We can analyze these functions by performing Log on both sides of the equeations. ( a relative comparision when n->$\infty$ ) 1. $\log f_{1} = \log (n!) = O(n\log n)$ 2. $\log f_{2} = \log (2n^{2}+n\log n) = O(\log n)$ 3. $\log f_{3} = \log (n^{2^{n}} + 6*2^{n}) = O(2^{n}\log n)$ comparing all these we find that $f_{3}$ is the biggest function. and $f_{2}$ is the smallest. or $f_{2} < f_{1} < f_{3}$ => (A) is False.
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