+1 vote
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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.

by Veteran (57.2k points)
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PS: this log comparision does not always work . we have to be little careful while applying this.

+1 vote