Question

Find the set of values of p for which the function f(x) = x³ – 2x² – px + 1 is one-one. 

Answer 1

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