Question

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

Answer 1

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