UGC NET CSE | June 2014 | Part 3 | Question: 61

Question

Given the following equalities : $E_{1}: n^{K+\in} + n^{K} \lg n = \theta (n^{K+\in})$ for all fixed $K$ and $\in, K \geq 0$ and $\in >0$. $E_{2}:n^{3}2^{n}+6n^{2}3^{n}=O (n^{3}2^{n})$. Which of the following is true ?

1. $E_{1}$ is correct and $E_{2}$ is correct.
2. $E_{1}$ is correct and $E_{2}$ is not correct.
3. $E_{1}$ is not correct and $E_{2}$ is correct.
4. $E_{1}$ is not correct and $E_{2}$ is not correct.

Answer 1

for E2:

FIND WHICH ONE IS LARGE

N^3*2^N=6N^2*3^N

N*2^N=6*3^N(NOW TAKE LOG ON BOTH SIDE)

LOG(N*2^N)=LOG(6*3^N)

LOG N +N LOG 2 =LOG 6 +N LOG 3

LOG N+ N =  1.5 N( LOG6=CONSTANT,,,LOG3=1.5)

LOG N  =0.5N

O(0.5N)=O(6*N^2*3^N)

SO E2 IS FALSE

FOR E1:

COMPARE N^E=logn

since E>0 N^E is large compared to logn

so E1 IS CORRECT

OPTION B...........E1 IS CORRECT AND E2 IS NOT CORRECT