The derivative of $f(x)=\cos ^{-1}\left[\sin \sqrt{\frac{1+x}{2}}\right]+x^x$ with respect to $x$ at $x=1$ is equal to

If $f(x) = \begin{cases} x^{3}-3x+2, & x < 2 \\ x^{3}-6x^{2}+9x+2, & x \geq 2 \end{cases}$, then:

Let $f(x) = x|x|$,$g(x) = \sin x$ and $h(x) = (g \circ f)(x)$. Then

Let $y(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$. Then $y'(x) - y''(x)$ at $x = -1$ is equal to:

Consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \begin{cases} (2 - \sin(\frac{1}{x}))|x|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f$ is

Let $[x]$ denote the greatest integer function,and let $m$ and $n$ respectively be the numbers of the points,where the function $f(x) = [x] + |x - 2|$,$-2 < x < 3$,is not continuous and not differentiable. Then $m + n$ is equal to: