If the Rolle's theorem holds for the function $$f(x) = 2x^3 + ax^2 + bx$$ in the interval $$[-1, 1]$$ for the point $$c = \frac{1}{2}$$, then the value of $$2a + b$$ is:

We are given that Rolle's theorem holds for the function $$ f(x) = 2x^3 + a x^2 + b x $$ in the interval $$[-1, 1]$$ at the point $$ c = \frac{1}{2} $$. Rolle's theorem requires three conditions: the function must be continuous on the closed interval $$[-1, 1]$$, differentiable on the open interval $$(-1, 1)$$, and $$ f(-1) = f(1) $$. Additionally, since the theorem holds at $$ c = \frac{1}{2} $$, we have $$ f'\left(\frac{1}{2}\right) = 0 $$.

First, since $$ f(x) = 2x^3 + a x^2 + b x $$ is a polynomial, it is continuous and differentiable everywhere, so the first two conditions of Rolle's theorem are automatically satisfied. We now use the third condition, $$ f(-1) = f(1) $$.

Compute $$ f(1) $$:

$$ f(1) = 2(1)^3 + a(1)^2 + b(1) = 2 + a + b $$

Compute $$ f(-1) $$:

$$ f(-1) = 2(-1)^3 + a(-1)^2 + b(-1) = 2(-1) + a(1) - b = -2 + a - b $$

Set $$ f(-1) = f(1) $$:

$$ -2 + a - b = 2 + a + b $$

Simplify this equation. Subtract $$ a $$ from both sides:

$$ -2 - b = 2 + b $$

Add $$ b $$ to both sides:

$$ -2 = 2 + 2b $$

Subtract 2 from both sides:

$$ -4 = 2b $$

Divide both sides by 2:

$$ b = -2 $$

So, we have $$ b = -2 $$.

Next, use the condition that $$ f'\left(\frac{1}{2}\right) = 0 $$. First, find the derivative of $$ f(x) $$:

$$ f'(x) = \frac{d}{dx}(2x^3 + a x^2 + b x) = 6x^2 + 2a x + b $$

Evaluate $$ f'\left(\frac{1}{2}\right) $$:

$$ f'\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^2 + 2a\left(\frac{1}{2}\right) + b = 6 \cdot \frac{1}{4} + a + b = \frac{6}{4} + a + b = \frac{3}{2} + a + b $$

Set this equal to zero:

$$ \frac{3}{2} + a + b = 0 $$

Substitute $$ b = -2 $$:

$$ \frac{3}{2} + a + (-2) = 0 $$

Simplify:

$$ \frac{3}{2} - 2 + a = 0 $$

$$ \frac{3}{2} - \frac{4}{2} + a = 0 $$

$$ -\frac{1}{2} + a = 0 $$

Solve for $$ a $$:

$$ a = \frac{1}{2} $$

Now, we have $$ a = \frac{1}{2} $$ and $$ b = -2 $$. The question asks for the value of $$ 2a + b $$:

$$ 2a + b = 2 \cdot \frac{1}{2} + (-2) = 1 - 2 = -1 $$

Thus, $$ 2a + b = -1 $$. Comparing with the options:

A. -1

B. 2

C. 1

D. -2

Hence, the correct answer is Option A.