Question

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

• A. 41
• B. 1
• C. 17/7
• D. 1/4

Answer 1

Answer: (A)

$\frac{3x^2+9x+17}{3x^2+9x+7}$
$=1+\frac{10}{3\left(x^2+3x+\frac{7}{3}\right)}$
$=1+\frac{10}{3\left[{\left(x+\frac{3}{2}\right)}^2+\frac{1}{12}\right]}$
$\Rightarrow$ Maximum value of $\frac{3x^2+9x+17}{3x^2+9x+7}$ occurs at $x=-\frac{3}{2}\cdot$
$\therefore$ Maximum value of $\frac{3x^2+9x+17}{3x^2+9x+7}$
$=1+\frac{10}{3\left(\frac{1}{12}\right)}$
= 1 + 40 = 41