The graph of the function $y = f(x)$ is symmetrical about the line $x = 2$, then

Let $f(x) = {(x + 1)^2} - 1,\;\;(x \ge - 1)$. Then the set $S = \{ x:f(x) = {f^{ - 1}}(x)\} $ is

Let $\quad E_1=\left\{x \in R : x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$ and $\quad E_2=\left\{x \in E_1: \sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$.

(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )

Let $f : E _1 \rightarrow R$ be the function defined by $f(x)=\log _c\left(\frac{x}{x-1}\right)$ and $g: E_2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)$

The correct option is:

Let $R$ be the set of all real numbers and $f(x)=\sin ^{10} x\left(\cos ^8 x+\cos ^4 x+\cos ^2 x+1\right)$ $x \in R$. Let  $S=\{\lambda \in R$ there exists a point $c \in(0,2 \pi)$ with $\left.f^{\prime}(c)=\lambda f(c)\right\}$ Then,

Range of $f(x) = \;[x]\; - x$ is

Let $R$ be the set of all real numbers and let $f$ be a function from $R$ to $R$ such that $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$, for all $x \in R$. Then $2 f(0)+3 f(1)$ is equal to