Functions

$ \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} $

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.

For real values of $f(x)$

$ \sqrt{9-\sqrt{x^2-144}} \geq 0 $

On squaring both sides, we get

$ \begin{aligned} 9-\sqrt{x^2-144} & \geq 0 \\ 9 & \geq \sqrt{x^2-144} \end{aligned} $

Again, squaring on both sides, we get

$ \begin{aligned} 9^2 & \geq x^2-144 \\ 81+144 & \geq x^2 \\ 225 & \geq x^2 \\ x^2 & \leq 225 \end{aligned} $

On taking principal square root

$ |x| \leq 15 $

i.e. $x \leq 15$ or $x \geq-15$

$ \begin{aligned}Now,\,\,\, x^2-144 & \geq 0 \\ x^2 & \geq 144 \end{aligned} $

On taking principal square root

$ \begin{gathered} \sqrt{x^2} \geq \sqrt{12^2} \\ |x| \geq 12 \end{gathered} $

i.e. $x \leq-12$ or $x \geq 12$

We can see on number line the domain of the given function will be cross shadow area.

i.e. $[-15,-12] \cup[12,15]$

Given function $f: R \rightarrow\left[\frac{5}{2}, \infty\right]$ is defined by

$ f(x)=|2 x+1|+|x-2| $

Consider the graph

There are two points, where the expressions inside the absolute values change signs.

$ x=-\frac{1}{2} \text { or }-0.5 \text { and } x=2 $

If $x<-0.5$

$ \begin{aligned} f(x) & =-2 x-1-x+2 \\ & =-3 x+1 \end{aligned} $

If -0.5 < x < 2

$ f(x)=2 x+1-x+2=x+3 $

If $x>2$

$ f(x)=2 x+1+x-2=3 x-1 $

Each vertical line should intersect the graph at exactly one point. If not then function is not onto.

If a horizontal line intersects the graph more than once, the function is not one-one.

Clearly, from graph, through horizontal line two cut points implies that not one-one and through vertical line one cut point implies that onto.

Hence, function is onto but not one-one.

We start with the equation:

$ 3 f(x) - 2 f\left(\frac{1}{x}\right) = x. $

Differentiating both sides with respect to $ x $, we have:

$ 3 f'(x) - 2 \cdot f'\left(\frac{1}{x}\right) \times \left(-\frac{1}{x^2}\right) = 1. $

This simplifies to:

$ 3 f'(x) + \frac{2}{x^2} f'\left(\frac{1}{x}\right) = 1. $

First, set $ x = 2 $:

$ 3 f'(2) + \frac{2}{(2)^2} f'\left(\frac{1}{2}\right) = 1. $

This results in:

$ 3 f'(2) + \frac{1}{2} f'\left(\frac{1}{2}\right) = 1. $

Multiply throughout by 2 to clear the fraction:

$ 6 f'(2) + f'\left(\frac{1}{2}\right) = 2. \quad (i) $

Next, set $ x = \frac{1}{2} $:

$ 3 f'\left(\frac{1}{2}\right) + \frac{2}{\left(\frac{1}{2}\right)^2} f'(2) = 1. $

This becomes:

$ 3 f'\left(\frac{1}{2}\right) + 8 f'(2) = 1. $

Solve for $ f'\left(\frac{1}{2}\right) $:

$ f'\left(\frac{1}{2}\right) = \frac{1 - 8 f'(2)}{3}. \quad (ii) $

Substitute the expression for $ f'\left(\frac{1}{2}\right) $ from equation (ii) into equation (i):

$ 6 f'(2) + \frac{1 - 8 f'(2)}{3} = 2. $

Multiply through by 3 to eliminate the fraction:

$ 18 f'(2) + 1 - 8 f'(2) = 6. $

Simplify:

$ 10 f'(2) = 5. $

So, we find:

$ f'(2) = \frac{5}{10} = \frac{1}{2}. $

Mathematics

(d) We have,

$ f(x)=\log _2 \log _3 \log _5\left(x^2-5 x+11\right) $

$\log _5$ must be positive.

$\log _5\left(x^2-5 x+11\right)$ is defined if

$ x^2-5 x+11>0 $

Now, $D=b^2-4 a c=(-5)^2-4(1)(11)$

$ =-19 $

$\because$ If the discriminant is negative and $a>0, x^2-5 x+11>0, \forall$ real $x$.

Now, $\log _3$ must be positive.

$\log _3\left(\log _5\left(x^2-5 x+11\right)\right)$ is defined if

$ \log _5\left(x^2-5 x+11\right)>0 $

Since, $\log _5\left(x^2-5 x+11\right)>0$

$ \begin{aligned} & \Rightarrow \quad x^2-5 x+11>5^{\circ} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad\left [\because \log _a x=b, \text { then } x=a^b\right] \\ & \Rightarrow \quad x^2-5 x+11>1 \\ & \Rightarrow \quad x^2-5 x+10>0 \\ & D=b^2-4 a c=(-5)^2-4(1)(10)=-15 \end{aligned} $

The discriminate is negative, $x^2-5 x+10$ has no real roots and is always positive. $\therefore x^2-5 x+10>0, \forall$ real $x$

Now, $\log _2$ must be positive.

$\log _2\left(\log _3\left(\log _5\left(x^2-5 x+11\right)\right)\right.$ is defined

If

$ \begin{aligned} & \log _3\left(\log _5\left(x^5-5 x+11\right)\right)>0 \\ & \log _3\left(\log _5\left(x^2-5 x+11\right)\right)>0 \\ & \log _5\left(x^2-5 x+11\right)>3^{\circ} \end{aligned} $

$ \begin{array}{lr} \Rightarrow & \log _5\left(x^2-5 x+11\right)>1 \\ \Rightarrow & x^2-5 x+11>5 \\ \Rightarrow & x^2-5 x+6>0 \\ & (x-2)(x-3)>0 \end{array} $

The quadratic $x^2-5 x+6$ changes sign at $x=2$ and $x=3$.

Thus, $x^2-5 x+6>0$ for $x \in(-\infty, 2) \cup(3, \infty)$.

Thus, the domain of $f(x)$ is $(-\infty, 2) \cup(3, \infty)$.

To find the range of the function $ f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15} $, we begin by factorizing the numerator and denominator.

Denominator Factorization:

$ 2x^2 + 13x + 15 = 0 $

$ 2x^2 + 10x + 3x + 15 = 0 $

$ 2x(x + 5) + 3(x + 5) = 0 $

$ (x + 5)(2x + 3) = 0 \quad \Rightarrow \quad x = -5, \frac{-3}{2} $

Numerator Factorization:

$ x^2 + 2x - 15 = 0 $

$ x^2 + (5x - 3x) - 15 = 0 $

$ x(x + 5) - 3(x + 5) = 0 $

$ (x + 5)(x - 3) = 0 \quad \Rightarrow \quad x = -5, 3 $

Domain and Range:

The domain is $(-\infty, -5) \cup (-5, \frac{-3}{2}) \cup (\frac{-3}{2}, \infty)$.

The range is $\left(-\infty, \frac{1}{2}\right) \cup \left(\frac{1}{2}, \frac{8}{7}\right) \cup \left(\frac{8}{7}, \infty\right)$, excluding $\frac{1}{2}$ and $\frac{8}{7}$, which do not correspond to any real $x$ values for $f(x)$.

Thus, the range of the function is:

$ R - \left\{\frac{1}{2}, \frac{8}{7}\right\} $

Given, $f(\Omega)=6$

and $f(x+y)=f(x)+12 y$

Now,

$ \begin{aligned} f(2) & =f(1+1)=f(1)+12(1) \\ & =6+12=18 \\ f(3) & =f(1+2)=f(1)+12(2) \\ & =6+24=30 \\ f(4) & =f(1+3)=f(1)+12(3) \\ & =6+36=42 \end{aligned} $

and, so on

$ \begin{aligned} \therefore \quad f(n) & =6+(n-1) 12 \\ & =6+12 n-12=12 n-6 \end{aligned} $

Now, $\sum\limits_{r=1}^n f(r)=f(1)+f(2)+f(3)+\ldots+f(n)$

$ \begin{aligned} & =6+18+30+42+\ldots+(12 n-6) \\ & =6[1+3+5+7+\ldots+(2 n-1)] \\ & =6 n^2 \end{aligned} $

To determine the domain of the function $ f(x) = \sqrt{2+x} + \sqrt{3-x} $, we need to ensure that all expressions under the square roots are non-negative.

For $\sqrt{2+x}$ to be defined, the expression inside the square root must be non-negative:

$ 2 + x \geq 0 \quad \Rightarrow \quad x \geq -2 $

For $\sqrt{3-x}$ to be defined, the expression inside the square root must also be non-negative:

$ 3 - x \geq 0 \quad \Rightarrow \quad x \leq 3 $

Combining both conditions, the value of $ x $ must satisfy:

$ -2 \leq x \leq 3 $

Therefore, the domain of the function $ f(x) $ is the closed interval $[-2, 3]$.

Let $ f(x) = 3 + 2x $ and $ g_n(x) $ be the composition of $ n $ functions $ f $ with itself. If all the lines $ y = g_n(x) $ pass through a fixed point $(\alpha, \beta)$, we need to find $\alpha + \beta$.

The function $ f: \mathbb{R} \to \mathbb{R} $ is defined as:

$ f(x) = 2x + 3 = g_1(x) $

Now, find $ f \circ f(x) = g_2(x) $:

$ \begin{aligned} g_2(x) &= 2(2x + 3) + 3 \\ &= 4x + 6 + 3 \\ &= 4x + 9 \end{aligned} $

Next, find $ f \circ f \circ f(x) = g_3(x) $:

$ \begin{aligned} g_3(x) &= 2(4x + 9) + 3 \\ &= 8x + 18 + 3 \\ &= 8x + 21 \end{aligned} $

Observing a pattern, we have:

$ \begin{aligned} g_1(x) &= 2x + 3 = 2^1(x + 3) - 3 \\ g_2(x) &= 4x + 9 = 2^2(x + 3) - 3 \\ g_3(x) &= 8x + 21 = 2^3(x + 3) - 3 \\ & \vdots \\ g_n(x) &= 2^n(x + 3) - 3 \end{aligned} $

Then, the expression becomes:

$ \begin{aligned} y &= g_n(x) = 2^n(x + 3) - 3 \\ y + 3 &= 2^n(x + 3) \end{aligned} $

Now, consider the equation of a line passing through $(\alpha, \beta)$:

$ y - \beta = m(x - \alpha) $

Comparing the equations,

$ \begin{aligned} \alpha &= -3 \\ \beta &= -3 \end{aligned} $

Thus, $\alpha + \beta = -3 + (-3) = -6$.

$\text { The graph of } f(x) \text { is shown below. }$

$\therefore$ The function is not onto.

The function is many one as image of $f(x)$ is for every value of $a$ and $b$.

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

We have,

$ \begin{array}{lc} & P(x)=x^5+a x^4+b x^3+c x^2+d x+e \\ \because & P(0)=1 \\ \Rightarrow & 0+0+0+0+0+e=1 \\ \Rightarrow & e=1 \\ \because & P(1)=2 \\ \Rightarrow & 1+a+b+c+d+1=2 \\ \Rightarrow & a+b+c+d=0 ......\text { (i) } \\ \Rightarrow & 32+16 a+8 b+4 c+2 d+1=5 \\ \Rightarrow & 16 a+8 b+4 c+2 d=-28 \\ \Rightarrow & 8 a+4 b+2 c+d=-14 ......\text { (ii) }\end{array} $

$ \begin{aligned} & \because P(3)=10 \\ & \Rightarrow \quad 243+81 a+27 b+9 c+3 d+1=10 \\ & \Rightarrow 81 a+27 b+9 c+3 d=-234 \\ & \Rightarrow \quad 27 a+9 b+3 c+d=-78 ......\text { (iii) }\end{aligned} $

$ \begin{aligned} & P(4)=17 \\ & \Rightarrow 1024+256 a+64 b+16 c+4 d+1=17 \\ & \Rightarrow 256 a+64 b+16 c+4 d=-1008 \\ & \Rightarrow 64 a+16 b+4 c+d=-252 ......\text { (iv) } \end{aligned} $

From Eqs. (i), (ii), (iii) and (iv), we get

$ \begin{aligned} & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{array}\right]\left[\begin{array}{l} a \\ b \\ c \\ d \end{array}\right]=\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \\ & R_2 \rightarrow R_2-8 R_1 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \\ & R_3 \rightarrow R_3-27 R_1 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & -18 & -24 & -26 \\ 64 & 16 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \\ & R_4 \rightarrow R_4-64 R_1 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & -18 & -24 & -26 \\ 0 & -48 & -60 & -63 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -78 \\ -252 \end{array}\right]} \end{aligned} $

$ \begin{aligned} & R_3 \rightarrow R_3-\frac{9}{2} R_2 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & 0 & 3 & \frac{11}{2} \\ 0 & -48 & -60 & -63 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -15 \\ -252 \end{array}\right]} \\ & R_4 \rightarrow R_4-12 R_2 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & 0 & 3 & \frac{11}{2} \\ 0 & 0 & 12 & 21 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -15 \\ -84 \end{array}\right]} \\ & R_4 \rightarrow R_4-4 R_3 \\ & {\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -7 \\ 0 & 0 & 3 & \frac{11}{2} \\ 0 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 0 \\ -14 \\ -15 \\ -24 \end{array}\right]} \end{aligned} $

Thus,

$ \begin{aligned} a+b+c+d & =0 \\ -4 b-6 c-7 d & =-14 \\ 3 c+\frac{11}{2} d & =-15 \\ -d & =-24 \end{aligned} $

On solving, we get

$ \begin{aligned} & a=-10, b=35, c=-49, d=24 \\ & \text { Now, } P(5)=3125+625 \times(-10) \\ & \quad+125 \times 35+25 \times(-49)+24 \times 5+1 \\ & =3125-6250+4375-1225+120+1 \\ & =146 \end{aligned} $

We are given a real-valued function $ f: [a, \infty) \rightarrow [b, \infty) $ defined by $ f(x) = 2x^2 - 3x + 5 $. This function is a quadratic and is stated to be a bijection.

First, let's find the vertex of the quadratic function, as it will help determine the minimum value of the function, which is crucial for it to be a bijection.

The vertex formula for a quadratic function $ ax^2 + bx + c $ is given by:

$ \left(\frac{-b}{2a}, \frac{-D}{4a}\right) $

where $ D = b^2 - 4ac $. For our function, $ a = 2 $, $ b = -3 $, and $ c = 5 $.

Calculating the vertex:

Calculate $ x $-coordinate of the vertex:

$ \frac{-(-3)}{2 \times 2} = \frac{3}{4} $

Calculate the discriminant $ D $:

$ D = (-3)^2 - 4 \times 2 \times 5 = 9 - 40 = -31 $

Calculate $ y $-coordinate of the vertex:

$ \frac{-(-31)}{8} = \frac{31}{8} $

Thus, the vertex of the quadratic function is $ \left(\frac{3}{4}, \frac{31}{8}\right) $.

Since $ f(x) = 2x^2 - 3x + 5 $ is a parabola opening upwards (as the coefficient of $ x^2 $ is positive), the minimum value of $ f(x) $ occurs at the vertex. Therefore, to achieve a bijective function, the domain should start from the $ x $-coordinate of the vertex, and the range should start from the $ y $-coordinate of the vertex.

This implies:

$ a = \frac{3}{4} $

$ b = \frac{31}{8} $

To find the value of $ 3a + 2b $:

$ 3 \times \frac{3}{4} + 2 \times \frac{31}{8} = \frac{9}{4} + \frac{62}{8} = \frac{9}{4} + \frac{31}{4} = \frac{40}{4} = 10 $

Thus, $ 3a + 2b = 10 $.

To find the domain of the real-valued function $ f(x) = \frac{1}{\sqrt{\log_{0.5}(2x-3)}} + \sqrt{4-9x^2} $, we analyze the conditions under which each part of the function is defined.

First Part: $ \frac{1}{\sqrt{\log_{0.5}(2x-3)}} $

For the square root in the denominator to be defined and non-zero, the expression inside must be positive: $\log_{0.5}(2x-3) > 0$.

The base of the logarithm is less than 1. This means:

$ 2x - 3 < 1 \quad \Rightarrow \quad x < 2 $

Additionally, the expression inside the logarithm must be positive:

$ 2x - 3 > 0 \quad \Rightarrow \quad x > \frac{3}{2} $

Second Part: $ \sqrt{4-9x^2} $

The expression under the square root must be non-negative:

$ 4 - 9x^2 > 0 \quad \Rightarrow \quad x^2 < \frac{4}{9} $

Solving for $ x $, we derive:

$ -\frac{2}{3} < x < \frac{2}{3} $

Intersection of Conditions:

From the first part, we need $ x $ to satisfy both conditions: $ \frac{3}{2} < x < 2 $.

From the second part, $ x $ must be in the interval $ -\frac{2}{3} < x < \frac{2}{3} $.

The intersection of these conditions is empty, as there is no number that simultaneously satisfies $ \frac{3}{2} < x < 2 $ and $ -\frac{2}{3} < x < \frac{2}{3} $.

Hence, the domain of the function is a null set.

$f(x)=x^3-x$ is not one-one function as $f(A)=f(0)=f(-1)=0$ $f(x)=x^3-x$ is a polynomial function and every polynomial fanction is continnous and differentiable in its domala.

If $x \rightarrow \infty$, then $f(x) \rightarrow \infty$

$[\because f(x)$ is a cubic function]

If $x \rightarrow-\infty$, then $f(x) \rightarrow-\infty$

So, $\rightarrow \infty

$f(x) \in R$ and $x \in R$

$\therefore f(x)$ is onto function.

To determine $ g(x) $, let's first understand the given functions:

We have $ f(x) = \sqrt{x - 1} $. This means $ f(x) $ is a function that transforms $ x $ into $ \sqrt{x - 1} $.

Next, the composition $ g(f(x)) = x + 2x^2 + 1 $ can be rewritten using substitution and simplification steps:

Begin by rewriting the expression for $ g(f(x)) $:

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

Recognize a perfect square identity in:

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

Here, $ (\sqrt{x-1} + 1)^2 = \sqrt{x-1}^2 + 2\sqrt{x-1} + 1 $, which simplifies to $ x - 1 + 2\sqrt{x-1} + 1 = x + 2\sqrt{x-1} $.

Notice then that:

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

Thus, we can deduce that $ g(x) $ must match the transformed output when $ x = f(x) $ plus an added constant:

$ g(x) = (x + 2)^2 $

Therefore, $ g(x) $ is the function that squares the input, transformed by adding $ 2 $.

Let's solve the problem such that the given expression assumes all real values.

We're looking at the expression:

$ y = \frac{x + a}{2x^2 - 3x + 1} $

To ensure that $ y $ can take all real values, the denominator must not be zero:

$ 2x^2 - 3x + 1 \neq 0 $

This can be factored as:

$ (2x - 1)(x - 1) \neq 0 $

Thus, $ x \neq 1 $ and $ x \neq \frac{1}{2} $.

For $ y $ to take all real values for real $ x $, the quadratic expression $ (3y+1)^2 - 4y(y-2) \geq 0 $ must hold:

$ (3y+1)^2 - 4[y(y-2) - a] \geq 0 $

Simplifying the terms:

$ 9y^2 + 6y + 1 - (4y^2 - 8y - 4a) \geq 0 $

This simplifies to:

$ 5y^2 + 14y + 1 + 4a \geq 0 $

As a quadratic in terms of $ y $, for it to be non-positive (invertible for any $ x $), the discriminant must be negative:

The discriminant:

$ 6^2 - 4 \times 5 \times (1 + 4a) < 0 $

$ 36 - 20(1 + 4a) < 0 $

$ 36 - 20 - 80a < 0 $

$ 16 < 80a $

Thus:

$ a > \frac{1}{2} $

Therefore, the condition for $ a $ is:

$ -1 < a < -\frac{1}{2} $

The correct understanding implies the feasible range for $ a $.

To find the value of $ h $ that removes the $ x^3 $ term from the polynomial $ f(x) = x^4 + 2x^3 - 19x^2 - 8x + 60 = 0 $ when transformed to $ f(x + h) = 0 $, follow these steps:

Polynomial Transformation: The transformation affects each term in the original polynomial. For the $ x^3 $ term in the polynomial, apply the transformation:

$ f(x+h) = (x+h)^4 + 2(x+h)^3 - 19(x+h)^2 - 8(x+h) + 60 $

Focus on the $ x^3 $ Terms: When expanding, consider only the terms involving $ x^3 $:

From $(x+h)^4$:

$ ^4C_1 \cdot x^3 \cdot h = 4hx^3 $

From $2(x+h)^3$:

$ 2 \times 1 \times x^3 = 2x^3 $

Collect $ x^3 $ Terms: Adding these terms together gives:

$ 4hx^3 + 2x^3 = (4h + 2)x^3 $

Set the Coefficient to Zero: To eliminate the $ x^3 $ term from the polynomial, the coefficient must be zero:

$ 4h + 2 = 0 $

Solve for $ h $:

$ 4h = -2 \quad \implies \quad h = \frac{-2}{4} = \frac{-1}{2} $

Thus, the value of $ h $ that removes the $ x^3 $ term is $ h = -\frac{1}{2} $.

$f(x)=\log \left(\left(\frac{2 x^2-3}{x}\right)+\sqrt{\frac{4 x^4-11 x^2+9}{|x|}}\right)$

$\begin{aligned} & f(-x)=\log \left(\frac{2(-x)^2-3}{(-x)}+\sqrt{\frac{4(-x)^4-11(-x)^2+9}{|-x|}}\right) \\ & f(-x)=\log \left[\frac{2 x^2-3}{-x}+\sqrt{\frac{4 x^4-11 x^2+9}{|x|}}\right] \end{aligned}$

We know that, if function $f(x)$ is an odd function. Then,

$\begin{aligned} & f(x)+f(-x)=0 \\ & \log \left[\left(\frac{2 x^2-3}{x}\right)+\sqrt{\frac{4 x^4-11 x^2+9}{|x|}}\right] \\ & +\log \left[\sqrt{\frac{4 x^4-11 x^2+9}{|x|}}-\frac{\left(2 x^2-3\right)}{x}\right] \\ & =\log \left[\frac{4 x^4-11 x^2+9}{x^2}-\frac{\left(2 x^2-3\right)^2}{x^2}\right] \\ & =\log \left[\frac{4 x^4-11 x^2+9-4 x^4-9+12 x^2}{x^2}\right] \\ & =\log 1=0 \end{aligned}$

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

$f(f(x))=-x$ [Given]

$\begin{aligned} & \Rightarrow \quad \frac{f(x)-2}{2(f(x))+1}=-x \\ & \Rightarrow \quad \frac{\frac{x-2}{2 x+1}-2}{\frac{2 x-4}{2 x+1}+1}=-x \\ & \Rightarrow \frac{x-2-4 x-2}{2 x-4+2 x+1}=-x \\ & \Rightarrow \quad \frac{-3 x-4}{4 x-3}=-x \Rightarrow \frac{3 x+4}{4 x-3}=x \\ & \Rightarrow \quad 3 x+4=4 x^2-3 x \\ & \Rightarrow \quad 4 x^2-6 x-4=0 \\ & \Rightarrow \quad 2 x^2-3 x-2=0 \\ \end{aligned}$

Now, $\alpha+\beta=\frac{3}{2}, \alpha \beta=-1$

Now, $(\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta$

$\Rightarrow \quad \frac{9}{4}=\alpha^2+\beta^2-2 \quad[\because \alpha \beta=-1]$

$\begin{aligned} & \Rightarrow \alpha^2+\beta^2=\frac{17}{4} \\ & \Rightarrow 4\left(\alpha^2+\beta^2\right)=17 \end{aligned}$

Given, function $f(x)=\sin \left(\log \left(\frac{\sqrt{4-x^2}}{1-x}\right)\right)$ is defined for $x \neq 1$ and $x>-2$ and also $x<1$. Thus, the domain of above function is $D(f)=\{x \in R ; x \in(-2,1)\}$

$f(x)=\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}$

Let $y=\frac{x^2+2 x+8}{x^2+2 x+4}\quad [\because f(x)=\sqrt{y}]$

$\begin{aligned} & \Rightarrow y x^2+2 x y+4 y=x^2+2 x+8 \\ & \Rightarrow(y-1) x^2+2 x(y-1)+4 y-8=0 \end{aligned}$

For $x$ to be real, discriminant of equation must be greater than equals to zero.

$\begin{array}{rlrl} & {[2(y-1)]^2-4(y-1)(4 y-8)} & \geq 0 \\ \Rightarrow & 4(y-1)^2-4(y-1) 4(y-2) & \geq 0 \\ \Rightarrow & (y-1)^2-(y-1) 4(y-2) & \geq 0 \\ \Rightarrow & (y-1)[y-1-4 y+8] & \geq 0 \\ \Rightarrow & (y-1)(-3 y+7) & \geq 0 \\ & y =1, \frac{7}{3} \end{array}$

$\begin{aligned} & \therefore \quad(y-1)(-3 y+7) \geq 0 \Rightarrow y \in\left(1, \frac{7}{3}\right] \\ & {\left[\because \text { At } y=1, x^2+2 x+4 \neq x^2+2 x+8\right]} \\ & \because \quad y \in\left(1, \frac{7}{3}\right] \\ & \therefore \quad f(x) \in\left(1, \sqrt{\frac{7}{3}}\right] \\ & \end{aligned}$

$\begin{aligned} & f(x)=\sqrt{2-x^2} \text { and } g(x)=\log (1-x) \\ & \begin{aligned} (f+g)(x)= & f(x)+g(x) \\ = & \sqrt{2-x^2}+\log (1-x) \end{aligned} \end{aligned}$

$\begin{aligned} 2-x^2 \geq 0 \\ \Rightarrow x^2 \leq 2 \\ \Rightarrow |x| \leq \sqrt{2} \\ \Rightarrow -\sqrt{2} \leq x \leq \sqrt{2} \end{aligned}$

Also, $1-x>0 \Rightarrow x<1$

Required domain : $-\sqrt{2} \leq x \leq \sqrt{2} \cap x<1$

$\Rightarrow \quad-\sqrt{2} \leq x<1$

Let $f(x)=y$, then $x=f^{-1}(y)$, here

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

or

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

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

$f(x)=[x], g(x)=|x|$

Now, $f\left(-\frac{5}{3}\right)=\left[-\frac{5}{3}\right]$

$=[-1.667]=-2$

Then, $\quad g\left(f\left(-\frac{5}{3}\right)\right)=g(-2)$

$(g \circ f)\left(-\frac{5}{3}\right)=|-2|=2$

$\therefore \quad(g \circ f)\left(-\frac{5}{3}\right)=2$

Given, $f(x)=a x+b$ and $g(x)=c x^3+d$

$\begin{aligned} f(g(x)) & =f\left(c x^3+d\right) \\ & =a\left(c x^3+d\right)+b \\ (f \circ g)(x) & =a c x^3+a d+b \quad .....(\text{i}) \end{aligned}$

Let $(f \circ g)(x)=y$, then

$x=(f \circ g)^{-1} y$

From Eq. (i), we get

$y=a c x^3+a d+b$

i.e.

$\begin{aligned} a c x^3 & =y-a d-b \\ x^3 & =(y-a d-b) / a c \\ x & =\left[\frac{(y-a d-b)}{a c}\right]^{1 / 3} \end{aligned}$

$\begin{array}{ll} \therefore & (f \circ g)^{-1}(y)=\left[\frac{(y-a d-b)}{a c}\right]^{1 / 3} \\ \text { or } & (f \circ g)^{-1}(x)=\left[\frac{(x-a d-b)}{a c}\right]^{1 / 3}\end{array}$

$f(10-x)=3 x^2+4 x-5$ and $f(x)=p x^2+q x+r$

Let $10-x=y$, then $x=10-y$

$\begin{aligned} f(y) & =3(10-y)^2+4(10-y)-5 \\ & =3\left(100+y^2-20 y\right)+40-4 y-5 \\ & =340+3 y^2-60 y-4 y-5 \\ & =3 y^2-64 y+335 \end{aligned}$

Replace $y$ by $x \Rightarrow f(x)=3 x^2-64 x+335$

compare with $f(x)=p x^2+q x+r$, we obtain

$\begin{aligned} & p=3, q=-64, r=335 \\ \Rightarrow \quad & p+q+r=274 \end{aligned}$

$\begin{aligned} f(x) & =\sin x+\cos x \\ g(x) & =x^2-1 \\ g[f(x)] & =(\sin x+\cos x)^2-1 \\ & =1+\sin 2 x-1=\sin 2 x \end{aligned}$

Among the given options, $\sin 2 x$ is monotonous (here strictly increasing) in $-\frac{\pi}{4} \leq 2 \leq \frac{\pi}{4}$.