Relations and Functions Test

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Relations and Functions Test - 72

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Weekly Quiz Competition

Question 1
1 / -0

If $$f (x) = x + 2, g (x) = 2 x +3,$$ then find gof

A
$$2x -7 $$

B
$$7x + 2$$

C
$$2x + 7$$

D
$$7 + 2x$$
Question 2
1 / -0

Let $$\displaystyle a > 1$$ be a real number and $$\displaystyle f\left ( x \right ) = \log _{a} x^{2}$$ for $$\displaystyle x > 0$$. If $$\displaystyle f^{-1}$$ is the inverse function of $$f$$ and $$b$$ and $$c$$ are real numbers then $$\displaystyle f^{-1} \left ( b+c \right )$$ is equal to

A
$$\displaystyle f^{-1}\left ( b \right ) . f^{-1}\left ( c \right )$$

B
$$\displaystyle f^{-1}\left ( b \right ) + f^{-1}\left ( c \right )$$

C
$$\displaystyle \frac {1}{f\left ( b+c \right )}$$

D
$$\displaystyle \frac {1}{f^{-1}\left ( b \right ) + f^{-1} \left ( c \right )}$$

Solution

$$y=\log_{a}(x^2)$$
$$a^{y}=x^{2}$$
$$a^{y/2}=x$$
Hence $$f^{-1}=a^{x/2}$$
$$f^{-1}(b+c)$$
$$=a^{(b+c)/2}$$
$$=a^{b/2}a^{c/2}$$
$$=f^{-1}(b)f^{-1}(c)$$
Hence, option 'A' is correct.
Question 3
1 / -0

Which one of the following functions is not one-one?

A
$$f:(-1,\infty )\rightarrow R$$ given by $$ f(x)={ x }^{ 2 }+2x\quad $$

B
$$g:(2,\infty )\rightarrow R$$ given by $$g(x)={ e }^{ { x }^{ 3 }-3x+2 }\quad $$

C
$$h:R\rightarrow R$$ given by $$h(x)={ 2 }^{ { x }(x-1) }\quad $$

D
$$\phi :(-\infty ,0)\rightarrow R$$ given by $$\phi (x)=\cfrac { { x }^{ 2 } }{ { x }^{ 2 }+1 } $$

Solution

$$ x(x-1)$$ is not a one -one function in whole $$R$$.
This function has a minima at $$x=\dfrac12$$, therefore repeats value after $$x=\dfrac12$$.
We can also see that $$x(x-1) = 0$$ for $$x=0,1$$ and therefore $$e^{x(x-1)} =1$$ for both $$x=0 \&\ 1$$, which rules it out as a one to one function.
Question 4
1 / -0

$$f(x)=x^3+3x^2+4x+b \sin x+c \cos x, \forall x\in R$$ is a one-one function, then the value of $$b^2+c^2$$ is

A
$$\geq 1$$

B
$$\geq 2$$

C
$$\leq 1$$

D
none of these

Solution

Here, $$f(x)=x^3+3x^2+4x+b \sin x+c \cos x$$
$$f'(x)=3x^2+6x+4+b \cos x-c \sin x$$
Now, for $$f(x)$$ to be one-one, the only possibility is
$$f'(x)\geq 0, \forall x\in R$$
ie, $$3x^2+6x+4+b \cos x-c \sin x\geq 0, \forall x\in R$$
ie, $$3x^2+6x+4 \geq c \sin x-b \cos x, \forall x\in R$$
As we are approximating lower bound for the function by a number instead of a function, hence we chose the number.
ie, $$3x^2+6x+4 \geq \sqrt {b^2+c^2}, \forall x\in R$$
ie, $$\sqrt {b^2+c^2} \leq 3(x^2+2x+1)+1,\forall x\in R$$
$$\sqrt {b^2+c^2} \leq 3(x+1)^2+1, \forall x\in R$$
$$\sqrt {b^2+c^2} \leq 1, \forall x \in R$$
$$\Rightarrow b^2+c^2 \leq 1, \forall x \in R$$
Hence, (c) is the correct answer.
Question 5
1 / -0

If $$f(x)=2x+|x|, g(x)=\dfrac {1}{3}(2x-|x|)$$ and $$h(x)=f(g(x))$$, then domain of $$\sin^{-1}\underset {\text {n times}}{\underbrace {(h(h(h(h.....h(x).....))))}}$$ is

A
$$[-1, 1]$$

B
$$\left [-1, -\dfrac {1}{2}\right ]\cup \left [\dfrac {1}{2}, 1\right ]$$

C
$$\left [-1, -\dfrac {1}{2}\right ]$$

D
$$\left [\dfrac {1}{2}, 1\right ]$$

Solution

Since, $$f(x)=\left\{\begin{matrix}2x+x, & x\geq 0\\ 2x-x, & x < 0\end{matrix}\right.=\left\{\begin{matrix}3x, & x\geq 0\\ x, & x < 0\end{matrix}\right.$$
and $$g(x)=\dfrac {1}{3}\left\{\begin{matrix}2x-x, & x\geq 0\\ 2x+x, & x < 0\end{matrix}\right.=\left\{\begin{matrix}\dfrac {x}{3}, & x\geq 0\\ x, & x < 0\end{matrix}\right.$$
$$\therefore f(g(x))=\left\{\begin{matrix}3\left (\dfrac {x}{3} \right ) & x\geq0\\ x & x < 0\end{matrix}\right.$$
$$\Rightarrow f(g(x))=x, \forall x\in R$$
$$\therefore h(x)=x$$
$$\Rightarrow \sin^{-1} (h(h(h....h(x)....)))=\sin^{-1}x$$
$$\therefore$$ Domain of $$\sin^{-1} (h(h(h(h.....h(x)....))))$$ is $$[-1, 1]$$
Hence, A is the correct answer.
Question 6
1 / -0

Which of the following functions is/are injective map(s) ?

A
$$f(x)=x^2+2, x \in (-\infty,\infty)$$

B
$$f(x)=|x+2|, x \in [-2,\infty)$$

C
$$f(x)=(x-4)(x-5), x \in (-\infty,\infty)$$

D
$$f(x)=\dfrac{4x^2 + 3x -5}{4+3x-5x^2}, x\in(-\infty, \infty)$$

Solution

The function $$f(x)=x^2 + 2, x \in (-\infty, \infty)$$ is not injective as
$$f(1)=f(-1) $$ but $$1 \neq -1.$$

The function $$f(x)=(x-4)(x-5), x \in (-\infty, \infty)$$ is not one-one as $$f(4)=f(5)$$ but $$4 \neq 5.$$

The functin $$f(x)=\dfrac{4x^2 + 3x -5}{4 + 3x - 5x^2}, x \in (-\infty,\infty)$$ is also not injective as $$f(1)=f(-1)$$ but $$1 \neq -1$$.

For the function , $$f(x)=-|x+2|, x\in [-2,\infty)$$

Let $$f(x)=f(y),x,y \in [-2,\infty) \Rightarrow |x+2| = |y+2| $$
$$\Rightarrow x+2 = y+2$$
$$\Rightarrow x=y$$
So , $$f$$ is an injective map.
Question 7
1 / -0

The function $$f$$ is one to one and the sum of all the intercepts of the graph is $$5$$. The sum of all the intercept of the graph $$\displaystyle y = f^{-1} \left ( x \right )$$ is

A
$$5$$

B
$$\dfrac15$$

C
$$\dfrac25$$

D
$$-5$$

Solution

Since the function $$f$$ is one-one there exist only one $$x$$-intercept and only one $$y$$-intercept for the function.
Let $$a$$ be the $$y$$-intercept of the function $$f$$. Hence $$f\left( 0 \right) = a$$, but then we have $${f^{ - 1}}\left( a \right) = 0$$.
Therefore $$a$$ is an $$x$$-intercept of the function $${f^{ - 1}}$$.
Similarly the $$x$$-intercept of the function$$f$$ is $$y$$-intercept of the function $${f^{ - 1}}$$, say $$b$$.
Hence the sum of the intercepts of the function $$f$$ is same as the sum of the intercepts of the function $${f^{ - 1}}$$ which is equal to 5.
Question 8
1 / -0

If $$\displaystyle f \left ( x \right ) = px + q$$ and $$\displaystyle f \left ( f\left ( f\left ( x \right ) \right ) \right ) = 8x + 21$$, where $$p$$ and $$q$$ are real numbers, the $$ p + q$$ equals

A
$$3$$

B
$$5$$

C
$$7$$

D
$$11$$

Solution

$$f(x)=px+q$$
$$f(f(x))=p(px+q)+q$$
$$=p^2x+q(p+1)$$
$$f(f(f(x)))=p(p^2x+q(p+1))+q$$
$$=p^{3}x+q(p(p+1)+1)$$
$$=8x+21$$
By comparing coefficients, we get
$$p^{3}=8$$
$$p=2$$
And $$q(2(3)+1)=21$$
$$7q=21$$
$$q=3$$
Hence
$$f(x)=2x+3$$
Therefore, $$p+q=2+3=5$$
Question 9
1 / -0

Let $$f\left( x \right) =\left\{ \begin{matrix} 1+|x|,\; x<-1 \\ \left[ x \right] ,\; x\ge -1 \end{matrix} \right. $$ where $$[\cdot]$$ denotes the greatest integer function. Then $$\displaystyle f\left \{f(-2.3) \right\}$$ is equal to

A
$$4$$

B
$$2$$

C
$$-3$$

D
$$3$$

Solution

$$f\left( x \right) =\left\{ \begin{matrix} 1+|x|,\; x<-1 \\ \left[ x \right] ,\; x\ge -1 \end{matrix} \right. $$
.

$$f\left( -2.3 \right) =1+\left| -2.3 \right| =3.3\dots ( \because x < -1$$)
.

hence, $$f\left( f\left( -2.3 \right) \right) =f\left( 3.3 \right) =\left[ 3.3 \right] =3$$

$$f\left( f\left( -2.3 \right) \right)=3$$
Question 10
1 / -0

The inverse function of the function $$\displaystyle f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$ is

A
$$\displaystyle \frac{1}{2}\log\frac{1+x}{1-x}$$

B
$$\displaystyle \frac{1}{2}\log\frac{2+x}{2-x}$$

C
$$\displaystyle \frac{1}{2}\log\frac{1-x}{1+x}$$

D
none of these

Solution

$$y=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$

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

$$y\left(e^{2x}+1\right)=e^{2x}-1$$

$$\Rightarrow e^{2x}\left(y\right)+y=e^{2x}-1$$

$$\Rightarrow y+1=e^{2x}\left(1-y\right)$$

$$e^{2x}=\left(\dfrac{1+y}{1-y}\right)$$

Taking logarithm on both the sides, we get

$$2x=\log\left(\dfrac{1+y}{1-y}\right)$$

$$x=\dfrac{1}{2}\log\left(\dfrac{1+y}{1-y}\right)$$

Hence $$f^{-1}=\dfrac{1}{2}\log\left(\dfrac{1+x}{1-x}\right)$$

Alternatively, $$f\left(x\right)$$ is $$\tan h\left(x\right)$$

So, it's inverse will be $$\dfrac{1}{2}\log\left(\dfrac{1+x}{1-x}\right)$$.

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Relations and Functions Test - 72

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Relations and Functions Test - 72

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SHARING IS CARING

Question 1

$$2x -7 $$

$$7x + 2$$

$$2x + 7$$

$$7 + 2x$$

Question 2

$$\displaystyle f^{-1}\left ( b \right ) . f^{-1}\left ( c \right )$$

$$\displaystyle f^{-1}\left ( b \right ) + f^{-1}\left ( c \right )$$

$$\displaystyle \frac {1}{f\left ( b+c \right )}$$

$$\displaystyle \frac {1}{f^{-1}\left ( b \right ) + f^{-1} \left ( c \right )}$$

Solution

Question 3

$$f:(-1,\infty )\rightarrow R$$ given by $$ f(x)={ x }^{ 2 }+2x\quad $$

$$g:(2,\infty )\rightarrow R$$ given by $$g(x)={ e }^{ { x }^{ 3 }-3x+2 }\quad $$

$$h:R\rightarrow R$$ given by $$h(x)={ 2 }^{ { x }(x-1) }\quad $$

$$\phi :(-\infty ,0)\rightarrow R$$ given by $$\phi (x)=\cfrac { { x }^{ 2 } }{ { x }^{ 2 }+1 } $$

Solution

Question 4

$$\geq 1$$

$$\geq 2$$

$$\leq 1$$

none of these

Solution

Question 5

$$[-1, 1]$$

$$\left [-1, -\dfrac {1}{2}\right ]\cup \left [\dfrac {1}{2}, 1\right ]$$

$$\left [-1, -\dfrac {1}{2}\right ]$$

$$\left [\dfrac {1}{2}, 1\right ]$$

Solution

Question 6

$$f(x)=x^2+2, x \in (-\infty,\infty)$$

$$f(x)=|x+2|, x \in [-2,\infty)$$

$$f(x)=(x-4)(x-5), x \in (-\infty,\infty)$$

$$f(x)=\dfrac{4x^2 + 3x -5}{4+3x-5x^2}, x\in(-\infty, \infty)$$

Solution

Question 7

$$5$$

$$\dfrac15$$

$$\dfrac25$$

$$-5$$

Solution

Question 8

$$3$$

$$5$$

$$7$$

$$11$$

Solution

Question 9

$$4$$

$$2$$

$$-3$$

$$3$$

Solution

Question 10

$$\displaystyle \frac{1}{2}\log\frac{1+x}{1-x}$$

$$\displaystyle \frac{1}{2}\log\frac{2+x}{2-x}$$

$$\displaystyle \frac{1}{2}\log\frac{1-x}{1+x}$$

none of these

Solution

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