Find the set of values of p for which the function f(x) = x3 – 2x2 – px + 1 is one-one.
Given f(x) = x3 – 2x2 – px + 1
f'(x) = 3x2 – 4x – p
Since f is one-one function
So, f is either strictly increasing or decreasing function
Thus, f'(x) > 0 (since, a = 3 > 0)
⇒ D < 0
⇒ 16 + 12p < 0
⇒ 4 + 3p < 0
⇒ p < – 4/3