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: Check the pair in the options
Test option D, $a = -5, b = 8$.

Step 2: f(1) and f(2)
$f(1) = 1 - 5 + 8 = 4$ and $f(2) = 8 - 20 + 16 = 4$, so they are equal.

Step 3: Derivative
$f'(\tfrac{4}{3}) = 3\cdot\tfrac{16}{9} - 10\cdot\tfrac{4}{3} + 8 = \tfrac{16}{3} - \tfrac{40}{3} + \tfrac{24}{3} = 0$.

Final Answer:
The pair -5 and 8 satisfies both conditions. \[ \boxed{\text{(D) }-5, 8} \]