Question

If $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{x}^p\cos \frac{1}{x},& x\ne 0\end{array}\\ \begin{array}{cc}0,& x=0\end{array}\end{array}$, then at $x=0,f\left(x\right)$ is

• A. continuous if $p>0$
• B. differentiable if $p>1$
• C. continuous if $p>1$
• D. differentiable if $p>0$

Answer 1

Answer: (A), (B)

The correct options are
A continuous if $p>0$
B differentiable if $p>1$

Continuity at $x=0$

$L.H.L.=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}{(-h)}^p\cos \frac{1}{h}=0$ if $p>0$

$R.H.L.=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}{h}^p\cos \frac{1}{h}=0$ if $p>0$

and,$f\left(0\right)=0.$

$\therefore f\left(x\right)$ is continuous at $x=0$ if $p>0$

$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$$=\underset{h\to 0}{lim}\frac{{(-h)}^p\cos \frac{1}{h}-0}{-h}$

$=\underset{h\to 0}{lim}{(-h)}^{p-1}\cos \frac{1}{h}=0$ if $p>0,$

and, $R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$$=\underset{h\to 0}{lim}\frac{h^p\cos \frac{1}{h}-0}{h}$

$=\underset{h\to 0}{lim}{h}^{p-1}\cos \frac{1}{h}=0$ if $p>0$

$\therefore f\left(x\right)$ is differentiable at $x=0$ if $p>1.$