Question

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

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

Range of the function $f\left(g\right(x\left)\right)$ is

• A. $[1,5]$
• B. $[2,3]$
• C. $[1,2]\cup (3,5]$
• D. $[1,5]-\left\{3\right\}$

Answer 1

The correct option is C $[1,2]\cup (3,5]$

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

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

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


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

For $-1\le x\le 1$
$x^2\in [0,1]\Rightarrow x^2+1\in [1,2]$
For $1<x\le \sqrt{2}$
$2x^2+1\in (3,5]$
$\therefore$ range is $[1,2]\cup (3,5]$