Let $f(x)$ be a polynomial of degree $6$ in $x$,in which the coefficient of $x^{6}$ is unity and it has extrema at $x=-1$ and $x=1$. If $\lim_{x \rightarrow 0} \frac{f(x)}{x^{3}}=1$,then $5 \cdot f(2)$ is equal to .............

The value of $f(4)-f(3)$ is

Let $k$ and $m$ be positive real numbers such that the function $f(x) = \begin{cases} 3x^2 + k\sqrt{x+1}, & 0 < x < 1 \\ mx^2 + k^2, & x \geq 1 \end{cases}$ is differentiable for all $x > 0$. Then $\frac{8f'(8)}{f'(\frac{1}{8})}$ is equal to $.............$.

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

Match the items of List-$I$ with those of List-$II$.

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+|x-1|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

In the following $[x]$ denotes the greatest integer less than or equal to $x$. Match the functions in Column $I$ with the properties in Column $II$.

Column $I$	Column $II$
$(A)$ $f(x) = x|x|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{|x|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = |x - 1| + |x + 1|$	$(s)$ not differentiable at least at one point in $(-1, 1)$