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Question

If n be a positive integer and (7+43)n=p+β, where p is a positive integer and β is a proper fraction, then

A
(1β)(p+β)=2
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B
(1β)1/n+(p+β)1/n=14
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C
(1β)(p+β)=1
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D
(1β)1/n+(p+β)1/n=83
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Solution

The correct option is C (1β)(p+β)=1
Given
p+β=(7+43)n
0β<1
Assuming
f=(743)n
Clearly,
0<f<1
Now,
p+β+f=(743)n+(7+43)n=2[7n+nC27n2(43)2+]=2k, kNβ+f=2kpβ+f=Integer
As 0<β+f<2, so
β+f=1

So,
(p+β)(1β)=(p+β)(f)=(7+43)n×(743)n=(4916×3)n(p+β)(1β)=1

(1β)1/n+(p+β)1/n=f1/n+(p+β)1/n=(743)+(7+43)=14

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