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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
At $x = 0$, The function $y = {e^{ - |x|}}$ is
  • A
    Continuous and differentiable at $x = 0$
  • B
    Neither continuous nor differentiable at $x = 0$
  • ✓
    Continuous but not differentiable at $x = 0$
  • D
    Not continuous but differentiable at $x = 0$
✓

Answer

Correct option: C.
Continuous but not differentiable at $x = 0$
(c)
We have, $f(x) = \left\{ {\begin{array}{*{20}{c}}{{e^{ - x}},}&{x \ge 0}\\{{e^x},}&{x < 0}\end{array}} \right.$Clearly, $f(x)$ is continuous and differentiable for all non zero $x.$
Now $\mathop {\lim }\limits_{x \to 0 - } f(x) = \mathop {\lim }\limits_{x \to 0} {e^x} = 1$,
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x){e^{ - x}} = 1$
Also, $f(0) = {e^0} = 1$. So, $f(x)$ is continuous for all $x$.
($LHD $ at $x = 0)$ $ = {\left( {\frac{d}{{dx}}({e^x})} \right)_{x = 0}} = 1$
( $RHD $  at $x = 0)$  $ = {\left( {\frac{d}{{dx}}({e^{-x}})} \right)_{x = 0}} = 1$
So, $\mathop {\lim }\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}$ is not differentiable at $L\,f'\,(1) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(1 - h) - f(1)}}{{ - h}}$.
Hence $f(x) = {e^{ - \,|\,x\,|}}$ is everywhere continuous but not differentiable at $x = 0$.

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