Question

Range of the function $f\left(x\right)=\frac{x^2+x+2}{x^2+x+1}$ is:

• A. $[1,2]$
• B. $[1,\infty ]$
• C. $[2,\frac{7}{3}]$
• D. $[1,\frac{7}{3}]$

Answer 1

Answer: (D)

$[1,\frac{7}{3}]$

$\frac{(x^2+x+2)}{(x^2+x+1)}=\frac{(x^2+x+1)+1}{(x^2+x+1)}$

$=1+\frac{1}{(x^2+x+1)}$

$=1+\frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}$

Minimum value of ${(x+\frac{1}{2})}^2=0$

Maximum value of ${(x+\frac{1}{2})}^2=\infty$

$\therefore$ Range $=(1,1+\frac{1}{3/4})$

$⟹$ Range $=[1,\frac{7}{3}]$