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

Explanation:

$y=\frac{2x^2-3x+8}{2x^2+3x+8}$

⇒ x²(2y – 2) + x(3y + 3) + 8y – 8 = 0

For real roots,

D ≥ 0

⇒ (3y + 3)² – 4(2y – 2) (8y – 8) ≥ 0

⇒ 9y² + 18y + 9 - 4(16y² - 16y - 16y + 16) ≥ 0

⇒ (11y – 5) (5y – 11) ≤ 0

$\Rightarrow y\in [\frac{5}{11},\frac{11}{5}]$

y = 1 is also included

So, y_max = 5/11 and y_min = 11/5

So, sum of max and min value = $\frac{5}{11}+\frac{11}{5}=\frac{146}{55}$

So, m = 146 and n = 55, gcd(m, n) = 1 

m + n = 146 + 55 = 201.

Option (4) is true.