Question

If Rolle's theorem holds for the function \(x^3+\mathrm{a} x^2+\mathrm{b} x, 1 \leq x \leq 2\) at the point \(\frac{4}{3}\), then the values of \(a\) and \(b\) are respectively

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

Hint:

Rolle theorem needs f(1) = f(2) and f prime equal to zero at the given point; solve these two equations.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
If Rolle theorem holds on \([1,2]\), then \(f(1) = f(2)\) and \(f'(c) = 0\) at \(c = \tfrac{4}{3}\).

Step 2: Use f(1) = f(2)
\(1 + a + b = 8 + 4a + 2b \Rightarrow 3a + b = -7\).

Step 3: Use f'(4/3) = 0
\(f'(x) = 3x^2 + 2ax + b\), so \(3\cdot\dfrac{16}{9} + \dfrac{8a}{3} + b = 0 \Rightarrow 8a + 3b = -16\).

Step 4: Solve
From the first equation \(b = -7 - 3a\). Substituting: \(8a - 21 - 9a = -16\), so \(a = -5\) and \(b = 8\).

Final Answer:
The values are a = -5 and b = 8. \[ \boxed{\text{(D) }-5, 8} \]