See Our team

Wondering how we keep quality?

Got unsolved questions?

Ask Questions

Engineering
GATE
CBSE
NCERT
Psychology
English
Computer
Constitution
Astrology
Yoga
Economics
Physics
Biology
Electronics
Microprocessor
Career
Interview
Anatomy
Botany

## Consider the following functions from positives integers to real numbers 10, √n, n, log2n, 100/n. The CORRECT arrangement of the above functions in increasing order of asymptotic complexity is: -gate computer science 2017

A) log2n, 100/n, 10, √n, n

B) 100/n, 10, log2n, √n, n

C) 10, 100/n ,√n, log2n, n

D) 100/n, log2n, 10 ,√n, n

By:satyashiromani

Taged users:

|satyashiromani|milan-ransingh|deepuckraj|milanyoyoyogmailcom

Likes:

|deepuckraj

Dislikes:

Be first to dislike this question

Talk about thisDelete|Like|Dislike|

## Answers

**D) 100/n, 10, log˅2n, √n, n**

__Explanation__:

For the large number, value of inverse of number is less than a constant and value of constant is less than value of square root. 10 is constant, not affected by value of n. √n Square root and log2n is logarithmic. So log2n is definitely less than √n n has linear growth and 100/n grows inversely with value of n. For bigger value of n, we can consider it 0, so 100/n is least and n is max. So the increasing order of asymptotic complexity will be :

100/n < 10 < log2n < √n < n

So, option (b) is true.

deepuckraj

Likes:

Be first to like this answer

Dislikes:

Be first to dislike this answer

Talk about this|Once you have earned teacher badge you can edit this questionDelete|Like|Dislike|

------------------------------------

#### Can you help us to add better answer here? Please see this

Loading...

## Use Me ?