Question

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

Answer 1

Let $y=\frac{3x^2+9x+17}{3x^2+9x+7}$
$=1+\frac{10}{3x^2+9x+7}$ ....(1)
Now, $3x^2+9x+7=3(x^2+3x)+7$
$\Rightarrow (x+\frac{3}{2}{)}^2+\frac{1}{4}$
Since, $(x+\frac{3}{2}{)}^2\ge 0$
$\Rightarrow (x+\frac{3}{2}{)}^2+\frac{1}{4}\ge \frac{1}{4}$ for all $xϵR$
$\Rightarrow 3x^2+9x+7\ge \frac{1}{4}$
Maximum value of $\frac{10}{3x^2+9x+7}$ is 40
Maximum value of y is $1+40=41$
$\therefore 5k+1=41$
$\Rightarrow k=8$.