Question

Consider the function $f\left(x\right)=\{\begin{array}{cc}{x}^m\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}$ then -

• A. $f\left(x\right)$ is continuous when $m<0$ as well as differentiable if $m<0$
• B. $f\left(x\right)$ is continuous when $m>0$ as well as defferentiable if $0<m<1$
• C. $f\left(x\right)$ is continuous when $m<0$ as well as differentiable if $m>0$
• D. $f\left(x\right)$ is continuous when $m>0$ as well as differentiable if $m>1$.

Answer 1

The correct option is D $f\left(x\right)$ is continuous when $m>0$ as well as differentiable if $m>1$.

${lim}_{x\to 0}f\left(x\right)=f\left(0\right)$
${lim}_{x\to 0}{x}^{m}sin\frac{1}{x}=0$
Which is possible only when $m>0$
Hence $f\left(x\right)$ is continuous when $m>0$
Now , for differentiability
${lim}_{x\to 0}\frac{f\left(x\right)-f\left(0\right)}{x}$ must exist finitely
$i.e{lim}_{x\to 0}\frac{x^{m}sin\frac{1}{x}-0}{x}$
${lim}_{x\to 0}{x}^{m-1}sin\frac{1}{x}$ must exist finitely
Which is possible if $m-1>0$ i.e. $m>1$