If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$,then $f'\left( \frac{\pi}{4} \right)$ is

Let $f : (0, \pi) \rightarrow \mathbb{R}$ be a twice differentiable function such that $\lim _{t \rightarrow x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x$ for all $x \in (0, \pi)$. If $f \left(\frac{\pi}{6}\right) = -\frac{\pi}{12}$,then which of the following statement$(s)$ is (are) $TRUE$?
$(A) f \left(\frac{\pi}{4}\right) = \frac{\pi}{4 \sqrt{2}}$
$(B) f(x) < \frac{x^4}{6} - x^2$ for all $x \in (0, \pi)$
$(C)$ There exists $\alpha \in (0, \pi)$ such that $f^{\prime}(\alpha) = 0$
$(D) f^{\prime \prime}\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = 0$

$A$ function $f(x) = \begin{cases} 1 + x, & x \le 2 \\ 5 - x, & x > 2 \end{cases}$ is

Let $f: R \rightarrow R$ be a function defined by $f(x) = \begin{cases} \max_{t \leq x} \{t^3 - 3t\} & x \leq 2 \\ x^2 + 2x - 6 & 2 < x < 3 \\ [x-3] + 9 & 3 \leq x \leq 5 \\ 2x + 1 & x > 5 \end{cases}$ where $[t]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I = \int_{-2}^{2} f(x) dx$. Then the ordered pair $(m, I)$ is equal to:

Match the items given in List-$I$ with those of the items of List-$II$:

List-$I$	List-$II$
$a$. If $y=|x|+|x-2|$, then at $x=2, \frac{dy}{dx}=$	$i$. $2$
$b$. If $f(x)=|\cos 2x|$, then $f^{\prime}(\frac{\pi}{4}+)=$	$ii$. $0$
$c$. If $f(x)=\sin(\pi[x])$, where $[\cdot]$ denotes the greatest integer function, then $f^{\prime}(1-)=$	$iii$. $-2$
$d$. If $f(x)=\log|x-1|, x \neq 1$, then $f^{\prime}(\frac{1}{2})=$	$iv$. Does not exist

Let $f : R \to R$ be a differentiable function satisfying $f''(3) + f'(2) = 0$. Then $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + f\left( {3 + x} \right) - f\left( 3 \right)}}{{1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)^{\frac{1}{x}}}$ is equal to