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Consider the following in respect of the function [2016-II]
$f(x)=\left\{\begin{array}{l}2+x, \quad x \geq 0 \\ 2-x, x < 0\end{array}\right.$
1- $\operatorname{limf}_{x \rightarrow 1}(x)$ does not exist.
2- $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$
3- $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ Which of the above statements is/are correct?
Options:
$f(x)=\left\{\begin{array}{l}2+x, \quad x \geq 0 \\ 2-x, x < 0\end{array}\right.$
1- $\operatorname{limf}_{x \rightarrow 1}(x)$ does not exist.
2- $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$
3- $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ Which of the above statements is/are correct?
Solution:
1295 Upvotes
Verified Answer
The correct answer is:
3 only
For $x \geq 1$
$\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} 2+x=2+1=3$
For $x < 1$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 2+x=2+1=3$
So, $\lim _{x \rightarrow 1} f(x)$ exist.
At $x=0$
RHL: $\lim _{h \rightarrow 0^{+}} f(0+h)=\lim _{h \rightarrow 0} 2+h=2$
LHL $\lim _{h \rightarrow 0^{-}} f(0-h)=\lim _{h \rightarrow 0} 2-h=2$
$f(0)=2+0=2 .$
So, $\mathrm{RHL}=\mathrm{LHL}=f(0)$
$\Rightarrow f(x)$ is continuous at $x=0$ Differentiability at $x=0$
LHD: $\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0^{-}} \frac{2+h-2}{-h}$
$=\frac{-h}{h}=-1$
RHD : $\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0^{+}} \frac{2+h-2}{h}=1$
Since $\mathrm{LHD} \neq \mathrm{RHD}$
So, $f(x)$ is not differentiable at $x=0$.
$\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} 2+x=2+1=3$
For $x < 1$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 2+x=2+1=3$
So, $\lim _{x \rightarrow 1} f(x)$ exist.
At $x=0$
RHL: $\lim _{h \rightarrow 0^{+}} f(0+h)=\lim _{h \rightarrow 0} 2+h=2$
LHL $\lim _{h \rightarrow 0^{-}} f(0-h)=\lim _{h \rightarrow 0} 2-h=2$
$f(0)=2+0=2 .$
So, $\mathrm{RHL}=\mathrm{LHL}=f(0)$
$\Rightarrow f(x)$ is continuous at $x=0$ Differentiability at $x=0$
LHD: $\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0^{-}} \frac{2+h-2}{-h}$
$=\frac{-h}{h}=-1$
RHD : $\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0^{+}} \frac{2+h-2}{h}=1$
Since $\mathrm{LHD} \neq \mathrm{RHD}$
So, $f(x)$ is not differentiable at $x=0$.
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