Question

If x is real, the maximum value of $\frac{3x^2+9x+17}{3x^2+9x+7}$ is

• A. $41$
• B. $1$
• C. $\frac{17}{7}$
• D. $40$

Answer 1

The correct option is A $41$

We have

$\frac{3x^2+9x+17}{3x^2+9x+7}$

$=\frac{3x^2+9x+10+7}{3x^2+9x+7}$

$=1+\frac{10}{3x^2+9x+7}$

$=1+\frac{10}{3(x^2+3x+\frac{7}{3})}$

$=1+\frac{10}{3}\times \frac{1}{x^2+2x\times \frac{3}{2}+{\left(\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2+\frac{7}{3}}$ by completing the square method

$=1+\frac{10}{3}\times \frac{1}{{(x+\frac{3}{2})}^2-\frac{9}{4}+\frac{7}{3}}$

$=1+\frac{1}{{(x+\frac{3}{2})}^2+\frac{1}{12}}$

Minimum value of ${(x+\frac{3}{2})}^2=0$

Minimum value of ${(x+\frac{3}{2})}^2+\frac{1}{12}=\frac{1}{12}$

Maximum value of $\frac{1}{{(x+\frac{3}{2})}^2+\frac{1}{12}}=\frac{1}{\frac{1}{12}}=12$

Maximum value of $\frac{10}{3}\times \frac{1}{{(x+\frac{3}{2})}^2+\frac{1}{12}}=\frac{10}{3}\times 12$

$=40$

Maximum value of

$1+\frac{10}{3}\times \frac{1}{{(x+\frac{3}{2})}^2+\frac{1}{12}}=1+40=41$