Question

If $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}x-3,& x<0\end{array}\\ \begin{array}{cc}{x}^2-3x+2,& x\ge 0\end{array}\end{array}$ and $g\left(x\right)=f\left(\left|x\right|\right)+\left|f\left(x\right)\right|$, then $g\left(x\right)$ is

• A. Continuous is $R-\left\{0\right\}$
• B. Continuous in $R$
• C. Differentiable in $R-\{0,1,2\}$
• D. Differentiable in $R-\{1,2\}$

Answer 1

Answer: (A), (C)

The correct options are
A Continuous is $R-\left\{0\right\}$
C Differentiable in $R-\{0,1,2\}$

We have, $f\left|x\right|=\{\begin{array}{c}\begin{array}{cc}\left|x\right|-3,& \left|x\right|<0\end{array}\\ \begin{array}{cc}{\left|x\right|}^2-3\left|x\right|+2,& \left|x\right|\ge 0\end{array}\end{array}$

Since, $|x|<0$ is not possible, so we get,

$f\left(\left|x\right|\right)={\left|x\right|}^2-3\left|x\right|+2,\left|x\right|\ge 0$

$=\{\begin{array}{c}\begin{array}{cc}{x}^2+3x+2,& x<0\end{array}\\ \begin{array}{cc}{x}^2-3x+2,& x\ge 0\end{array}\end{array}$ ...(1)

Again, $\left|f\left(x\right)\right|=\{\begin{array}{c}\begin{array}{cc}|x-3|& x<0\end{array}\\ \begin{array}{cc}|x^2-3x+2|& x\ge 0\end{array}\end{array}$

$=\{\begin{array}{c}\begin{array}{cc}(x^2-3x+2),& 0\le x<1\end{array}\\ \begin{array}{cc}-(x^2-3x+2),& 1\le x<2\end{array}\\ \begin{array}{cc}(x^2-3x+2),& 2\le x\end{array}\end{array}$ ...(2)

From (1) and (2), we get

$g\left(x\right)=f\left(\left|x\right|\right)=\left|f\left(x\right)\right|$

$=\{\begin{array}{c}\begin{array}{cc}{x}^2+2x+5,& x<0\end{array}\\ \begin{array}{cc}2x^2-6x+4,& 0\le x<1\end{array}\\ \begin{array}{cc}0,& 1\le x<2\end{array}\\ \begin{array}{cc}2x^2-6x+4,& x\ge 2\end{array}\end{array}$

and $g^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}2x+2,& x<0\end{array}\\ \begin{array}{cc}4x-6,& 0<x<1\end{array}\\ \begin{array}{cc}0,& 1<x<2\end{array}\\ \begin{array}{cc}4x-6,& x>2\end{array}\end{array}$

Clearly, $g\left(x\right)$ is continuous in $R-\left\{0\right\}$ and differentiable in $R-\{0,1,2\}$