Question

Consider the functions $f\left(x\right)=\{\begin{array}{c}x+1,x\le 1\\ 2x+1,1<x\le 2\end{array}$

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

The number of roots of the equation $f\left(g\right(x\left)\right)=2$ is

Answer 1

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


Now from graph of $g\left(x\right),$
$g\left(x\right)\le 1$
$\Rightarrow x\in [-1,1]$
and $g\left(x\right)\in (1,2]\Rightarrow x\in (1,\sqrt{2}]$

$\therefore f\left(g\left(x\right)\right)=\{\begin{array}{cc}g\left(x\right)+1,& x\in [-1,1]\\ 2g\left(x\right)+1,& x\in (1,\sqrt{2}]\end{array}$


$=\{\begin{array}{cc}{x}^2+1,& x\in [-1,1]\\ 2x^2+1,& x\in (1,\sqrt{2}]\end{array}$

$f\left(g\right(x\left)\right)=2$
$x^2+1=2\Rightarrow x=\pm 1$
and
$2x^2+1=2$
$x=\pm \frac{1}{\sqrt{2}}\notin (1,\sqrt{2}]$
So only two roots are there.