Question

The range of the function $f\left(x\right)=\left(\frac{x^2+8}{x^2+4}\right),x\in R$ is

• A. $[-1,\frac{3}{2}]$
• B. $(1,2]$
• C. $(1,2)$
• D. $[1,2]$
• E. $[\frac{3}{2},2]$

Answer 1

The correct option is B $(1,2]$

Given,

$f\left(x\right)=\frac{x^2+8}{x^2+4},x\in R$
$\Rightarrow x^2(y-1)+4y=8$
$\Rightarrow x^2=\frac{8-4y}{y-1}$
$\Rightarrow x=\sqrt{\frac{8-4y}{y-1}}$

$\because x$ is real.
Thus

$y-1>0$ and $8-4y\ge 0$
$\Rightarrow y>1$ and $\le 2$
$\Rightarrow y\in (1,2]$