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 $x = 0$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f_1(x)=\left\{\begin{array}{lll}|x| & \text { if } & x<0, \\ e^x & \text { if } & x \geq 0 ;\end{array}\right.$
$f_2(x)=x^2$
$f_3(x)=\left\{\begin{array}{ccc}\sin x & \text { if } & x < 0, \\ x & \text { if } & x \geq 0\end{array}\right.$ and
$f_4(x)=\left\{\begin{array}{ccc}f_2\left(f_1(x)\right) & \text { if } & x < 0, \\ f_2\left(f_1(x)\right)-1 & \text { if } & x \geq 0\end{array}\right.$
| List $I$ | List $II$ |
| $P.$ $ f_4$ is | $1.$ onto but not one-one |
| $Q.$ $f_3$ is | $2.$ neither continuous nor one-one |
| $R.$ $f _2 \circ f _1$ is | $3.$ differentiable but not one-one |
| $S.$ $ f_2$ is | $4.$ continuous and one-one |
Codes: $ \quad P \quad Q \quad R \quad S $