SSC IMD SA Mock 04 – Question 141

Question:

Arrange the following in the order of growth:
\[ O(n), O(n^2), O(2^n), O(\log n), O(n \log n), O(n^2 \log n) \]

निम्नलिखित को वृद्धि दर (order of growth) के क्रम में व्यवस्थित कीजिए:
\[ O(n), O(n^2), O(2^n), O(\log n), O(n \log n), O(n^2 \log n) \]

Options:

  • O(n), O(n^2), O(log n), O(2^n), O(n log n), O(n^2 log n)
  • O(n), O(2^n), O(2^n), O(log n), O(n^2 log n), O(n log n)
  • O(log n), O(n), O(n log n), O(n^2), O(n^2 log n), O(2^n)
  • O(n^2), O(n), O(2^n), O(log n), O(n log n), O(n^2 log n)
  • O(n), O(n^2), O(log n), O(2^n), O(n log n), O(n^2 log n)
  • O(n), O(2^n), O(2^n), O(log n), O(n^2 log n), O(n log n)
  • O(log n), O(n), O(n log n), O(n^2), O(n^2 log n), O(2^n)
  • O(n^2), O(n), O(2^n), O(log n), O(n log n), O(n^2 log n)

Correct Answer: c)

Explanation:

The standard asymptotic order of growth in increasing order is:
\[ O(\log n) < O(n) < O(n \log n) < O(n^2) < O(n^2 \log n) < O(2^n) \]
Hence, Option 3 is the correct sequence.

समय जटिलता की स्पर्शोन्मुख वृद्धि दर (order of growth) का बढ़ता हुआ सही क्रम निम्नलिखित है:
\[ O(\log n) < O(n) < O(n \log n) < O(n^2) < O(n^2 \log n) < O(2^n) \]
अतः विकल्प 3 सही क्रम दर्शाता है।

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