Q54 / 120 · VITEEE 2018 · Mathematics

$f(x)=\left\{\begin{array}{cc}x \sin 1 / x & , \quad x \neq 0 \\ 0 & , x=0\end{array}\right.$ at $x=0$ is

  • A
    continuous as well as differentiable
  • B
    differentiable but not continuous
  • Correct answer
    C
    continuous but not differentiable
  • D
    neither continuous nor differentiable

Solution

For function to be continuous : $$ \begin{array}{l} \mathrm{f}(0+\mathrm{h})=\mathrm{f}(0-\mathrm{h})=\mathrm{f}(0) \\ \mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \sin 1 / \mathrm{h}=0 \times(\text { a finite } \\ \text { quantity })=0 \\ \mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}-\mathrm{h} \sin (1 /-\mathrm{h})=0 \times(\text { a finite } \\ \text { quantity })=0 \end{array} $$ Also, $\lim _{x \rightarrow 0} x \sin 1 / x=0 \times($ a finite quantity $)$ $=0$ $\Rightarrow$ function is continuous at $x=0$ For function to be differentiable : $$ \begin{array}{l} \mathrm{f}^{\prime}(0+\mathrm{h})=\mathrm{f}^{\prime}(0-\mathrm{h}) \\ \mathrm{f}^{\prime}(0+\mathrm{h})=\frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}} \\ =\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \sin \frac{1}{\mathrm{~h}}-0}{\mathrm{~h}}=\lim _{\mathrm{h} \rightarrow 0} \sin \left(\frac{1}{\mathrm{~h}}\right) \end{array} $$ which does not exist.

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

Relevant Questions