Let fx be a function defined on -0- 1- such that fx - x- if x - Q 1-x- if x - Q -Then- for all x - -0- 1- ffx-

If x ϵ Q, f(x)=x f{f(x)}=f(x)=x If x ϵ̸ Q, f(x)=1 x f{f(x)}= f(1 x) = 1 (1 x) = x ⇒ f{f(x)} = x ∀ x ϵ [0,1] ...

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Byju's Answer

Standard XIII

Mathematics

Definition of Functions

Let fx be a f...

Question

Let f(x) be a function defined on [0, 1] such that $f (x) = {\begin{matrix} x, & i f x ϵ Q 1 - x, & i f x / ϵ Q \end{matrix}$
Then, for all $x ϵ [0, 1], f (f (x)) =$

constant

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$1 + x$

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$x$

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none of these

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Solution

The correct option is C
$x$

$I f x ϵ Q, f (x) = x f {f (x)} = f (x) = x I f x / ϵ Q, f (x) = 1 - x f {f (x)} = f (1 - x) = 1 - (1 - x) = x \Rightarrow f {f (x)} = x \forall x ϵ [0, 1]$

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Similar questions

Let f(x) be a function defined on [0, 1] such that $f (x) = {\begin{matrix} x, & i f x ϵ Q 1 - x, & i f x / ϵ Q \end{matrix}$
Then, for all $x ϵ [0, 1], f (f (x)) =$

Q. If

f (x)

is defined on

[0, 1]

by the rule

f (x) = {\begin{matrix} x & , & i f & x \in Q 1 - x & , & i f & x \notin Q \end{matrix}

Then, for all

x \in [0, 1]

prove that

f (f (x)) = x

Q. Let f(x) be a function defined on [0, 1] such that

f (x) = {\begin{matrix} x, & i f x ϵ Q 1 - x, & i f x / ϵ Q \end{matrix}

Then, for all

x ϵ [0, 1], f (f (x)) =

Q. Let

f (x)

be defined for all

x

ϵ

R

and continuous. Let

f (x + y) - f (x - y) = 4 x y

for all

x, y \in R

and

f (0) = 0

, then

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Let f(x) be a function defined on [0, 1] such that f(x)={x, if x ϵ Q1−x, if x /ϵ Q Then, for all x ϵ [0,1],f(f(x))=

The correct option is C x If x ϵ Q,f(x)=xf{f(x)}=f(x)=xIf x /ϵ Q,f(x)=1−xf{f(x)}=f(1−x)=1−(1−x)=x⇒f{f(x)}=x ∀ x ϵ [0,1]

Let f(x) be a function defined on [0, 1] such that $f (x) = {\begin{matrix} x, & i f x ϵ Q 1 - x, & i f x / ϵ Q \end{matrix}$
Then, for all $x ϵ [0, 1], f (f (x)) =$

The correct option is C
$x$

$I f x ϵ Q, f (x) = x f {f (x)} = f (x) = x I f x / ϵ Q, f (x) = 1 - x f {f (x)} = f (1 - x) = 1 - (1 - x) = x \Rightarrow f {f (x)} = x \forall x ϵ [0, 1]$