Question

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

• A. $1$
• B. $\frac{17}{7}$
• C. $\frac{1}{4}$
• D. $41$

Answer 1

The correct option is D $41$

$\frac{3x^2+9x+17}{3x^2+9x+7}=\frac{3x^2+9x+7+10}{3x^2+9x+7}$

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

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

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

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

$=1+\frac{10/3}{(x+3/2{)}^2+\frac{28-27}{12}}$

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

Maximum value of $(x+3/2{)}^2=0$

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

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

Maximum value of $\frac{10/3}{(x+3/2{)}^2+1/12}=\frac{10}{3}\times 12=40$

Maximum value of $1+\frac{10/3}{(x+3/2{)}^2+1/12}=1+40=41$.