Continuity And Differentiability Question 1
Question 1 - 24 January - Shift 1
Let $[ f(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} ]$ Then at ( x = 0 )
(1) $f$ is continuous but not differentiable
(2) $f$ is continuous but $f^{\prime}$ is not continuous
(3) $f$ and $f^{\prime}$ both are continuous
(4) $f^{\prime}$ is continuous but not differentiable
Show Answer
Answer: (2)
Solution:
Formula: Continuity of a Function at a Point, Differentiability at a point
Continuity of $f(x): f(0^{+})=h^{2} \cdot \sin \frac{1}{h}=0$
$f(0^{-})=(-h)^{2} \cdot \sin (\frac{-1}{h})=0$
$f(0)=0$
$f(x)$ is continuous
$f^{\prime}(0^{+})=\lim _{h \to 0} \frac{f(0+h)-f(0)}{h}=\frac{h^{2} \cdot \sin (\frac{1}{h})-0}{h}=0$
$f^{\prime}(0^{-})=\lim _{h \to 0} \frac{f(0-h)-f(0)}{-h}=\frac{h^{2} \cdot \sin (\frac{1}{-h})-0}{-h}=0$
$f(x)$ is differentiable.
$f^{\prime}(x)=2 x \cdot \sin (\frac{1}{x})+x^{2} \cdot \cos (\frac{1}{x}) \cdot \frac{-1}{x^{2}}$
$f^{\prime}(x)=\begin{array}{cc} 2 x \cdot \sin (\frac{1}{x})-\cos (\frac{1}{x}), & x \neq 0 \\ 0, & x=0\end{array}$
$\Rightarrow f^{\prime}(x)$ is not continuous (as $\cos (\frac{1}{x})$ is highly)
