Question

\(m\) is the smallest positive integer such that for any integer \(n \ge m\), the quantity \(n^3 - 7n^2 + 11n - 5\) is positive. What is the value of \(m\)?

• A. 4
• B. None of these
• C. 8
• D. 5

Hint:

Factorise the cubic; it has a repeated root at n = 1 and another root at n = 5.

Answer 1

The Correct Option is B

Step 1: Factorise:
\(n^3 - 7n^2 + 11n - 5 = (n-1)^2(n-5)\). (Check: \(n = 1\) makes the cubic \(1-7+11-5 = 0\), and dividing out gives \((n-1)(n^2-6n+5) = (n-1)^2(n-5)\).)

Step 2: Sign of the expression:
\((n-1)^2 \ge 0\) always, so the sign depends on \(n-5\). The expression is positive exactly when \(n>5\) (and \(n \ne 1\)). At \(n = 5\) it equals 0, which is not positive.

Step 3: Find \(m\):
We need every integer \(n \ge m\) to give a positive value. \(m = 5\) fails because \(n = 5\) gives 0. \(m = 6\) works because every \(n \ge 6\) gives a positive value. So \(m = 6\).

Step 4: Compare with the options:
6 is not among 4, 5 and 8, so the answer is "None of these".

Final Answer:
The correct option is (2): None of these