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.