Let the sum of the maximum and the minimum values of the function [f(x)=(2 x^2-3 x+8)/(2 x…

Question

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 :

Accepted Answer

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

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

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

Solutions · 1