If $f(x)=\left\{\begin{array}{cc}x^\alpha \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, &…
If $f(x)=\left\{\begin{array}{cc}x^\alpha \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{array}\right.$; Which of the following is true?
- $f(x)$ is continuous and differentiable if $0 \leq \alpha \lt 1$
- $f(x)$ is discontinuous and not differentiable if $0 \leq \alpha \lt 1$
- $f(x)$ is continuous and differentiable for $\alpha\gt1$
- $f(x)$ is discontinuous and differentiable for $\alpha\gt1$
Solution
Given, $f(x)=\left\{\begin{array}{cc}x^\alpha \sin \left(\frac{1}{x}\right) ; & x \neq 0 \\ 0 ; & x=0\end{array}\right.$
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^\alpha \sin \left(\frac{1}{x}\right)=0 \text { if } \alpha\gt0$
So, $f(x)$ is continuous for $\alpha\gt0$
and $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0} \frac{h^\alpha \sin \left(\frac{1}{h}\right)-0}{h}=\lim _{x \rightarrow 0} h^{\alpha-1} \sin \left(\frac{1}{h}\right)$
So $f^{\prime}(0)$ is finite and exists if $\alpha\gt1$
$\Rightarrow f(x)$ is differentiable at $x=0$ if $\alpha\gt1$.
Asked in: AP EAMCET 2024 (19 May Shift 2)
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