Q67 / 170 · Manipal (MET) 2019 · Mathematics
Let $f(x)=x^p \cos \left(\frac{1}{x}\right)$, when $x \neq 0$ and $f(x)=0$, when $x=0$. Then $f(x)$ will be differentiable at $x=0$, if
- A$p\gt0$
- Correct answerB$p\gt1$
- C$0 \lt p \lt 1$
- D$\frac{1}{2} \lt p \lt 1$
Solution
Given that, $f(x)=\left\{\begin{aligned} x^p \cos \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{aligned}\right.$
$f(x)$ will be differentiable at $x=0$, if $\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$ exists finitely.
$\Rightarrow \quad \lim _{x \rightarrow 0} \frac{x^p \cos \left(\frac{1}{x}\right)}{x}$ exists finitely
$\Rightarrow \lim _{x \rightarrow 0} x^{p-1} \cos \left(\frac{1}{x}\right)$ exists finitely
$\Rightarrow \quad p-1\gt0 \Rightarrow p\gt1$
Hence, the correct answer is option: 'B'.