The function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x) = \begin{cases} \frac{1}{x} \log \left(\frac{1+3x}{1-2x}\right), & x \neq 0 \\ k, & x=0 \end{cases}$ is continuous at $x=0$. Then the value of $k$ is:

Let $f(x) = [2x^2 + 1]$ and $g(x) = \begin{cases} 2x - 3, & x < 0 \\ 2x + 3, & x \geq 0 \end{cases}$,where $[t]$ denotes the greatest integer function $\leq t$. Then,in the open interval $(-1, 1)$,the number of points where $f(g(x))$ is discontinuous is equal to:

For what value of $k$ is the function $f(x) = \begin{cases} \frac{\text{log}(1+2x) \sin x^{\circ}}{x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}$ continuous at $x = 0$?

If $f$ is a continuous real-valued function defined on a closed interval $[a, b]$,then the range of the function is . . . . . .

Find the values of $k$ so that the function $f$ is continuous at the indicated point.
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$

Let $f(x) = \min \{x, x^2\}$ for every real number $x$. Then: