Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^{p}sin\frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}$ then f(x) is continuous but not differential at x = 0 if

• A. $0<p\le 1$
• B. $0\le p<\infty$
• C. $-\infty <p<0$
• D. $p=0$

Answer 1

The correct option is A $0<p\le 1$

$f\left(x\right)=x^{p}sin\frac{1}{x},x\ne 0andf\left(x\right)=0,x=0$
Since at x = 0 , f(x) is a continuous function
$\therefore {lim}_{x\to 0}f\left(x\right)=f\left(0\right)=0\Rightarrow {lim}_{x\to 0}{x}^{p}sin\frac{1}{x}=0\to p>0$
f (x) is differentiable at x = 0, if ${lim}_{x\to 0}\frac{f\left(x\right)-f\left(0\right)}{x-0}$ exists
$\Rightarrow {lim}_{x\to 0}\frac{x^{p}sin\frac{1}{x}-0}{x-0}exists$
$\Rightarrow {lim}_{x\to 0}{x}^{p-1}sin\frac{1}{x}exists$
$\Rightarrow p-1>0orp>1$
$Ifp\le 1,then{lim}_{x\to 0}{x}^{p-1}sin\left(\frac{1}{x}\right)$ does not exist and at x = 0 f (x) is not differentiable.
$\therefore$ for $0<p\le 1$ f (x) is a continuous function at x = 0 but not differentiable.