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in Continuity and Differentiability by (28.7k points)
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The value of e in Rolle’s theorem show f(x) = 2x3 – 5x2 – 4x + 3, is x ∈[1/3, 3]

A. 2

B. \(-\frac{1}{3}\)

C. -2

D. \(\frac{2}{3}\)

1 Answer

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Best answer

Correct answer is A.

f(x) = 2x3 – 5x2 – 4x + 3

f’(x) = 6x2 – 10x – 4

f’(c) = 6c2 – 10c – 4

\(\therefore\) f’(c) = 0

\(\therefore\) 6c2 – 10c – 4 = 0

3c2 – 5c – 2 = 0

3c2 + c – 6c – 2 = 0

c (3c + 1) – 2 (3c + 1) = 0

(3c + 1) (c – 2) = 0

c = 2 or c = \(-\frac{1}{3}\)

\(\therefore\) c = 2 Є \(\big(\frac{1}{3}, 3 \big)\)

Thus, as per Rolle’s Theorem, c =  2 Є \(\big(\frac{1}{3}, 3 \big).\)

So, the required value of c = 2

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