If the function $f(x) = \frac{e^{x}(e^{\tan x-x}-1)+\log_{e}(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$, then the value of $f(0)$ is equal to

Let $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\ 2x \cot x+b, & \frac{\pi}{4} \leq x < \frac{\pi}{2} \\ a \cos 2x-b \sin x, & \frac{\pi}{2} \leq x \leq \pi \end{cases}$. If $f(x)$ is continuous for $0 \leq x \leq \pi$,then:

The value of $f(0)$ so that the function $f(x) = \frac{1 - \cos(1 - \cos x)}{x^4}$ is continuous everywhere is

Let $f(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \ge 0 \end{cases}$,then for all values of $x$

Let $f(x) = \frac{g(x)}{h(x)}$,where $g$ and $h$ are continuous functions on the open interval $(a, b)$. Which of the following statements is true for $a < x < b$?

If $f(x) = \begin{cases} x - 1, & x < 0 \\ \frac{1}{4}, & x = 0 \\ x^2, & x > 0 \end{cases}$,then