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Official PYQ · 2003

GATE CS 2003 Q20 — Asymptotic Worst Case Time and Space Complexity

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Consider the following three claims:

  1. (n+k)m=Θ(nm)(n+k)^m = \Theta(n^m) where kk and mm are constants
  2. 2n+1=O(2n)2^{n+1} = O(2^n)
  3. 22n+1=O(2n)2^{2n+1} = O(2^n)

Which of the following claims are correct?

Choose one option
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Answer: A

We analyze each claim individually using the definitions of asymptotic notations:

  1. Claim I: (n+k)m=Θ(nm)(n+k)^m = \Theta(n^m) where k,mk, m are constants.
    Expanding using the binomial theorem, the leading term is nmn^m. Since kk and mm are constants, all other terms are of lower order. Thus, (n+k)m=Θ(nm)(n+k)^m = \Theta(n^m). This claim is correct.
  2. Claim II: 2n+1=O(2n)2^{n+1} = O(2^n).
    We can rewrite 2n+12^{n+1} as 2⋅2n2 \cdot 2^n. Since 2⋅2n≤c⋅2n2 \cdot 2^n \le c \cdot 2^n holds for any constant c≥2c \ge 2, we have 2n+1=O(2n)2^{n+1} = O(2^n). This claim is correct.
  3. Claim III: 22n+1=O(2n)2^{2n+1} = O(2^n).
    We can rewrite 22n+12^{2n+1} as 2⋅4n2 \cdot 4^n. There is no constant cc such that 2⋅4n≤c⋅2n2 \cdot 4^n \le c \cdot 2^n for all sufficiently large nn, because the ratio 2⋅4n2n=2⋅2n\frac{2 \cdot 4^n}{2^n} = 2 \cdot 2^n grows infinitely. This claim is incorrect.

Therefore, only claims I and II are correct.

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