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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 $f(x)=\left\{\begin{array}{ll}x^2\left|\cos \frac{\pi}{x}\right|, & x \neq 0 \\ 0, & x=0\end{array}, x \in I R\right.$, then $f$ is
  • A
    differentiable both at $x=0$ and at $x=2$
  • ✓
    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 $x=2$
✓

Answer

Correct option: B.
differentiable at $x=0$ but not differentiable at $x=2$
b
$(I)$ for derivability at $x=0$

$L.H.D.\quad =f^{\prime}\left(0^{-}\right) $$ =\lim _{h \rightarrow 0^{+}} \frac{f(0-h)-f(0)}{-h} $

$ =\lim _{h \rightarrow 0^{+}} \frac{\left.h^2 \cdot \cos \left(-\frac{\pi}{h}\right) \right\rvert\,-0}{-h} $

$ =\lim _{h \rightarrow 0^{+}}-h .\left|\cos \frac{\pi}{h}\right|=0$

$R H D \quad f^{\prime}\left(0^{+}\right)=\operatorname{\ell im}_{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}$

$=\lim _{h \rightarrow 0^{+}} \frac{h^2 \cdot\left|\cos \left(\frac{\pi}{h}\right)\right|-0}{h}=0$

So $f(x)$ is derivable at $x=0$

$(ii)$ check for derivability at $x=2$

$R H D=f^{\prime}\left(2^{+}\right) $$ =\lim _{h \rightarrow 0^{-}} \frac{f(2+h)-f(2)}{h} $

$ =\lim _{h \rightarrow 0^{-}} \frac{\left.(2+h)^2 \cdot \cos \left(\frac{\pi}{2+h}\right) \right\rvert\,-0}{h} $

$ =\lim _{h \rightarrow 0^{-}} \frac{(2+h)^2 \cdot \cos \left(\frac{\pi}{2+h}\right)}{h}$

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

$=\operatorname{Cim}_{h \rightarrow 0^{-}} \frac{(2+h)^2 \cdot \sin \left(\frac{\pi h}{2(2+h)}\right)}{\left(\frac{\pi}{2(2+h)}\right) h} \cdot \frac{\pi}{2(2+h)} $

$=(2)^2 \cdot \frac{\pi}{2(2)}=\pi $

$\text { LHD }=\underset{h \rightarrow 0^{-}}{ } \frac{f(2-h)-f(2)}{-h} $

$=\lim _{h \rightarrow 0^{+}} \frac{\left.(2-h)^2 \cdot \cos \left(\frac{\pi}{2-h}\right) \right\rvert\,-0}{-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 \cdot \sin \left(\frac{\pi}{2}-\frac{\pi}{2-h}\right)}{h} $

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

$=-\pi $

So $f(x)$ is not derivable at $x=2$

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