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Let f be a function defined on [0, 1] such that $f (x) = \{\begin{cases} x x \in Q \\ 1 - x, x \notin Q \end{cases}$ Then for all x ∈ [0, 1], fof (x) is

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Correct option is C

We have, f(x)=x x∈Q1−x, x∉Q∴fof(x)=f(x), f(x)∈Q1−f(x), f(x)∉Q=x,x∈Q,x∈Q1−x,x∉Q,x∈Q1−x,1−x∈Q,x∉Q1−(1−x),1−x∉Q,x∉Q=x, x∈Q x, x∉Q ∴fof(x)=x,x∈[0,1].

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Let f be a function defined on [0, 1] such that f(x)=x x∈Q1−x, x∉Q Then for all x ∈ [0, 1], fof (x) is

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Let f be a function defined on [0, 1] such that $f (x) = \{\begin{cases} x x \in Q \\ 1 - x, x \notin Q \end{cases}$ Then for all x ∈ [0, 1], fof (x) is