Question

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

• A. $p<0$
• B. $0<p<1$
• C. $p=1$
• D. $p>1$

Answer 1

Answer: (D)

Given, $f\left(x\right)=\{\begin{array}{l}{x}^p\cos \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}$
Since, $f\left(x\right)$ is differentiable at $x=0$ , therefore it is continuous at $x=0$
$\therefore \underset{x\to 0}{\lim }f(x)=f(0)=0$
$\Rightarrow \underset{x\to 0}{\lim }{x}^p\cos \left(\frac{1}{x}\right)=0\Rightarrow p>0$
As $f(x)$ is differentiable at $x=0$
$\therefore \underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x-0}$ exists finitely
$\Rightarrow \underset{x\to 0}{\lim }\frac{x^p\cos \frac{1}{x}-0}{x}$ exists finitely
$\Rightarrow \underset{x\to 0}{\lim }{x}^{p-1}\cos \frac{1}{x}$ exists finitely
$\Rightarrow p-1>0\Rightarrow p>1$