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If(5+26)n=I+f;n,I∈N and 0≤f<1, then I is equal to

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a

1f−f

b

11+f−f

c

11+f+f

d

11−f−f

answer is D.

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Detailed Solution

∵I+f+f′=(5+26)n+(5−26)n=2k [even integer]∴ f+f′=1Now, (I+f)f′=(5+26)n(5−26)n=(1)n=1⇒ (I+f)(1−f)=1 or I=1(1−f)−f

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If(5+26)n=I+f;n,I∈N and 0≤f<1, then I is equal to