Question:
The smallest integer n such that n3 - 11n2 + 32n - 28 > 0 is
Correct Answer: 8
Given, $n^{3}-11 n^{2}+32 n-28>0$
When $n=2, n^{3}-11 n^{2}+32 n-28=0$
$\Rightarrow(n-2)\left(n^{2}-9 n+14\right)>28$
$\Rightarrow(n-2)(n-7)(n-2)>28$
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.
Given, $n^{3}-11 n^{2}+32 n-28>0$
When $n=2, n^{3}-11 n^{2}+32 n-28=0$
$\Rightarrow(n-2)\left(n^{2}-9 n+14\right)>28$
$\Rightarrow(n-2)(n-7)(n-2)>28$
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.
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