Functions Test 34

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Functions Test 34

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Weekly Quiz Competition

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

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

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

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

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

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

$$\therefore g(x) = \dfrac{e{^y}- e^{-y}}{2} \Rightarrow g (2)= \dfrac{e^{2}-e^{-2}}{2}$$
Question 3
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 4
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 5
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 6
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 7
1 / -0

if $$f\left( x \right) = 3x + 2$$ , $$g\left( x \right) = {x^2} + 1$$,then the values of $$\left( {f_og} \right)\left( {{x^2} - 1} \right)$$

A
$$3{x^4} - 6{x^2} + 8$$

B
$$3{x^4} + 3x + 4$$

C
$$6{x^4} + 3{x^2} - 2$$

D
$$6{x^4} + 3{x^2} + 2$$

Solution

$$f\left(x\right)=3x+2$$

$$g\left(x\right)={x}^{2}+1$$
$$f.g\left(x\right)=f\left({x}^{2}+1\right)$$
$$f.g\left(x\right)=3\left({x}^{2}+1\right)+2=3{x}^{2}+5$$

$$f.g\left(x\right)=3{x}^{2}+5$$
$$f.g\left({x}^{2}-1\right)=3{\left({x}^{2}-1\right)}^{2}+5$$

$$=3\left({x}^{4}-2{x}^{2}+1\right)+5$$

$$=3{x}^{4}-6{x}^{2}+3+5$$

$$=3{x}^{4}-6{x}^{2}+8$$
Question 8
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Let A = {1,2,3,4,5} and B={1,2,3,4,5}. If $$f:A\rightarrow B$$ is an one-one function and $$f(x)=x$$ holds only for one value of $$x\epsilon \{ 1,2,3,4,5\} ,$$ then the number of such possible function is

A
$$120$$

B
$$36$$

C
$$45$$

D
$$44$$

Solution
Question 9
1 / -0

The domain function $$\sqrt{log_{10}(\frac{5x-x^{2}}{4})}$$ is

A
[1,4]

B
[-1,4]

C
[0,5]

D
[-1,5]

Solution
Question 10
1 / -0

$$f(x) = \mid |x|^2 - 2 |x| -3 \mid$$ is non differentiable at $$k$$ points in its domain , the value of $$k$$ is

A
$$3$$

B
$$2$$

C
$$4$$

D
$$1$$

Solution