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

If $f(x) = \cos (\log x)$,then $f(x)f(y) - \frac{1}{2}[f(x/y) + f(xy)] = $

Match the functions given in List-$I$ with their relevant characteristics from List-$II$.

List-$I$	List-$II$
$(A)$ $\sinh x$	$(I)$ Domain is $(-1, 1)$, even function
$(B)$ $\text{sech } x$	$(II)$ Domain is $[1, \infty)$, neither even nor odd function
$(C)$ $\tanh x$	$(III)$ Even function
$(D)$ $\text{cosech}^{-1} x$	$(IV)$ Range is $\mathbb{R}$, odd function
	$(V)$ Range is $(-1, 1)$, odd function

The correct answer is

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:

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 $f: R \rightarrow (0,1)$ be a continuous function. Then,which of the following function$(s)$ has(have) the value zero at some point in the interval $(0,1)$?