Q57 / 150 · MHT CET 2024 (15 May Shift 1) · Mathematics

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

  • A
    only two points
  • Correct answer
    B
    only three points
  • C
    four or more points
  • D
    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 $x=0,1,3$.

Hence, the correct answer is option: 'B'.

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