SQL
February 12, 2024Question 9541 – SQL
February 12, 2024GATE 2003
Question 20 |
Consider the following three claims
I. (n + k)m = Θ(nm), where k and m are constants II. 2n+1 = O(2n) III. 22n+1 = O(2n)
Which of these claims are correct?
I and II
| |
I and III | |
II and III | |
I, II, and III |
Question 20 Explanation:
I) (n+k)m = Θ(nm)
Which is true by considering leading ordered term present in polynomial expression.
II) 2n+1 = Θ(nm) → True

2n×2n can’t be written as Θ(2n)
So, this is False.
Which is true by considering leading ordered term present in polynomial expression.
II) 2n+1 = Θ(nm) → True
2n×2n can’t be written as Θ(2n)
So, this is False.
Correct Answer: A
Question 20 Explanation:
I) (n+k)m = Θ(nm)
Which is true by considering leading ordered term present in polynomial expression.
II) 2n+1 = Θ(nm) → True

2n×2n can’t be written as Θ(2n)
So, this is False.
Which is true by considering leading ordered term present in polynomial expression.
II) 2n+1 = Θ(nm) → True
2n×2n can’t be written as Θ(2n)
So, this is False.
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