1. Tardigrade
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  4. If f(x)= begincases xp cos((1/x)), x≠ 0 0, x=0 endcases is differentiable at x=0 , then

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Q. If $f\left(x\right)=\begin{cases} x^{p}\cos\left(\frac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{cases}$ is differentiable at $x=0$ , then

NTA AbhyasNTA Abhyas 2022

Solution:

Given, $f\left(x\right)=\begin{cases} x^{p}\cos\left(\frac{1}{x}\right), \, x\neq 0 \\ 0, \, \, \, x=0 \end{cases}$
Since, $f\left(x\right)$ is differentiable at $x=0$ , therefore it is continuous at $x=0$
$\therefore \displaystyle\lim _{x \rightarrow 0} f(x)=f(0)=0 $
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} x^{p} \cos \left(\frac{1}{x}\right)=0 \Rightarrow p>0$
As $f(x)$ is differentiable at $x=0$
$\therefore \displaystyle\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$ exists finitely
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{x^{p} \cos \frac{1}{x}-0}{x}$ exists finitely
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} x^{p-1} \cos \frac{1}{x}$ exists finitely
$\Rightarrow p-1>0 \Rightarrow p>1$