In the usual notation,the value of $\Delta \nabla$ is equal to

Let $f_1: R \rightarrow R, f_2:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R, f_3:\left(-1, e^{\frac{\pi}{2}}-2\right) \rightarrow R$ and $f_4: R \rightarrow R$ be functions defined by:
$(i)$ $f_1(x)=\sin \left(\sqrt{1-e^{-x^2}}\right)$
$(ii)$ $f_2(x)=\begin{cases} \frac{|\sin x|}{\tan^{-1} x} & \text{if } x \neq 0 \\ 1 & \text{if } x=0 \end{cases}$,where the inverse trigonometric function $\tan^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$(iii)$ $f_3(x)=\left[\sin \left(\log_e(x+2)\right)\right]$,where,for $t \in R, [t]$ denotes the greatest integer less than or equal to $t$.
$(iv)$ $f_4(x)=\begin{cases} x^2 \sin \left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$

$LIST-I$	$LIST-II$
$P$. The function $f_1$ is	$1$. $NOT$ continuous at $x=0$
$Q$. The function $f_2$ is	$2$. Continuous at $x=0$ and $NOT$ differentiable at $x=0$
$R$. The function $f_3$ is	$3$. Differentiable at $x=0$ and its derivative is $NOT$ continuous at $x=0$
$S$. The function $f_4$ is	$4$. Differentiable at $x=0$ and its derivative is continuous at $x=0$

The correct option is:

Let $f(x) = \begin{cases} x^{3/5} & \text{if } x \le 1 \\ -(x - 2)^3 & \text{if } x > 1 \end{cases}$. Then the number of critical points on the graph of the function is:

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

Which of the following is not true?

Given the following properties of a function $f(x)$:
$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given by the following number line:
$f'(x)$ is positive for $x < -5$,negative for $-5 < x < 2$,positive for $2 < x < 4$,and negative for $x > 4$.
On the possible graph of $y = f(x)$,we have: