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Let $f : [0, 1] \rightarrow [0, 1]$ be defined by f(x) = { x, if is rational
1-x , if is irrational}.
Then (fof)x is

(a) constant

(b) 1 + x

(c) x

(d) None of these

Answers (1)

(c)

Here, $f : [0, 1] \rightarrow [0, 1]\\$

$\\f = f\textsuperscript{-1} \\ Therefore, (fof)x = x\\$

Posted by

infoexpert22

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