Continuity And Differentiability Question 1

Question 1 - 24 January - Shift 1

Let $f(x)=\left{\begin{array}{cc}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{array} ;\right.$ 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.

$f^{\prime}(x)=2 x \cdot \sin (\frac{1}{x})+x^{2} \cdot \cos (\frac{1}{x}) \cdot \frac{-1}{x^{2}}$

$f^{\prime}(x)=\left{\begin{array}{cc} 2 x \cdot \sin (\frac{1}{x})-\cos (\frac{1}{x}), & x \neq 0 \\ 0, & x=0\end{array} ;\right.$

$\Rightarrow f^{\prime}(x)$ is not continuous (as $\cos (\frac{1}{x})$ is highly