Let $f(x) = |x|$ and $g(x) = |x| + a$,where $a > 0$. For $0 \leq x \leq b$,the set $\{(x, y) \mid g(x) \leq y \leq f(x)\}$ represents all the points in the interior of:

Match the following:

List-$I$	List-$II$
$A$. $\frac{x}{e^x-1} + \frac{x}{2} + 4; x \neq 0$	$I$. is neither odd nor even function
$B$. $\tan^{-1}(\log|x+\sqrt{x^2+1}|), x > 0$	$II$. is an even function
$C$. For $3 < x < 5, |x-2|+|x-3|+|x-5|$	$III$. is an odd function
$D$. $\sin 2x + \sin^2 x + \cos 3x, \forall x \in \mathbb{R}$	$IV$. is the identity function
	$V$. is a constant function

Let $c, k \in R$. If $f(x)=(c+1) x^{2}+(1-c^{2}) x+2 k$ and $f(x+y)=f(x)+f(y)-x y$,for all $x, y \in R$,then the value of $|2(f(1)+f(2)+f(3)+\ldots+f(20))|$ is equal to

Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be two functions defined by $f(x)=\log _{e}(x^{2}+1)-e^{-x}+1$ and $g(x)=\frac{1-2e^{2x}}{e^{x}}$. Then,for which of the following range of $\alpha$,the inequality $f(g(\frac{(\alpha-1)^{2}}{3})) > f(g(\alpha-\frac{5}{3}))$ holds?

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be functions defined by
$f(x)=\begin{cases} x|x| \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{cases}$ and $g(x)=\begin{cases} 1-2x, & 0 \leq x \leq \frac{1}{2} \\ 0, & \text{otherwise} \end{cases}$
Let $a, b, c, d \in R$. Define the function $h: R \rightarrow R$ by
$h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in R$
Match each entry in $List-I$ to the correct entry in $List-II$.

$List-I$	$List-II$
$(P)$ If $a=0, b=1, c=0$ and $d=0$,then	$(1)$ $h$ is one-one
$(Q)$ If $a=1, b=0, c=0$ and $d=0$,then	$(2)$ $h$ is onto
$(R)$ If $a=0, b=0, c=1$ and $d=0$,then	$(3)$ $h$ is differentiable on $R$
$(S)$ If $a=0, b=0, c=0$ and $d=1$,then	$(4)$ the range of $h$ is $[0,1]$
	$(5)$ the range of $h$ is $\{0,1\}$

The correct option is

Let $f(x) = \frac{\sqrt{x - 2\sqrt{x - 1}}}{\sqrt{x - 1} - 1}$. Then: