Q95 / 180 · KCET 2010 · Mathematics

The function $f(x)=|x-2|+x$ is

  • A
    differentiable at both $x=2$ and $x=0$
  • B
    differentiable at $x=2$ but not at $x=0$
  • C
    continuous at $\mathrm{x}=2$ but not at $\mathrm{x}=0$
  • Correct answer
    D
    continuous at both $\mathrm{x}=2$ and $\mathrm{x}=0$

Solution

$$ f(x)=|x-2|+x $$ First we check the continuity At $\mathrm{x}=0$, $\begin{aligned} \text { RHL } f(0+h) &=\lim _{h \rightarrow 0}|0+h-2|+(0+h) \\ &=|-2|=2 \\ \text { LHL } f(0-h) &=\lim _{h \rightarrow 0}|0-h-2|+(0-h) \\ &=|-2|=2 \end{aligned}$ At $\mathrm{x}=2$, $$ \begin{aligned} \operatorname{RHL} \mathrm{f}(2+\mathrm{h}) &=\lim _{\mathrm{h} \rightarrow 0}|2+\mathrm{h}-2|+(2+\mathrm{h}) \\ &=0+2+0=2 \end{aligned} $$ LHL $f(2-h)=\lim _{h \rightarrow 0}|2-h-2|+(2+h)$ $$ =0+2-0=2 $$ and $\quad f(0)=2, f(2)=2$ Hence, $f(x)$, is continuous at $x=0,2$ Now, we check differentiability $$ f(x)=\left\{\begin{array}{c} (-x+2-x), x < 0 \\ (-x+2+x), 0 \leq x \leq 2 \\ (x-2+2), 2 \leq x \end{array}\right. $$ $$ f(x)=\left\{\begin{array}{c} 2-2 x, x < 0 \\ 2,0 \leq x \leq 2 \\ 2 x-2,2 \leq x \end{array}\right. $$ Now,$f^{\prime}(x)=\left\{\begin{array}{c}-2, x < 0 \\ 0,0 \leq x \leq 2 \\ 2,2 \leq x\end{array}\right.$ Hence, $f(x)$ is not differentiable at $x=0,2$.

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

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