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Let $f(x)=\left\{\begin{array}{ll}-2, & -3 \leq x \leq 0 \\ x-2, & 0 < x \leq 3\end{array}\right.$ and $g(x)=f(|x|)+|f(x)|$
Which of the following statements is/are correct?
1- $\mathrm{g}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$.
2- $\mathrm{g}(\mathrm{x})$ is differentiable at $\mathrm{x}=2$. Select the correct answer using the code given below:
Options:
Which of the following statements is/are correct?
1- $\mathrm{g}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$.
2- $\mathrm{g}(\mathrm{x})$ is differentiable at $\mathrm{x}=2$. Select the correct answer using the code given below:
Solution:
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Verified Answer
The correct answer is:
Neither 1 nor 2
$f(x)=\left\{\begin{array}{ll}-2, & -3 \leq x \leq 0 \\ x-2, & 0 < x \leq 3 \text { and }\end{array}\right.$
$g(x)=f(|x|)+|f(x)|$
At $x=0$
For LHD $: g(x)=-2+|-2|=-2+2=0 \Rightarrow g(x)=0$
$\begin{array}{l}\text { LHD }=\lim _{x \rightarrow 0^{-}} \frac{g(x)-g(0)}{x-0}=\lim _{h \rightarrow 0} \frac{g(-h)-g(0)}{-h} \\ \quad= & \lim _{h \rightarrow 0} \frac{0-0}{-h}=\lim _{h \rightarrow 0} 0 \\ \text { LHD }=0\end{array}$
For RHD : $g(x)=|x|-2+|x-2|$
$g(x)=x-2-(x-2) \quad x>0$ (and just greater
than zero)
$g(x)=x-2-x+2=0$
Now $g(x)$ is not continuous at $x=0$, hence $g(x)$ is not differentiable at $\mathrm{x}=0$ At $x=2$ For LHD :
$g(x)=|x|-2+|x-2|=x-2-(x-2)$
$\quad=x-2-x+2=0$
$\therefore \mathrm{LHD}=\lim _{x \rightarrow 2^{-}} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{0}{x-2}=0$
For RHD:
$g(x)=|x|-2+|x-2|=x-2+x-2=2 x-4$
$\begin{aligned} \Rightarrow \mathrm{RHD} &=\lim _{x \rightarrow 2^{+}} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{2 x-4-2(2)+4}{x-2} \\ &=\lim _{x \rightarrow 2} \frac{2(x-2)}{x-2}=2 \\ \Rightarrow \mathrm{LHD} & \neq \mathrm{RHD} \end{aligned}$
Thus $g(x)$ is not differentiable at $x=2$.
$g(x)=f(|x|)+|f(x)|$
At $x=0$
For LHD $: g(x)=-2+|-2|=-2+2=0 \Rightarrow g(x)=0$
$\begin{array}{l}\text { LHD }=\lim _{x \rightarrow 0^{-}} \frac{g(x)-g(0)}{x-0}=\lim _{h \rightarrow 0} \frac{g(-h)-g(0)}{-h} \\ \quad= & \lim _{h \rightarrow 0} \frac{0-0}{-h}=\lim _{h \rightarrow 0} 0 \\ \text { LHD }=0\end{array}$
For RHD : $g(x)=|x|-2+|x-2|$
$g(x)=x-2-(x-2) \quad x>0$ (and just greater
than zero)
$g(x)=x-2-x+2=0$
Now $g(x)$ is not continuous at $x=0$, hence $g(x)$ is not differentiable at $\mathrm{x}=0$ At $x=2$ For LHD :
$g(x)=|x|-2+|x-2|=x-2-(x-2)$
$\quad=x-2-x+2=0$
$\therefore \mathrm{LHD}=\lim _{x \rightarrow 2^{-}} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{0}{x-2}=0$
For RHD:
$g(x)=|x|-2+|x-2|=x-2+x-2=2 x-4$
$\begin{aligned} \Rightarrow \mathrm{RHD} &=\lim _{x \rightarrow 2^{+}} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{2 x-4-2(2)+4}{x-2} \\ &=\lim _{x \rightarrow 2} \frac{2(x-2)}{x-2}=2 \\ \Rightarrow \mathrm{LHD} & \neq \mathrm{RHD} \end{aligned}$
Thus $g(x)$ is not differentiable at $x=2$.
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