If f (x) is defined on [0, 1] by the rule f(x) = begincases x &: textx text is rational 1 - x &: textx text is irrational endcases then for all x ∈ R, f (f(x)) is

Question

If f (x) is defined on [0, 1] by the rule f(x) = begincases x &: textx text is rational 1 - x &: textx text is irrational endcases then for all x ∈ R, f (f(x)) is (A) constant (B) 1 + x (C) x (D) none of these

Tardigrade Admin · Accepted Answer

f (f (x)) = f (x) = x if x is rational = f (1 - x) = 1 - (1 - x) = x if x is irrational. Hence f (f (x)) = x ∀ x ∈ R

Q. If f (x) is defined on [0, 1] by the rule $f (x) = {x 1 - x : x is rational : x is irrational$
then for all x $\in$ R, f (f(x)) is

constant

1 + x

x

none of these

f (f (x)) = f (x) = x if x is rational
= f (1 - x) = 1 - (1 - x) = x if x is irrational.
Hence f (f (x)) = x $\forall$ x $\in$ R

Q. If f (x) is defined on [0, 1] by the rule f(x)={x1−x​:x is rational:x is irrational​ then for all x ∈ R, f (f(x)) is