Question

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. Then the values of $a$ and $b$ are respectively \dots

• A. 11, -6
• B. -6, 11
• C. -11, 6
• D. 6, -11

Hint:

For polynomial functions, adding or subtracting a standalone constant $b$ merely shifts the graph perfectly up or down without altering its endpoints' relative heights or the location of its horizontal tangents ($f'(c)=0$). Thus, $b$ cannot be uniquely determined by Rolle's Theorem alone.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are told a polynomial function satisfies all the conditions of Rolle's Theorem on the closed interval $[1, 3]$. We must determine the constants $a$ and $b$.

Step 2: Key Formula or Approach:
Rolle's Theorem states that for a function to be applicable on $[a, b]$, three conditions must be met:
1. $f(x)$ is continuous on $[a, b]$. (Always true for polynomials).
2. $f(x)$ is differentiable on $(a, b)$. (Always true for polynomials).
3. The values at the endpoints must be exactly equal: $f(a) = f(b)$.

Step 3: Detailed Explanation:
Apply the crucial third condition of Rolle's theorem to our interval endpoints $x=1$ and $x=3$:
$$f(1) = f(3)$$
Evaluate $f(1)$:
$$f(1) = (1)^3 - 6(1)^2 + a(1) + b = 1 - 6 + a + b = a + b - 5$$
Evaluate $f(3)$:
$$f(3) = (3)^3 - 6(3)^2 + a(3) + b = 27 - 54 + 3a + b = 3a + b - 27$$
Set them equal to each other:
$$a + b - 5 = 3a + b - 27$$
The constant $b$ cancels out on both sides:
$$a - 5 = 3a - 27$$
Rearrange to solve for $a$:
$$27 - 5 = 3a - a$$
$$22 = 2a \implies a = 11$$
Because the constant $b$ cancelled out during the evaluation, Rolle's theorem technically holds for any real value of $b$.
However, scanning the multiple-choice options, only option (a) provides $a = 11$. Therefore, we can confidently select it.

Step 4: Final Answer:
The values are 11 and -6, matching option (a).