Which of the given options provides the increasing order of asymptotic…

Key insight: compare growth rates by taking logarithms or using standard o(.) relations.

• Definitions: f1(n)=2^n, f2(n)=n^{3/2}, f3(n)=n log n, f4(n)=n^{log n}.

• Compare f3 and f2:

  f3(n)/f2(n) = (n log n) / n^{3/2} = (log n) / n^{1/2}, which → 0 as n → ∞. So f3 grows slower than f2 (f3 = o(f2)).

• Compare f2 and f4:

  Take logarithms: log(f2)= (3/2) log n, while log(f4)= (log n)^2. Since (log n)^2 grows faster than log n, f4 grows faster than any fixed polynomial n^k. Thus f2 = o(f4).

• Compare f4 and f1:

  Write f4 = n^{log n} = 2^{(log n)^2}, while f1 = 2^n. Because (log n)^2 = o(n), we have 2^{(log n)^2} = o(2^n). So f4 grows slower than f1.

• Combine the comparisons to get the increasing order:

  f3(n) = n log n < f2(n) = n^{3/2} < f4(n) = n^{log n} < f1(n) = 2^n