Question

If Rolle's theorem holds for the function f(x) = x³ - ax² + bx - 4, x ∈ [1, 2] with f'(4/3) = 0, then ordered pair (a, b) is equal to :

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

Hint:

Rolle's theorem states that for a differentiable function where $f(a)=f(b)$, there must be at least one $c \in (a, b)$ such that $f'(c)=0$.

Answer 1

The Correct Option is A

Step 1: Rolle's theorem requires $f(1) = f(2)$. $f(1) = 1 - a + b - 4 = b - a - 3$. $f(2) = 8 - 4a + 2b - 4 = 2b - 4a + 4$. $b - a - 3 = 2b - 4a + 4 \Rightarrow 3a - b = 7 \quad \cdots (1)$.
Step 2: $f'(x) = 3x^2 - 2ax + b$. Given $f'(4/3) = 0$: $3(16/9) - 2a(4/3) + b = 0 \Rightarrow 16/3 - 8a/3 + b = 0 \Rightarrow 8a - 3b = 16 \quad \cdots (2)$.
Step 3: Solve (1) and (2). From (1), $b = 3a - 7$. Substitute in (2): $8a - 3(3a - 7) = 16 \Rightarrow 8a - 9a + 21 = 16 \Rightarrow -a = -5 \Rightarrow a = 5$.
Step 4: $b = 3(5) - 7 = 8$. (a, b) = (5, 8).