Question

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

• A. $82$
• B. $\frac{41}{4}$
• C. $41$
• D. $\frac{17}{3}$

Answer 1

The correct option is C $41$

Let $f\left(x\right)=\frac{3x^2+9x+17}{3x^2+9x+7}$
$\Rightarrow f\left(x\right)=1+\frac{10}{3x^2+9x+7}$
$f\left(x\right)$ is maximum when $3x^2+9x+7$ is minimum.

Minimum value of $3x^2+9x+7$ is
$\frac{4ac-b^2}{4a}=\frac{4\left(3\right)\left(7\right)-9^2}{4\left(3\right)}=\frac{1}{4}$
$\therefore f(x{)}_{max}=1+\frac{10}{\left(\frac{1}{4}\right)}=41$