(4)

We have, $[x+3]+[x+4]\le 3$

$\Rightarrow \left[x\right]+3+\left[x\right]+4\le 3\Rightarrow 2\left[x\right]\le -4\Rightarrow \left[x\right]\le -2$

$\Rightarrow x\in (-\infty ,-1)$

Now,
$3^x(\sum _{r=1}^{\infty }{\left(\frac{{\displaystyle 3}}{{\displaystyle {10}^r}}\right))}^{x-3}<3^{-3x}\Rightarrow 3^x{\left(\frac{{\displaystyle 1}}{{\displaystyle 3}}\right)}^{x-3}<3^{-3x}$

$\Rightarrow 3^x·3^{3-x}<3^{-3x}\Rightarrow 3^3<3^{-3x}\Rightarrow 3<-3x\Rightarrow x<-1$

$\Rightarrow x\in (-\infty ,-1)$  $\therefore \text{Both sets }A\text{ and }B\text{ are equal.}$

(1)

$f\left(x\right)=\frac{1}{\sqrt{\lceil x\rceil -x}}$

If $x\in I$, $\lceil x\rceil =\left[x\right]$ (greatest integer function) 

If $x\notin I$, $\lceil x\rceil =\left[x\right]+1$

$\Rightarrow f\left(x\right)=$$\{\begin{array}{cc}\frac{1}{\sqrt{\left[x\right]-x}},& x\in I\\ \frac{1}{\sqrt{\left[x\right]+1-x}},& x\notin I\end{array}$

$\Rightarrow f\left(x\right)=$$\{\begin{array}{ccc}\frac{1}{\sqrt{-\left\{x\right\}}},& x\in I & \text{(does not exist)}\\ -\frac{1}{\sqrt{1-\left\{x\right\}}},& x\notin I& \end{array}$

$\Rightarrow \text{Domain of }f\left(x\right)=R-I$

Now, $f\left(x\right)=\frac{1}{\sqrt{1-\left\{x\right\}}}, x\notin I$

$\Rightarrow 0<\left\{x\right\}<1\Rightarrow 0<\sqrt{1-\left\{x\right\}}<1 \Rightarrow \frac{1}{\sqrt{1-\left\{x\right\}}}>1$

$\Rightarrow \text{Range}=(1,\infty )\Rightarrow A=R-I\text{ and }B=(1,\infty )$

$\text{So, }A\cap B=(1,\infty )-N \Rightarrow A\cup B\ne (1,\infty )$

$\Rightarrow S1\text{ is true.}$

(1)

Let $f\left(x\right)=\frac{Ax+B}{Cx-A}$

$f\left(f\right(x\left)\right)= \frac{A\left(\frac{Ax+B}{Cx-A}\right)+B}{C\left(\frac{Ax+B}{Cx-A}\right)-A}=\frac{A^{2}x+BCx}{BC+A^2}=x \dots \left(i\right)$

$f\left(f\right(\frac{{\displaystyle 4}}{{\displaystyle x}}\left)\right)=\frac{4}{x} \text{[From (i)]}$

$\therefore f\left(f\right(x\left)\right)+f\left(f\right(\frac{{\displaystyle 4}}{{\displaystyle x}}\left)\right)=x+\frac{4}{x}\ge 4 \text{[Using A.M.}\ge \text{G.M.]}$

(1)

(2)

${\left[x\right]}^2-3\left[x\right]-10>0$

$\Rightarrow {\left[x\right]}^2-5\left[x\right]+2\left[x\right]-10>0 \Rightarrow \left(\right[x]-5)\left(\right[x]+2)>0$

$\Rightarrow \left[x\right]<-2\text{ or }\left[x\right]>5 \Rightarrow x<-2\text{ or }x\ge 6$

$\Rightarrow x\in (-\infty ,-2)\cup [6,\infty )$

(*)

For $x\in R$, $f\left(x\right)$ and $g\left(x\right)$ are real-valued functions.

$g\left(x\right)=\sqrt{x}+1 \text{and} fog\left(x\right)=x+3-\sqrt{x}$

$fog\left(x\right)=x+3-\sqrt{x}\Rightarrow f\left(g\right(x\left)\right)=x+3-\sqrt{x}$

$={(\sqrt{x}+1)}^2-3(\sqrt{x}+1)+5$

$={\left[g\right(x\left)\right]}^2-3\left[g\right(x\left)\right]+5 (\because g(x)=\sqrt{x}+1)$

So, $f\left(x\right)=x^2-3x+5$

$\therefore f\left(0\right)=0-3\times 0+5=5$

Note: But if we consider the domain of the composite function $fog\left(x\right)$, then $f\left(0\right)$ will not be defined as $g\left(x\right)$ can’t be equal to zero.

(3)

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

$x=2\Rightarrow f\left(2\right)+f(-1)=3$             $\dots \left(i\right)$

$x=-1\Rightarrow f(-1)+f\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)=0$       $\dots \left(ii\right)$

$x=\frac{1}{2}\Rightarrow f\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)+f\left(2\right)=\frac{3}{2}$            $\dots \left(iii\right)$

$\left(i\right)+\left(iii\right)-\left(ii\right)\Rightarrow 2f\left(2\right)=\frac{9}{2}$     $\therefore$  $f\left(2\right)=\frac{9}{4}$

(4)

$x^2-4x+\left[x\right]+3=x\left[x\right]$

$\Rightarrow x^2-x\left[x\right]-(x-[x\left]\right)+3(x-1)=0$

$\Rightarrow x(x-[x\left]\right)-1(x-[x\left]\right)-3(x-1)=0$

$\Rightarrow (x-1)(x-[x\left]\right)-3(x-1)=0$

$\Rightarrow (x-1)(x-[x]-3)=0$

$\Rightarrow (x-1)\left(\right\{x\}-3)=0\Rightarrow x=1\text{ as }\left\{x\right\}\ne 3$

(1)

Given, $f\left(x\right)=\frac{2^{2x}}{2^{2x}+2}=\frac{4^x}{4^x+2}$

$f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}$

Now, $f\left(x\right)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{(1-x)}}{4^{(1-x)}+2}$

$f\left(x\right)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}=\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)}$

                         $=\frac{4^x}{4^x+2}+\frac{2}{2+4^x}=1$

$f\left(\frac{{\displaystyle 1}}{{\displaystyle 2023}}\right)+f\left(\frac{{\displaystyle 2}}{{\displaystyle 2023}}\right)+\dots +f\left(\frac{{\displaystyle 2022}}{{\displaystyle 2023}}\right)$

$=f\left(\frac{{\displaystyle 1}}{{\displaystyle 2023}}\right)+f\left(\frac{{\displaystyle 2}}{{\displaystyle 2023}}\right)+f\left(\frac{{\displaystyle 3}}{{\displaystyle 2023}}\right) +\dots +f(1-\frac{{\displaystyle 3}}{{\displaystyle 2023}})+f(1-\frac{{\displaystyle 2}}{{\displaystyle 2023}})+f(1-\frac{{\displaystyle 1}}{{\displaystyle 2023}})$

$=1+1+1+\dots +1 \left(1011\text{ times}\right)=1011$

(3)

$f(x+y)=f\left(x\right)·f\left(y\right) \forall x,y\in N$

Put $x=y=1, f\left(2\right)=3^2; \text{Put }x=2,y=1$

$f\left(3\right)=3^3, f\left(4\right)=3^4$ and so on.

Now, $\sum _{k=1}^{n}f\left(k\right)=3279\Rightarrow 3+3^2+3^3+\dots +3^n=3279$

$\Rightarrow \frac{3(3^n-1)}{3-1}=3279\Rightarrow 3^n=2187$   $\therefore$  $n=7$