Question

$\text{Range of the function f(x)}=\frac{x^2+x+2}{x^2+x+1};xϵRis$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\text{We have, f(x)}=\frac{x^2+x+2}{x^2+x+1}=\frac{(x^2+x+1)}{x^2+x+1}=1+\frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}$

$\text{We can see here that as x}\to \infty ,f\left(x\right)\to 1\text{which is the min value of f(x)}$.

$\text{Also f(x) is max when}{(x+\frac{1}{2})}^2+\frac{3}{4}$

$\text{is min which is so when x}=-\frac{1}{2}\text{and then}\frac{3}{4}.$

$\therefore f_{max}=1+\frac{1}{\frac{3}{4}}=\frac{7}{3}\therefore R_i=(1,\frac{7}{3}]$