Question

The smallest integer n such that n³ - 11n² + 32n - 28 > 0 is

Answer 1

Calculation:

Given, n³ – 11n² + 32n – 28 > 0

⇒ (n - 2)(n² – 9n + 14) > 0

⇒ (n - 2)(n - 7)(n - 2) > 0

⇒ For n < 2, (n – 2)(n – 7)(n – 2) is negative.
⇒ For 2 < n < 7, (n – 2)(n – 7)(n – 2) is negative.
⇒ For n > 7, (n – 2)(n – 7)(n – 2) is positive.
When n = 8, (n – 2)(n – 7)(n – 2) = 36, which is greater than 0.

The least integral value of n which satisfies the inequation is 8.