A) 4
B) 5
C) 2
D) 3
Correct Answer: D
Solution :
We know that \[e=\underset{n\to \infty }{\mathop{\lim }}\,\,{{\left( 1+\frac{1}{n} \right)}^{n}}\] and \[2<e<3\]. \ \[{{(1+0.0001)}^{10000}}<3\] (By putting n = 10000) Also \[{{(1+0.0001)}^{10000}}=1+10000\times {{10}^{-4}}\]\[+\frac{10000\times 9999}{2!}\times {{10}^{-8}}+......\]upto10001 terms \[\Rightarrow {{(1+0.0001)}^{10000}}>2\]. Hence 3 is the positive integer just greater than \[{{(1+0.0001)}^{10000}}>2\]. Hence D is the correct option.
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