Relations and Functions Test

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Relations and Functions Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

The inverse of the function $$\log_{e}x$$ is

A
$$10^{x}$$.

B
$$e ^{x}$$

C
$$10^{e}$$.

D
$$x^{e}$$.

Solution

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

$$ \Rightarrow x=f\left( y \right) =\log _{ e }{ y } $$

$$\Rightarrow y={ e }^{ x }$$
Question 2
1 / -0

If $$\displaystyle A= \left \{ a,b,c,d \right \}, B= \left \{ 1,2,3 \right \}$$ find whether or not the following sets of ordered pairs are relations from $$A$$ to $$B$$ or not.
$$\displaystyle R_{1}= \left \{ \left ( a,1 \right ), \left ( a,3 \right ) \right \}$$
$$\displaystyle R_{2}= \left \{ \left ( a,1 \right ), \left ( c,2 \right ), \left ( d,1 \right ) \right \}$$
$$\displaystyle R_{3}= \left \{ \left ( a,1 \right ), \left ( b,2 \right ), \left ( 3,c \right ) \right \}.$$

A
$$R_{1}$$ $$R_{2}$$ are relations but $$R_{3}$$ is not a relation.

B
$$R_{1}$$ $$R_{3}$$ are relations but $$R_{2}$$ is not a relation.

C
All are relations

D
none of these

Solution

Given, $$ A= \left \{ a,b,c,d \right \}, B= \left \{ 1,2,3 \right \}$$ and
$$ R_{1}= \left \{ \left ( a,1 \right ), \left ( a,3 \right ) \right \} $$
$$ R_{2}= \left \{ \left ( a,1 \right ), \left ( c,2 \right ), \left ( d,1 \right ) \right \} $$
$$ R_{3}= \left \{ \left ( a,1 \right ), \left ( b,2 \right ), \left ( 3,c \right ) \right \}.$$
We know that every relation from $$A$$ to $$B$$ will be a subset of $$A \times B$$.
Thus, both $$ R_{1} $$ and $$R_{2} $$ are subsets of $$ A\times B$$ and hence are relations but $$ R_{3}$$ is not a relation because $$R_{3}$$ is not a subset of $$A\times B$$ because the element $$ (3,c) \notin (A\times B).$$ It belongs to $$B\times A.$$
Question 3
1 / -0

If $$\displaystyle X= \left \{ 1,2,3,4,5 \right \}, Y= \left \{ 1,3,5,7,9 \right \}$$ determine which of the following sets are mappings, relations or neither from A to B:
(i)$$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \in X, y \in Y \right \}$$

A
It is clearly a one-one onto mapping i.e. a bijection. It is also a relation.

B
It is clearly a many-one onto mapping. It is also a relation.

C
It is clearly a one-one but not onto mapping. It is also a relation.

D
It is not a mapping but a relation

Solution

Given def of F as
$$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \epsilon X, y \epsilon Y \right \}$$
where $$\displaystyle X= \left \{ 1,2,3,4,5 \right \}, Y= \left \{ 1,3,5,7,9 \right \}$$

So, we get $$\left \{ \left ( 1,3 \right ), \left ( 2,4 \right ), \left ( 3,5 \right ), \left ( 4,6 \right ), \left ( 5,7 \right ) \right \} $$
Since, $$\displaystyle y \in Y$$ and in the above 4,6 do not belong to Y.
$$\displaystyle \therefore $$ we exclude the ordered pair (2,4) and (4,6)
$$\displaystyle \therefore \displaystyle F=\left \{ \left ( 1,3 \right ), \left ( 3,5 \right ), \left ( 5,7 \right ) \right \}$$
F is not a mapping because the elements 2,4 $$\displaystyle \epsilon X$$ do not have any image. $$\displaystyle F\subset X\times Y$$ and hence F is a relation from X to Y.
Question 4
1 / -0

Let $$f(x)=x^{2}-2x$$ and $$g(x)=f(f(x)-1)+f(5-f(x)),$$ then

A
$$g(x)<0,\forall x\in R$$

B
$$g(x)<0$$ for some $$x\in R$$

C
$$g(x)\leq 0$$ for some $$x\in R$$

D
$$g(x)\geq 0,\forall x\in R$$

Solution

Given, $$ f(x) = x^2 - 2x $$
Now, $$ f(f(x) - 1) \\ = f(x^2 - 2x - 1 ) \\ = (x^2 - 2x - 1 )^2 - 2 (x^2 - 2x - 1 ) $$
And, $$ f(5 - f(x) ) \\ = f(5 - x^2 + 2x ) \\ = (5 - x^2 + 2x )^2 - 2 (5 - x^2 + 2x ) $$
Hence, $$ g(x) = f(f(x) - 1) + f (5 - f(x)) \\ \therefore g(x) = (x^2 - 2x - 1)^2 - 2 (x^2 - 2x -1 ) + (5 - x^2 + 2x )^2 - 2(5 - x^2 + 2x ) \\ \Rightarrow g(x) = (x^2 - 2x - 1)^2 + (5- x^2 + 2x)^2 - 2(x^2 - 2x - 1 + 5 - x^2 + 2x) \\ \therefore g(x) = (x^2 + 2x - 1 )^2 + ( 5 - 2x ^2 + 2 x^2)^2 - 8$$
Since the first two terms are in square,
$$\therefore $$ it can not be negative and if x = 9 then we also get positive value.
$$\therefore g(x) \geq 0 \, \, \forall x \in R $$
$$ \therefore $$ option D is correct
Question 5
1 / -0

Let $$R$$ be a relation from a set $$A$$ to a set $$B$$,then

A
$$\displaystyle R=A\cup B$$

B
$$\displaystyle R=A\cap B$$

C
$$\displaystyle R\subseteq A\times B$$

D
$$\displaystyle R\subseteq B\times A$$

Solution

If $$R$$ is a relation from $$A$$ to $$B$$ then $$R$$ is the subset of $$A \times B$$ because it contains all the possible mappings from $$A$$ to $$B$$.
Question 6
1 / -0

If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ are functions defined by $$f(x)=3x-1; g(x)=\sqrt{x+6}$$, then the value of $$(g\circ f^{-1})(2009)$$ is

A
$$26$$

B
$$29$$

C
$$16$$

D
$$15$$

Solution

Given $$f(x)=3x-1, g(x)=\sqrt{x+6}$$
Let $$f(x)=y$$
$$ \Rightarrow y=3x-1\Rightarrow y+1-3x\Rightarrow x=\cfrac{y+1}{3}$$
Now, $$ f^{-1}(y)=x=\cfrac{y+1}{3}$$
$$\Rightarrow f^{-1}(x)=\cfrac{x+1}{3}$$ and $$g(x)=\sqrt {x+6}$$
Consider, $$(g\circ f^{-1})(2009)=g[{f^{-1}(2009)}]$$
$$ =g\left(\cfrac{2009+1}{3}\right)=g\left(\cfrac{2010}{3}\right)$$
$$=\sqrt{\cfrac{2010}{3}+6}=\sqrt{\cfrac{2028}{3}}=\sqrt{676}=26$$
Hence, option $$A$$ is correct.
Question 7
1 / -0

Find inverse $$f(x)=\log_{e}(x+\sqrt{x^{2}+1})$$

A
$$\sinh(x) $$

B
$$\cosh(x)$$

C
$$\tanh(x) $$

D
$$\coth(x)$$

Solution

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

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

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

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

$$\Rightarrow e^{2y}=1+2xe^{y}$$

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

$$\Rightarrow x=\dfrac{e^{y}-e^{-y}}{2}$$
Replacing x with y
$$f^{-1}(x)=\dfrac{e^{x}-e^{-x}}{2}$$

$$=\sinh(x)$$
Question 8
1 / -0

Let $$f:[2,\infty)\rightarrow [1,\infty)$$defined by $$f(x)=2^{x^{4}-4x^{2}}$$ and $$\displaystyle g:\left[ \frac{\pi}{2},\pi \right] \rightarrow A$$ defined by $$ \displaystyle g(x)=\frac {\sin x+4}{\sin x-2}$$ be two invertible functions, then
$$f^{-1}(x)$$ is equal to

A
$$\sqrt{2+\sqrt{4-\log_{2}x}}$$

B
$$\sqrt{2+\sqrt{4+\log_{2}x}}$$

C
$$\sqrt{2-\sqrt{4+\log_{2}x}}$$

D
None of these

Solution

$$\because ff^{-1}(x)=x$$
$$2^{(f^{-1}(x))^{4}}-4^{(f^{-1}(x))^{2}}=x$$
$$\Rightarrow (f^{-1}(x))^{4}-4(f^{-1}(x))^{2}=\log _{2}x=0$$
$$\therefore (f^{-1}(x))^{2}=2+\sqrt{4+\log _{2}x}$$
$$\therefore $$ Range of $$f^{-1}(x)$$ is $$(2, \infty )$$.
$$\therefore f^{-1}(x)=\sqrt{2+\sqrt{4+\log _{2}x}}$$
Question 9
1 / -0

If $$f_{0}(x)\, =\, \dfrac{x}{(x\, +\, 1)}$$ and $$f_{n\, +\, 1}\, =\, f_{0}\circ f_{n}(x)$$ for $$n = 0, 1, 2,\cdots$$ then $$f_{n}(x)$$ is

A
$$\displaystyle \frac{x}{(n\, +\, 1) x\, +\, 1}$$

B
$$f_{0}(x)$$

C
$$\displaystyle \frac{nx}{nx\, +\, 1}$$

D
$$\displaystyle \frac{x}{nx\, +\, 1}$$

Solution

Given $$f_{0}\left(x\right)=\dfrac{x}{ \left(x\, +\, 1\right)}\cdots\cdots\left(1\right)$$

Also given $$f_{n\, +\, 1}\, =\, f_{0}\, o\, f_{n}$$ for $$(n = 0, 1, 2,\cdots)\ldots\left(2\right)$$

Put $$n=0$$ in eqn (2)

$$f_1=f_{0} o f_0$$

$$f_1\left(x\right)=f_{0}[f_{0}\left(x\right)]$$

$$\Rightarrow f_{1}\left(x\right)=f_{0}\left(\dfrac{x}{x+1}\right)$$

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

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

$$\Rightarrow f_1\left(x\right)=\dfrac{x}{2x+1}$$

Put $$n=1$$ in eqn $$(2)$$

$$f_2=f_{0} o f_1$$

$$f_2\left(x\right)=f_{0}[f_{1}\left(x\right)]$$

$$\Rightarrow f_{2}\left(x\right)=f_{0}\left(\dfrac{x}{2x+1}\right)$$

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

$$=\dfrac{x}{3x+1}$$

$$\Rightarrow f_2\left(x\right)=\dfrac{x}{3x+1}$$

Hence, $$f_n\left(x\right)=\dfrac{x}{\left(n+1\right)x+1}$$
Question 10
1 / -0

If the function $$f : R \rightarrow$$ A given by $$f(x)\, =\, \displaystyle \frac{x^{2}}{x^{2}\, +\, 1}$$ is a surjection, then A is

A
$$R$$

B
$$[0, 1]$$

C
$$(0, 1]$$

D
$$[0, 1)$$

Solution

As 'f' is surjective,
Range of f = co-domain of 'f'
$$\implies$$ A = range of 'f'
Since, $$f(x)=\dfrac{x^2}{x^2+1}$$
$$\implies$$ $$y=\dfrac{x^2}{x^2+1}$$
$$\implies$$ $$y(x^2+1)=x^2$$
$$\implies$$ $$(y-1)x^2+y=0$$
$$\implies$$ $$x^2=\dfrac{-y}{y-1}$$
$$\implies$$ $$x=\sqrt{\dfrac{y}{1-y}}$$
$$\implies$$ $$\dfrac{y}{1-y} \geq 0$$
$$\implies$$ $$y \epsilon [0,1)$$
$$\implies$$ $$A=[0,1)$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 47

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Relations and Functions Test - 47

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SHARING IS CARING

Question 1

$$10^{x}$$.

$$e ^{x}$$

$$10^{e}$$.

$$x^{e}$$.

Solution

Question 2

$$R_{1}$$ $$R_{2}$$ are relations but $$R_{3}$$ is not a relation.

$$R_{1}$$ $$R_{3}$$ are relations but $$R_{2}$$ is not a relation.

All are relations

none of these

Solution

Question 3

It is clearly a one-one onto mapping i.e. a bijection. It is also a relation.

It is clearly a many-one onto mapping. It is also a relation.

It is clearly a one-one but not onto mapping. It is also a relation.

It is not a mapping but a relation

Solution

Question 4

$$g(x)<0,\forall x\in R$$

$$g(x)<0$$ for some $$x\in R$$

$$g(x)\leq 0$$ for some $$x\in R$$

$$g(x)\geq 0,\forall x\in R$$

Solution

Question 5

$$\displaystyle R=A\cup B$$

$$\displaystyle R=A\cap B$$

$$\displaystyle R\subseteq A\times B$$

$$\displaystyle R\subseteq B\times A$$

Solution

Question 6

$$26$$

$$29$$

$$16$$

$$15$$

Solution

Question 7

$$\sinh(x) $$

$$\cosh(x)$$

$$\tanh(x) $$

$$\coth(x)$$

Solution

Question 8

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

$$\sqrt{2+\sqrt{4+\log_{2}x}}$$

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

None of these

Solution

Question 9

$$\displaystyle \frac{x}{(n\, +\, 1) x\, +\, 1}$$

$$f_{0}(x)$$

$$\displaystyle \frac{nx}{nx\, +\, 1}$$

$$\displaystyle \frac{x}{nx\, +\, 1}$$

Solution

Question 10

$$R$$

$$[0, 1]$$

$$(0, 1]$$

$$[0, 1)$$

Solution

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