Question

If ${}^{\prime }{x}^{\prime }$ 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. $\frac{1}{4}$

Answer 1

The correct option is A $41$

$f\left(x\right)=\left(\frac{(3x^2+9x+7)+10}{3x^2+9x+7}\right)=1+\left(\frac{10}{3x^2+9x+7}\right)nowf\left(x\right)willbemaximumwhen3x^2+9x+7willbeminimumanditwillbe=\left(\frac{-D}{4a}\right)=\left(\frac{-(9^2-4\cdot 3\cdot 7)}{4\cdot 3}\right)=\left(\frac{-(81-84)}{4\cdot 3}\right)=\left(\frac{1}{4}\right)\therefore valueoff\left(x\right)=1+\left(\frac{10}{\left(\frac{1}{4}\right)}\right)=1+40=41$