Limit Continuity And Differentiability Ques 54
- Let $f(x)$ be defined in the interval $[-2,2]$ such that $ f(x)=\left\{\begin{array}{cc} -1, & -2 \leq x \leq 0 \\ x-1, & 0<x \leq 2 \end{array}\right. $
and $ g(x)=f(|x|)+|f(x)| $
Test the differentiability of $g(x)$ in $(-2,2)$.
(1986, 5M)
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Answer:
Correct Answer: 54.$g(x)$ is differentiable for all $x \in(-2,2)-\{0,1)$
Solution: Given that, $f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x \leq 0 \\ (x-1), & 0<x \leq 2\end{array}\right.$
Since, $x \in[-2,2]$.
Therefore, $|x| \in[0,2]$
$\Rightarrow \quad f(|x|)=|x|-1, \forall x \in[-2,2]$
$\Rightarrow \quad f(|x|)= \begin{cases}x-1, & 0 \leq x \leq 2 \\ -x-1, & -2 \leq x \leq 0\end{cases}$
Also, $\quad|f(x)|=\left\{\begin{array}{cc}1, & -2 \leq x<0 \\ 1-x, & 0 \leq x<1 \\ x-1, & 1 \leq x \leq 2\end{array}\right.$
Also,
$ \begin{aligned} g(x) & =f(|x|)+|f(x)| \\ & = \begin{cases}-x-1+1, & -2 \leq x \leq 0 \\ x-1+1-x, & 0 \leq x<1 \\ x-1+x-1, & 1 \leq x \leq 2\end{cases} \end{aligned} $
$ g(x)=\left\{\begin{array}{cc} -x, & -2 \leq x \leq 0 \\ 0, & 0 \leq x<1 \\ 2(x-1), & 1 \leq x \leq 2 \end{array}\right. $
$\therefore \quad g^{\prime}(x)=\left\{\begin{array}{cc}-1, & -2 \leq x \leq 0 \\ 0, & 0 \leq x<1 \\ 2, & 1 \leq x \leq 2\end{array}\right.$
$\therefore \quad$ RHD (at $x=1$ ) $=2, \operatorname{LHD}$ (at $x=1$ ) $=0$
$\Rightarrow \quad g(x)$ is not differentiable at $x=1$.
Also, RHD (at $x=0)=0$, LHD at $(x=0)=-1$
$\Rightarrow \quad g(x)$ is not differentiable at $x=0$.
Hence, $g(x)$ is differentiable for all $x \in(-2,2)-\{0,1\}$
BETA
