Let f(x)=
Question:

Let $f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x<0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.$ and

$g(x)=|f(x)|+f(|x|)$. Then, in the interval $(-2,2), g$ is :

 

  1. (1) differentiable at all points

  2. (2) not continuous

  3. (3) not differentiable at two points

  4. (4) not differentiable at one point


Correct Option: , 4

Solution:

$f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x<0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.$

Then, $f(|x|)=\left\{\begin{array}{cc}-1, & -2 \leq|x|<0 \\ |x|^{2}-1, & 0 \leq|x| \leq 2\end{array}\right.$

$\Rightarrow \quad f(|x|)=x^{2}-1,-2 \leq x \leq 2$

$\Rightarrow \quad g(x)=\left\{\begin{array}{cc}1+x^{2}-1, & -2 \leq x<0 \\ \left(x^{2}-1\right)+\left|x^{2}-1\right|, & 0 \leq x \leq 2\end{array}\right.$

$=\left\{\begin{array}{cc}x^{2}, & -2 \leq x<0 \\ 0, & 0 \leq x<1 \\ 2\left(x^{2}-1\right), & 1 \leq x \leq 2\end{array}\right.$

$g^{\prime}\left(0^{-}\right)=0, g^{\prime}\left(0^{+}\right)=0, g^{\prime}\left(1^{-}\right)=0, g^{\prime}\left(1^{+}\right)=4$

$\Rightarrow g(x)$ is non-differentiable at $x=1$

$\Rightarrow g(x)$ is not differentiable at one point.

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