madeeasy test series

Question

Consider the following functions:

• $f1=n^{\log n}$
• $f2=2^n$
• $f3=2^{\sqrt{n}}$

Which of the following is true with regard to asymptotic growth?

A. $f1\leq f3\leq f2$

B. $f3\leq f1\leq f2$

Comments

B. f3 <= f1 <= f2

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Its A.

f1 <= f3 <= f2

try with n = 2²⁰ or any bigger value

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$1)\ n^{logn} = \Large 2^{log\ n^{(logn)}}$ $=2^{log^2n}$

$2)\  2^n$

$3)2^{\sqrt{n}}$

$\therefore n>\sqrt{n}>log^2n \Rightarrow f_2>f_3>f_1$

Answer 1

Option A. Put n as 2^2^128...