If Rolle's theorem holds for the function `f(x)=x^(3)-ax^(2)+bx-4, x in [1,2]` with `f'((4)/(3))=0`, then ordered pair (a, b) is equal to :

Options

A. (5, 8)
B. (-5, 8)
C. (5, -8)
D. (-5, -8)

Answer

To solve the problem, we will follow these steps:

### 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.

Given the function \( f(x) = x^3 - ax^2 + bx - 4 \), we need to check the conditions of Rolle's theorem on the interval [1, 2].

### Step 2: Calculate \( f(1) \) and \( f(2) \)
First, we calculate \( f(1) \) and \( f(2) \):

\[ f(1) = 1^3 - a(1^2) + b(1) - 4 = 1 - a + b - 4 = b - a - 3 \]

\[ f(2) = 2^3 - a(2^2) + b(2) - 4 = 8 - 4a + 2b - 4 = 2 - 4a + 2b \]

### Step 3: Set \( f(1) = f(2) \)
According to Rolle's theorem, we set \( f(1) = f(2) \):

\[ b - a - 3 = 2 - 4a + 2b \]

### Step 4: Rearranging the equation
Rearranging the equation gives:

\[ b - a - 3 = 2 - 4a + 2b \]

Subtract \( b \) from both sides:

\[ -a - 3 = 2 - 4a + b \]

Now, rearranging gives:

\[ -a + 4a + b = 2 + 3 \]

This simplifies to:

\[ 3a + b = 5 \quad \text{(Equation 1)} \]

### Step 5: Find \( f' \) and set \( f'(\frac{4}{3}) = 0 \)
Next, we find the derivative of \( f(x) \):

\[ f'(x) = 3x^2 - 2ax + b \]

We need to evaluate \( f'(\frac{4}{3}) \):

\[ f'\left(\frac{4}{3}\right) = 3\left(\frac{4}{3}\right)^2 - 2a\left(\frac{4}{3}\right) + b \]

Calculating \( \left(\frac{4}{3}\right)^2 = \frac{16}{9} \):

\[ f'\left(\frac{4}{3}\right) = 3 \cdot \frac{16}{9} - \frac{8a}{3} + b = \frac{48}{9} - \frac{8a}{3} + b = \frac{16}{3} - \frac{8a}{3} + b \]

Setting \( f'(\frac{4}{3}) = 0 \):

\[ \frac{16}{3} - \frac{8a}{3} + b = 0 \]

### Step 6: Rearranging the derivative equation
Rearranging gives:

\[ b = \frac{8a}{3} - \frac{16}{3} \quad \text{(Equation 2)} \]

### Step 7: Solve the system of equations
Now we have two equations:

1. \( 3a + b = 5 \)
2. \( b = \frac{8a}{3} - \frac{16}{3} \)

Substituting Equation 2 into Equation 1:

\[ 3a + \left(\frac{8a}{3} - \frac{16}{3}\right) = 5 \]

Multiplying through by 3 to eliminate the fraction:

\[ 9a + 8a - 16 = 15 \]

This simplifies to:

\[ 17a = 31 \quad \Rightarrow \quad a = \frac{31}{17} \]

### Step 8: Substitute back to find \( b \)
Now substituting \( a \) back into Equation 1 to find \( b \):

\[ 3\left(\frac{31}{17}\right) + b = 5 \]

Calculating \( 3 \cdot \frac{31}{17} = \frac{93}{17} \):

\[ b = 5 - \frac{93}{17} = \frac{85}{17} - \frac{93}{17} = -\frac{8}{17} \]

### Final Ordered Pair
Thus, the ordered pair \( (a, b) \) is:

\[ \left(\frac{31}{17}, -\frac{8}{17}\right) \]