Relations and Functions Test

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Relations and Functions Test - 73

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Question 1
1 / -0

$$K(x)$$ is a function such that $$K(f(x))=a+b+c+d$$,
Where,
$$a=\begin{cases}
0 & \text{ if f(x) is even} \\
-1 & \text{ if f(x) is odd} \\
2 & \text{ if f(x) is neither even nor odd}
\end{cases}$$
$$b=\begin{cases}
3 & \text{ if f(x) is periodic} \\
4 & \text{ if f(x) is aperiodic}
\end{cases}$$
$$c=\begin{cases}
5 & \text{ if f(x) is one one} \\
6 & \text{ if f(x) is many one}
\end{cases}$$
$$d=\begin{cases}
7 & \text{ if f(x) is onto} \\
8 & \text{ if f(x) is into}
\end{cases}$$
$$h:R\rightarrow R,h(x)=\left ( \displaystyle \frac{e^{2x}+e^{x}+1}{e^{2x}-e^{x}+1} \right )$$

On the basis of above information, answer the following questions.$$K(\phi(x)) $$

A
$$15$$

B
$$16$$

C
$$17$$

D
$$18$$

Solution

$$\phi(x)\rightarrow \left(\dfrac{-\pi}{2},\dfrac{\pi}{2}\right)$$
Now we know that for the given domain, $$\tan(x)$$ is one-one, periodic, odd and an into function.
Hence $$f(\phi(x))$$
$$=-1+3+5+8$$
$$=16-1$$
$$=15$$
Question 2
1 / -0

Let $$f:{x, y, z}\rightarrow (a, b, c)$$ be a one-one function. It is known that only one of the following statements is true:
(i) $$f(x)\neq b$$
(ii)$$f(y)=b$$
(iii)$$f(z)\neq a$$

A
$$f=\{(x, a), (y, b), (z, c)\}$$

B
$$f=\{(x, b), (y, a), (z, c)\}$$

C
$$f=\{(x, b), (y, c), (z, c)\}$$

D
$$f=\{(x, b), (y, c), (z, a)\}$$

Solution

When (i) is true, then $$f(x) \neq b, f(y) \neq b , f(z) = a $$

$$\Rightarrow$$ Two ordered pair function is possible $$ f(x) = a, f(y) = c, f(z) = a$$ or $$f(x) = c, f(y) = a, f(z) = a$$

But given $$f$$ is one-one and only one such function is possible. Hence (i) can't be true.

When (ii) is true, then $$f(y) = b, f(z) =a , f(x) = b$$. This is also not possible.

Clearly if (iii) is true then it is satisfying every condition.
Hence ordered pair of $$f$$ is $$\{(x,a), (y,b), (z,c)\}$$
Question 3
1 / -0

If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ are given by $$f(x)=|x|$$ and $$g(x)=[x]$$ for each $$x\in R,$$ then $$\left\{ x\in R:g\left( f\left( x \right) \right) \le f\left( g\left( x \right) \right) \right\} =$$

A
$$Z\cup \left( -\infty ,0 \right) $$

B
$$\left( -\infty ,0 \right) $$

C
$$Z$$

D
$$R$$

Solution

$$g\left( f\left( x \right) \right) \le f\left( g\left( x \right) \right) \Rightarrow g\left( \left| x \right| \right) \le f\left( \left[ x \right] \right) \Rightarrow \left[ \left| x \right| \right] \le \left[ \left| x \right| \right] $$
This is true for each $$x\in R$$
Question 4
1 / -0

Let $$\displaystyle f:\left[-\frac{\pi}{3},\frac{2\pi}{3}\right]\rightarrow [0,4]$$ be a function defined by $$\displaystyle f(x)=\sqrt 3 \sin x-\cos x+2$$ then $$\displaystyle f^{-1}(x)$$ equals

A
$$\displaystyle \frac{2\pi}{3}-\cos ^{-1}\left(\frac{x-2}{2}\right)$$

B
$$\displaystyle \sin ^{-1}\left ( \frac{x-2}{2} \right )+\frac{\pi}{3}$$

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

D
$$\displaystyle \sin ^{-1}\left ( \frac{x+2}{2} \right )+\frac{\pi}{6}$$

Solution

$$y=2\left(\dfrac{\sqrt{3}}{2}\sin\left(x\right)-\dfrac{1}{2}\cos\left(x\right)\right)+2$$

$$\Rightarrow \dfrac{y}{2}-1=\dfrac{\sqrt{3}}{2}\sin\left(x\right)-\dfrac{1}{2}\cos\left(x\right)$$

$$\Rightarrow \dfrac{y-2}{2}=-\cos\left(\dfrac{\pi}{3}+x\right)$$

$$\Rightarrow \dfrac{y-2}{2}=\cos\left(\pi-\left(\dfrac{\pi}{3}+x\right)\right)$$

$$\Rightarrow \dfrac{y-2}{2}=\cos\left(\dfrac{2\pi}{3}-x\right)$$

$$\Rightarrow \cos^{-1}\left(\dfrac{y-2}{2}\right)=\dfrac{2\pi}{3}-x$$

$$x=\dfrac{2\pi}{3}-\cos^{-1}\left(\dfrac{y-2}{2}\right)$$
Therefore,
$$f^{-1}\left(x\right)=\dfrac{2\pi}{3}-\cos^{-1}\left(\dfrac{x-2}{2}\right)$$
Question 5
1 / -0

Suppose $$\displaystyle f\left ( x \right )=\left ( x+1 \right )^{2}$$ for $$\displaystyle x\geq -1$$. If $$\displaystyle g(x)$$ is the function whose graph is the reflection of the graph of $$\displaystyle f(x)$$ with respect to the line $$y=x $$, then $$\displaystyle g(x)$$ is equal to

A
$$\displaystyle -\sqrt{x}-1$$ for $$\displaystyle x\geq 0$$

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

C
$$\displaystyle \sqrt{x+1}$$ for $$\displaystyle x > -1$$

D
$$\displaystyle \sqrt{x}-1$$ for $$\displaystyle x\geq 0$$

Solution

$$g(x)=f^{-1}(x)$$
$$y=(x+1)^{2}$$
$$\sqrt{y}=x+1$$
$$\sqrt{y}-1=x$$
$$g(x)=f^{-1}(x)=\sqrt{x}-1$$ for $$x\geq 0$$
Question 6
1 / -0

If $$\displaystyle f\left ( x \right )= \frac{3x+2}{5x-3}$$ , then

A
$$\displaystyle f\left ( f{}'\left ( x \right ) \right )=f^{-1}\left ( x \right )$$

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

C
$$\displaystyle f\left ( f{}'\left ( x \right ) \right )=f'\left ( x \right )$$

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

Solution

$$\displaystyle y=\frac{3x+2}{5x-3}$$

$$\displaystyle 5xy-3y=3x+2$$

$$\displaystyle x(5y-3)=3y+2$$

$$\displaystyle x=\frac{3y+2}{5y-3}$$

Hence $$f(x)$$ is its own inverse.

$$\displaystyle f^{-1}(x)=\frac{3x+2}{5x-3}=f(x)$$
Question 7
1 / -0

If $$\displaystyle f\left ( x \right )+f\left ( \dfrac1x \right )=0,\; f(e)=1; g\left ( x \right )=f^{-1}\left ( x \right )$$ then $$\displaystyle g{}'\left ( x \right )$$ equals

A
$$\displaystyle e^{x}$$

B
$$\displaystyle x $$

C
$$\displaystyle x^{2}$$

D
$$\displaystyle e^{-x}$$

Solution

$$f(x)+f\left(\dfrac{1}{x}\right)=0$$ is satisfied by $$f(x)=k \ln x$$, where $$k$$ is a constant.
Given that $$f(e)=1 \Rightarrow k=1$$
$$f(x)= \ln x$$
$$ \Rightarrow f^{-1}x=e^x$$
$$ \Rightarrow g(x)=e^x$$
$$ \Rightarrow g'(x)=e^x$$
Question 8
1 / -0

If $$f:[1,\infty )\rightarrow [2,\infty )$$ is given by $$\displaystyle f\left( x \right)=x+\frac { 1 }{ x } ,$$ then $$f^{-1}(x)$$ equals

A
$$\displaystyle \frac { x+\sqrt { { x }^{ 2 }-4 } }{ 2 } $$

B
$$\displaystyle \frac { x }{ 1+{ x }^{ 2 } } $$

C
$$\displaystyle \frac { x-\sqrt { { x }^{ 2 }-4 } }{ 2 } $$

D
$$1+\sqrt{x^2-4}$$

Solution

Let $$\displaystyle y=x+\frac { 1 }{ x } $$

$$\displaystyle \Rightarrow { x }^{ 2 }-xy+1=0\quad \Rightarrow x=\frac { y\pm \sqrt { { y }^{ 2 }-4 } }{ 2 } $$

$$\because \quad x\in \left( 2,\infty \right) $$

$$\displaystyle \therefore x=\frac { y+\sqrt { { y }^{ 2 }-4 } }{ 2 } $$

Hence, $$\displaystyle f^{ -1 }\left( x \right) =\frac { x+\sqrt { { x }^{ 2 }-4 } }{ 2 } $$
Question 9
1 / -0

Let $$\displaystyle f\left ( x \right )=\frac{ax^{2}+2x+1}{2x^{2}-2x+1}$$, the value of $$a$$ for which $$\displaystyle f:R\rightarrow \left [ -1,2 \right ]$$ is onto , is

A
$$\displaystyle \left [ 2,5 \right ]$$

B
$$\displaystyle \left [ -5,-2 \right ]$$

C
$$\displaystyle \left [ 0,5 \right ]$$

D
None of these.

Solution

Given $$\displaystyle f\left ( x \right )=\frac{ax^{2}+2x+1}{2x^{2}-2x+1}$$

Since, $$\displaystyle f:R\rightarrow \left [ -1,2 \right ]$$ is onto

$$R(f)=Co-domain [-1,2]$$

$$-1 \le \dfrac{ax^{2}+2x+1}{2x^{2}-2x+1}\le 2$$

$$\Rightarrow -(2x^{ 2 }-2x+1)\le ax^{ 2 }+2x+1\le 2(2x^{ 2 }-2x+1)$$

$$\Rightarrow -2x^{ 2 }+2x-1\le ax^{ 2 }+2x+1\le 4x^{ 2 }-4x+2$$ ....(1)

$$\Rightarrow -2x^{ 2 }+2x-1\le ax^{ 2 }+2x+1$$

$$\Rightarrow (a+2)x^2+2\ge 0$$

So for all $$x\in R$$ , $$a+2\ge 0$$

$$\Rightarrow a\ge -2$$ .....(2)

From the inequality (1), it follows that

$$ax^{ 2 }+2x+1\le 4x^{ 2 }-4x+2$$

$$\Rightarrow (a-4)x^2+6x-1 \le 0$$

$$\Rightarrow a-4 < 0$$ and $$D\le 0$$

$$\Rightarrow a<4 $$ and $$36+4a-16 \le 0$$

$$\Rightarrow a<4 $$ and $$a\le -5$$ ....(3)

From (2) and (3), we get

$$a\in (-\infty,-5]\cup [-2,4)$$
Question 10
1 / -0

The total number of injective mappings from a set with $$m$$ elements to a set with $$n$$ elements, $$m \leq n $$ is

A
$$\displaystyle m^{n}$$

B
$$\displaystyle n^{m}$$

C
$$\displaystyle \frac{n!}{(n-m)!}$$

D
$$n!$$

Solution

$$a_{1} \in $$A can have $$n$$ images in $$B$$, but the element $$a_{2}$$ will have only $$(n-1)$$ images as the mappings are to be one-one (injective).

Similarly the elements $$a_{3}$$ will have $$(n-2)$$ images.

Hence the total number of mappings will be,

$$n(n-1)(n-2)...(n-\overline{m-1})=n(n-1)(n-2)...(n-m+1)$$

Multiply above and below by $$(n-m)(n-m-1)...3.2.1$$

$$\therefore$$ Required numbers is $$\displaystyle \frac{n!}{(n-m)!} $$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 73

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Relations and Functions Test - 73

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SHARING IS CARING

Question 1

$$15$$

$$16$$

$$17$$

$$18$$

Solution

Question 2

$$f=\{(x, a), (y, b), (z, c)\}$$

$$f=\{(x, b), (y, a), (z, c)\}$$

$$f=\{(x, b), (y, c), (z, c)\}$$

$$f=\{(x, b), (y, c), (z, a)\}$$

Solution

Question 3

$$Z\cup \left( -\infty ,0 \right) $$

$$\left( -\infty ,0 \right) $$

$$Z$$

$$R$$

Solution

Question 4

$$\displaystyle \frac{2\pi}{3}-\cos ^{-1}\left(\frac{x-2}{2}\right)$$

$$\displaystyle \sin ^{-1}\left ( \frac{x-2}{2} \right )+\frac{\pi}{3}$$

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

$$\displaystyle \sin ^{-1}\left ( \frac{x+2}{2} \right )+\frac{\pi}{6}$$

Solution

Question 5

$$\displaystyle -\sqrt{x}-1$$ for $$\displaystyle x\geq 0$$

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

$$\displaystyle \sqrt{x+1}$$ for $$\displaystyle x > -1$$

$$\displaystyle \sqrt{x}-1$$ for $$\displaystyle x\geq 0$$

Solution

Question 6

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

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

$$\displaystyle f\left ( f{}'\left ( x \right ) \right )=f'\left ( x \right )$$

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

Solution

Question 7

$$\displaystyle e^{x}$$

$$\displaystyle x $$

$$\displaystyle x^{2}$$

$$\displaystyle e^{-x}$$

Solution

Question 8

$$\displaystyle \frac { x+\sqrt { { x }^{ 2 }-4 } }{ 2 } $$

$$\displaystyle \frac { x }{ 1+{ x }^{ 2 } } $$

$$\displaystyle \frac { x-\sqrt { { x }^{ 2 }-4 } }{ 2 } $$

$$1+\sqrt{x^2-4}$$

Solution

Question 9

$$\displaystyle \left [ 2,5 \right ]$$

$$\displaystyle \left [ -5,-2 \right ]$$

$$\displaystyle \left [ 0,5 \right ]$$

None of these.

Solution

Question 10

$$\displaystyle m^{n}$$

$$\displaystyle n^{m}$$

$$\displaystyle \frac{n!}{(n-m)!}$$

$$n!$$

Solution

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