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