Let α₀ and β₀ be the distinct roots of 2x² + (cosθ)x - 1 = 0, θ ∈ (0, 2π). If m and M are the minimum and the maximum values of α₀⁴ + β₀⁴, then 16(M + m) equals:

(1) 24 

(2) 25 

(3) 27 

(4) 17

Given: 2x² + (cosθ)x - 1 = 0, θ ∈ (0, 2π)

Let α₀ and β₀ be the roots.

By Vieta's formulas:

• α₀ + β₀ = -cosθ/2
• α₀·β₀ = -1/2

We need to find: α₀⁴ + β₀⁴

α₀² + β₀² = (α₀ + β₀)² - 2α₀β₀ = cos²θ/4 - 2(-1/2) = cos²θ/4 + 1

α₀⁴ + β₀⁴ = (α₀² + β₀²)² - 2(α₀β₀)²

= (cos²θ/4 + 1)² - 2(-1/2)²

= (cos²θ/4 + 1)² - 2(1/4)

= (cos²θ/4 + 1)² - 1/2

= cos⁴θ/16 + cos²θ/2 + 1 - 1/2

= cos⁴θ/16 + cos²θ/2 + 1/2

Let y = cos²θ, where 0 ≤ y ≤ 1

f(y) = y²/16 + y/2 + 1/2 = (y² + 8y + 8)/16

f'(y) = (2y + 8)/16 = (y + 4)/8 > 0 for all y ∈ [0,1]

So f is increasing on [0,1]

Minimum at y = 0: M = 8/16 = 1/2

Maximum at y = 1: M = (1 + 8 + 8)/16 = 17/16

Therefore: m = 1/2, M = 17/16

16(M + m) = 16(17/16 + 1/2) = 16(17/16 + 8/16) = 16(25/16) = 25

Answer: (2) 25