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

Options

A. 1
B. `-1`
C. 2
D. `-2`

Answer

To solve the problem step by step, we need to apply Rolle's Theorem to the function \( f(x) = 2x^3 + ax^2 + bx \) over the interval \([-1, 1]\) and find the values of \( a \) and \( b \) such that the theorem holds at the point \( c = \frac{1}{2} \).

### Step 1: Apply Rolle's Theorem
Rolle's Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), and if \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

### Step 2: Evaluate \( f(-1) \) and \( f(1) \)
We need to find \( f(-1) \) and \( f(1) \) and set them equal to each other.

1. **Calculate \( f(-1) \)**:
\[ f(-1) = 2(-1)^3 + a(-1)^2 + b(-1) = 2(-1) + a(1) - b = -2 + a - b \]

2. **Calculate \( f(1) \)**:
\[ f(1) = 2(1)^3 + a(1)^2 + b(1) = 2(1) + a(1) + b = 2 + a + b \]

### Step 3: Set \( f(-1) = f(1) \)
Setting the two expressions equal gives:
\[ -2 + a - b = 2 + a + b \]

### Step 4: Simplify the equation
Subtract \( a \) from both sides:
\[ -2 - b = 2 + b \]
Now, add \( b \) to both sides:
\[ -2 = 2 + 2b \]
Subtract 2 from both sides:
\[ -4 = 2b \]
Divide by 2:
\[ b = -2 \]

### Step 5: Find \( f' \) and evaluate at \( c = \frac{1}{2} \)
Next, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 + ax^2 + bx) = 6x^2 + 2ax + b \]

Now, substitute \( x = \frac{1}{2} \):
\[ f'\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^2 + 2a\left(\frac{1}{2}\right) + b = 6\left(\frac{1}{4}\right) + a + b \]
\[ = \frac{3}{2} + a + b \]

### Step 6: Set \( f'\left(\frac{1}{2}\right) = 0 \)
Now, we set this equal to zero:
\[ \frac{3}{2} + a + b = 0 \]
Substituting \( b = -2 \):
\[ \frac{3}{2} + a - 2 = 0 \]
\[ \frac{3}{2} + a - \frac{4}{2} = 0 \]
\[ a - \frac{1}{2} = 0 \implies a = \frac{1}{2} \]

### Step 7: Calculate \( 2a + b \)
Now that we have \( a \) and \( b \):
\[ 2a + b = 2\left(\frac{1}{2}\right) + (-2) = 1 - 2 = -1 \]

### Final Answer
Thus, the value of \( 2a + b \) is:
\[ \boxed{-1} \]