Relations and Functions Test

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Relations and Functions Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$f(x)=\frac{x}{\sqrt{1-x^{2}}}$$

B
$$f(x)=\frac{x}{\sqrt{1+x^{2}}}$$

C
$$x^{2}$$

D
x

Solution

Given:-
$$f{\left( x \right)} = \cfrac{x}{\sqrt{1 - {x}^{2}}}$$
$$g{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}$$
To find:-
$$fog{\left( x \right)} = ?$$
$$fog{\left( x \right)}$$
$$= f{\left( g{\left( x \right)} \right)}$$
$$= f{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{1 - {\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}^{2}}}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{1 - \left( \cfrac{{x}^{2}}{1 + {x}^{2}} \right)}}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{\left( \cfrac{1 + {x}^{2} - {x}^{2}}{1 + {x}^{2}} \right)}}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\left( \cfrac{1}{1 + {x}^{2}} \right)}$$
$$= \cfrac{x}{\sqrt{1 + {x}^{2}}}$$
$$\Rightarrow fog{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}$$
Hence the correct answer is $$\cfrac{x}{\sqrt{1 + {x}^{2}}}$$.
Question 2
1 / -0

If $$ f ( x ) = \left( a - x ^ { n } \right) ^ { 1 / n }$$ where $$ a > 0$$ and } $$n$$ is a positive integer then$$( f o f ) ( x )$$ is

A
$$f ( x )$$

B
x

C
0

D
1

Solution

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

if $$f\left( x \right) = \log \left( {\dfrac{{1 +x}}{{1 - x}}} \right)$$ and $$g\left( x \right) = \dfrac{{3x + {x^3}}}{{1 + 3{x^2}}}$$ then $$\left( {f(g(x)))} \right)$$ is equal to

A
$$ - f\left( x \right)$$

B
$$3f\left( x \right)$$

C
$${\left( {f\left( x \right)} \right)^3}$$

D
$$f\left( {3x} \right)$$

Solution

$$f(x)=\log\bigg(\dfrac{1+x}{1-x}\bigg)\ and\ g(x)=\dfrac{3x+x^{3}}{1+3x^{2}}$$

$$f(g(x))=\log\bigg(\dfrac{1+\dfrac{3x+x^{3}}{1+3x^{2}}}{1-\dfrac{3x+x^{3}}{1+3x^{2}}}\bigg)$$

$$f(g(x))=\log\bigg(\dfrac{\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}}}{\dfrac{1+3x^{2}-3x-x^{3}}{1+3x^{2}}}\bigg)$$

$$f(g(x))=\log\bigg(\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}-3x-x^{3}}\bigg)$$

$$f(g(x))=\log\bigg(\dfrac{(1+x)^3}{(1-x)^{3}}\bigg)$$

$$f(g(x))=\log\bigg(\dfrac{1+x}{1-x}\bigg)^{3}$$

$$f(g(x))=3\log\bigg(\dfrac{1+x}{1-x}\bigg)$$

$$f(g(x))=3f(x)$$
Question 4
1 / -0

Let $$f:R\rightarrow R$$ is a function satisfying $$f(2-x)=f(2+x)$$ and $$f(20-x)=f(x)\forall x\in R$$
If $$f(0)=5$$ then the minimum possible no. of values of $$x$$ satisfying $$f(x)=5$$ for $$x=[0.,70]$$, is

A
$$21$$

B
$$12$$

C
$$11$$

D
$$22$$

Solution

Given, $$f(2 - x) = f(2 + x) .. (i)$$
$$\therefore f(x)$$ is symmetric about $$x = 2$$
$$f(20 - x) = f(x) ..(ii)$$
$$x \rightarrow x + 10$$
$$f(10 - x) = f(10 + x)$$
$$f(x)$$ is symmetric about $$x = 10$$
$$x \rightarrow x + 2$$
$$f(18 - x) = f(x + 2) .. (ii)$$
$$f(2 - x) = f(18 - x)$$
$$x \rightarrow -x$$
$$f(2 + x) = f(18 + x)$$
$$\therefore x + 2 \rightarrow x$$
$$\therefore f(x) = f(x + 16)$$
$$f(x)$$ has period = $$16$$
$$f(x) = 5 , \, x \in [0, 170]$$
put $$x = 2$$ in (i) $$f(0) = f(4)$$
put $$x = 4$$ in (ii) $$f(4) = f(16)$$
when $$x \in [0, 16] \rightarrow f(x) $$ has two solution
when $$x \in [0, 160]$$ has $$20$$ solution.
When $$x in [0, 170]$$ has $$'21'$$ solution
Question 5
1 / -0

Let $$f\left( x \right) = {x^2}$$ and $$g\left( x \right) = {2^x}$$. Then the solution of the equation $$fog\left( x \right) = gof\left( x \right)$$ is

A
$$R$$

B
$$\left\{ {0} \right\}$$

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

D
none

Solution

$$f\left(x\right)={x}^{2}$$

$$f{g\left(x\right)}=f{\left({2}^{x}\right)}={\left({2}^{x}\right)}^{2}={2}^{2x}$$

$$g\left(x\right)={2}^{x}$$

$$g{f\left(x\right)}=g{\left({x}^{2}\right)}={2}^{{x}^{2}}$$

Given:$$f{g\left(x\right)}=g{f\left(x\right)}$$

$$\Rightarrow\,{2}^{2x}={2}^{{x}^{2}}$$

Since bases are same we can equate the powers
$$\Rightarrow\,2x={x}^{2}$$
$$\Rightarrow\,{x}^{2}-2x=0$$
$$\Rightarrow\,x\left(x-2\right)=0$$
$$\Rightarrow\,x=0,2$$

When $$x=0,\,f{g\left(x\right)}=g{f\left(x\right)}\Rightarrow\,{2}^{0}={2}^{0}=1$$
When $$x=2,\,f{g\left(x\right)}=g{f\left(x\right)}\Rightarrow\,{2}^{2\times 2}={2}^{{2}^{2}}\Rightarrow\,16=16$$
Hence $$x=\left\{0,2\right\}$$
Question 6
1 / -0

All values of a for which f : R $$ \to R$$ defined by f(x)= $${x^3} + a{x^2} + 3x + 100$$ is a one one functions, are

A
$$( - \infty , - 2)$$

B
$$( - \infty ,4)$$

C
$$(4, - 4)$$

D
$$( - 3,3)$$

Solution

$$f ^ { \prime } ( x ) = 3 x ^ { 2 } + 2 a x + 3$$
$$f ^ { \prime } ( x ) = 0$$ and f ( x ) > 0 $$f ^ { \prime } ( x ) < 0$$
$$ 3 x ^ { 2 } + 2 a x + 3 = 0$$
$$ x = \dfrac { - 2 a \pm \sqrt { 4 a ^ { 2 } - 36 } } { 6 }$$
$$ 4 a ^ { 2 } < 36$$
$$ a \in ( - 3,3 ) $$
Question 7
1 / -0

If $$f ( x ) = \left( a - x ^ { n } \right) ^ { 1 / n }$$ where $$a > 0$$ and $$n$$ is a positive integer then $$( f o f ) ( x )$$ is

A
$$f ( x )$$

B
$$x$$

C
$$0$$

D
$$1$$

Solution

$$f(n)= (a-x^{n})^{1/n}$$
$$(fof)^{(n)}=(a-((a-x^{n})^{1/n})^{n})^{1/n}$$
$$=(a-a+x^{n})^{1/n}$$
$$fof(n)=x$$
Question 8
1 / -0

Let $$A = \{ 1,2,3 \}$$ . Which of the following functions on A is invertible?

A
$$f = \{ ( 1,1 ) , ( 2,1 ) , ( 3,1 ) \}$$

B
$$f = \{ ( 1,2 ) , ( 2,3 ) , ( 3,1 ) \}$$

C
$$f = \{ ( 1,2 ) , ( 2,3 ) , ( 3,2 ) \}$$

D
$$f = \{ ( 1,1 ) , ( 2,2 ) , ( 3,1 ) \}$$

Solution

$$A=\left\{ 1,2,3 \right\} $$
$$\therefore$$ $$A$$ is invertible when obviously
$$f=\left\{ \left( 1,1 \right) ,\left( 2,1 \right) ,\left( 3,1 \right) \right\} $$
Question 9
1 / -0

If $$f ( x ) = \sin ^ { - 1 } ( \sin x ) + \cos ^ { - 1 } ( \sin x ) \text { and } \phi ( x ) = f ( f ( f ( x ) ) )$$ then $$\phi ^ { \prime } ( x )$$

A
1

B
$$\sin x$$

C
0

D
none of these

Solution

$$f(x)=\sin ^{ -1 }{ \left( \sin { x } \right) } +\cos ^{ -1 }{ \left( \sin { x } \right) } =\cfrac { \pi }{ 2 } \left[ \because \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\cfrac { \pi }{ 2 } \right] $$
$$f(f(x))=f\left(\cfrac { \pi }{ 2 }\right )=\cfrac { \pi }{ 2 } $$
$$\phi (x)=f(f(f(x)))=f\left(\cfrac { \pi }{ 2 }\right )=\cfrac { \pi }{ 2 } $$
$$\phi (x)=\cfrac { \pi }{ 2 } $$
$$\phi '(x)=0$$
Question 10
1 / -0

$$f:R \rightarrow R$$ such that $$f(x)=\ell n(x+\sqrt {x^{2}+1})$$. Another function $$g(x)$$ is defined such that $$gof(x)=x\ \forall\ x \in\ R$$. Then $$g(2)$$ is -

A
$$\dfrac {e^{2}+e^{-2}}{2}$$

B
$$e^{2}$$

C
$$\dfrac {e^{2}-e^{-2}}{2}$$

D
$$e^{-2}$$

Solution

$$f(x)= ln (x+ \sqrt{x^{2}+1})$$

$$g(x)= g(f(x))$$ $$ g(2)=?$$

ln $$(x+ \sqrt{x^{2}+1})=y$$

$$\sqrt{x^2+1}= e^y-{x}$$

$$x^{2}+1= e^{2y}-2e^{y}x+x^{2}$$

$$e^{2y}-2e^{y}x-1=o \Rightarrow$$

$$x =\dfrac{e^{2y}-1}{2e^{y}}$$

$$=\dfrac{(e^{y}- e^{-y})}{2}$$

$$\therefore g(x) = \dfrac{e{^y}- e^{-y}}{2} \Rightarrow g (2)= \dfrac{e^{2}-e^{-2}}{2}$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 66

Result

Relations and Functions Test - 66

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SHARING IS CARING

Question 1

$$f(x)=\frac{x}{\sqrt{1-x^{2}}}$$

$$f(x)=\frac{x}{\sqrt{1+x^{2}}}$$

$$x^{2}$$

x

Solution

Question 2

$$f ( x )$$

x

0

1

Solution

Question 3

$$ - f\left( x \right)$$

$$3f\left( x \right)$$

$${\left( {f\left( x \right)} \right)^3}$$

$$f\left( {3x} \right)$$

Solution

Question 4

$$21$$

$$12$$

$$11$$

$$22$$

Solution

Question 5

$$R$$

$$\left\{ {0} \right\}$$

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

none

Solution

Question 6

$$( - \infty , - 2)$$

$$( - \infty ,4)$$

$$(4, - 4)$$

$$( - 3,3)$$

Solution

Question 7

$$f ( x )$$

$$x$$

$$0$$

$$1$$

Solution

Question 8

$$f = \{ ( 1,1 ) , ( 2,1 ) , ( 3,1 ) \}$$

$$f = \{ ( 1,2 ) , ( 2,3 ) , ( 3,1 ) \}$$

$$f = \{ ( 1,2 ) , ( 2,3 ) , ( 3,2 ) \}$$

$$f = \{ ( 1,1 ) , ( 2,2 ) , ( 3,1 ) \}$$

Solution

Question 9

1

$$\sin x$$

0

none of these

Solution

Question 10

$$\dfrac {e^{2}+e^{-2}}{2}$$

$$e^{2}$$

$$\dfrac {e^{2}-e^{-2}}{2}$$

$$e^{-2}$$

Solution

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