Question

Let $f\left(x\right)$ be defined on $[-2,2]$ and is given by $f\left(x\right)=\{\begin{array}{cc}-1,& -2\le x\le 0\\ x-1,& 0<x\le 2\end{array},g\left(x\right)=|x|$. Then $fog\left(x\right)+gof\left(x\right)=$

• A. $\{\begin{array}{cc}-x,& -2\le x\le 0\\ 0,& 0<x\le 1\\ 2(x-1),& 1<x\le 2\end{array}$
• B. $\{\begin{array}{cc}1,& -2\le x\le 0\\ -x+1,& 0<x\le 1\\ x-1,& 1<x\le 2\end{array}$
• C. $\{\begin{array}{cc}-x+2,& -2\le x\le 0\\ 0,& 0<x\le 1\\ x-1,& 1<x\le 2\end{array}$
• D. $\{\begin{array}{cc}2& -2\le x\le 0\\ 0& 0<x\le 1\\ 2(x-1)& 1<x\le 2\end{array}$

Answer 1

The correct option is A $\{\begin{array}{cc}-x,& -2\le x\le 0\\ 0,& 0<x\le 1\\ 2(x-1),& 1<x\le 2\end{array}$

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

From graph of $g\left(x\right)$




$0<g\left(x\right)\le 2⟺x\in [-2,2]-\left\{0\right\}$
$-2\le g\left(x\right)\le 0⟺x=0$

So, $f\left(g\right(x\left)\right)=\{\begin{array}{cc}-1,& x=0\\ g\left(x\right)-1,& x\in [-2,2]-\left\{0\right\}\end{array}$
$f\left(g\right(x\left)\right)=\{\begin{array}{cc}-1,& x=0\\ |x|-1,& x\in [-2,2]-\left\{0\right\}\end{array}\cdots \left(i\right)$
and
$g\left(f\right(x\left)\right)=|f\left(x\right)|=\{\begin{array}{cc}f\left(x\right),& f\left(x\right)\ge 0\\ -f\left(x\right),& f\left(x\right)<0\end{array}$

From graph of $f\left(x\right)$


$f\left(x\right)\ge 0⟺x\in [1,2]$
$f\left(x\right)<0⟺x\in [-2,1)$
$g\left(f\right(x\left)\right)=\{\begin{array}{cc}f\left(x\right),& x\in [1,2]\\ -f\left(x\right),& x\in [-2,1)\end{array}$


$g\left(f\right(x\left)\right)=\{\begin{array}{cc}1,& x\in [-2,0)\\ -x+1,& x\in [0,1)\\ x-1,& x\in [1,2]\end{array}\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$fog\left(x\right)+gof\left(x\right)=\{\begin{array}{cc}-x-1+1=-x,& x\in [-2,0)\\ -1-x+1=-x,& x=0\\ x-1-x+1=0,& x\in (0,1)\\ x-1+x-1=2(x-1),& x\in [1,2]\end{array}$
$=\{\begin{array}{cc}-x,& -2\le x\le 0\\ 0,& 0<x\le 1\\ 2(x-1),& 1<x\le 2\end{array}$