof(1) = g{f(1)} = g(2) = 3
gof(3) = g{f(3)} = g(5) = 1
gof(4) = g{f(4)} = g(1) = 3
$\therefore$ gof = {(1, 3), (3, 1), (4, 3)}
Now, fog(2) = f{g(2)} = f(3) = 5
fog(5) = f{g(5)} = f(1) = 2
fog(1) = f{g(1)} = f(3) = 5
fog = {(2, 5), (5, 2), (1, 5)}

View full question & answer→

Question 31 Mark

Fill in the blank.
Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ______.

Answer

Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = {(3, 8), (6, 6), (9, 4), (12, 2)}.
Solution:
Given that, 2a + 3b = 30
3b = 30 - 2a
$\text{b}=\frac{30-2\text{a}}{3}=10-\frac{2\text{a}}{3}$
Since 'a' and 'b' are natural numbers, 'a' must be multiple of '3'
For a = 3, b = 8
a = 6, b = 6
a = 9, b = 4
a = 12, b = 2
R = {(3, 8), (6, 6), (9, 4), (12, 2)}

View full question & answer→

Question 41 Mark

Fill in the blank.
Let the relation R be defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(a, b): |a^2 – b^2| < 8$. Then R is given by _______.

Answer

Let the relation R be defined on the set A = {1, 2, 3, 4, 5} by
$R = \{(a, b): |a^2 – b^2| < 8$. Then R is given by $\{(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (4, 3), (3, 4), (4, 4), (5, 5)\}$.
Solution:
Given, $A = \{1, 2, 3, 4, 5\}$
$R = \{(a, b): |a^2 - b^2| < 8\}$
$R = \{(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (4, 3), (3, 4), (4, 4), (5, 5)\}$

View full question & answer→

Question 51 Mark

Fill in the blank.
Let f : R → R be defined by $\text{f}(\text{x})=\frac{\text{x}}{\sqrt{1+\text{x}^2}}.$ Then (fofof)(x) = _______.

Answer

Let f : R → R be defined by $\text{f}(\text{x})=\frac{\text{x}}{\sqrt{1+\text{x}^2}}.$ Then $(\text{fofof})(\text{x})=\frac{\text{x}}{\sqrt{1+3\text{x}^2}}.$ Solution: Given that, $\text{f}(\text{x})=\frac{\text{x}}{\sqrt{1+\text{x}^2}}.$ $\therefore\ (\text{fofof})(\text{x})=\text{f}[\text{f}(\text{f}(\text{x}))]$ $=\text{f}\bigg[\text{f}\bigg(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\bigg)\bigg]$ $=\text{f}\begin{pmatrix}\frac{\frac{\text{x}}{\sqrt{1+\text{x}^2}}}{\sqrt{1+\frac{\text{x}^2}{1+\text{x}^2}}} \end{pmatrix}$ $=\text{f}\Big(\frac{\text{x}}{\sqrt{1+2\text{x}^2}}\Big)$ $=\frac{\frac{\text{x}}{\sqrt{1+2\text{x}^2}}}{\sqrt{1+\frac{\text{x}^2}{1+2\text{x}^2}}}$$=\frac{\text{x}}{\sqrt{1+3\text{x}^2}}$

View full question & answer→

Question 61 Mark

Fill in the blank.
If $\text{f}(\text{x})=[4-(\text{x}-7)^3],$ then $f^{-1}(x) =$ _______.

Answer

If $\text{f}(\text{x})=[4-(\text{x}-7)^3],$ then $\text{f}^{-1}(\text{x})=7+(4-\text{x})^\frac{1}{3}$ Solution: Given that, $\text{f}(\text{x})=[4-(\text{x}-7)^3],$Let $\text{y}=[4-(\text{x}-7)^3]$
$(\text{x}-7)^3=4-\text{y}$
$\Rightarrow\ (\text{x}-7)=(4-\text{y})^\frac{1}{3}$
$\Rightarrow\ \text{x}=7+(4-\text{y})^\frac{1}{3}$
$\text{f}^{-1}(\text{x})=7+(4-\text{x})^\frac{1}{3}$

View full question & answer→

Maths Fill In The Blanks[1 Marks ]

Take a timed test