Relations and Functions Test

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Relations and Functions Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

Let $$\displaystyle f:(-\infty,1]\rightarrow (-\infty,1]$$ such that $$\displaystyle f(x)=x(2-x).$$ Then $$f^{-1}(x)$$ is

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

B
$$1-\sqrt{1-x}$$

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

D
none of these

Solution

We have, $$f(x)=x(2-x)$$

$$y=x(2-x)$$

$$-y=x(x-2)$$

$$-y=x^{2}-2x$$

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

$$1-y=(x-1)^{2}$$

$$\pm\sqrt{1-y}=x-1$$

Now $$f:(-\infty,1] \rightarrow (-\infty,1]$$

$$x-1=-\sqrt{1-y}$$

$$x=1-\sqrt{1-y}$$

Hence $$f^{-1}(x)=1-\sqrt{1-x}$$
Question 2
1 / -0

If $$\displaystyle f(x)=\frac{1}{1-x},x\neq 0,1$$ then the graph of the function $$\displaystyle y=f\left \{ f(f(x)) \right \},x> 1,$$ is

A
a circle

B
an ellipse

C
a straight line

D
a pair of straight lines

Solution

$$f(f(x))=\dfrac{1}{1-\dfrac{1}{1-x}}$$

$$=-\dfrac{1-x}{x} =g(x)$$

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

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

$$=x$$

$$y=x$$ is an equation of straight line.

Hence, 'C' is correct.
Question 3
1 / -0

Let $$\displaystyle f:\left \{ x,y,z \right \}\rightarrow \left \{ a,b,c \right \}$$ be a one-one function and only one of the conditions $$(i)f(x)\neq b, (ii)f(y)=b,(iii)f(z)\neq a$$ is true then the function $$f$$ is given by the set

A
$$\displaystyle \left \{ (x,a),(y,b),(z,c)\right \}$$

B
$$\displaystyle \left \{ (x,a),(y,c),(z,b)\right \}$$

C
$$\displaystyle \left \{ (x,b),(y,a),(z,c)\right \}$$

D
$$\displaystyle \left \{ (x,c),(y,b),(z,a)\right \}$$

Solution

$$f:\left\{ x,y,z \right\} \rightarrow \left\{ a,b,c \right\} $$ is a one-one function

$$ \Rightarrow$$ each element in $$ \left\{ x,y,z \right\} $$ will have exactly one image in $$\left\{ a,b,c \right\}$$

and no two elements of $$ \left\{ x,y,z \right\}$$ will have same image in $$\left\{ a,b,c \right\} $$

Coming to the given 3 conditions, only one is true.

1) if $$f\left( x \right) \neq b$$ is true then $$f\left( y \right) =b$$ is false which makes $$f\left( z \right) \neq a$$ true $$\Longrightarrow f\left( x \right) \neq b$$ is false.

2) if $$f\left( y \right) =b$$ is true then $$f\left( x \right) \neq b$$ will also be true $$\Longrightarrow f\left( y \right) =b$$ is false

$$ \therefore f\left( z \right) \neq a$$ is the true condition and remainig two are false conditions.

$$\therefore f\left( x \right) =b, f\left( y \right) =a, f\left( z \right) =c$$

hence $$f=\left\{ \left( x,b \right) ,\left( y,a \right) ,\left( z,c \right) \right\} $$
Question 4
1 / -0

If $$\displaystyle f(x)=3x-5$$ then $$f^{-1}(x)=$$

A
$$\displaystyle \frac{1}{3x-5}$$

B
$$\displaystyle \frac{x+5}{3}$$

C
does not exist because $$f$$ is not one-one

D
does not exist because $$f$$ is not onto

Solution

$$f(x)=3x-5$$

Since $$f(x)$$ is a purely linear function (straight line), it is therefore a one-one function.

Hence $$f(x)$$ is revertible for all real $$x.$$

$$y=3x-5$$

$$y+5=3x$$

$$x=\dfrac{y+5}{3}$$

Replacing $$x$$ by $$y$$, gives us

$$f^{-1}(x)=\dfrac{x+5}{3}$$
Question 5
1 / -0

Which of the following function is one-one?

A
$$f:R\rightarrow R$$ given by$$ f(x)=|x-1|$$ for all $$x\in R$$

B
$$g:\left[ -\dfrac{\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \rightarrow R$$ given by $$g(x)=|sinx|$$ for all $$ x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] $$

C
$$h:\left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \in R$$ given by $$ h(x)=sinx$$ for all $$ x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] $$

D
$$\phi :R\rightarrow R$$ given by $$f(x)={ x }^{ 2 }-4$$ for all $$ x\in R$$

Solution

Option A - Consider $$x = 2 $$ and $$ x =0 $$ , the value of $$f(x)$$ is same. Hence it is not one-one
Option B - If we replace $$x$$ by $$-x$$, then the value of $$g(x)$$ remains the same. Hence it is not one-one
Option C - $$h(x)$$ is an increasing function for the given values of $$x$$. Hence it is one-one function
Option D -$$f(x)$$ is an even function. So it is not one-one for the given values of $$x$$
Question 6
1 / -0

If $$\displaystyle f(x)=x^{n},n\in N$$ and $$(gof)(x)=ng(x)$$ then $$g(x)$$ can be

A
$$n\:|x|$$

B
$$3.\sqrt[3]{x}$$

C
$$e^{x}$$

D
$$\log\:|x|$$

Solution

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

$$g(f(x))=ng(x)$$ ...$$(i)$$

$$\log(f(x))=n\log(x)$$ ...$$(ii)$$

Taking $$\log(|x|)$$ as $$g(x)$$ the above expression is reduced to eq $$i$$.

Hence

$$g(x)=\log(|x|)$$.
Question 7
1 / -0

If $$f:R^+\rightarrow A$$, Where $$A=\{x:-5<x<\infty\}$$ be defined by $$f(x)=x^2-5$$. Then $$f^{-1}(7)=$$

A
only $$-2{\sqrt3}$$

B
only $$2{\sqrt3}$$

C
both options A and B

D
none

Solution

The value $$f^{-1}(7)$$ will be the roots of the equation

$$x^2-5= 7\Rightarrow x^2 = 12\Rightarrow x = \pm 2\sqrt{3}$$

But only positive. $$\therefore x = 2\sqrt{3}$$
Question 8
1 / -0

If the function $$f:[1,\infty)\rightarrow [1,\infty)$$ is defined by $$\displaystyle f(x)=3^{x(x-1)}$$ then $$f^{-1}(x) $$ is

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

B
$$\displaystyle \frac{1}{2}\left ( 1+\sqrt{1+4\log_{3}x}\right)$$

C
$$\displaystyle \frac{1}{2}\left ( 1-\sqrt{1+4\log_{3}x}\right)$$

D
not defined

Solution

For the given domain, the function is one one.
Thus,
$$f(x)=2^{x(x-1)}$$

$$y=2^{x^{2}-x}$$

$$lny=(x^{2}-x)ln2$$

$$log_{2}y=x^{2}-x$$

$$log_{2}y=(x-\dfrac{1}{2})^{2}-\dfrac{1}{4}$$

$$log_{2}y+\dfrac{1}{4}=(x-\dfrac{1}{2})^{2}$$

$$\dfrac{\pm 1}{2}(\sqrt{1+4log_{2}y})=x-\dfrac{1}{2}$$

Since $$f:[1,\infty)\rightarrow[1,\infty)$$
Thus domain is $$[1,\infty)$$
Hence domain is positive so negative sign will be ignored. $$So$$

$$x-\dfrac{1}{2}=\dfrac{1}{2}(\sqrt{1+4log_{2}y})$$

$$x=\dfrac{1}{2}[1+\sqrt{1+4log_{2}y}]$$
Replacing x and y, gives us

$$f^{-1}(x)=\dfrac{1}{2}[1+\sqrt{1+4log_{2}x}]$$
Question 9
1 / -0

The composite mapping $$fog$$ of the map $$f: R\rightarrow R,f(x)=\sin x$$ and $$g: R\rightarrow R, g(x)=x^2$$ is

A
$$x^2 \sin x$$

B
$$(\sin x)^2$$

C
$$\sin x^2$$

D
$$\dfrac{ \sin x}{x^2}$$

Solution

Composite mapping will be $$f(g(x))$$

$$=f(x^2)$$

$$=\sin(x^2)$$

Hence option $$'C'$$ is the answer.
Question 10
1 / -0

If the function $$f:R\rightarrow R$$ be such that $$\displaystyle f(x)=x-[x],$$ where $$[y]$$ denotes the greatest integer less than or equal to $$y$$, then $$f^{-1}(x)$$ is

A
$$\displaystyle \frac{1}{x-[x]}$$

B
$$\displaystyle [x]-x$$

C
not defined

D
none of these

Solution

$$f\left( x \right) = x - \left[ x \right] = \left\{ x \right\}$$ is not one-one function from R to R. Hence $${f^{ - 1}}\left( x \right)$$ can not be defined.

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Relations and Functions Test - 43

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Relations and Functions Test - 43

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SHARING IS CARING

Question 1

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

$$1-\sqrt{1-x}$$

$$\sqrt{1-x}$$

none of these

Solution

Question 2

a circle

an ellipse

a straight line

a pair of straight lines

Solution

Question 3

$$\displaystyle \left \{ (x,a),(y,b),(z,c)\right \}$$

$$\displaystyle \left \{ (x,a),(y,c),(z,b)\right \}$$

$$\displaystyle \left \{ (x,b),(y,a),(z,c)\right \}$$

$$\displaystyle \left \{ (x,c),(y,b),(z,a)\right \}$$

Solution

Question 4

$$\displaystyle \frac{1}{3x-5}$$

$$\displaystyle \frac{x+5}{3}$$

does not exist because $$f$$ is not one-one

does not exist because $$f$$ is not onto

Solution

Question 5

$$f:R\rightarrow R$$ given by$$ f(x)=|x-1|$$ for all $$x\in R$$

$$g:\left[ -\dfrac{\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \rightarrow R$$ given by $$g(x)=|sinx|$$ for all $$ x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] $$

$$h:\left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \in R$$ given by $$ h(x)=sinx$$ for all $$ x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] $$

$$\phi :R\rightarrow R$$ given by $$f(x)={ x }^{ 2 }-4$$ for all $$ x\in R$$

Solution

Question 6

$$n\:|x|$$

$$3.\sqrt[3]{x}$$

$$e^{x}$$

$$\log\:|x|$$

Solution

Question 7

only $$-2{\sqrt3}$$

only $$2{\sqrt3}$$

both options A and B

none

Solution

Question 8

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

$$\displaystyle \frac{1}{2}\left ( 1+\sqrt{1+4\log_{3}x}\right)$$

$$\displaystyle \frac{1}{2}\left ( 1-\sqrt{1+4\log_{3}x}\right)$$

not defined

Solution

Question 9

$$x^2 \sin x$$

$$(\sin x)^2$$

$$\sin x^2$$

$$\dfrac{ \sin x}{x^2}$$

Solution

Question 10

$$\displaystyle \frac{1}{x-[x]}$$

$$\displaystyle [x]-x$$

not defined

none of these

Solution

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