If Rolle's theorem holds for the function f(x) = x^3 - ax^2 + bx - 4, x [1, 2] with f'(4/3) = 0 , then the ordered pair (a, b) is equal to: - Mathematics MCQ Mathematics Mock Test - 4 2026 | ExamTest.live

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If Rolle's theorem holds for the function f (x) = x^{3} - a x^{2} + b x - 4, x \in [1, 2] with f^{'} (4/3) = 0, then the ordered pair (a, b) is equal to:

Solution & Step-by-step Explanation

Rolle's Theorem conditions on [1, 2] : f (1) = f (2) $ 1^{3} - a (1)^{2} + b (1) - 4 = 2^{3} - a (2)^{2} + b (2) - 4 $ 1 - a + b - 4 = 8 - 4 a + 2 b - 4 3 a - b = 7 \dots (i) f^{'} (4/3) = 0 $ f^{'} (x) = 3 x^{2} - 2 a x + b $ f^{'} (4/3) = 3 (16/9) - 2 a (4/3) + b = 0 \frac{16}{3} - \frac{8 a}{3} + b = 0 \Rightarrow 8 a - 3 b = 16 \dots (ii) Solve (i) and (ii):From (i), b = 3 a - 7 . Substitute in (ii): 8 a - 3 (3 a - 7) = 16 8 a - 9 a + 21 = 16 \Rightarrow - a = - 5 \Rightarrow a = 5 Then b = 3 (5) - 7 = 8 .Ordered pair (a, b) = (5, 8) .

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If Rolle's theorem holds for the function f (x) = x^{3} - a x^{2} + b x - 4, x \in [1, 2] with f^{'} (4/3) = 0, then the ordered pair (a, b) is equal to:

(5, 8)

(-5, 8)

(5, -8)

(-5, -8)

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