$f\left(x\right)=ln\left(\frac{x^2+e}{x^2+1}\right)=ln\left(\frac{x^2+1-1+e}{x^2+1}\right)=ln(1+\frac{e-1}{x^2+1})$
$0<\frac{e-1}{x^2+1}\le (e-1)\Rightarrow 1<(1+\frac{e-1}{x^2+1})\le e\Rightarrow 0<ln\left(\frac{x^2+e}{x^2+1}\right)\le 1$
Hence range is $(0,1]$
 Hence (B) is correct answer.

$f\left(x\right)=x^2+1+\frac{1}{x^2+1}-1x^2+1+\frac{1}{x^2+1}\ge 2[\because AM\ge GM]x^2+\frac{1}{x^2+1}\ge 1\therefore f\left(x\right)ϵ[1,\infty )$

$2f\left(x\right)+f(1-x)=x^2....\left(1\right)$

Replacing x with (1-x), we get

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

$2\times \left(1\right)-\left(2\right)\Rightarrow 3f\left(x\right)=2x^2-(1-x{)}^2$

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

Every relation from A to B is a subset of A $\times$ B. Every relation from A to B is an element of  P(A $\times$ B).

$n(A\times B)=5\times 2=10$

Hence, the total number of relations,

$N=2^{10}=1024$

$f\left(x\right)=\frac{tan\left(\pi [x^2-x]\right)}{1+sin\left(cosx\right)}=\left\{0\right\}\text{because of}[x^2-x]is\text{integer}.$

$f\left(x\right)=2x^2+bx+c$

$f\left(0\right)=0+0+c=3$

$\Rightarrow c=3$

$f\left(2\right)=2\times 2^2+2b+3=1$

$\Rightarrow 11+2b=1\Rightarrow b=-5$

$f\left(1\right)=2-5+3=0$

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

$x+1\ge 0\Rightarrow x\ge -1$

$Also,2x-3\ne 1$

$\Rightarrow x\ne \frac{3}{2}$

Let $y=f\left(x\right)=\frac{x^2+x+2}{x^2+x+1};xϵR$
$\therefore y=\frac{x^2+x+2}{x^2+x+1},$
$y=1+\frac{1}{x^2+x+1}[i.ey>1]$             . . . (i)
$\Rightarrow y{x}^2+yx+y=x^2+x+2\Rightarrow x^2(y-1)+x(y-1)+(y-2)=0,\because xϵR\Rightarrow \text{D}\ge 0\Rightarrow (y-1{)}^2-4(y-1\left)\right(y-2)\ge 0\Rightarrow (y-1\left)\right\{(y-1)-4(y-2)\}\ge 0\Rightarrow (y-1\left)\right(-3y+7)\ge 0$
$\Rightarrow 1\le y\le \frac{7}{3}$                          . . . (ii)
From (i) and (ii), we get
$1<y\le \frac{7}{3}$

option (a) is not a function as 2 has multiple images and 3 has no image.

option (b) is a function from B to A.

option (c) is not a function as 2 has multiple images.

option (d) is a function from A to B.

$|x-3|=\{\begin{array}{cc}x-3,& x\ge 3\\ 3-x,& x<3\end{array}$

$\therefore \frac{|x-3|}{x-3}=1>0,x>3$

The function is not defined at x = 3.