Question

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

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

Answer 1

The correct option is D None of these

Let $y=n^3-7n^2+11n-5$
$=(n-1)(n^2-6n+5)=(n-1{)}^2(n-5)$
Now, $(n-1{)}^2$ is always positive. And for n < 5, the expression gives a negative quantity. Therefore, the least value of n will be 6. Hence m = 6.