TheCATExam
Sign in

CAT 2001 — QA Question 15

Higher Degree PolynomialsEasy

m is the smallest positive integer such that for any integer n > m, the quantity n3 – 7n2 + 11n – 5 is positive. What is the value of m?

Answer & solution

  • A

    4

  • 5

  • C

    8

  • D

    None of these

Solution

n3 − 7n2 + 11n − 5

For n = 1, the expression reduces to zero

∴ (n − 1) is a factor.

n3 − 7n2 + 11n − 5 = n3 − n2 − 6n2 + 6n + 5n − 5

                               = n2(n − 1) − 6n(n − 1) + 5(n − 1)

                               = (n − 1)(n2 − 6n + 5)

                               = (n − 1)(n − 1)(n − 5)

                               = (n − 1)2(n − 5)

Since, (n − 1)2 is always positive.

∴ The expression is positive only when n > 5.

∴ n = 6 and m = 5

Hence, option (b).

Alternatively,

Check using options. If we take m = 4 then n = 5, then we do not get a positive value of the given equation. Therefore, check the next higher value.

n = 6 and m = 5 gives the positive value.

Hence, option (b).

CAT 2001 QA Q15: m is the smallest positive integer such that for any integer n > m, the quantity n 3 – 7n 2 + 11n &ndash — Solution | TheCATExam