MCQ
The function $y = {e^{ - |x|}}$ is
- AContinuous and differentiable at $x = 0$
- BNeither continuous nor differentiable at $x = 0$
- ✓Continuous but not differentiable at $x = 0$
- DNot continuous but differentiable at $x = 0$
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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| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| $P(X)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |