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UGC-NET | UGC NET CS 2018 July – II | Question 21

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  • Last Updated : 03 Jul, 2020

The solution of the recurrence relation
T(m) = T(3m / 4) + 1 is :
(A) θ (lg m)
(B) θ (m)
(C) θ (mlg m)
(D) θ (lglg m)


Answer: (A)

Explanation: Given,

T(m) 
= T(3m/4) + 1
= T(m/(4/3)) + 1 

In this problem we use Master’s Theorem:

T(m) = aT(m/b) + n^k logn 

Compare both equation,

we get a = 1 and b = 4/3 and k=0 

Now find (log a base b), i.e., (log 1 base 4/3), which is greater than (4/3)0
And whenever, (logb(a)) and bk is greater than answer is f(m)* log(m)
Here, f(m) = 1
Hence, correct answer 1*log(m) = log(m)

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