If Rolle's theorem holds for the function f(x) = x³ - ax² + bx - 4, x ∈ [1,2] with f'( 4/3 ) = 0, then ordered pair (a, b) is equal to :

Solution

Maths sir · 4 steps

Rolle's theorem gives you two equations, f(1) = f(2) and f'(4/3) = 0. Solve them together for a and b.

1. Equal end values. Rolle's theorem needs f(1) = f(2). Here f(1) = 1 - a + b - 4 = b - a - 3 and f(2) = 8 - 4a + 2b - 4 = 4 - 4a + 2b. Setting them equal gives 3a - b = 7.
2. Derivative condition. f'(x) = 3x² - 2ax + b, so f'(4/3) = 16/3 - 8a/3 + b = 0. Multiplying by 3 gives 8a - 3b = 16.

   3a - b = 7, 8a - 3b = 16

3. Solve together. From the first equation, b = 3a - 7. Substituting gives 8a - 9a + 21 = 16, so a = 5 and b = 8. Check: f(1) = 8 - 5 - 3 = 0 and f(2) = 4 - 20 + 16 = 0, so the end values match.
4. So. (a, b) = (5, 8), which is option (d).

The idea to keep. Rolle's theorem on [p, q] always gives you f(p) = f(q) for free. Pair that with f'(c) = 0 and you have two equations for two unknowns.

Answer from the NTA final answer key. Solution by PrepWiser.