The function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$

$f(x) = \begin{cases} \frac{\sin(x-[x])}{x-[x]} & , x \in (-2, -1) \\ \max \{2x, 3[|x|]\} & , |x| < 1 \\ 1 & , \text{otherwise} \end{cases}$ where $[t]$ denotes the greatest integer $\leq t$. If $m$ is the number of points where $f$ is not continuous and $n$ is the number of points where $f$ is not differentiable,then the ordered pair $(m, n)$ is

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

Let $f: R \rightarrow (0, \infty)$ be a twice differentiable function such that $f(3) = 18$, $f'(3) = 0$, and $f''(3) = 4$. Then $\lim_{x \rightarrow 1} \left( \log_{e} \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-1)^{2}}} \right)$ is equal to:

Match the following:
In the following,$[x]$ denotes the greatest integer less than or equal to $x$.

$(a)$ $x|x|$	$(i)$ continuous in $(-1, 1)$
$(b)$ $\sqrt{|x|}$	$(ii)$ differentiable in $(-1, 1)$
$(c)$ $x+[x]$	$(iii)$ strictly increasing in $(-1, 1)$
$(d)$ $|x-1|+|x+1|$	$(iv)$ not differentiable at,at least one point in $(-1, 1)$

If $f(x)=\sin \left(\cos ^{-1}\left(\frac{1-2^{2 x}}{1+2^{2 x}}\right)\right)$ and its first derivative with respect to $x$ is $-\frac{ b }{ a } \log _{ e } 2$ when $x =1,$ where $a$ and $b$ are integers,then the minimum value of $\left| a ^{2}- b ^{2}\right|$ is.........