For $\theta \in [0, \pi]$,let $f(\theta) = \sin(\cos \theta)$ and $g(\theta) = \cos(\sin \theta)$. Let $a = \max_{0 \leq \theta \leq \pi} f(\theta)$,$b = \min_{0 \leq \theta \leq \pi} f(\theta)$,$c = \max_{0 \leq \theta \leq \pi} g(\theta)$,and $d = \min_{0 \leq \theta \leq \pi} g(\theta)$. The correct inequalities satisfied by $a, b, c, d$ are:

The graph of the function $f(x) = x + \frac{1}{8} \sin(2 \pi x)$,$0 \leq x \leq 1$ is shown below. Define $f_1(x) = f(x)$,$f_{n+1}(x) = f(f_n(x))$,for $n \geq 1$.
Which of the following statements are true?
$I.$ There are infinitely many $x \in [0, 1]$ for which $\lim_{n \rightarrow \infty} f_n(x) = 0$
$II.$ There are infinitely many $x \in [0, 1]$ for which $\lim_{n \rightarrow \infty} f_n(x) = \frac{1}{2}$
$III.$ There are infinitely many $x \in [0, 1]$ for which $\lim_{n \rightarrow \infty} f_n(x) = 1$
$IV.$ There are infinitely many $x \in [0, 1]$ for which $\lim_{n \rightarrow \infty} f_n(x)$ does not exist.

Suppose $f:[-2, 2] \to R$ is defined by $f(x) = \begin{cases} -1 & \text{for } -2 \le x \le 0 \\ x - 1 & \text{for } 0 < x \le 2 \end{cases}$,then find the set $\{ x \in (-2, 2) : x \le 0 \text{ and } f(|x|) = x \}$.

Consider the following statements:
$(i)$ $A$ relation is a special case of a function.
$(ii)$ $A$ function is a special case of a relation.
$(iii)$ Both relation and function are the same.

Let $f:[0,1] \rightarrow \mathbb{R}$ be an injective continuous function that satisfies the condition $-1 < f(0) < f(1) < 1$. Then,the number of functions $g:[-1,1] \rightarrow [0,1]$ such that $(g \circ f)(x) = x$ for all $x \in [0,1]$ is

If $f(x) = \log_{e}\left(\frac{1-x}{1+x}\right)$,$|x| < 1$,then $f\left(\frac{2x}{1+x^2}\right)$ is equal to