Question

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

• A. $41$
• B. $\frac{-3}{2}$
• C. $1$
• D. $\frac{17}{7}$

Answer 1

The correct option is A $41$

Given,

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

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

$3x^2+9x+7=(\sqrt{3}x{)}^2+2\times \sqrt{3}\times 3\frac{\sqrt{3}}{2}\times x+\frac{27}{4}+(7-\frac{27}{4})$

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

Thus the minimum value is $\frac{1}{4}$

Hence total value, maximum value

$=1+\frac{10}{\frac{1}{4}}=1+40=41$