Let $\mathrm{f}:[-1,3] \rightarrow \mathbb{R}$ be defined as $\begin{cases}|x|+[x], & -1 \leqslant x \lt 1…
Let $\mathrm{f}:[-1,3] \rightarrow \mathbb{R}$ be defined as
$\begin{cases}|x|+[x], & -1 \leqslant x \lt 1 \\ x+|x|, & 1 \leqslant x \lt 2 \\ x+[x], & 2 \leqslant x \leqslant 3\end{cases}$
where $[t]$ denotes the greatest integer function.
Then f is discontinuous at
only two points
only three points
four or more points
only one point
Solution
$\begin{aligned}
& \mathrm{f}(x)=\left\{\begin{array}{cc}
|x|+[x], & -1 \leq x \lt 1 \\
x+|x|, & 1 \leq x \lt 2 \\
x+[x], & 2 \leq x \leq 3
\end{array}\right. \\
& \therefore \quad \mathrm{f}(x)=\left\{\begin{array}{cc}
-(x+1), & -1 \leq x \lt 0 \\
x, & 0 \leq x \lt 1 \\
2 x, & 1 \leq x \lt 2 \\
x+2, & 2 \leq x \lt 3 \\
x+3, & x=3
\end{array}\right.
\end{aligned}$
$\therefore \quad \mathrm{f}(x)$ is discontinuous at $x=0,1,3$.