Question

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

Answer 1

Correct Answer: 8

Given n3-11n2+32n-28>0
When n = 2, n3-11n2+32n-28 = 0
⇒ (n-2)(n2-9n+14)>28
⇒ (n-2)(n-7)(n-2)>28
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 28. Least integral value of n which satisfies the inequation is 8.