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

Question

Aicharya · Accepted Answer

$y=\frac{3 x^2+9 x+17}{3 x^2+9 x+7}$ $3 x^2(y-1)+9 x(y-1)+7 y-17=0$ $D \geq 0 \quad \because x$ is real $81(y-1)^2-4 x 3(y-1)(7 y-17) \geq 0$ $\Rightarrow(y-1)(y-41) \leq 0 \Rightarrow 1 \leq y \leq 41$

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

Solution