If $f(x) = \begin{cases} \frac{x^4 - 16}{x - 2}, & x \neq 2 \\ 16, & x = 2 \end{cases}$,then:

Let $f: R \to R$ be a function defined by $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi \right)$,where $[x]$ denotes the greatest integer function. Then $f$ is:

The function $f(x) = |x| + \frac{|x|}{x}$ is

If $f(x) = \begin{cases} \frac{(e^{ax}-1) \log(1+x)}{\sin^2 x} & \text{if } x > 0 \\ 2 & \text{if } x = 0 \\ \frac{\cos 4x - \cos bx}{\tan^2 x} & \text{if } x < 0 \end{cases}$ is continuous at $x = 0$,then $\sqrt{b^2 - a^2} = $

Let $f$ be a differentiable function on the open interval $(a, b)$. Which of the following statements must be true?
$I$. $f$ is continuous on the closed interval $[a, b]$
$II$. $f$ is bounded on the open interval $(a, b)$
$III$. If $a < a_1 < b_1 < b$,and $f(a_1) < 0 < f(b_1)$,then there is a number $c$ such that $a_1 < c < b_1$ and $f(c) = 0$

Let $f(x) = \begin{cases} \frac{1 + \cos 2\pi x}{1 - \sin \pi x}, & x < \frac{1}{2} \\ p, & x = \frac{1}{2} \\ \frac{\sqrt{2x - 1}}{\sqrt{4 + \sqrt{2x - 1}} - 2}, & x > \frac{1}{2} \end{cases}$. If $f(x)$ is discontinuous at $x = \frac{1}{2}$,then: