CAT 2001 | Question: 15

Question

$m$ 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 $m?$

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

Comments

given conditon in question is $n\geq m$

Answer 1

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.