Let the sum of the maximum and the minimum values of the function f(x)=(2 x²-3 x+8)/(2 x²+3 x+8) be m/n, where gcd(m, n)=1. Then m+n is equal to:

Limit, Continuity and Differentiability: JEE Main 2024 Mathematics question

JEE Main 2024 · 4 Apr, Morning shift · Q2685

JEE Main 20244 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

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}

\frac{m}{n}

, where

\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1

. Then

\mathrm{m}+\mathrm{n}

is equal to :