Question

If $x$ is real, the 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

Answer: (D)

$41$

$f\left(x\right)=\frac{3x^2+9x+17}{3x^2+9x+7}=\frac{3x^2+9x+7+10}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}$

$f^{{}^{\prime }}\left(x\right)=0$ for maximum value

$f^{{}^{\prime }}\left(x\right)=\frac{10(6x+9)}{{(3x^2+9x+7)}^2}6x+9=0x=-\frac{3}{2}$

As $x$ is real so maximum value is $f\left(x\right)=1+\frac{10}{3{(-\frac{3}{2})}^2+9{(-\frac{3}{2})}^2+7}=1+\frac{10}{\frac{24}{4}-\frac{27}{2}+7}=1+\frac{10}{\frac{27-54+28}{4}}=1+\frac{40}{1}=41$

Answer