If $f : R \rightarrow R$ is defined by $f(x) = x^{2} - 3x + 2$,find $f(f(x))$.

$f: R \rightarrow R$ and $g:[0, \infty) \rightarrow R$ are defined by $f(x)=x^2$ and $g(x)=\sqrt{x}$. Which one of the following is not true?

Let $R$ be a relation '$ < $' from $A$ to $B$,where $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$ such that $(a, b) \in R \iff a < b$. Then $R \circ R^{-1}$ is:

For $x \in R - \{0, 1\}$,let ${f_1}(x) = \frac{1}{x}$,${f_2}(x) = 1 - x$,and ${f_3}(x) = \frac{1}{1 - x}$ be three given functions. If a function $J(x)$ satisfies $(f_2 \circ J \circ f_1)(x) = f_3(x)$,then $J(x)$ is equal to:

Let $f(x) = \sin \left(\frac{\pi}{6} \sin \left(\frac{\pi}{2} \sin x\right)\right)$ for all $x \in R$ and $g(x) = \frac{\pi}{2} \sin x$ for all $x \in R$. Let $(f \circ g)(x)$ denote $f(g(x))$ and $(g \circ f)(x)$ denote $g(f(x))$. Then which of the following is (are) true?
$(A)$ Range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(B)$ Range of $f \circ g$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(C)$ $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)} = \frac{\pi}{6}$
$(D)$ There is an $x \in R$ such that $(g \circ f)(x) = 1$

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x > 0 \end{cases}$. Then for all values of $x$,the value of $f(g(x))$ is: