Relations and Functions Test

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Relations and Functions Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

If $$f:(-\infty ,\infty )\rightarrow (-\infty ,\infty )$$ is defined by $$f(x)=5x-6$$, then $$f^{-1}(x)=$$

A
$$\dfrac{x+5}{6}$$

B
$$\dfrac{x-5}{6}$$

C
$$\dfrac{x-6}{5}$$

D
$$\dfrac{x+6}{5}$$

Solution

$$f(x)=5x-6=y$$
$$\Rightarrow 5x=y+6$$
$$x=\dfrac{y+6}{5}$$
$$\therefore f\left ( x \right )=y\Rightarrow x=f^{-1}\left ( y \right )$$
$$\therefore f^{-1}\left ( y \right )=\dfrac{y+6}{5}$$
Question 2
1 / -0

If $$f:R\rightarrow R $$ is defined by $$\displaystyle f(x)={\dfrac{2x+1}{3}}$$ then $$f^{-1}(x)=$$

A
$$\dfrac{3x-1}{2}$$

B
$$\dfrac{x-3}{2}$$

C
$$\dfrac{2x-1}{3}$$

D
$$\dfrac{x-4}{3}$$

Solution

$$f\left ( x \right )=\dfrac{2x+1}{3}$$
Let $$f\left ( x \right )=y=\dfrac{2x+1}{3}$$
$$\Rightarrow 2x+1=3y$$
$$2x=3y-1$$
$$x=\dfrac{3y-1}{2}$$
$$\therefore f\left ( x \right )=y\Rightarrow f^{-1}\left ( y \right )=x$$
$$\Rightarrow f^{-1}\left ( y \right )=\dfrac{3y-1}{2}$$
$$\Rightarrow f^{-1}\left ( x \right )=\dfrac{3x-1}{2}$$
Question 3
1 / -0

If $$f(x)=\dfrac{5x+6}{7x+9}$$ then $$f^{-1}(x)=$$

A
$$\dfrac{y+6}{7y+9}$$

B
$$\dfrac{7y+9}{5y+6}$$

C
$$\dfrac{9y-6}{-7y+9}$$

D
$$\dfrac{9y-6}{-7y+5}$$

Solution

Let $$f\left ( x \right )=y=\dfrac{5x+6}{7x+9}$$
$$7xy+9y=5x+6$$
$$\Rightarrow x\left ( 7y-5 \right )=6-9y$$
$$x=\dfrac{9y-6}{-7y+5}$$
$$x=f^{-1}\left ( y \right )=\dfrac{9y-6}{-7y+5}$$
Question 4
1 / -0

If $$f$$ from $$R$$ into $$R$$ is defined by $$f(x)=x^{3}-1$$, then $$f^{-1}\left \{ -2,0,7 \right \}=$$

A
$$\left \{ -1,1,2 \right \}$$

B
$$\left \{ 0,1,2 \right \}$$

C
$$\left \{ \pm 1,\pm 2 \right \}$$

D
$$\left \{ 0,\pm 2 \right \}$$

Solution

$$f$$ is invertible throughout $$R$$

$$\therefore f\left ( x \right )=x^{3}-1=y$$

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

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

$$f^{-1}\left ( 0 \right )=1$$

$$f^{-1}\left ( 7 \right )=2$$

$$\therefore f^{-1}\left \{ -2,0,7 \right \}=\left \{ -1,1,2 \right \}$$
Question 5
1 / -0

If $$f(x)=e^{5x+13}$$ then $$f^{-1}(x)=$$

A
$$\dfrac{13-\log y}{5}$$

B
$$\dfrac{-13+\log y}{5}$$

C
$$\dfrac{5+\log y}{13}$$

D
$$\dfrac{5-\log y}{13}$$

Solution

Let $$f\left ( x \right )=e^{5x+13}=y$$
$$f\left ( x \right )=y\Rightarrow x=f^{-1}\left ( y \right )$$
$$5x+13=\ln y$$
$$5 x=-13+\log y$$
$$ x=\dfrac{-13+\log y}{5}$$
$$\therefore f^{-1}\left ( y \right )=\dfrac{-13+\log y}{5}$$
Question 6
1 / -0

If $$f(x)=3x-1$$ and $$g(x)=5x+6$$ then $$(g^{-1}of^{-1})(2)=$$

A
$$10$$

B
$$-1$$

C
$$11$$

D
$$12$$

Solution

$$f\left ( x \right )=3x-1=y$$
$$\Rightarrow x=\displaystyle\frac{y+1}{3}$$
$$\therefore \displaystyle f^{-1}\left ( y \right )=\frac{y+1}{3}$$
$$\Rightarrow \displaystyle f^{-1}\left ( x \right )=\frac{x+1}{3}$$

$$g\left ( x \right )=5x+6=z$$
$$\Rightarrow x=\displaystyle \frac{z-6}{5}$$
$$g^{-1}\left ( z \right )=\displaystyle \frac{z-6}{5}$$
$$\Rightarrow g^{-1}\left ( x \right )=\displaystyle \frac{x-6}{5}$$

$$\displaystyle g^{-1}\left ( f^{-1} \left ( x \right )\right )=g^{-1}\left ( \frac{x+1}{3} \right )$$

$$=\displaystyle \frac{\dfrac{x+1}{3}-6}{5}$$

$$\displaystyle =\frac{x-17}{15}$$

$$\displaystyle \therefore g^{-1}\left ( f^{-1}\left ( 2 \right ) \right )=\frac{2-17}{15}=-1$$
Question 7
1 / -0

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

A
$$\dfrac{x+\sqrt{x^{2}-4}}{2}$$

B
$$\dfrac{x}{1+x^{2}}$$

C
$$\dfrac{x-\sqrt{x^{2}-4}}{2}$$

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

Solution

Let $$f\left ( x \right )=x+\dfrac{1}{x}=y$$
$$\Rightarrow x=f^{-1}\left ( y \right )$$ & $$x^{2}-yx+1=0$$
Solving $$x^{2}-yx+1$$, we get
$$x^{2}-yx+1=0$$
$$x=\dfrac{y\pm \sqrt{y^{2}-4}}{2}$$
$$\therefore f^{-1}=\dfrac{x+\sqrt{x^{2}-4}}{2}$$
$$\because$$ $$f$$ is defined from $$\left ( 1,\infty \right )\rightarrow \left ( 2,\infty \right )$$ negative part is discarded.
Question 8
1 / -0

The function $$f:(0,\infty )\rightarrow (-\infty ,\infty )$$ is defined by $$ f(x)=\log_{3} x $$ then $$ f^{-1}(x)=$$

A
$$3^{x}$$

B
$$3^{-x}$$

C
$$-3^{x}$$

D
$$-3x^{-x}$$

Solution

$$f\left ( x \right )=\log _{3}x$$
$$f^{-1}:\left ( -\infty ,\infty \right )\rightarrow \left ( 0, \infty \right )$$
Let $$\log _{3}x=y=f\left ( x \right )$$
$$\Rightarrow x=3^{y}$$
$$f\left ( x \right )=y\Rightarrow f^{-1}\left ( y \right )=x=3^{y}$$
$$\therefore f^{-1}\left ( x \right )=3^{x}$$
Question 9
1 / -0

If $$f:\left \{ 1,2,3,..... \right \}\rightarrow \left \{ 0,\pm 1,\pm 2,..... \right \}$$ is defined by $$f(n)=\begin{cases} \dfrac{n}{2} & \text{ if } n \space is \space even \\-\left (\dfrac{n-1}{2} \right ) & \text{ if } n \space is \space odd \end{cases}$$ then $$f^{-1}(-100)$$ is

A
Function is not invertible.

B
$$199$$

C
$$201$$

D
$$200$$

Solution

$$f(n)$$ is positive if $$n$$ is even & negative if $$n$$ is odd.
$$\therefore f^{-1}\left ( -100 \right )=-2x+1$$
$$=-2\left ( -100 \right )+1$$
$$=200+1$$
$$=201$$
Question 10
1 / -0

$$f:R\rightarrow R$$ is defined by $$f(x)=x^{2}+4$$ then $$f^{-1}(13)=$$

A
$$\left \{ -3,3 \right \}$$

B
$$\left \{ -2,2 \right \}$$

C
$$\left \{ -1,1 \right \}$$

D
Not invertible

Solution

$$f\left ( x \right )=x^{2}+4=y$$
$$\Rightarrow x^{2}=y-4$$
$$x=\pm \sqrt{y-4}$$
$$\therefore f^{-1}\left ( 13 \right )=\pm \sqrt{13-4}=\pm 3 $$
$$f(3)=13\,\&\,f(-3)=13$$
Thus image $$13$$ has two pre-images i.e, $$3$$ and $$-3$$
$$\therefore$$ $$f$$ is not invertible

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Relations and Functions Test - 37

Result

Relations and Functions Test - 37

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SHARING IS CARING

Question 1

$$\dfrac{x+5}{6}$$

$$\dfrac{x-5}{6}$$

$$\dfrac{x-6}{5}$$

$$\dfrac{x+6}{5}$$

Solution

Question 2

$$\dfrac{3x-1}{2}$$

$$\dfrac{x-3}{2}$$

$$\dfrac{2x-1}{3}$$

$$\dfrac{x-4}{3}$$

Solution

Question 3

$$\dfrac{y+6}{7y+9}$$

$$\dfrac{7y+9}{5y+6}$$

$$\dfrac{9y-6}{-7y+9}$$

$$\dfrac{9y-6}{-7y+5}$$

Solution

Question 4

$$\left \{ -1,1,2 \right \}$$

$$\left \{ 0,1,2 \right \}$$

$$\left \{ \pm 1,\pm 2 \right \}$$

$$\left \{ 0,\pm 2 \right \}$$

Solution

Question 5

$$\dfrac{13-\log y}{5}$$

$$\dfrac{-13+\log y}{5}$$

$$\dfrac{5+\log y}{13}$$

$$\dfrac{5-\log y}{13}$$

Solution

Question 6

$$10$$

$$-1$$

$$11$$

$$12$$

Solution

Question 7

$$\dfrac{x+\sqrt{x^{2}-4}}{2}$$

$$\dfrac{x}{1+x^{2}}$$

$$\dfrac{x-\sqrt{x^{2}-4}}{2}$$

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

Solution

Question 8

$$3^{x}$$

$$3^{-x}$$

$$-3^{x}$$

$$-3x^{-x}$$

Solution

Question 9

Function is not invertible.

$$199$$

$$201$$

$$200$$

Solution

Question 10

$$\left \{ -3,3 \right \}$$

$$\left \{ -2,2 \right \}$$

$$\left \{ -1,1 \right \}$$

Not invertible

Solution

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