If $f(x) = \frac{1}{\sqrt{x + 2\sqrt{2x - 4}}} + \frac{1}{\sqrt{x - 2\sqrt{2x - 4}}}$ for $x > 2$,then $f(11) = $

The number of solutions of the equation $|x^2 - 2|x|| = 2^x$ is:

Let $f_1: R \rightarrow R$,$f_2:[0, \infty) \rightarrow R$,$f_3: R \rightarrow R$ and $f_4: R \rightarrow [0, \infty)$ be defined by:
$f_1(x) = \begin{cases} |x| & \text{if } x < 0 \\ e^x & \text{if } x \geq 0 \end{cases}$
$f_2(x) = x^2$
$f_3(x) = \begin{cases} \sin x & \text{if } x < 0 \\ x & \text{if } x \geq 0 \end{cases}$ and
$f_4(x) = \begin{cases} f_2(f_1(x)) & \text{if } x < 0 \\ f_2(f_1(x)) - 1 & \text{if } x \geq 0 \end{cases}$

List $I$	List $II$
$P. f_4$ is	$1. \text{onto but not one-one}$
$Q. f_3$ is	$2. \text{neither continuous nor one-one}$
$R. f_2 \circ f_1$ is	$3. \text{differentiable but not one-one}$
$S. f_2$ is	$4. \text{continuous and one-one}$

Codes: $P \quad Q \quad R \quad S$

$f: R \rightarrow R$ is a function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$. Assertion $(A): f(c) = \frac{1}{3}$ for some values of $c \in R$. Reason $(R): 0 < f(x) \leq \frac{1}{2\sqrt{2}}$ for all $x \in R$. Then which of the following options is correct?

Let $[x]$ denote the integral part of $x \in R$. Let $g(x) = x - [x]$. Let $f(x)$ be any continuous function with $f(0) = f(1)$. Then the function $h(x) = f(g(x))$:

Consider a function $f: R \to R$ such that $f(x + a) = \frac{1}{2} + \sqrt{f(x) - f^2(x)}$,where $a$ is a real constant. Then $f(x)$ must be