Let the sum of the maximum and the minimum values of the function f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8} be \frac{m}{n}, where \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1. Then \mathrm{m}+\mathrm{n} is equal to :

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Answer: C. 201 \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}ight)\left(y-\frac{11}{5}ight) \leq 0 \ & \Rightarrow y \in\left[\frac{5}{11}, \frac{11}{5}ight] \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}

Let the sum of the maximum and the minimum values of the function $$f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}$$ be $$\frac{m}{n}$$, where $$\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$$. Then $$\mathrm{m}+\mathrm{n}$$ is equal to :

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