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=`1/2`,then the value of 2a+b is :

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Verify the conditions of Rolle's Theorem
According to Rolle's theorem, for the function \( f(x) = 2x^3 + ax^2 + bx \) to satisfy the theorem on the interval \([-1, 1]\), it must hold that:
1. \( f(-1) = f(1) \)
2. There exists at least one \( c \) in the interval \((-1, 1)\) such that \( f'(c) = 0 \).

### Step 2: Calculate \( f(1) \) and \( f(-1) \)
First, we calculate \( f(1) \):
\[ f(1) = 2(1)^3 + a(1)^2 + b(1) = 2 + a + b \]

Next, we calculate \( f(-1) \):
\[ f(-1) = 2(-1)^3 + a(-1)^2 + b(-1) = -2 + a - b \]

### Step 3: Set up the equation from the condition \( f(-1) = f(1) \)
Setting \( f(1) \) equal to \( f(-1) \):
\[ 2 + a + b = -2 + a - b \]

### Step 4: Simplify the equation
Subtract \( a \) from both sides:
\[ 2 + b = -2 - b \]

Adding \( b \) to both sides gives:
\[ 2 + 2b = -2 \]

Now, solving for \( b \):
\[ 2b = -2 - 2 \implies 2b = -4 \implies b = -2 \]

### Step 5: Use \( c = \frac{1}{2} \) to find \( a \)
Now we need to find \( a \) using the derivative \( f'(x) \):
\[ f'(x) = 6x^2 + 2ax + b \]
Substituting \( c = \frac{1}{2} \):
\[ f'\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^2 + 2a\left(\frac{1}{2}\right) + b = 0 \]
Calculating \( f'\left(\frac{1}{2}\right) \):
\[ f'\left(\frac{1}{2}\right) = 6 \cdot \frac{1}{4} + a + b = \frac{3}{2} + a + b \]
Substituting \( b = -2 \):
\[ \frac{3}{2} + a - 2 = 0 \]
This simplifies to:
\[ \frac{3}{2} + a - \frac{4}{2} = 0 \implies a - \frac{1}{2} = 0 \implies a = \frac{1}{2} \]

### Step 6: Calculate \( 2a + b \)
Now we can find \( 2a + b \):
\[ 2a + b = 2\left(\frac{1}{2}\right) + (-2) = 1 - 2 = -1 \]

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