If $f(x) = |x|$,then $f(x)$ is

Consider the piecewise defined function $f(x) = \begin{cases} \sqrt{-x} & \text{if } x < 0 \\ 0 & \text{if } 0 \leqslant x \leqslant 4 \\ x - 4 & \text{if } x > 4 \end{cases}$. Choose the answer which best describes the continuity of this function.

Find all points of discontinuity of the function $f$ defined by $f(x) = |x| - |x + 1|$.

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f(x)=x-[x]$,$g(x)=1-x+[x]$,and $h(x)=\min \{f(x), g(x)\}$ for $x \in [-2, 2]$. Then $h$ is :

If $f(x) = \begin{cases} \frac{x - 1}{2}, & 0 \leqslant x < 1 \\ 1/2, & 1 \leqslant x < 2 \end{cases}$ and $g(x) = (2x + 1)(x - k) + 3$ for $0 \leqslant x < \infty$,then $g(f(x))$ will be continuous at $x = 1$ if $k$ is equal to:

The function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1} & x \neq 2 \\ k & x = 2 \end{cases}$ is continuous from the right at the point $x = 2$. Then $k$ is equal to: