Q104 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

In the interval $[0,3]$, the function $f(x)=|x-1|+|x-2|$ is

  • A
    Discontinuous
  • Correct answer
    B
    Differentiable
  • C
    Continuous but not differentiable at $x=2$ only
  • D
    Continuous but not differentiable at $x=1$ and $x=2$

Solution

$f(x)=|x-1|+|x-2|, x \in[0,3]$ $\begin{array}{l} f(x)=\left\{\begin{array}{cc} -2 x+3, & x \leq 1 \\ 1, & 1 \lt x \lt 2 \\ 2 x-3, & x \geq 2 \end{array}\right. \\ \lim _{x \rightarrow 1^{-}} f(x)=1 \text { and } \lim _{x \rightarrow 1^{+}} f(x)=1, f(1)=1 \\ \Rightarrow f(x) \text { is continous at } x=1 \end{array}$ $\therefore f(x)$ is not differentiable at $x=1$ Similarly, $\lim _{x \rightarrow 2^{+}} f^{\prime}(x)=2, \lim _{x \rightarrow 2^{-}} f^{\prime}(x)=0$ $\therefore f(x)$ is not differentiable at $x=2$

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

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