Relations and Functions Test

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Relations and Functions Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle f:[1,+\infty ]\rightarrow [2,+\infty )$$ is given by $$f(x)=x+\dfrac{1}{x}$$ then $$f^{-1}(x)$$ equals

A
$$\displaystyle \frac{x+\sqrt{x^{2}+4}}{2}$$

B
$$\displaystyle \frac{x}{1+x^{2}}$$

C
$$\displaystyle \frac{x+\sqrt{x^{2}-4}}{2}$$

D
$$\displaystyle 1+\sqrt{x^{2}-4}$$

Solution

Let $$y = x+\cfrac{1}{x}$$
$$\Rightarrow x^2-yx+1=0\Rightarrow x = \cfrac{y+\sqrt{y^2-4}}{2}=f^{-1}(y)$$
Hence $$f^{-1}(x) = \cfrac{x+\sqrt{x^2-4}}{2}$$
Note: -ve term in step two is discarded using given domain and co-domain.
Question 2
1 / -0

If $$f : \{1,2,3,...\} \rightarrow \{0, \pm 1, \pm 2,...\}$$ is defined by
$$\displaystyle y=f(x)=\begin{cases} \displaystyle \frac { x }{ 2 } \quad \quad \text{ if x is even } \\ -\displaystyle \frac { (x-1) }{ 2 } \quad ,\text{ if x is odd } \end{cases}$$, then $$f^{-1}(100)$$ is

A
Function is not invertible as it is not onto

B
$$199$$

C
$$201$$

D
$$200$$

Solution

Since $$100$$ is even $$f^{-1}(x)=2x$$

Hence $$f^{-1}(100)=200$$
Question 3
1 / -0

If $$\displaystyle f(y)=\frac{y}{\sqrt{1-y^2}}$$; $$\displaystyle g(y)=\frac{y}{\sqrt{1+y^2}}$$ then $$(fog)y$$ is equal to

A
$$\displaystyle \frac{y}{\sqrt{1-y^2}}$$

B
$$\displaystyle \frac{y}{\sqrt{1+y^2}}$$

C
$$y$$

D
$$2f(x)$$

Solution

Given $$\displaystyle f(y)=\frac{y}{\sqrt{1-y^2}}$$ and $$\displaystyle g(y)=\frac{y}{\sqrt{1+y^2}}$$
$$\therefore \displaystyle (fog)y = f(g(y))=\dfrac{\dfrac{y}{\sqrt{1+y^2}}}{\sqrt{1-\dfrac{y^2}{1+y^2}}}=y$$
Question 4
1 / -0

$$f:R\rightarrow R$$ is a function defined by $$f(x)=10x-7$$. If $$g=f^{-1}$$, then $$g(x)$$ is equals

A
$$\dfrac{1}{10x-7}$$

B
$$\dfrac{1}{10x+7}$$

C
$$\dfrac{x+7}{10}$$

D
$$\dfrac{x-7}{10}$$

Solution

$$y=10x-7$$

It is a one one function over the given domain.

$$y+7=10(x)$$

$$x=\dfrac{y+7}{10}$$

Hence

$$f^{-1}$$ will be $$g(x)$$

$$g(x)=\dfrac{x+7}{10}$$

Hence, option 'C' is correct.
Question 5
1 / -0

Let $$f:[4,\infty )\rightarrow [4,\infty )$$ be a function defined by $$f\left( x \right)={ 5 }^{ x\left( x-4 \right) }$$, then $$f^{ -1 }\left( x \right)$$ is

A
$$2-\sqrt { 4+\log _{ 5 }{ x } } $$

B
$$2+\sqrt { 4+\log _{ 5 }{ x } } $$

C
$$\displaystyle { \left( \frac { 1 }{ 5 } \right) }^{ x\left( x-4 \right) }$$

D
None of these

Solution

Let $$y={ 5 }^{ x\left( x-4 \right) }\Rightarrow x\left( x-4 \right) =\log _{ 5 }{ y } \Rightarrow { x }^{ 2 }-4x-\log _{ 5 }{ y } =0$$

$$\displaystyle \Rightarrow x=\frac { 4\pm \sqrt { 16+4\log _{ 5 }{ y } } }{ 2 } =\left( 2\pm \sqrt { 4+\log _{ 5 }{ y } } \right) $$

But $$x\ge 4$$, so $$x=\left( 2+\sqrt { 4+\log _{ 5 }{ y } } \right) $$

$$\therefore f^{ -1 }\left( y \right) =2+\sqrt { 4+\log _{ 5 }{ y } } $$

$$\therefore f^{ -1 }\left( x \right) =2+\sqrt { 4+\log _{ 5 }{ x } } $$
Question 6
1 / -0

The inverse of $$\displaystyle f\left ( x \right )=\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$$ is

A
$$\displaystyle \frac{1}{6}\log_{10}\left ( \frac{1+x}{1-x} \right )$$

B
$$\displaystyle \frac{1}{6}\log_{10}\left ( \frac{x}{1-x} \right )$$

C
$$\displaystyle \frac{1}{6}\log_{e}\left ( \frac{1+x}{1-x} \right )$$

D
$$\displaystyle \frac{1}{6}\log_{e}\left ( \frac{1-x}{1+x} \right )$$

Solution

$$\displaystyle y=\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$$

$$\Rightarrow \displaystyle y=\frac{e^{6x}-1}{e^{6x}+1}$$

$$\Rightarrow \displaystyle \frac{y+1}{y-1}=\frac{e^{6x}-1+e^{6x}+1}{e^{6x}-1-e^{6x}-1}$$

$$\Rightarrow \displaystyle \frac{y+1}{y-1}=\frac{2e^{6x}}{-2}$$

$$\Rightarrow \displaystyle \frac{1+y}{1-y}=e^{6x}$$

Taking $$\log_e$$ we get $$\displaystyle \log_{e}\left(\frac{1+y}{1-y}\right)=6x$$

$$\displaystyle x=\frac{1}{6}\left(\log_{e}\left(\frac{1+y}{1-y}\right)\right)$$

Hence $$f^{-1}$$$$\displaystyle =\frac{1}{6}\left(\log_{e}\left(\frac{1+x}{1-x}\right)\right)$$
Question 7
1 / -0

If $$\displaystyle f\left ( x \right )=\left\{\begin{matrix}
x^{2} x \geq 0\\
x x < 0
\end{matrix}\right.$$
then $$\displaystyle (f o f)(x)$$ is given by

A
$$x^{2}$$ for $$x\geq 0$$ and $$x$$ for $$ x< 0$$

B
$$\displaystyle x^{4}$$ for $$\displaystyle x\geq 0$$ and $$x^{2}$$ for $$x< 0$$

C
$$ \displaystyle x^{4}$$ for $$ \displaystyle x\geq 0$$ and $$-x^{2} $$ for $$x < 0$$

D
$$\displaystyle x^{4}$$ for $$ x\geq 0$$ and $$x $$ for $$ x< 0$$

Solution

For $$x\ge0$$,we have $$\displaystyle f \circ f\left( x \right)= {\left( {{x^2}} \right)^2}=\left( {{x^4}} \right)$$
For $$x<0$$,we have $$\displaystyle f \circ f\left( x \right)=x$$
Question 8
1 / -0

Let $$\displaystyle g(x)=1+x-[x]$$ and $$\displaystyle f(x)=\left\{\begin{matrix}{-1}\quad {x< 0} \\ {0} \quad {x=0}\\{1} \quad {x> 0} \end{matrix}\right.$$ Then for all $$\displaystyle x, f\left \{ g\left ( x \right ) \right \}$$ is equal to

A
$$x$$

B
$$1$$

C
$$\displaystyle f(x)$$

D
$$\displaystyle g(x)$$

Solution

Let $$g(x) = 1+x-[x] = 1+\{x\}> 0$$

since $$\{x\}\in [0,1) \forall x\in R$$

Hence $$f\{g(x)\} = 1$$
Question 9
1 / -0

If $$\displaystyle f(x)= \frac{3x+2}{5x-3}$$ then

A
$$\displaystyle f^{-1}(x)= -f(x)$$

B
$$\displaystyle f^{-1}(x)= f(x)$$

C
$$\displaystyle fo(f(x))= -x $$

D
$$\displaystyle f^{-1}(x)= -\frac{1}{19}f(x)$$

Solution

Let $$y=\displaystyle \frac{3x+2}{5x-3}$$

$$\Rightarrow \displaystyle x=\frac{3y+2}{5y-3}=f^{-1}(y)$$

$$\displaystyle \therefore f^{-1}(x)=\frac{3x+2}{5x-3}=f(x)$$
Question 10
1 / -0

The inverse of the function $$\displaystyle f(x) = \log_{2}(x+\sqrt{x^{2}+1}) $$ is

A
$$\displaystyle 2^{x}+2^{-x} $$

B
$$\displaystyle \frac{2^{x}+2^{-x}}{2}$$

C
$$\displaystyle \frac{2^{-x}-2^{x}}{2}$$

D
$$\displaystyle \frac{2^{x}-2^{-x}}{2}$$

Solution

We have to find the inverse of $$f(x)$$ which means $$f$$ is one one and onto function
Let $$\displaystyle f\left ( x \right )= y= \log _{2}\left (

x+\sqrt{x^{2}+1} \right )$$
$$\displaystyle \therefore 2^{y}=

x+\sqrt{x^{2}+1}$$ ...(i)
$$\displaystyle 2^{-y}=

\frac{1}{x+\sqrt{x^{2}+1}}= \frac{\sqrt{x^{2}+1}-x}{1}$$ ...(ii)

By subtracting (i) and (ii) we have
$$ \therefore 2^{y}-2^{-y}=

2x$$
or $$\displaystyle x= \frac{1}{2}\left ( 2^{y}-2^{-y} \right )$$
$$\displaystyle

\therefore f^{-1}\left ( y \right )= \frac{1}{2}\left ( 2^{y}-2^{-y}

\right )$$
Replace $$\displaystyle y\rightarrow x$$
$$\displaystyle \therefore f^{-1}\left ( x \right )= \frac{1}{2}\left ( 2^{x}-2^{-x} \right )$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 45

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Relations and Functions Test - 45

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SHARING IS CARING

Question 1

$$\displaystyle \frac{x+\sqrt{x^{2}+4}}{2}$$

$$\displaystyle \frac{x}{1+x^{2}}$$

$$\displaystyle \frac{x+\sqrt{x^{2}-4}}{2}$$

$$\displaystyle 1+\sqrt{x^{2}-4}$$

Solution

Question 2

Function is not invertible as it is not onto

$$199$$

$$201$$

$$200$$

Solution

Question 3

$$\displaystyle \frac{y}{\sqrt{1-y^2}}$$

$$\displaystyle \frac{y}{\sqrt{1+y^2}}$$

$$y$$

$$2f(x)$$

Solution

Question 4

$$\dfrac{1}{10x-7}$$

$$\dfrac{1}{10x+7}$$

$$\dfrac{x+7}{10}$$

$$\dfrac{x-7}{10}$$

Solution

Question 5

$$2-\sqrt { 4+\log _{ 5 }{ x } } $$

$$2+\sqrt { 4+\log _{ 5 }{ x } } $$

$$\displaystyle { \left( \frac { 1 }{ 5 } \right) }^{ x\left( x-4 \right) }$$

None of these

Solution

Question 6

$$\displaystyle \frac{1}{6}\log_{10}\left ( \frac{1+x}{1-x} \right )$$

$$\displaystyle \frac{1}{6}\log_{10}\left ( \frac{x}{1-x} \right )$$

$$\displaystyle \frac{1}{6}\log_{e}\left ( \frac{1+x}{1-x} \right )$$

$$\displaystyle \frac{1}{6}\log_{e}\left ( \frac{1-x}{1+x} \right )$$

Solution

Question 7

$$x^{2}$$ for $$x\geq 0$$ and $$x$$ for $$ x< 0$$

$$\displaystyle x^{4}$$ for $$\displaystyle x\geq 0$$ and $$x^{2}$$ for $$x< 0$$

$$ \displaystyle x^{4}$$ for $$ \displaystyle x\geq 0$$ and $$-x^{2} $$ for $$x < 0$$

$$\displaystyle x^{4}$$ for $$ x\geq 0$$ and $$x $$ for $$ x< 0$$

Solution

Question 8

$$x$$

$$1$$

$$\displaystyle f(x)$$

$$\displaystyle g(x)$$

Solution

Question 9

$$\displaystyle f^{-1}(x)= -f(x)$$

$$\displaystyle f^{-1}(x)= f(x)$$

$$\displaystyle fo(f(x))= -x $$

$$\displaystyle f^{-1}(x)= -\frac{1}{19}f(x)$$

Solution

Question 10

$$\displaystyle 2^{x}+2^{-x} $$

$$\displaystyle \frac{2^{x}+2^{-x}}{2}$$

$$\displaystyle \frac{2^{-x}-2^{x}}{2}$$

$$\displaystyle \frac{2^{x}-2^{-x}}{2}$$

Solution

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