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Question

Let f(n)=[12+n100] where [x] denotes the integral part of x. Then the value of 100n=1f(n) is?

A
50
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B
51
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C
1
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D
None of these
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Solution

The correct option is B 51
For all the values of n < 50, f(n) = 0
And for all the n50,f(n)=1. Hence, 51 such values are there.

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