$f(x) = \cos x - 1 + \frac{x^2}{2!}, x \in R$. Then $f(x)$ is

If $f''(x)$ is a positive function for all $x \in R$,$f'(3) = 0$ and $g(x) = f(\tan^2 x - 2 \tan x + 4)$ for $0 < x < \frac{\pi}{2}$,then the interval in which $g(x)$ is increasing is

If $f(x) = x e^{x(1 - x)}$,then $f(x)$ is

In which of the following intervals does $f(x) = 2x^3$ increase less rapidly than $g(x) = 9x^2 - 12x + 6$?

Let $f$ be a function defined on $[a, b]$ such that $f^{\prime}(x) > 0$ for all $x \in (a, b)$. Prove that $f$ is an increasing function on $(a, b)$.

For the function $f(x) = \cos x - x + 1, x \in R$,consider the following two statements:
$(S1)$ $f(x) = 0$ for only one value of $x$ in $[0, \pi]$.
$(S2)$ $f(x)$ is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$.