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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
Let $f(x)=\left\{\begin{array}{cl}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{array} ;\right.$ Then at $x=0$
  • A
    $f$ is continuous but not differentiable
  • ✓
    $f$ is continuous but $f ^{\prime}$ is not continuous
  • C
    $f$ and $f$ ' both are continuous
  • D
    $f ^{\prime}$ is continuous but not differentiable
✓

Answer

Correct option: B.
$f$ is continuous but $f ^{\prime}$ is not continuous
b
$\text { Continuity of } f(x): f\left(0^{+}\right)=h^2 \cdot \sin \frac{1}{h}=0$

$f\left(0^{-}\right)=(-h)^2 \cdot \sin \left(\frac{-1}{h}\right)=0$

$f(0)=0$

$f(x)$ is continuous

$f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\frac{h^2 \cdot \sin \left(\frac{1}{h}\right)-0}{h}=0$

$f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\frac{h^2 \cdot \sin \left(\frac{1}{-h}\right)-0}{-h}=0$

$f(x)$ is differentiable.

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

$\Rightarrow f^{\prime}(x)$ is not continuous (as $\cos \left(\frac{1}{x}\right)$ is highly oscillating at $x =0$ )

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