Relations and Functions Test

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Relations and Functions Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=\displaystyle \dfrac{x}{\sqrt{1-x^{2}}},g(x)=\displaystyle \dfrac{x}{\sqrt{1+x^{2}}} $$, then $$(fog)(x)=$$

A
$$x$$

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

C
$$\sqrt{1+x^{2}}$$

D
$$2x$$

Solution

$$fog\left(x\right)=f\left(g\left(x\right)\right)$$
$$\displaystyle =f\left(\dfrac{x}{\sqrt{1+x^{2}}}\right)$$
$$\displaystyle =\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\sqrt{1-\dfrac{x^{2}}{1+x^{2}}}}=x$$
Question 2
1 / -0

If $$f(x)=1+x+x^{2}+x^{3}+\ldots\ldots $$ for $$\left | x \right |<1$$ then $$f^{-1}(x)=$$

A
$$\dfrac{x-1}{x+1}$$

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

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

D
$$\dfrac{x-1}{x}$$

Solution

$$f(x)=1+x+x^{2}+\ldots\ldots$$
$$=\dfrac{1}{1-x}$$
Let $$y=\dfrac{1}{1-x}=f(x)$$
$$1-x=\dfrac{1}{y}$$
$$x=1-\dfrac{1}{y}=\dfrac{y-1}{y}$$
But $$x=f^{-1}(y)=\dfrac{y-1}{y}$$
Question 3
1 / -0

If the function is $$f:R\rightarrow R, g:R\rightarrow R$$ are defined as $$f(x)=2x+3, g(x)=x^{2}+7$$ and $$f[g(x)]=25$$ then $$x=$$

A
$$f(x)$$

B
$$\pm 2$$

C
$$\pm 3$$

D
$$\pm 4$$

Solution

$$f(g(x))=f(x^{2}+7) =25$$
$$=2(x^{2}+7)+3$$
$$=2x^{2}+17$$
$$\Rightarrow 2x^{2}=8$$
$$x^{2}=4 \Rightarrow x=\pm 2$$
Question 4
1 / -0

If $$f(x)=\displaystyle \frac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$$, then $$f^{-1}(x)=$$

A
$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-1}{x+1} \right )$$

B
$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-1} \right )$$

C
$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-2} \right )$$

D
$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-2}{x-1} \right )$$

Solution

Let $$f\left(x\right)=y=\displaystyle \frac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$$
$$\Rightarrow 2^{x}\left(y-1\right)=2^{-x}\left(1+y\right)$$
$$2^{2x}=\displaystyle\frac{y+1}{y-1}$$
$$\displaystyle 2x=\log_{2}\left(\frac{y+1}{y-1}\right)$$
$$\Rightarrow \displaystyle x=f^{-1}\left(y\right)=\frac{1}{2}\log_{2}\left(\frac{y+1}{y-1}\right)$$
So, $$\displaystyle f^{-1}x=\frac{1}{2}\log_{2}\left(\frac{x+1}{x-1}\right)$$
Question 5
1 / -0

I: If $$f:A\rightarrow B$$ is a bijection only then does $$f$$ have an inverse function
II: The inverse function $$f:R^{+}\rightarrow R^{+}$$ defined by $$f(x)=x^{2}$$ is $$f^{-1}(x)=\sqrt{x}$$

A
only I is true

B
only II is true

C
both I and II are true

D
neither I nor II true

Solution

$$f$$ is a bijection $$\Rightarrow$$ $$f$$ is one one and onto.
$$\therefore$$ there exists on inverse value for every value in co-domain.
$$\therefore$$ $$f$$ is invertible if and only if $$f$$ is a bijection
$$f:R^{+}\rightarrow R^{+} ; f(x)=x^{2}=y$$
$$\Rightarrow x=\sqrt{y}=f^{-1}(y)$$
Question 6
1 / -0

If $$F(n)=(-1)^{k-1}(n-1), G(n)=n-F(n)$$ then $$ (GoG)(n)=$$ (where $$k$$ is odd)

A
$$1$$

B
$$n$$

C
$$2$$

D
$$n-1$$

Solution

$$G(n)=n-(-1)^{k-1}(n-1)$$
$$GoG(n)=G(n-(-1)^{k-1}(n-1))$$
$$=n-(-1)^{k-1}(n-1)-(-1)^{k-1}((n-1)-(-1)^{k-1}(n-1))$$
$$=n-(n-1)$$
$$=1$$
Question 7
1 / -0

If $$f(x)=\dfrac{x}{\sqrt{1-x^{2}}}$$, then $$ (fof)(x)=$$

A
$$\dfrac{x}{\sqrt{1-x^{2}}}$$

B
$$\dfrac{x}{\sqrt{1-2x^{2}}}$$

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

D
$$x$$

Solution

$$fof\left ( x \right )=f\left ( f\left ( x \right ) \right )=\left ( x/\sqrt{1-x^{2}} \right )/\left ( \sqrt{1-\left ( x^{2}/\left ( 1-x^{2} \right ) \right )} \right )$$
$$=\left ( x/\sqrt{1-x^{2}} \right )/\sqrt{\left ( 1-x^{2}-x^{2} \right )/\left ( 1-x^{2} \right )}$$
$$=\left ( x/\sqrt{1-x^{2}} \right )/\left ( \sqrt{1-2 x^{2}} /\sqrt{1-x^{2}}\right)$$
$$=x/\sqrt{1-2x^{2}}$$
Question 8
1 / -0

If $$f(x)=\sin^{-1}\left \{ 3-(x-6)^{4} \right \}^{1/3}$$ then $$ f^{-1}(x)=$$

A
$$6+\sqrt[4]{3+\sin^{3}x}$$

B
$$6+\sqrt[4]{3-\sin^{3}x}$$

C
$$6+\sqrt[4]{3+\sin x}$$

D
$$6+\sqrt[4]{3-\sin x}$$

Solution

Let $$f(x)=y=\sin^{-1}(3-(x-6)^{4})^{\frac{1}{3}}$$
$$\sin y=(3-(x-6)^{4})^{\frac{1}{3}}$$
$$\sin^{3}y=3-(x-6)^{4}$$
$$(x-6)^{4}=3-\sin^{3}y$$
$$x-6=(3-\sin^{3}y)^{\frac{-1}{4}}$$
$$x=(3-\sin^{3}y)^{\frac{1}{4}}+6$$
$$x=f^{-1}(y)=(3-\sin^{3}y)^{\frac{1}{4}}+6.$$
Question 9
1 / -0

Which of the following functions defined from $$(-\infty ,\infty )$$ to $$ (-\infty ,\infty )$$ is invertible ?

A
$$f(x) = \sin (2x+3)$$

B
$$f(x) = x^{2} + 4$$

C
$$ f (x) =x^{3}$$

D
$$f (x) = \cos x$$

Solution

$$f(x)=\sin(2x+3)$$ lies only from $$[-1,1]$$ and hence is not onto.
So, $$f(x)$$ is not bijective
$$\therefore$$ it is not invertible

$$f(x)=x^{2}+4$$ is always positive and so not onto.
So, $$f(x)$$ is not bijective
$$\therefore$$ It is also not invertible.

$$f(x)=x^{3}$$ varies from $$(-\infty ,\infty )$$ as $$x$$ varies from $$(-\infty ,\infty ) $$
So, $$f(x)$$ is bijective
$$\therefore$$ It is invertible.

$$f(x)=\cos x$$ always lies between $$[-1,1]$$ and so is into function.
So, $$f(x)$$ is not bijective
Hence, not invertible.
Question 10
1 / -0

If $$f(x)=\dfrac{x}{\sqrt{1+x^{2}}}$$ then $$fofof(x)=$$

A
$$\dfrac{x}{\sqrt{1+3x^{2}}}$$

B
$$\dfrac{x}{\sqrt{1-x^{2}}}$$

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

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

Solution

$$fofof\left(x\right)=fof\left(\dfrac{x}{\sqrt{1+x^{2}}}\right)$$
$$=f\left(\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+x^{2}}}}\right)$$
$$=f\left(\dfrac{x}{\sqrt{1+2x^{2}}}\right)$$
$$=\left(\dfrac{\dfrac{x}{\sqrt{1+2x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+2x^{2}}}}\right)$$
$$=\dfrac{x}{\sqrt{1+3x^{2}}}$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 39

Result

Relations and Functions Test - 39

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SHARING IS CARING

Question 1

$$x$$

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

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

$$2x$$

Solution

Question 2

$$\dfrac{x-1}{x+1}$$

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

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

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

Solution

Question 3

$$f(x)$$

$$\pm 2$$

$$\pm 3$$

$$\pm 4$$

Solution

Question 4

$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-1}{x+1} \right )$$

$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-1} \right )$$

$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-2} \right )$$

$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-2}{x-1} \right )$$

Solution

Question 5

only I is true

only II is true

both I and II are true

neither I nor II true

Solution

Question 6

$$1$$

$$n$$

$$2$$

$$n-1$$

Solution

Question 7

$$\dfrac{x}{\sqrt{1-x^{2}}}$$

$$\dfrac{x}{\sqrt{1-2x^{2}}}$$

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

$$x$$

Solution

Question 8

$$6+\sqrt[4]{3+\sin^{3}x}$$

$$6+\sqrt[4]{3-\sin^{3}x}$$

$$6+\sqrt[4]{3+\sin x}$$

$$6+\sqrt[4]{3-\sin x}$$

Solution

Question 9

$$f(x) = \sin (2x+3)$$

$$f(x) = x^{2} + 4$$

$$ f (x) =x^{3}$$

$$f (x) = \cos x$$

Solution

Question 10

$$\dfrac{x}{\sqrt{1+3x^{2}}}$$

$$\dfrac{x}{\sqrt{1-x^{2}}}$$

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

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

Solution

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