Question

The range of the function $f\left(x\right)=\frac{x^2+x+2}{x^2+x+1},x\in R$ is

• A. $(1,\infty )$
• B. $(1,\frac{11}{7})$
• C. $(1,\frac{7}{3}]$
• D. $(1,\frac{7}{5})$

Answer 1

The correct option is C $(1,\frac{7}{3}]$

Let $y=\frac{x^2+x+2}{x^2+x+1}=1+\frac{1}{x^2+x+1}$

$x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}$
As $x\in R$,
$0\le {(x+\frac{1}{2})}^2<\infty$
$\Rightarrow \frac{3}{4}\le x^2+x+1<\infty$
$\Rightarrow 0<\frac{1}{x^2+x+1}\le \frac{4}{3}$
$\Rightarrow 1<1+\frac{1}{x^2+x+1}\le \frac{7}{3}$
$\Rightarrow 1<y\le \frac{7}{3}$

Hence, range of $f$ is $(1,\frac{7}{3}]$