f(n) = n (n log n)
g(n) = (n log n) $(log n)^{9}$
cancelling n log n on both sides we have f(n)= n, and g(n) = $(log n)^{9}$
n < (log n)9
example: 1. n = 1 , then f(n) = 1 and g(n) = 0.
2. n = 4 , then f(n) = 4 and g(n) = $2^{9}$
3. n = 128 / $2^{7}$, then f(n) = 128 and g(n) = $7^{9}$
4. n= 1024 / $2^{10}$ , then f(n) = 1024 and g(n) = $10^{9}$
Therefore f(n) < c g(n) where $n_{o}$ = 4. Hence f(n) = O(g(n)).
The answer is a.