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 {\gcd}( {\m}, {\n})=1. Then {\m}+ {\n} is equal to : [4 Apr 2024 Shift 1]

Application of Derivatives Part 2

Section: Mathematics

Question : 5 of 100

Marks: +1, -0

Let the sum of the maximum and the minimum values of the function f(x)=‌

2x2−3x+8

2x2+3x+8

be ‌

, where gcd(m,n)=1. Then m+n is equal to :

[4 Apr 2024 Shift 1]

Go to Question:

Prev Question

Next Question