Let f x be a function defined on -0-1- such that f x -left begin matrix 1-x- if - X - hot epsilon - Q -lend matrix right- Then- for all x -0-1- f f x -A- constantB- 1-xC- XD- none of these

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Definition of Functions

Let f x be a ...

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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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) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x^{3} + x^{2} - 16 x + 20}{(x - 2)^{2}}, & i f x \neq 2 k, i f x = 2 \end{matrix}

f f(x)be continuous for all x, then k =

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

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