Q27 / 60 · JEE Advanced 2012 (Paper 1) · Mathematics

Let $\int(x)=\left\{\begin{array}{r}x^{2}\left|\cos \frac{\pi}{x}\right|, \quad x \neq 0 \\ 0, \quad x=0\end{array}, x \in R\right.$ then $\int$ is

  • A
    differentiable both at $x=0$ and at $x=2$
  • Correct answer
    B
    differentiable at $x=0$ but not differentiable at $x=2$
  • C
    not differentiable at $x=0$ but differentiable at $x=2$
  • D
    differentiable neither at $x=0$ nor at $x=2$

Solution

$f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$ $\begin{aligned} =\lim _{h \rightarrow 0} \frac{h^{2}\left|\cos \frac{\pi}{h}\right|}{h} &=\lim _{h \rightarrow 0} h\left|\cos \frac{\pi}{h}\right| \\ &=0 \times \text { some finite value }=0 \end{aligned}$ $\begin{array}{c} =0 \times \text { some finite value }=0 \\ \text { and } f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{h^{2}\left|\cos \frac{\pi}{-h}\right|}{-h} \\ =\lim _{h \rightarrow 0}-h\left|\cos \frac{\pi}{h}\right|=0 \times \text { some finite value }=0 \end{array}$ $\begin{array}{l} \because f^{\prime}\left(0^{+}\right)=f^{\prime}\left(0^{-}\right) \therefore f \text { is differentiable at } x=0 \\ \text { Now } f^{\prime}\left(2^{+}\right)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h} \\ \quad=\lim _{h \rightarrow 0} \frac{(2+h)^{2}\left|\cos \frac{\pi}{2+h}\right|-4\left|\cos \frac{\pi}{2}\right|}{h} \end{array}$ $=\lim _{h \rightarrow 0} \frac{(2+h)^{2}\left(\cos \frac{\pi}{2+h}\right)}{h}$ $=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \sin \left(\frac{\pi}{2}-\frac{\pi}{2+h}\right)$ $=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \sin \left(\frac{\pi h}{2(2+h)}\right)$ $=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \times \frac{\sin \left(\frac{\pi h}{2(2+h)}\right)}{\left(\frac{\pi h}{2(2+h)}\right)} \times \frac{\pi h}{2(2+h)}=\pi$ and $f^{\prime}\left(2^{-}\right)=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$ $=\lim _{h \rightarrow 0} \frac{(2-h)^{2}\left|\cos \left(\frac{\pi}{2-h}\right)\right|-0}{-h}$ $=\lim _{h \rightarrow 0} \frac{-(2-h)^{2} \cos \left(\frac{\pi}{2-h}\right)}{-h}$ $=\lim _{h \rightarrow 0} \frac{(2-h)^{2} \sin \left(\frac{\pi}{2}-\frac{\pi}{2-h}\right)}{h}$ $=\lim _{h \rightarrow 0} \frac{(2-h)^{2}}{h} \times \frac{\sin \left(\frac{-\pi h}{2(2-h)}\right)}{\left(\frac{-\pi h}{2(2-h)}\right)} \times\left(\frac{-\pi h}{2(2-h)}\right)=-\pi$ $\because f^{\prime}\left(2^{+}\right) \neq f^{\prime}\left(2^{-}\right), \therefore f$ is not differentiable at $x=2 .$

Hence, the correct answer is option: 'B'.

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