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Question

Given f(1) = 2 and f(n + 1) =f(n)1f(n)+1nϵN then which of the following is/ are correct?


A

f(2015)=12

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B

(f(2012))f(2013)

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C

f(1001) = 2

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D

f(2015) = – 3

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Solution

The correct options are
A

f(2015)=12


B

(f(2012))f(2013)


C

f(1001) = 2


f(n+1)=f(n)1f(n)+1=f(n1)1f(n1)+11f(n1)1f(n1)+1+1=1f(n1) f(n+1)=1f(n1) (i)
In a similar way f(n1)=1f(n3) . . . (ii)
from (i) and (ii)
f(n+1)=11f(n3)=f(n3)or f(n+4)=f(n)nϵN
hence f(n) is a periodic sequence with period 4
Put n = 1, 2, 3
we get f(2) =13, f(3)=12, f(4)=3
Now f(2012) = f(4) = –3
f(2013) = f(1) = 2
f(2015) = f(3) = 12
f(1001) = f(1) = 2


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