Question

If the value of c prescribed in the Rolle’* theorem for the function f(x) = 2x(x – 3)n, n ∈ N on [0, 3] is 3/4, then find the value of n.

Answer 1

Given f(x) = 2x (x – 3)n

⇒ f'(x) = 2 (x – 3)n + 2nx (x – 3)n–1

Now, f'(c) = 0 gives

⇒ 2(c – 3)n + 2nc(c – 3)n–1 = 0

⇒ 2(c – 3)n = –2nc(c – 3)n–1

⇒ (c – 3)n = –nc(c – 3)n–1

⇒ (c – 3) = –nc

⇒ c (1 + n) = 3

⇒ (1 + n) = 3/c = 3/(3/4) = 4

⇒ n = 3

Hence, the value of n is 3