QJEEM24Ma3d6f: JEE Main 2024 Mathematics - Class 12 JEEM24Ma3d6f Solution: Let the sum of the maximum and the minimum values of the…

$\begin{aligned}& f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}=y, 2 x^2+3 x+8>0 \quad \forall x \in \mathbb{R} \\& \Rightarrow \quad x^2(2 y-2)+x(3 y+3)+8 y-8=0\end{aligned}$

Since $x \in \mathbb{R}$ , the equation has real roots

$\begin{aligned}& \Rightarrow \quad D \geq 0 \\& \Rightarrow(3 y+3)^2-4(2 y-2)(8 y-8) \geq 0 \\& \Rightarrow 9(y+1)^2-64 y(y-1)^2 \geq 0 \\& \Rightarrow(3 y+3)^2-(8 y-8)^2 \geq 0 \\& \Rightarrow(11 y-5)(-5 y+11) \geq 0 \\& \Rightarrow\left(y-\frac{5}{11}\right)\left(y-\frac{11}{5}\right) \leq 0 \\& \Rightarrow y \in\left[\frac{5}{11}, \frac{11}{5}\right]\end{aligned}$

Sum of maximum and minimum value

$\begin{aligned}& y_{\max }+y_{\min }=\frac{5}{11}+\frac{11}{5}=\frac{25+121}{55} \\& =\frac{146}{55}=\frac{m}{n} \Rightarrow m+n=201\end{aligned}$