Question

Let $f\left(x\right)=f_1\left(x\right)-2f_2\left(x\right)$,

where, $f_1\left(x\right)=\{\begin{array}{cc}min\{x^2,|x|\},& |x|\le 1\\ max\{x^2,|x|\},& |x|>1\end{array}$

and, $f_2\left(x\right)=\{\begin{array}{cc}min\{x^2,|x|\},& |x|>1\\ max\{x^2,|x|\},& |x|\le 1\end{array}$

and, $g\left(x\right)=\{\begin{array}{c}min\left\{f\right(t):-3\le t\le x,-3\le x<0\}\\ max\left\{f\right(t):0\le t\le x,0\le x\le 3\}\end{array}$

The graph of $y=g\left(x\right)$ in its domain is broken at

• A. $1$ point
• B. $2$ points
• C. $3$ points
• D. None of these

Answer 1

The correct option is A $1$ point

$f_1\left(x\right)=x^2$ and $f_2\left(x\right)=|x|$
or, $f\left(x\right)=f_1\left(x\right)-2f_2\left(x\right)=x^2-2|x|$

Graph of $f\left(x\right)$ is

$g\left(x\right)=\{\begin{array}{cc}f\left(x\right),& -3\le x<-1\\ & \\ -1,& -1\le x<0\\ & \\ 0,& 0\le x\le 2\\ & \\ f\left(x\right),& 2<x\le 3\end{array}$

$\Rightarrow g\left(x\right)=\{\begin{array}{cc}{x}^2+2x,& -3\le x<-1\\ & \\ -1,& -1\le x<0\\ & \\ 0,& 0\le x\le 2\\ & \\ x^2-2x,& 2<x\le 3\end{array}$

The graph is broken at $x=0$