The value of e in Rolle’s theorem show f(x) = 2x3 – 5x2 – 4x + 3

1.	The value of e in Rolle’s theorem show f(x) = 2x3 – 5x2 – 4x + 3, is x ∈[1/3, 3]A. 2B. \(-\frac{1}{3}\)C. -2D. \(\frac{2}{3}\)
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

The value of e in Rolle’s theorem show f(x) = 2x3 – 5x2 – 4x + 3, is x ∈[1/3, 3]A. 2B. \(-\frac{1}{3}\)C. -2D. \(\frac{2}{3}\)

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