# CAT 2001 | Question: 15

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

1. $4$
2. $5$
3. $8$
4. None of these

Given the equation is $n^3-7n^2+11n-5$, put $n=1$ we get: $1-7+11-5=0$, which means $n=1$ is one of the roots of the given cubic equation.

Now we can rewrite the equation as:

$(n-1)(n^2-6n+5)\implies(n-1)(n-1)(n-5)$

For $n<5$ it will give a negative value. This root is positive for $n> 5$  $n\geq m$ which is $m=6$

Option (D) is correct.

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