If $f(x) = \begin{cases} \frac{(8-2x)^{\frac{1}{3}} - 2}{3 - (243+5x)^{\frac{1}{5}}}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$, then $k =$

Consider the function $f(x) = \frac{x^3}{4} - \sin(\pi x) + 3$. Which of the following statements is true regarding the values attained by $f(x)$ in the interval $[-2, 2]$?

Let $f(x) = \begin{cases} \frac{(1 + \tan x)^{\frac{1}{x}} - e}{x} & x \neq 0 \\ k & x = 0 \end{cases}$ be continuous at $x = 0$,then the value of $k$ is:

If the function $f(x)$ is defined as:
$f(x) = \begin{cases} 1 + \sin \frac{\pi x}{2}, & -\infty < x \leq 1 \\ ax + b, & 1 < x < 3 \\ 6 \tan \frac{x \pi}{12}, & 3 \leq x < 6 \end{cases}$
and is continuous in $(-\infty, 6)$,then the values of $a$ and $b$ are respectively.

Examine the following function for continuity: $f(x) = \frac{1}{x-5}, x \neq 5$.

If $f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & \text{for } -1 \leq x < 0 \\ 2x^2+3x-2, & \text{for } 0 \leq x \leq 1 \end{cases}$ is continuous at $x=0$,then $k$ is equal to