Question

The maximum value of
$f\left(x\right)=\frac{3x^2+9x+17}{3x^2+9x+7}$ is $5k+1$ , then the value of $k$ is

Answer 1

Answer: 8

Given that,
$f(x)=\frac{3x^2+9x+7+10}{3x^2+9x+7}$
$\Rightarrow f(x)=1+\frac{10}{3x^2+9x+7}$
Now, $3x^2+9x+7\ge \frac{1}{4}$
if $a>0$ then $a{x}^2+bx+c\ge \frac{-D}{4a}$
$\Rightarrow 0<\frac{1}{3x^2+9x+7}\le 4$
$\Rightarrow 0<\frac{10}{3x^2+9x+7}\le 40$
$\Rightarrow 1<\frac{10}{3x^2+9x+7}+1\le 41$
Maximum value of $f(x)$ is 41 .
$5k+1=41$
$\Rightarrow k=8$