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CAT 2018 Slot 2 — QA Question 28

Higher Degree PolynomialsEasy

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

Answer & solution

Answer: 8

Solution

Easy

Factor the cubic by finding an integer root, then analyse the sign of the factored form. A squared factor never changes sign, so the sign is controlled by the remaining linear factor.

1

Find a root. Test small divisors of 2828. At n=2n=2:

23−11(2)2+32(2)−28=8−44+64−28=0⇒ (n−2) is a factor\begin{aligned} &2^3 - 11(2)^2 + 32(2) - 28 = 8 - 44 + 64 - 28 = 0\\ &\Rightarrow\ (n-2)\ \text{is a factor} \end{aligned}
2

Factor completely. Divide by (n−2)(n-2) and factor the resulting quadratic.

n3−11n2+32n−28=(n−2)(n2−9n+14)⇒ =(n−2)(n−2)(n−7)⇒ =(n−2)2(n−7)\begin{aligned} &n^3 - 11n^2 + 32n - 28 = (n-2)(n^2 - 9n + 14)\\ &\Rightarrow\ = (n-2)(n-2)(n-7)\\ &\Rightarrow\ = (n-2)^2(n-7) \end{aligned}
3

Solve the inequality. Since (n−2)2≥0(n-2)^2\ge 0 always, the expression is positive only when (n−7)>0(n-7)>0 and n≠2n\ne 2.

(n−2)2(n−7)>0⇒ n−7>0(square factor is positive for n≠2)⇒ n>7⇒ smallest integer n=8\begin{aligned} &(n-2)^2(n-7) > 0\\ &\Rightarrow\ n-7 > 0 \quad\text{(square factor is positive for }n\ne 2\text{)}\\ &\Rightarrow\ n > 7\\ &\Rightarrow\ \text{smallest integer } n = 8 \end{aligned}
n=8n = \textbf{8}
CAT 2018 Slot 2 QA Q28: The smallest integer n such that n 3 – 11n 2 + 32n – 28 > 0 is — Solution | TheCATExam