Finding Integral Part of Numbers of the Form a^b Where a Is Irrational
If n be a pos...
Question
If n be a positive integer and (7+4√3)n=p+β, where p is a positive integer and β is a proper fraction, then
A
(1−β)(p+β)=2
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B
(1−β)1/n+(p+β)1/n=14
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C
(1−β)(p+β)=1
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D
(1−β)1/n+(p+β)1/n=8√3
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Solution
The correct option is C(1−β)(p+β)=1 Given p+β=(7+4√3)n 0≤β<1
Assuming f=(7−4√3)n
Clearly, 0<f<1
Now, p+β+f=(7−4√3)n+(7+4√3)n=2[7n+nC2⋅7n−2⋅(4√3)2+⋯]=2k,k∈N⇒β+f=2k−p⇒β+f=Integer
As 0<β+f<2, so ⇒β+f=1
So, (p+β)(1−β)=(p+β)(f)=(7+4√3)n×(7−4√3)n=(49−16×3)n∴(p+β)(1−β)=1