Functions

$ \begin{aligned} & \text { We have, } y=\frac{10^x-10^{-x}}{10^x+10^{-x}}+1 \\ & \Rightarrow \quad y-1=\frac{10^x-10^{-x}}{10^x+10^{-x}} \\ & \text { Let } \quad a=10^x \\ & \therefore \quad 10^{-x}=\frac{1}{a} \\ & \therefore \quad y-1=\frac{a-\frac{1}{a}}{a+\frac{1}{a}}=\frac{a^2-1}{a^2+1} \\ & \Rightarrow \quad(y-1)\left(a^2+1\right)=a^2-1 \Rightarrow(y-1) a^2+(y-1)=a^2-1 \\ & \Rightarrow \quad a^2(y-1-1)=-y \Rightarrow a^2(y-2)=-y \\ & \Rightarrow \quad a^2=\frac{-y}{y-2} \Rightarrow 10^{2 x}=\frac{-y}{y-2} \\ & \Rightarrow \quad 2 x=\log _{10}\left(\frac{-y}{y-2}\right) \Rightarrow 2 x=\log _{10}\left(\frac{y}{2-y}\right) \\ & \Rightarrow \quad x=\frac{1}{2} \log _{10}\left(\frac{y}{2-y}\right) \end{aligned} $

Given, $f(x)=\tan \left(\frac{\pi}{\sqrt{x+1}+4}\right)$

For $\sqrt{x+1}$ to be defined,

$ \begin{array}{ll} \therefore & \sqrt{x+1}+4 \geq 0+4=4 \\ \Rightarrow & 0<\frac{1}{\sqrt{x+1}+4} \leq \frac{1}{4} \\ \Rightarrow & 0<\frac{\pi}{\sqrt{x+1}+4} \leq \frac{\pi}{4} \\ \therefore & f(x)=\tan \left(\frac{\pi}{\sqrt{x+1}}+4\right), \text { where } \\ & 0<\frac{\pi}{\sqrt{x+1}+4} \leq \frac{\pi}{4} \end{array} $

So, the function $f(x)$ is strictly increasing in the interval $\left(0, \frac{\pi}{4}\right]$

And the range of the given function is $\left(\tan (0), \tan \left(\frac{\pi}{4}\right)\right]$, which is $(0,1]$

$\therefore$ The range of $f(x)$ is $(0,1]$

Given,

$ \frac{x^3+3}{(x-3)^3}=a+\frac{b}{x-3}+\frac{c}{(x-3)^2}+\frac{d}{(x-3)^3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let $y=x-3$

$ \Rightarrow x=y+3 $

Then, $\frac{x^3+3}{(x-3)^3}=\frac{(y+3)^3+3}{y^3}$

$ \begin{aligned} & =\frac{y^3+9 y^2+27 y+27+3}{y^3} \\ & =\frac{y^3+9 y^2+27 y+30}{y^3} \\ & =1+\frac{9}{y}+\frac{27}{y^2}+\frac{30}{y^3} \\ & =1+\frac{9}{x-3}+\frac{27}{(x-3)^2}+\frac{30}{(x-3)^3} \end{aligned} $

Comparing this with Eq. (i), we get

$ a=1, b=9, c=27, d=30 $

So, $(a+d)-(b+c)=(1+30)-(9+27)$

$ =31-36=-5 $

$ \begin{aligned} &\begin{aligned} & \text { } f(x)=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right) \\ & \text { for } \frac{3}{4-x^2}, 4-x^2 \neq 0 \\ & \Rightarrow \quad x \neq \pm 2 \end{aligned}\\ &\text { and for } \log _{10}\left(x^3-x\right) \text {, }\\ &x^3-x>0 \Rightarrow x\left(x^2-1\right)>0 \end{aligned} $

and $x \neq 2$

Here, $x \in(-1,0) \cup(1,2) \cup(2, \infty)$

$\because f(x)=\frac{4-x^2}{4+x^2}, \forall x \in A$ and $f(x)$ is bijective So, Domain of $f(x) 4+x^2 \neq 0$ which is always true. therefore, Domain of $f(x) \in R$ Lets check if $f(x)$ be injective

$ \begin{aligned} & f\left(x_1\right)=f\left(x_2\right) \\ & \Rightarrow \frac{4-x_1^2}{4+x_1^2}=\frac{4-x_2^2}{4+x_2^2} \\ & \Rightarrow 16+4 x_2^2-4 x_1^2-x_1^2 x_2^2 \\ & \quad=16+4 x_1^2-4 x_2^2-x_1^2 x_2^2 \\ & \Rightarrow x_1^2=x_2^2 \Rightarrow x_2= \pm x_1 \end{aligned} $

for $f(x)$ to be injective.

We must have $x_1=x_2$ and $-4 \in A$ therefore, domain must be ( $-\infty, 0$ ] Now, determine the range

Let, $y=\frac{4-x^2}{4+x^2}$

$ \begin{aligned} & \Rightarrow y\left(4+x^2\right)=4-x^2 \Rightarrow 4 y+y x^2=4-x^2 \\ & \Rightarrow x^2=\frac{4(1-y)}{1+y} \\ & \because x^2 \geq 0 \end{aligned} $

So, $\frac{4(1-y)}{1+y} \geq 0$

So, $1-y \geq 0$ and $1+y>0$

therefore, range of $f(x) \in(-1,1]$

that means, $A \subseteq(-\infty, 0]$

and $B=(-1,1]$

Hence, $A \cap B=(-1,0]$

$f(x)=x^2+2 b x+2 c^2$ is a quadratic, opening upward

So, minmum at $(x)=\frac{-2 b}{2}=-b$

So, maximum of $g(x)$ at $x=\frac{-(-2 c)}{2(-1)}=-c$

then, $g(-c)=-c^2+2 c^2+b^2=c^2+b^2$

and given that

$\min f(x)>\max g(x)$

$\Rightarrow \quad-b^2+2 c^2>c^2+b^2$

$\Rightarrow \quad c^2>2 b^2 \Rightarrow\left(\frac{c}{b}\right)^2>2$

$\Rightarrow \quad\left|\frac{c}{b}\right|>\sqrt{2}$

Hence, $\left|\frac{c}{b}\right| \in(\sqrt{2}, \infty)$

We have, $f: R \rightarrow A$

$ f(x)=\cos x+\sqrt{3} \sin x-1 $

is onto.

∴ Range $=$ codomain

$ \begin{aligned} & \text { Now, range of } f(x) \\ & \begin{aligned} \because \quad & -\sqrt{1^2+(\sqrt{3})^2} \leq \cos x \\ & +\sqrt{3} \sin x \leq \sqrt{1^2+(\sqrt{3})^2} \\ & -2 \leq \cos x+\sqrt{3} \sin x \leq 2 \\ & -3 \leq \cos x+\sqrt{3} \sin x-1 \leq 1 \\ & -3 \leq f(x) \leq 1 \end{aligned} \\ & \therefore A=[-3,1] \end{aligned} $

$ \begin{aligned} & \text { Given, } g(x)=1+x-[x] \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =1+\{x\} \\ & \begin{aligned} \because \quad 0 & \{x\}<1 \\ \Rightarrow \quad 1 & \leq g(x)<2 \end{aligned} \\ & f(x)= \begin{cases}-1, & x<0 \\ 0, & x=0 \\ 1, & x>0\end{cases} \\ & \text { Now, } f\{g(x)\}= \begin{cases}-1, & g(x)<0 \\ 0, & g(x)=0 \\ 1, & g(x)>0\end{cases} \end{aligned} $

$\therefore g(x)$ is always $[1,2)$

$ \therefore \quad f\{g(x)\}=1 $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & f(x)=\left\{x+\left[\frac{x}{1+x^2}\right]\right\} \\ & f(\sqrt{2})=\left\{\sqrt{2}+\left[\frac{\sqrt{2}}{5}\right]\right\}=\{\sqrt{2}\}=\sqrt{2}-1 \\ & \text { and } f(-\sqrt{3})=\left\{-\sqrt{3}+\left[\frac{-\sqrt{3}}{4}\right]\right\} \\ & \quad=\{-\sqrt{3}-1\} \\ & \Rightarrow \quad f(-\sqrt{3})=2-\sqrt{3} \\ & \begin{aligned} \therefore f(\sqrt{2}) & +f(-\sqrt{3})=\sqrt{2}-1+2-\sqrt{3} \\ & =1+\sqrt{2}-\sqrt{3}=1+1.414-1.732 \\ & =2.414-1.732=0.682 \end{aligned} \end{aligned} \end{aligned} $

The function given is $f(x) = -3x - 3$.

The range of the function is given as $\{3, -6, -9, -18\}$.

To find the elements in the domain corresponding to this range, we need to find the $x$ values for which $f(x)$ equals each value in the range.

We set $f(x)$ equal to each value in the range and solve for $x$:

1. Set $f(x) = 3$:

   $-3x - 3 = 3$

   $-3x = 3 + 3$

   $-3x = 6$

   $x = \frac{6}{-3}$

   $x = -2$

2. Set $f(x) = -6$:

   $-3x - 3 = -6$

   $-3x = -6 + 3$

   $-3x = -3$

   $x = \frac{-3}{-3}$

   $x = 1$

3. Set $f(x) = -9$:

   $-3x - 3 = -9$

   $-3x = -9 + 3$

   $-3x = -6$

   $x = \frac{-6}{-3}$

   $x = 2$

4. Set $f(x) = -18$:

   $-3x - 3 = -18$

   $-3x = -18 + 3$

   $-3x = -15$

   $x = \frac{-15}{-3}$

   $x = 5$

So, the domain of $f$ corresponding to the given range $\{3, -6, -9, -18\}$ is the set of $x$ values $\{-2, 1, 2, 5\}$.

Now we need to check the given options to see which one is not in this domain:

A) -1: Is -1 in $\{-2, 1, 2, 5\}$? No.

B) -2: Is -2 in $\{-2, 1, 2, 5\}$? Yes. ($f(-2) = 3$)

C) 2: Is 2 in $\{-2, 1, 2, 5\}$? Yes. ($f(2) = -9$)

D) 5: Is 5 in $\{-2, 1, 2, 5\}$? Yes. ($f(5) = -18$)

The value -1 is not in the calculated domain.

The final answer is $\boxed{\text{-1}}$.

$ \begin{aligned} \text { } \because[2 x-2] & =[(2 x-1)-1] \\ & =[2 x-1]-1 \\ \text { let, } \quad[2 x-1] & =a \\ \Rightarrow \quad[2 x-2] & =a-1 \\ \text { so, } 3(a-1)+1 & =2 a-1 \\ \Rightarrow \quad 3 a-3+1 & =2 a-1 \Rightarrow a=1 \end{aligned} $

thus $[2 x-1]=1$

$ \begin{aligned} & \Rightarrow \quad 1 \leq(2 x-1)<2 \Rightarrow 2 \leq 2 x<3 \\ & \Rightarrow \quad 1 \leq x<1.5 \end{aligned} $

so, $x \in[1,1.5)$

Now, $k=2[2 x-1]-1=2(1)-1=1$

for $x \in[1,1.5)$

$ \begin{array}{rlrl} f(x) & =[k+5 x]=[1+5 x] & \\ \text { at } & x =1, f(x)=[6]=6 \quad(\text { minimum }) \\ \text { at } & x =1.5, f(x)=[1+5 \times 1.5] \\ & =[8.5]=8 \quad \text { (maximum) } \end{array} $

so, $f(x)$ can be $6,7,8$

Hence, the range of $f(x)=\{6,7,8\}$

$\because \quad y=f(x)=(x+1)^2-1$

$ \begin{aligned} & \Rightarrow y+1=(x+1)^2 \\ & \Rightarrow \quad x=\sqrt{y+1}-1 \end{aligned} $

thus, $f^{-1}(x)=\sqrt{x+1}-1$, defined for

$ x \geq-1 $

Now, $f(x)=f^{-1}(x)$

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

let $x+1=t$

$ \Rightarrow \quad t^4=t \Rightarrow t\left(t^3-1\right)=0 $

therefore, $t=0$

$ \Rightarrow \quad x+1=0 \Rightarrow x=-1 $

or $\quad t=1$

$ \begin{aligned} & \Rightarrow \quad x+1=1 \Rightarrow x=0 \\ & \text { Hence, }\left\{x \mid f(x)=f^{-1}(x)\right\} \\ & =\{0,-1\} \end{aligned} $

It is obvious. Since, $f$ is said to be one-one if $f(x)=f(y) \Rightarrow x=y$ or $f(x) \neq f(y) \Rightarrow x \neq y$. Also, in a function, one image may have more than one pre-image.

$f(x)=\sqrt{\frac{|x|-2}{|x|-3}}$

$ \Rightarrow \frac{|x|-2}{|x|-3} \geq 0 $

Case I $|x|-2 \geq 0$ and $|x|-3>0$

$ \begin{array}{ll} \Rightarrow & |x| \geq 2 \text { and }|x|>3 \\ \Rightarrow & |x|>3 \\ \Rightarrow & x>3 \text { or } x<-3 \end{array} $

Case II $|x|-2 \leq 0$ and $|x|-3<0$

$ \begin{aligned} &|x| \leq 2 \text { and }|x|<3 \\ & \Rightarrow \quad|x| \leq 2 \\ &-2 \leq x \leq 2 \\ &|x|-3 \neq 0 \Rightarrow|x| \neq 3 \\ & \Rightarrow x \neq 3 \text { and } x \neq-3 \end{aligned} $

Combining these, we get

$ (-\infty,-3) \cup[-2,2] \cup(3, \infty) $

Finally we write

$ R-[-3,-2) \cup(2,3] $

$f(x+y)=f(x)+f(y)+x y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ x, y \in N $

From Eq. (i),

$ \begin{aligned} & f(1)=2=\frac{1(1+3)}{2} \\ & x=1, y=1 \\ & f(2)=f(1)+f(1)+1 \\ & f(2)=5=\frac{2(2+3)}{2} \end{aligned} $

From Eq. (i), $x=1, y=2$

$ \begin{aligned} f(3) & =f(1)+f(2)+2 \\ & =2+5+2=9=\frac{3(3+3)}{2} \\ f(4) & =f(2)+f(2)+4 \\ & =14=\frac{4(4+3)}{2} \\ \Rightarrow f(x)= & \frac{n(n+3)}{2}=\frac{n^2}{2}+\frac{3 n}{2} \\ \sum_{k=0}^{10} f(k)= & \sum_{k=1}^{10} \frac{k(k+3)}{2}=\frac{1}{2} \sum_{k=1}^{10} k^2+\frac{3}{2} \sum_{k=1}^{10} k \\ = & \frac{1}{2}(385+165)=275 \end{aligned} $

$ \text { } \begin{aligned} f(x) & = \begin{cases}1-x, & x \in[-1,1] \\ x-1, & x \in(1,2]\end{cases} \\ & -1 \leq x \leq 1 \\ & +1 \geq-x \geq-1 \\ & +2 \geq 1-x \geq+0 \\ x & \in[-1,1] \\ f(x) & \in[0,2] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now $x \in(1,2], f(x) \in(0,1]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii) range $\equiv[0,2]$

Given, $f(x)=\frac{x^2-5 x+7}{x^2-5 x-7}=y$

$ \Rightarrow \quad(y-1) x^2-5(y-1) x-(7 y+7)=0 $

for $x \in R$

$ \begin{aligned} & D \geq 0 \\ & \Rightarrow \quad 25(y-1)^2+4(y-1)(7 y+\pi \geq 0 \\ & \Rightarrow \quad(y-1)[25 y-25+28 y+28] \geq 0 \\ &(y-1)(53 y+3) \geq 0 \\ & y \in\left(-\infty, \frac{-3}{53}\right] \cup(1, \infty) \text { at } y=1, x \text { is not } \end{aligned} $

possible

$\therefore$ Required sum $=2+(-1)=1$

$f(x)=2 x+\log \left(\frac{x}{2+x}\right), \frac{x}{2+x}>0$ and $\frac{x}{2+x} \neq 0 \Rightarrow x \neq 0$

$ \begin{aligned} &\begin{aligned} & x \in(-\infty,-2) \cup(0, \infty) \\ & \begin{aligned} f^{\prime}(x) & =2+\left(\frac{2+x}{x}\right) \cdot\left(\frac{2}{(2+x)^2}\right) \\ & =2+\frac{2}{x(x+2)}, x \neq-2 \\ f^{\prime}(x) & =0,2=\frac{-2}{x(x+2)} \\ & x^2+2 x+1=0 \\ & (x+1)^2=0 \Rightarrow x=-1 \\ & x=-1 \notin(-\infty,-2) \cup(0, \infty) \\ & f^{\prime}(x)>0 \text { for } x \in(-\infty,-2) \cup(0, \infty) \end{aligned} \end{aligned}\\ &\text { So, required interval, }\\ &(-\infty,-2) \cup(0, \infty) \end{aligned} $

$ \begin{aligned} &\text { } f(x)=\sqrt{\frac{[x]-1}{[x]^2-[x]-6}}\\ &\begin{aligned} & f(x) \text { is defined when } \frac{[x]-1}{[x]^2-[x]-6} \geq 0 \\ & \quad \frac{[x]-1}{([x]-3)([x]+2)} \geq 0 \\ & \Rightarrow \quad-2<[x] \leq 1 \text { or }[x] \geq 3 \\ & \Rightarrow \quad-1 \leq x<2 \text { or } x \geq 4 \text { or } x \in[-1,2) \cup[4, \infty) \end{aligned} \end{aligned} $

$f(x)=x-(-1)^x= \begin{cases}x-1, & x \in \text { even } \\ x+1, & x \in \text { odd }\end{cases}$

$f(x)$ is a linear function. Hence, it is one-one and its range is integer for integral value of $x$ so it is onto.

$\because$ It is both one-one and onto.

$f(x)=\log \left(x^2-1\right)+x \cot h^{-1} x \log \left(x^2-1\right)$ is defined when $x^2-1 > 0 x^2 > 1$

$-\infty < x < -1$ or $1 < x < \infty$ And $\cot h^{-1} x$ is defined when $|x|>1 \Rightarrow-\infty < x < -1$ or $1 < x < \infty$

$\because$ Domain of $f(x)$ is $x \in(-\infty,-1) \cup(1, \infty)$ or $R-[-1,1]$

$ \begin{aligned} &\text {We have } f(x)=\cos x-3\\ &\begin{aligned} & \text { Domain }=R \\ & \text { Range }-1 \leq \cos x \leq 1 \\ & -1-3 \leq \cos x-3 \leq 1-3 \\ & -4 \leq f(x) \leq-2 \\ & \text { Range }=[-4,-2] \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } f(x)=2 x-3 \text { and } g(x)=5 x^2-2 \\ & g \circ f(x)=5[f(x)]^2-2=5(2 x-3)^2-2 \\ & \because \text { Least value of } g \circ f(x)=-2 \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & f(x)=a \sin \left(x+\frac{\pi}{4}\right)+b \cos x+c \\ & f(x)=\frac{a \sin x}{\sqrt{2}}+\left(\frac{a}{\sqrt{2}}+b\right) \cos x+c=r \sin (x+\theta)+c \\ & \text { where } r=\sqrt{\left(\frac{a}{\sqrt{2}}\right)^2+\left(\frac{a}{\sqrt{2}}+b\right)^2} \text { and } \\ & \theta=\tan ^{-1}\left(\frac{a+\sqrt{2} b}{a}\right) \end{aligned} \end{aligned} $

For $f$ to be one-one, we must have

$ \begin{aligned} & -\frac{\pi}{2} \leq x+\theta \leq \frac{\pi}{2} \\ & \Rightarrow-\frac{\pi}{2}-\theta \leq x \leq \frac{\pi}{2}-\theta \Rightarrow x \in\left[-\frac{\pi}{2}-\theta, \frac{\pi}{2}-\theta\right] \end{aligned} $

Hence, $X=\left[\frac{-\pi}{2}-\theta, \frac{\pi}{2}-\theta\right]$, where

$ \theta=\tan ^{-1}\left(\frac{a+\sqrt{2} b}{a}\right) $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } f^{-1}(9)=x \\ & \Rightarrow \quad f(x)=9 \\ & \Rightarrow \quad x^2+5 x+9=9 \\ & \Rightarrow \quad x^2+5 x=0 \\ & \Rightarrow \quad x(x+5)=0 \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow & & x & =0,-5 \text { which are in } R . \\ \therefore & & f^{-1}(9) & =\{0,-5\} \end{aligned} $

$\begin{aligned} & F(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R} \\ & \sin 3 x+\cos 3 x \in[-\sqrt{2}, \sqrt{2}] \\ & 2+\sin 3 x+\cos 3 x \in[2-\sqrt{2}, 2+\sqrt{2}] \\ & \Rightarrow \frac{1}{2+\sin 3 x+\cos 3 x} \in\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right] \\ & \Rightarrow a=\frac{1}{2+\sqrt{2}}, b=\frac{1}{2-\sqrt{2}} \\ & \alpha=\frac{a+b}{2}=\frac{\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}}}{2} \\ & =\frac{4}{2 \times 2}=1 \\ & \beta=\sqrt{a b}=\sqrt{\left(\frac{1}{2+\sqrt{2}}\right) \times\left(\frac{1}{2-\sqrt{2}}\right)} \\ & =\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \\ & \Rightarrow \frac{\alpha}{\beta}=\sqrt{2} \\ \end{aligned}$

$ \begin{array}{ll} f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right) & \\ -1 \leq \frac{x-1}{2 x+3} \leq 1 & \frac{x-1}{2 x+3}+1 \geq 0 \\ \frac{x-1}{2 x+3}-1 \leq 0 & \frac{x-1+2 x+3}{2 x+3} \geq 0 \\ \frac{x-1-2 x-3}{2 x+3} \leq 0 & \frac{3 x+2}{2 x+3} \geq 0 \end{array} $

$\begin{aligned} & \frac{-x-4}{2 x+3} \leq 0 \\ & \frac{x+4}{2 x+3} \geq 0 \end{aligned}$

$\begin{aligned} & x \in(-\infty,-4] \cup\left[\frac{-2}{3}, \infty\right) \\ & \text { Domain : } R-\left(-4, \frac{-2}{3}\right) \\ & \alpha=-4 \\ & \beta=\frac{-2}{3} \\ & 12 \times \alpha \beta=12 \times 4 \times \frac{2}{3}=32 \end{aligned}$

$\begin{aligned} & f(x)=\left\{\begin{array}{l} -a \quad \text { if }-a \leq x \leq 0 \\ x+a \quad \text { if } 0< x \leq a \end{array}\right. \\ & f(|x|)=\left\{\begin{array}{cc} -a & -a \leq|x| \leq 0 \\ |x|+a & \text { if } 0 < |x| \leq a \end{array}\right. \end{aligned}$

$|x|<0$ is not possible, so

$\begin{aligned} & f(|x|)= \begin{cases}x+a & -a \leq x \leq 0 \\ -x+a & 0 < x \leq a\end{cases} \\ & |f(x)|= \begin{cases}a & -a \leq x \leq 0 \\ x+a & 0 < x \leq a\end{cases} \\ & h(x)= \begin{cases}-\frac{x}{2} & -a \leq x \leq 0 \\ 0 & 0 < x \leq a\end{cases} \end{aligned}$

Neither one-one nor onto.

$f\left(\frac{1}{\pi}\right)< f\left(\frac{1}{e}\right) \quad \text { as } \frac{1}{\pi}<\frac{1}{e}$

$\begin{aligned} & \Rightarrow\left(\frac{1}{1}\right)^{\frac{2}{\pi}}<\left(\frac{1}{\frac{1}{e}}\right)^{\frac{2}{e}} \\ & \Rightarrow(\pi)^{\frac{2}{\pi}}<(e)^{\frac{2}{e}} \\ & \Rightarrow \pi^e < e^\pi \end{aligned}$

$\begin{aligned} & f(x)=\frac{1}{7-\sin 5 x} \\\\ & -1 \leq \sin 5 x \leq 1 \\\\ & -1 \leq-\sin 5 x \leq 1 \\\\ & -1+7 \leq 7-\sin 5 x \leq 1+7 \\\\ & 6 \leq 7-\sin 5 x \leq 8 \\\\ & \frac{1}{8} \leq \frac{1}{7-\sin 5 x} \leq \frac{1}{6} \\\\ & \frac{1}{8} \leq f(x) \leq \frac{1}{6} \\\\ & \text { Range }=\left[\frac{1}{8}, \frac{1}{6}\right] \end{aligned}$

The function $ f(x)=\frac{x^2+2x-15}{x^2-4x+9}, x \in \mathbb{R} $ can be simplified to $ f(x)=\frac{(x-3)(x+5)}{x^2-4x+9} $.

For $ x=3 $ and $ x=-5 $, $ f(x) $ equals 0. Therefore, $ f(x) $ is not one-one as it yields the same output for different input values.

The range of $ f(x) $ is $ [-2, 1.6] $, indicating that $ f(x) $ does not cover all possible real values. Consequently, $ f(x) $ is not onto.

Thus, the function is neither one-one nor onto.

$\begin{aligned} & f(x)= \begin{cases}x-1, & x \geq 1 \\ 1-x & x<0\end{cases} \\ & g(x)=\left\{\begin{array}{cc} e^x & ; \quad x \geq 0 \\ x+1 & ; \quad x \leq 0 \end{array}\right. \end{aligned}$

$\begin{aligned} f(g(x)) & = \begin{cases}g(x)-1, & g(x) \geq 1 \\ 1-g(x) & g(x)<1\end{cases} \\ & =\left\{\begin{array}{cl} e^x-1 & x \geq 0 \\ -x & x<0 \end{array}\right. \end{aligned}$

$f(1)+f(3)=14$

Case I

$\begin{aligned} & f(1)=2, f(3)=12 \\ & f(1)=12, f(3)=2 \end{aligned}$

Total one-one function

$\begin{aligned} & =2 \times 5 \times 4 \times 3 \\ & =120 \end{aligned}$

Case II

$\begin{aligned} & f(1)=4, f(3)=10 \\ & f(1)=10, f(3)=4 \end{aligned}$

Total one-one function

$\begin{aligned} & =2 \times 5 \times 4 \times 3 \\ & =120 \\ & \text { Total cases }=120+120=240 \end{aligned}$

To find the domain of the function $f(x) = \frac{\sqrt{x^2-25}}{(4-x^2)}+\log_{10}(x^2+2x-15),$ we need to consider the domain conditions for both the square root function and the logarithmic function.

The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, so $x^2 - 25 \geq 0.$ This inequality is satisfied when $x \leq -5 \quad \text{or} \quad x \geq 5.$

The denominator of the rational part of $f(x)$, $(4-x^2)$, cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have $4 - x^2 \neq 0.$ This inequality is violated when $x = \pm2.$

Combining these conditions gives us the domain for the rational part of the function: $x \in (-\infty, -5] \cup (5, \infty) \quad \text{and} \quad x \neq 2,-2.$

Moving on to the logarithmic function, $\log_{10}(x^2+2x-15)$, the argument must be positive: $x^2 + 2x - 15 > 0.$ This is a quadratic inequality, which we can factor to find the solution: $(x+5)(x-3) > 0.$ From this, we see that the inequality is satisfied for $x < -5 \quad \text{or} \quad x > 3.$

The overall domain of $f(x)$ is the intersection of the domains for each piece. Taking the intersection of the two sets gives us: $x \in (-\infty, -5) \cup (5, \infty),$

Since the question states that the domain is of the form $(-\infty, \alpha) \cup [\beta, \infty)$, we can infer that $\alpha = -5 \quad \text{and} \quad \beta = 5.$

We calculate $\alpha^2 + \beta^3$ as follows: $\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150.$

So the correct answer, representing the sum of $\alpha^2$ and $\beta^3$, is: Option D $150$.

To find $(g \circ g \circ g)(4),$ we first need to understand the composition of $f$ with itself, i.e., $(f \circ f)(x) = f(f(x)) = g(x).$ We can then repeatedly apply $g$ to get the given expression.

First, let's calculate $(f \circ f)(x) = g(x):$

$g(x) = (f \circ f)(x) = f(f(x))$

$= f\left(\frac{4x+3}{6x-4}\right)$

To evaluate this expression, we substitute $\frac{4x+3}{6x-4}$ for $x$ in the function $f(x):$

$g(x) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}$

Now, we simplify the expression:

$g(x) = \frac{4(4x+3) + 3(6x-4)}{6(4x+3) - 4(6x-4)}$

$= \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16}$

$= \frac{34x}{34}$

$= x$

So, $g(x) = x$ for all $x$ in the domain of $g$, which is $\mathbb{R}-\left\{\frac{2}{3}\right\}$. It's important to note that the domain restriction is preserved through the composition because $f(x)$ has a vertical asymptote at $x = \frac{2}{3}$ which doesn't intersect the graph.

So, $g(x)$ is the identity function on its domain, which means that applying $g$ any number of times will result in the same input for $x$ in the given domain. Hence, we have:

$ (g \circ g \circ g \circ g)(4) = g(g(g(g(4)))) = g(g(g(4))) = g(g(4)) = g(4) = 4 $

This corresponds to option D, which is $4$.

$\begin{aligned} & \frac{2 x+3}{4 x^2+x-3}>0 \text { and }-1 \leq \frac{2 x-1}{x+2} \leq 1 \\ & \frac{2 x+3}{(4 x-3)(x+1)}>0 \quad \frac{3 x+1}{x+2} \geq 0 \& \frac{x-3}{x+2} \leq 0 \end{aligned}$

$(-\infty,-2) \cup\left[\frac{-1}{3}, \infty\right)$ ..... (1)

$(-2,3]$ ..... (2)

$\left[\frac{-1}{3}, 3\right]$ ..... (3) $\quad (1) \cap(2) \cap(3)$

$\begin{aligned} & \left(\frac{3}{4}, 3\right] \\ & \alpha=\frac{3}{4} \beta=3 \\ & 5 \beta-4 \alpha=15-3=12 \end{aligned}$

$\begin{aligned} & -1 \leq\left|\frac{2-|x|}{4}\right| \leq 1 \\ & \Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1 \\ & -4 \leq 2-|x| \leq 4 \\ & -6 \leq-|x| \leq 2 \\ & -2 \leq|x| \leq 6 \\ & |x| \leq 6 \end{aligned}$

$\Rightarrow x \in[-6,6]$ .... (1)

Now, $3-x\ne 1$

And $x\ne2$ .... (2)

and $3-x>0$

$x<3$ .... (3)

$\begin{aligned} & \text { From (1), (2) and (3) } \\ & \Rightarrow x \in[-6,3)-\{2\} \\ & \alpha=6 \\ & \beta=3 \\ & \gamma=2 \\ & \alpha+\beta+\gamma=11 \end{aligned}$

$f(g(x)) = \left\{ {\matrix{ {2 + 2g(x),} & { - 1 \le g(x) < 0} & {.....(1)} \cr {1 - {{g(x)} \over 3},} & {0 \le g(x) \le 3} & {.....(2)} \cr } } \right.$

$\text { By (1) } x \in \phi$

And by (2) $x \in[-3,0]$ and $x \in[0,1]$

Range of $\mathrm{f(g(x))}$ is $[0,1]$

$\begin{aligned} & f(x)=\frac{2 x+3}{2 x+1} ; x \neq-\frac{1}{2} \\ & g(x)=\frac{|x|+1}{2 x+5}, x \neq-\frac{5}{2} \end{aligned}$

Domain of $f(g(x))$

$f(g(x))=\frac{2 g(x)+3}{2 g(x)+1}$

$x \neq-\frac{5}{2}$ and $\frac{|x|+1}{2 x+5} \neq-\frac{1}{2}$

$x \in R-\left\{-\frac{5}{2}\right\}$ and $x \in R$

$\therefore$ Domain will be $\mathrm{R}-\left\{-\frac{5}{2}\right\}$

$\begin{aligned} & \mathrm{f}: \mathrm{N}-\{1\} \rightarrow \mathrm{N} \\ & \mathrm{f}(\mathrm{n})=\text { The highest prime factor of } \mathrm{n} . \\ & \mathrm{f}(2)=2 \\ & \mathrm{f}(4)=2 \\ & \Rightarrow \text { many one } \\ & 4 \text { is not image of any element } \\ & \Rightarrow \text { into } \end{aligned}$

Hence many one and into

Neither one-one nor onto.

Given two functions, $ f(x) = ax^2 + bx + c $, which is even, and $ g(x) = px^3 + qx^2 + rx $, which is odd, we need to analyze the combined function $ h(x) = f(x) + g(x) $. We know that $ h(-2) = 0 $.

Properties of Even and Odd Functions:

Even Function: $ f(x) = f(-x) $.

Odd Function: $ g(x) = -g(-x) $.

Since $ f(x) $ is even:

$ f(x) = f(-x) \quad \Rightarrow \quad f(2) = f(-2) $

For the odd function $ g(x) $:

$ g(-x) = -g(x) \quad \Rightarrow \quad g(-2) = -g(2) $

Using the Given Information:

Given $ h(-2) = 0 $, we have:

$ h(-2) = f(-2) + g(-2) = 0 $

From the properties mentioned:

$ f(-2) = f(2) \quad \text{and} \quad g(-2) = -g(2) $

Substitute these into $ h(-2) = 0 $:

$ f(2) - g(2) = 0 $

This implies:

$ f(2) = g(2) $

Expressing $ f(2) $ and $ g(2) $:

Calculate $ f(2) $:

$ f(2) = a(2)^2 + b(2) + c = 4a + 2b + c $

And $ g(2) $:

$ g(2) = p(2)^3 + q(2)^2 + r(2) = 8p + 4q + 2r $

Equating the two expressions from $ f(2) = g(2) $:

$ 4a + 2b + c = 8p + 4q + 2r $

Thus, the expression for $ 8p + 4q + 2r $ is:

$ 8p + 4q + 2r = 4a + 2b + c $