If the Rolle's theorem holds for the function f(x) = 2x³ + ax² + bx in interval [-1, 1] for the point c = 1/2, then find the value of 2a + b?

f(-1) = f(1)
-2 + a - b = 2 + a + b
=> 2a + 2b = -4
=> a + b = -2
f'(c) = 6c² + 2ac + b
f'(c) = 0 at c = 1/2
=> 6(1/2)² + 2a(1/2) + b = 0
=> 6(1/4) + a + b = 0
=> 3/2 + a + b = 0
=> a + b = -3/2
This is a contradiction. Let's re-examine.
f(-1) = -2 + a - b
f(1) = 2 + a + b
Since Rolle's theorem holds, f(-1) = f(1)
-2 + a - b = 2 + a + b
-4 = 2b
b = -2
f'(x) = 6x² + 2ax + b
f'(1/2) = 6(1/4) + a + b = 0
3/2 + a - 2 = 0
a = 1/2
2a + b = 2(1/2) + (-2) = 1 - 2 = -1