Question

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

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

Answer 1

Answer: (B)

$f\left(x\right)=\frac{x^2+8}{x^2+4}$
$\Rightarrow f\left(x\right)=\frac{x^2+4}{x^2+4}+\frac{4}{x^2+4}$
$\Rightarrow f\left(x\right)=1+\frac{4}{x^2+4}>1$
$\Rightarrow f\left(x\right)$ is maximum at $x=0$.
Hence $R\left(f\left(x\right)\right)=(1,2]$