Question

If the function \(f(x) = ax^3-bx^2-8x-4\) satisfies Roll's theorem in \([1,3]\), if \(f^'(2) = 0\) then \(a-b\) is equal to...

• A. \(2\)
• B. \(-2\)
• C. \(0\)
• D. \(1\)

Hint:

Use f(1) = f(3) and f'(2) = 0 together.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Rolle's theorem needs \(f(1)=f(3)\) along with continuity and differentiability. The point \(c\) with \(f'(c)=0\) is given as \(c=2\).

Step 2: Use f(1) = f(3):
\(f(1)=a-b-8-4=a-b-12\). \(f(3)=27a-9b-24-4=27a-9b-28\). Equate:
\[ a-b-12=27a-9b-28\ \Rightarrow\ 13a-4b=8 \]

Step 3: Use f'(2) = 0:
\(f'(x)=3ax^2-2bx-8\), so \(f'(2)=12a-4b-8=0\), which gives \(3a-b=2\).

Step 4: Solve:
From \(b=3a-2\): \(13a-4(3a-2)=8\), so \(a+8=8\) and \(a=0\). Then \(b=-2\).

Step 5: Find a - b:
\(a-b=0-(-2)=2\), option (A).

Final Answer:
Solving the two conditions gives a = 0, b = -2, so a - b = 2. \[ \boxed{2} \]