Question

Find the set of values of a for which the function f:R → R is given by f(x) = x³ + (a + 2)x² + 3ax + 5 is one-one.

Answer 1

Given f(x) = x³ + (a + 2)x² + 3ax + 5

Since the function is one-one, so f is monotonic.

Now, f'(x) = 3x² + 2(a + 2) x + 3a > 0

⇒ 4(a + 2)² – 36 < 0

⇒ (a + 2)² – 9 < 0

⇒ (a + 2)² – 3² < 0

⇒ (a + 2 + 3)(a + 2 – 3) < 0

⇒ (a + 5)(a – 1) < 0

⇒ –5 < a < 1