Relations and Functions Test

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Relations and Functions Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x) = x^3 $$ and $$g(x) = sin2x$$, then

A
$$g[f(1)] = 1$$

B
$$f(g(\pi/12) = 1/8$$

C
$$g{f(2)} = \sin 2$$

D
none of these

Solution

Given,
$$ f(x)=x^3 $$ and $$g(x)=\sin 2x$$

Clearly,

$$ f(1)=1 \Rightarrow g[f(1)]=\sin 2 $$

$$ g(\cfrac{\pi}{12}) =\dfrac{1} {2} \Rightarrow f[g(\dfrac{\pi}{12})]=\dfrac{1}{8}$$

and,$$ f(2)=8 \Rightarrow g[f(2)]= \sin 16 $$

So, the correct option is (B).
Question 2
1 / -0

If $$f(x) = (a x^n)^{1/n},$$ where $$\ n \in N$$, then $$f\{f(x)\}$$ equals

A
$$0$$

B
$$x$$

C
$$x^n$$

D
none of these

Solution

We have, $$f(x) = (a x^n)^{1/n}=a^{1/n}x=kx$$ where $$k = a^{1/n}$$

$$\therefore f\{f(x)\} = kf(x) = k^2x =a^{2/n} x$$
Question 3
1 / -0

Let $$X = \left\{1,2,3,4\right\}$$ and $$Y = \left\{1,3,5,7,9\right\}$$. Which of the following is relations from $$X$$ to $$Y$$

A
$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$

B
$$R_2 = \left\{(1,1),(2,1),(3,3),(4,3),(5,5)\right\}$$

C
$$R_3 = \left\{(1,1),(1,3),(3,5),(3,7),(5,7)\right\}$$

D
$$R_4 = \left\{(1,3),(2,5), (2,4), (7,9)\right\}$$

Solution

We have
$$X = \left\{1,2,3,4\right\}$$ and $$Y = \left\{1,3,5,7,9\right\}$$.
$$X \times Y=\{(1,1),(1,3),(1,5),(1,7),(1,9), (2,1),(2,3),(2,5),$$
$$(2,7),(2,9), (3,1),(3,3),(3,5),(3,7),(3,9),(4,1),(4,3),(4,5),(4,7),(4,9)\}$$

Let's take option A,
$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$
$$R_{1}= \{(1,3),(2,5),(3,7), (4,9)\}$$
Since, $$R_1 \subseteq X\times Y$$
So, $$R_1$$ is a relation from $$X$$ to $$Y$$

Clearly, $$R_2$$ is not a relation from $$X$$ to $$Y$$ as $$(5,5) \notin X\times Y$$
Similarly, $$R_3$$ is not a relation from $$X$$ to $$Y$$ as $$(5,7) \notin X\times Y$$
In the same manner, $$R_4$$ is also not a relation from $$X$$ to $$Y$$ as $$(7,9) \notin X\times Y$$
Question 4
1 / -0

If $$f(x) =\dfrac {x+2}{x-1}=y$$, then.

A
$$x = f(y)$$

B
$$f(1) = 3$$

C
$$f(y) = 2f(x)$$

D
$$f(y) = 2 + f(x)$$

Solution

Given, $$ f(x) = \dfrac {x+2}{x-1}=y $$

$$ =>x+2 = y (x-1) $$

$$ => y +2 = x y - x $$

$$ => x (y-1) = (y +2) $$

$$ => x = \dfrac {y+2}{y-1} $$

$$ => x = f (y) $$
Question 5
1 / -0

If $$f(x) =\ln {\displaystyle \frac { 1+x }{ 1-x } } $$ and $$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$$, then $$f[g(x)]$$ equals.

A
$$f(x)$$

B
$$[f(x)]^3$$

C
$$3f(x)$$

D
$${f(x)}^2$$

Solution

Given $$f(x)=\ln {\displaystyle \frac { 1+x }{ 1-x } } $$ and $$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$$

then $$f[g(x)]=\ln {\displaystyle \frac { 1+\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } }{ 1-\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } } } $$

$$=\ln {\displaystyle \frac { 1+3x^{ 2 }+3x+x^{ 3 } }{ 1+3x^{ 2 }-3x-x^{ 3 } } } $$

$$=\ln\left[ {\displaystyle \frac { 1+x }{ 1-x } }\right]^3 $$

$$=3\ln {\displaystyle \frac { 1+x }{ 1-x } } $$

$$\Rightarrow f[g(x)]=3f(x)$$

Hence, option C.
Question 6
1 / -0

Let $$f(x) = e^{3x}, g(x) = \log_ex, x > 0$$, then $$fog (x)$$ is

A
$$3x$$

B
$$x^3$$

C
$$\log_{10}3x$$

D
$$\log3x$$

Solution

Given $$f(x) = e^{3x}, g(x) = \log_ex, x > 0$$

$$fog(x)=f(\log_ex)=e^{3\log_ex}=x^3$$

Hence, option B.
Question 7
1 / -0

If $$f(x) = x^3- 1 $$ and domain of $$f = \{0, 1, 2, 3\}$$, then domain of $$f^{-1}(x)$$ is

A
$$\{0, 1, 2, 3\}$$

B
$$\{1, 0, 7, 26\}$$

C
$$\{0, -1, 7, 26\}$$

D
$$\{ 1, 2, 3,4\}$$

Solution

We know domain of $$f^{-1}(x)$$ will be the range of $$f(x)$$
which are $$ = \{0^3-1, 1^3-1, 2^3-1, 3^3-1\} = \{-1,0,7,26\}$$
Question 8
1 / -0

Let $$f : R \rightarrow R, g : R \rightarrow R$$ be two function such that
$$f(x) = 2x- 3, g(x) = x^3 + 5$$
The function $$(fog)^{1}(x)$$ is equal to.

A
$$\left ( \dfrac {x+7}{2} \right )^{1/3}$$

B
$$\left (x- \dfrac {7}{2} \right )^{1/3}$$

C
$$\left ( \dfrac {x-2}{7} \right )^{1/3}$$

D
$$\left ( \dfrac {x-7}{2} \right )^{1/3}$$

Solution

Given $$f(x) = 2x-3$$ and $$g(x) = x^3+ 5$$
$$(fog)(x) = fg(x) = f(x^3+5) = 2(x^3+5) -3 = 2x^3 +7$$
Let $$(fog)(x) = y = 2x^3+7$$
$$y-7 = 2x^3$$
$$x = (\dfrac{y-7}{2})^{\dfrac{1}{3}}$$
$$\therefore (fog)^{-1}(x) = (\dfrac{x-7}{2})^{\dfrac{1}{3}}$$
Question 9
1 / -0

If $$f(x) = 3x - 5, \,\,then\,\, f^{-1}(x)$$.

A
is given by $$\dfrac {1}{3x-5}$$

B
is given by $$\dfrac {x+5}{3}$$

C
does not exist because f is not one-one

D
does not exist because f is not onto

Solution

Given, $$f(x)= 3x-5$$

Let $$y=f^{ -1 }\left( x \right) $$

$$ \Rightarrow x=f\left( y \right) =3y-5$$

$$ \Rightarrow y=\dfrac { x+5 }{ 3 } $$
Question 10
1 / -0

The inverse of the function $$f(x)= [-1+ (x -3)^5]^{\tfrac 17}$$ is

A
$$3 + (1 + x^7)^{\tfrac 15}$$

B
$$3 - (1- x^7)^{\tfrac 15}$$

C
$$3 - (1 + x^7)^{\tfrac 15}$$

D
none of these

Solution

$$f\left( x \right) =[-1+(x-3)^{ 5 }]^{\frac17}$$

Let,

$$ y=f^{ -1 }\left( x \right) $$

$$ \Rightarrow x=f\left( y \right) =[-1+(y-3)^{ 5 }]^{\frac17}$$

$$\Rightarrow y=3+(1+x^{ 7 })^{\frac15}$$

Hence, option A.

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Relations and Functions Test - 50

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Relations and Functions Test - 50

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SHARING IS CARING

Question 1

$$g[f(1)] = 1$$

$$f(g(\pi/12) = 1/8$$

$$g{f(2)} = \sin 2$$

none of these

Solution

Question 2

$$0$$

$$x$$

$$x^n$$

none of these

Solution

Question 3

$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$

$$R_2 = \left\{(1,1),(2,1),(3,3),(4,3),(5,5)\right\}$$

$$R_3 = \left\{(1,1),(1,3),(3,5),(3,7),(5,7)\right\}$$

$$R_4 = \left\{(1,3),(2,5), (2,4), (7,9)\right\}$$

Solution

Question 4

$$x = f(y)$$

$$f(1) = 3$$

$$f(y) = 2f(x)$$

$$f(y) = 2 + f(x)$$

Solution

Question 5

$$f(x)$$

$$[f(x)]^3$$

$$3f(x)$$

$${f(x)}^2$$

Solution

Question 6

$$3x$$

$$x^3$$

$$\log_{10}3x$$

$$\log3x$$

Solution

Question 7

$$\{0, 1, 2, 3\}$$

$$\{1, 0, 7, 26\}$$

$$\{0, -1, 7, 26\}$$

$$\{ 1, 2, 3,4\}$$

Solution

Question 8

$$\left ( \dfrac {x+7}{2} \right )^{1/3}$$

$$\left (x- \dfrac {7}{2} \right )^{1/3}$$

$$\left ( \dfrac {x-2}{7} \right )^{1/3}$$

$$\left ( \dfrac {x-7}{2} \right )^{1/3}$$

Solution

Question 9

is given by $$\dfrac {1}{3x-5}$$

is given by $$\dfrac {x+5}{3}$$

does not exist because f is not one-one

does not exist because f is not onto

Solution

Question 10

$$3 + (1 + x^7)^{\tfrac 15}$$

$$3 - (1- x^7)^{\tfrac 15}$$

$$3 - (1 + x^7)^{\tfrac 15}$$

none of these

Solution

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