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UGC NET CS 2020 Official Paper

Option 2 : \({2^{{{10}^6}}}and\;{10^{{12}}}\)

Official Paper 1: Held on 24 Sep 2020 Shift 1

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50 Questions
100 Marks
60 Mins

The correct answer is **option 2.**

**Explanation:**

A microsecond is 10^{-6} seconds. Hence, one second =10^{6 }microseconds

One hour=60x60x106= 3.6x10^{9 }microseconds

One month=2.592x10^{12}microseconds

One century=3.1104x1015microseconds

f(n)=logn in this case,the largest value n such that logn <=10^{6}.

We rewrite as, 2^{logn }<= 2^{(10)^6}

thus, 2^{1000000 }= (2^{10})^{100000 }= (10^{3})100000 = 10^{300000}

Hence, log2(n) and √n, In one second

\({2^{{{10}^6}}}and\;{10^{{12}}}\) the value will be **10 ^{6} and 10^{6}**

Hence the correct answer is **option 2.**