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Question

Let f(x) be a function defined by f(x) = x - [x], 0 x ϵ R where [x] is the greatest integer less than or equal to x. Then the number of solutions of f(x)+f(1x)=1 are :



Your Answer
A

0

Correct Answer
B

Infinite

Your Answer
C

1

Your Answer
D

2


Solution

The correct option is B

Infinite


Given f(x) = x - [x]. 0 x ϵ R.

f(x) + f(1x) = x - [x] + 1x[1x]=1 

x + 1x1=[x]+[1x]

x2+1xx = ( integer) k (Say).

x2(k+1)x+1=0

x is real so (k+1)2 - 4 0      [ b24ac0

or k22k30     (k+3)(k-1)

k -3       or    k 1 .

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