Functions

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} $

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

$ \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 $

To determine the range of the function $ f(x) = \log_3(5 + 4x - x^2) $, we need to analyze the behavior of the quadratic expression $ 5 + 4x - x^2 $.

Analysis of $ 5 + 4x - x^2 $

The expression $ 5 + 4x - x^2 $ is a quadratic function of the form $ ax^2 + bx + c $, where $ a = -1 $, $ b = 4 $, and $ c = 5 $.

Determining the Range of the Quadratic

Since $ a = -1 < 0 $, the parabola opens downwards, and the quadratic function achieves its maximum value at its vertex. The maximum (or minimum for upward-opening parabolas) value of a quadratic function $ ax^2 + bx + c $ is given by $ -\frac{D}{4a} $, where $ D = b^2 - 4ac $ is the discriminant.

Calculate the discriminant:

$ D = 4^2 - 4 \times (-1) \times 5 = 16 + 20 = 36 $

The maximum value of the quadratic $ 5 + 4x - x^2 $ is:

$ -\frac{D}{4a} = -\frac{36}{4 \times (-1)} = 9 $

Hence, the range of $ 5 + 4x - x^2 $ is $ (-\infty, 9] $.

Translating to the Logarithmic Function

Considering the function $ f(x) = \log_3(5 + 4x - x^2) $, the possible values of $ 5 + 4x - x^2 $ determine the input values for the logarithm.

Since $ 5 + 4x - x^2 \leq 9 $, we take the logarithm base 3 of each side:

$ \log_3(5 + 4x - x^2) \leq \log_3(9) $

Given that $ \log_3(9) = 2 $, we conclude:

$ f(x) = \log_3(5 + 4x - x^2) \leq 2 $

Thus, the range of $ f(x) $ is $ (-\infty, 2] $.

To find the sum of the maximum and minimum values of the function $ f(x) = \frac{x^2 - x + 1}{x^2 + x + 1} $, we start by setting:

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

Rewriting, we have:

$ y(x^2 + x + 1) = x^2 - x + 1 $

This simplifies to:

$ (y - 1)x^2 + (y + 1)x + (y - 1) = 0 $

To ensure real values of $ x $, the discriminant $ D $ of this quadratic in $ x $ must be non-negative. The discriminant is given by:

$ (y + 1)^2 - 4(y - 1)^2 \geq 0 $

Simplifying the inequality:

$ (y + 1 - 2(y - 1))(y + 1 + 2(y - 1)) \geq 0 $

$ (3 - y)(3y - 1) \geq 0 $

Reversing the inequality:

$ (y - 3)(3y - 1) \leq 0 $

This implies $ y $ lies within the interval:

$ y \in \left[\frac{1}{3}, 3\right] $

Therefore, the sum of the maximum and minimum values of $ y $ is:

$ \frac{1}{3} + 3 = \frac{10}{3} $

Given,

$ f(x)=\frac{1}{\sqrt{|x-|x||}} $

For

$ \begin{array}{r} x>0,|x|=x \\\\ x-|x|=0 \\\\ |x-|x||=0 \end{array} $

$\therefore f(x)$ is not possible for $x>0$.

For

$ \begin{aligned} x<0,|x| & =-x \\\\ x-|x| & =-2 x \\\\ |x-|x|| & =2 x \\\\ f(x) & =\frac{1}{\sqrt{-2 x}} \end{aligned} $

$\therefore A$ is set of negative numbers and $B$ is a set of positive numbers.

$ \therefore A \cap B=\phi $

We have,

$ \begin{aligned} & f(x)=\sqrt[3]{\frac{x-2}{2 x^2-7 x+5}}+\log \left(x^2-x-2\right) \\ \therefore & \quad 2 x^2-7 x+5 \neq 0 \text { and } x^2-x-2>0 \\ \Rightarrow & (2 x-5)(x-1) \neq 0 \text { and }(x+1)(x-2)>0 \\ \Rightarrow & x \neq 1, \frac{5}{2} \text { and } x \in(-\infty,-1) \cup(2, \infty) \\ \Rightarrow & x \in(-\infty,-1) \cup\left(2, \frac{5}{2}\right) \cup\left(\frac{5}{2}, \infty\right) \end{aligned} $

$\therefore$ Domain of $f$ is

$ (-\infty,-1) \cup\left(2, \frac{5}{2}\right) \cup\left(\frac{5}{2}, \infty\right) $

We have, $f\left(3 x+\frac{1}{2 x}\right)=9 x^2+\frac{1}{4 x^2}$

$ \therefore \quad f\left(3 x+\frac{1}{2 x}\right)=\left(3 x+\frac{1}{2 x}\right)^2-3 $

Now, $f\left(x+\frac{1}{x}\right)=1$

$ \begin{array}{cc} \Rightarrow & \left(x+\frac{1}{x}\right)^2-3=1 \Rightarrow\left(x+\frac{1}{x}\right)^2=4 \\ \Rightarrow & x^2+\frac{1}{x^2}+2=4 \Rightarrow x^2+\frac{1}{x^2}-2=0 \\ \Rightarrow & \left(x-\frac{1}{x}\right)^2=0 \Rightarrow x-\frac{1}{x}=0 \\ \Rightarrow & x^2-1=0 \Rightarrow x^2=1 \\ \Rightarrow & x= \pm 1 \end{array} $

We have,

$ \begin{aligned} & f(x)=\sqrt{\cos (\sin x)}+\cos ^{-1}\left(\frac{1+x^2}{2 x}\right) \\\\& \Rightarrow \quad \cos (\sin x)>0 \\\\ & \because \quad \sin x \in[-1,1] \\\\ & \therefore \quad \cos (\sin x)>0, \forall x \in R \\\\ & \text { and } \quad-1 \leq \frac{1+x^2}{2 x} \leq 1 \\\\ & \quad-1 \leq \frac{1+x^2}{2 x} \text { and } \frac{1+x^2}{2 x} \leq 1 \\\\ & \Rightarrow \quad x \in\{-1,1\} \end{aligned} $

We have, $f(x)=\frac{1}{\sqrt{64-(0.5)^{24+x-x^2}}}$

$f(x)$ will be defined, if $64-(0.5)^{24+x-x^2}$ $>0$

$ \begin{aligned} & =64 > (0.5)^{24+x-x^2} \\ & =64 > 2^{x^2-x-24} \\ & =2^6 > 2^{x^2-x-24} \\ & =x^2-x-24 < 6 \\ & =x^2-x-30 < 0 \\ & =(x+5)(x-6) < 0 \\ & =-5 < x < 6 \end{aligned} $

$ \because x=-4,-3,-2-1,0,1,2,3,4,5 $

Then, $A=\{-4,-3,-2,-1,0,1,2,3,4,5\}$

Sum of all absolute values of $A$ is

$ \begin{aligned} & =4+3+2+1+0+1+2+3+4+5 \\ & =25 \end{aligned} $

We have, $f(x)=\log \left(x+\sqrt{x^2+1}\right)$

$ \begin{aligned} \Rightarrow \quad & f(-x)=\log \left(-x+\sqrt{x^2+1}\right) \\ & =\log \left[\sqrt{x^2+1}-x \times \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right] \\ & =\log \left[\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right] \\ & =\log \left(\frac{1}{\sqrt{x^2+1}+x}\right) \\ & =-\log \left(x+\sqrt{x^2+1}\right) \\ & =-f(x) \end{aligned} $

$\therefore f(x)$ is an odd function.

We need to determine the range of the function:

$ f(x) = \frac{15}{3 \sin x + 4 \cos x + 10} $

Start with the expression inside the denominator:

$ 3 \sin x + 4 \cos x $

The range of this expression is given by:

$ -\sqrt{3^2 + 4^2} \leq 3 \sin x + 4 \cos x \leq \sqrt{3^2 + 4^2} $

Calculating the bounds:

$ -\sqrt{9 + 16} \leq 3 \sin x + 4 \cos x \leq \sqrt{9 + 16} $

$ -5 \leq 3 \sin x + 4 \cos x \leq 5 $

Adding 10 to the entire inequality, we have:

$ 5 \leq 3 \sin x + 4 \cos x + 10 \leq 15 $

The function $ f(x) $ can then be rewritten as:

$ f(x) = \frac{15}{3 \sin x + 4 \cos x + 10} $

Therefore, the minimum value of the denominator is 5, and the maximum is 15.

The range of $ f(x) $ can be calculated as follows:

Minimum value of $ f(x) $:

$ f(x)_{\min} = \frac{15}{15} = 1 $

Maximum value of $ f(x) $:

$ f(x)_{\max} = \frac{15}{5} = 3 $

Hence, the range of the function $ f(x) $ is $[1, 3]$.

Given the functions:

$ f(x) = x^2 - 1 $

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

$ h(x) = \begin{cases} 0, & \text{if } x \leq 0 \\ x, & \text{if } x \geq 0 \end{cases} $

Let's analyze the statements:

Function Composition:

$ f \circ g = f(g(x)) = f(\sqrt{x^2 + 1}) = (\sqrt{x^2 + 1})^2 - 1 = x^2 + 1 - 1 = x^2 $.

Since $ f \circ g = x^2 $, it is clear that this function is not one-to-one because different values of $ x $ (such as $ x $ and $-x$) yield the same $ x^2 $. Therefore, $ f \circ g $ is not invertible.

Composition involving $ h $:

$ h \circ f \circ g = h(f(g(x))) = h(x^2) $.

The function $ h(x^2) $ has the rule:

$ h(x^2) = \begin{cases} 0, & \text{if } x \leq 0 \\ x^2, & \text{if } x \geq 0 \end{cases} $

Since $ x^2 $ is always non-negative and $ h(x^2) = x^2 $ for $ x \geq 0 $, this simplifies to $ h(x^2) = x^2 $ across non-negative ranges, reinforcing that $ h(x^2) $ is equivalent to $ x^2 $.

Considering these observations, statements (III) and (IV) are true:

Statement (III): $ f \circ g $ is not invertible.

Statement (IV): $ h \circ f \circ g = x^2 $.

Thus, statements (III) and (IV) are correct.