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Question

If p=(8+37)n and f=p[p], then the value of p(1f) is
(where [.] denotes the greatest integer function)

A
1
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B
2
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C
2n
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D
22n
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Solution

The correct option is A 1
Given p=(8+37)n
Assuming p1=(837)n
Clearly, 0<p1<1
Now,
p+p1=(8+37)n+(837)n=2[ nC08n+ nC28n2(37)2+]=2k, kZ
Given
p=[p]+f[p]+f+p1=2k
f+p1=2k[p](1)
Which is integer
We know that,
p1(0,1), f(0,1)0<p1+f<2
Using equation (1), the only interger possible
f+p1=1p1=1f

Therefore,
p(1f)=pp1=[(8+37)n(837)n]=(649×7)n=1

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