Question

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

• A. $1$
• B. $\frac{17}{7}$
• C. $\frac{1}{4}$
• D. $41$

Answer 1

The correct option is D

$41$

Explanation for the correct answer:

Solve for maximum value

The given function is

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

$f\left(x\right)$ will have maximum value when $f\text{'}\left(x\right)=0$

Differentiating with respect to $x$ we get

$f\text{'}\left(x\right)=\frac{d}{dx}\left(1+10{\left(3x^2+9x+7\right)}^{-1}\right)$

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

Equating $f\text{'}\left(x\right)$ with $0$ we get

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

$\Rightarrow$ $6x+9=0$

$\Rightarrow$ $x=\frac{-3}{2}$

Thus, $f\left(x\right)$ is maximum when $x=\frac{-3}{2}$

Substituting value of $x=\frac{-3}{2}$ in $f\left(x\right)$ we get,

$\Rightarrow$$f{\left(x\right)}_{maximum}=1+\frac{10}{3{\left({\displaystyle \frac{-3}{2}}\right)}^2+9\left({\displaystyle \frac{-3}{2}}\right)+7}$

$\Rightarrow$$f{\left(x\right)}_{maximum}=1+\frac{10}{{\displaystyle \frac{27}{4}}-\left({\displaystyle \frac{27}{2}}\right)+7}$

$\Rightarrow$$f{\left(x\right)}_{maximum}=1+\frac{10}{{\displaystyle \frac{1}{4}}}$

$\Rightarrow$$f{\left(x\right)}_{maximum}=41$

Hence, option (D) is correct.