Question

Range of the function $f\left(x\right)=\frac{x^2+x+2}{x^2+x+1};xϵR$ 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 $(1,\frac{7}{3}]$

We have, $f\left(x\right)=\frac{x^2+x+2}{x^2+x+1}=\frac{(x^2+x+1)+1}{x^2+x+1}=1+\frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}$
We can see here that as $x\to \infty ,f\left(x\right)\to 1$, which is the min value of f(x). Also f(x) is max when ${(x+\frac{1}{2})}^2+\frac{3}{4}$ is min which is so when $x=-\frac{1}{2}$ and then $\frac{3}{4}$.
$\therefore$ $f_{max}=1+\frac{1}{\frac{3}{4}}=\frac{7}{3}$
$\therefore$ $R_i=(1,\frac{7}{3}]$