Question

Let $f\left(x\right)=\{\begin{array}{cc}\left[x\right];& -2\le x\le -\frac{1}{2}\\ 2x^2-1;& -\frac{1}{2}<x\le 2\end{array}$ and $g\left(x\right)=f\left(|x|\right)+|f\left(x\right)|$, where $[.]$ represents the greatest integer function. The number of points where $g\left(x\right)$ is non-differentiable is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$f\left(x\right)=\{\begin{array}{cc}\left[x\right];& -2\le x\le -\frac{1}{2}\\ 2x^2-1;& -\frac{1}{2}<x\le 2\end{array}$

$f\left(x\right)=\{\begin{array}{cc}-2;& -2\le x\le -1\\ -1;& -1\le x\le -\frac{1}{2}\\ 2x^2-1;& -\frac{1}{2}<x\le 2\end{array}$

$|f\left(x\right)|=\{\begin{array}{cc}2;& -2\le x\le -1\\ 1;& -1\le x\le -\frac{1}{2}\\ |2x^2-1|;& -\frac{1}{2}<x\le 2\end{array}$

$|f\left(x\right)|=\{\begin{array}{cc}2;& -2\le x\le -1\\ 1;& -1\le x\le -\frac{1}{2}\\ 1-2x^2;& -\frac{1}{2}<x\le \frac{1}{\sqrt{2}}\\ 2x^2-1;& \frac{1}{\sqrt{2}}<x\le 2\end{array}$

$f\left(|x|\right)=\{\begin{array}{cc}-2;& -2\le |x|\le -1\\ -1;& -1\le |x|\le -\frac{1}{2}\\ 2|x{|}^2-1;& -\frac{1}{2}<x\le 2\end{array}$
Hence,
$f\left(|x|\right)=2x^2-1;-2\le x\le 2$
Now, $g\left(x\right)=|f\left(x\right)|+f\left(|x|\right)$
$g\left(x\right)=\{\begin{array}{cc}2x^2+1;& -2\le x<-1\\ 2x^2;& -1\le x\le -\frac{1}{2}\\ 0;& -\frac{1}{2}<x<\frac{1}{\sqrt{2}}\\ 4x^2-2;& \frac{1}{\sqrt{2}}\le x\le 2\end{array}$

$g(-1^{-})=\underset{x\to -1}{lim}(2x^2+1)=3$ and $g(-1^{+})=\underset{x\to -1}{lim}2x^2=2$

$g(-{\frac{1}{2}}^{-})=\underset{x\to -\frac{1}{2}}{lim}\left(2x^2\right)=\frac{1}{2}$ and $g(-{\frac{1}{2}}^{+})=\underset{x\to -\frac{1}{2}}{lim}0=0$

$g\left({\frac{1}{\sqrt{2}}}^{-}\right)=\underset{x\to \frac{1}{\sqrt{2}}}{lim}0=0$ and $g\left({\frac{1}{\sqrt{2}}}^{+}\right)=\underset{x\to \frac{1}{\sqrt{2}}}{lim}(4x^2-2)=0$

Hence, $g\left(x\right)$ is discontinuous at $x=-1,-\frac{1}{2}$
$\Rightarrow g\left(x\right)$ is non-differentiable at $x=-1,-\frac{1}{2}$
Now, $g^{\prime }\left({\frac{1}{\sqrt{2}}}^{-}\right)=0,g^{\prime }\left({\frac{1}{\sqrt{2}}}^{+}\right)=\frac{8}{\sqrt{2}}$
Hence, $g\left(x\right)$ is non-differentiable at $x=\frac{1}{\sqrt{2}}$