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 Concept:
Rolle's Theorem requires: (i) $f(x)$ is continuous in $[1, 3]$, (ii) $f(x)$ is differentiable in $(1, 3)$, and (iii) $f(1) = f(3)$. Also, there exists $c \in (1, 3)$ such that $f'(c) = 0$.

Step 2: Formula Application:
Set $f(1) = f(3)$: $(1)^3 - 6(1)^2 + a(1) + b = (3)^3 - 6(3)^2 + a(3) + b$.

Step 3: Explanation:
$1 - 6 + a + b = 27 - 54 + 3a + b$ $-5 + a = -27 + 3a$ $22 = 2a \implies a = 11$. While $b$ can technically be any real number for the theorem to apply, looking at the options, if we assume the function has a root at $x=1$ or fits a standard polynomial form, $b = -6$ is the consistent choice provided.

Step 4: Final Answer:
The values are $a = 11$ and $b = -6$.