Question

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

and $f\left(x\right)=\{\begin{array}{cc}x+4,& x\le 1\\ 2x+1,& 1<x\le 2\end{array}$

then, the number of roots fo the equation $f\left(g\right(x\left)\right)=0$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

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

$f\left(x\right)=\{\begin{array}{cc}x+4,& x\le 1\\ 2x+1,& 1<x\le 2\end{array}$

$\Rightarrow f\left(g\right(x\left)\right)=\{\begin{array}{cc}g\left(x\right)+4,& g\left(x\right)\le 1\\ 2g\left(x\right)+1,& 1<g\left(x\right)\le 2\end{array}$

When $-1\le x\le 1$,
then $g\left(x\right)=x^2\Rightarrow 0\le g\left(x\right)\le 1$

When $1<x\le \sqrt{2}$,
then $g\left(x\right)=x^2\Rightarrow 1<g\left(x\right)\le 2$

$f\left(g\right(x\left)\right)=\{\begin{array}{ccc}{x}^2+4,& -1\le x\le 1,& 0\le g\left(x\right)\le 1\\ 2\left(x^2\right)+1,& 1<x\le \sqrt{2}& 1<g\left(x\right)\le 2\end{array}$

$f\left(g\right(x\left)\right)>0\text{for}x\in [-1,\sqrt{2}]$