If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then the function $f(x)$ at $x = 0$ is:

If the function $f(x) = \begin{cases} x + a \sqrt{2} \sin x & \text{if } 0 \leq x \leq \frac{\pi}{4} \\ 2x \cot x + b & \text{if } \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous in $[0, \pi]$,then $a - b = $

Let $a, b \in \mathbb{R}$ $(a \neq 0)$. If the function $f$ is defined as $f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \leq x < 1 \\ a, & 1 \leq x < \sqrt{2} \\ \frac{2b^2-4b}{x}, & \sqrt{2} \leq x < \infty \end{cases}$ is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is

If $f(x) = \begin{cases} \sin x, & \text{if } x \leq 0 \\ x^2+a^2, & \text{if } 0 < x < 1 \\ bx+2, & \text{if } 1 \leq x \leq 2 \\ 0, & \text{if } x > 2 \end{cases}$ is continuous on $\mathbb{R}$, then $a+b+ab = $

Examine the continuity of the function $f(x) = 2x^{2} - 1$ at $x = 3$.

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 $f(c) = 0$