You should never evaluate relative growth of functions by plugging in values.

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$$f(n) = n^2 \log n, \qquad g(n) = n \log^{10} n$$

Dividing both these functions by $(n \log n)$, we get:

$$F(n) = n, \qquad G(n) = \log^9 n$$

Since * any polynomial function grows faster than any logarithmic function*, we have:

$F(n)$ grows faster than $G(n)$

$\implies$ $f(n)$ grows faster than $g(n)$

$\implies n \log^{10} n = O(n^2 \log n)$

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These function have not uniformity in comparison actually .... ans is C

If F(n) = O(g(n)) then F(n) <= C g(n) which after simplfying gives

n < c (logn)^9 , now put n = 2^10 , 2^20 this will be true but put 2^60 , 2^80 it will fail the identity for same constant which we derived in for 2^10 or 2^20 .....

If F(n) = O(g(n)) then F(n) <= C g(n) which after simplfying gives

n < c (logn)^9 , now put n = 2^10 , 2^20 this will be true but put 2^60 , 2^80 it will fail the identity for same constant which we derived in for 2^10 or 2^20 .....

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