If the functions f ( x ) and g ( x ) are defined on such that f ( x ) = { 0 , x rational x , x irrational ; g ( x ) = { 0 , x irrational x , x rational then ( f - g ) ( x ) is [2005]

Question

If the functions $f (x)$ and $g (x)$ are defined on $ℝ \to ℝ$ such that

$f (x) =$ ${\begin{matrix} 0, & x \in rational \\ x, & x \in irrational \end{matrix};$ $g (x) =$ ${\begin{matrix} 0, & x \in irrational \\ x, & x \in rational \end{matrix}$

then $(f - g) (x)$ is [2005]

Smart Achievers · Accepted Answer

(1)

$Given f (x) and g (x) defined on ℝ \to ℝ$

and $f (x) =$ ${\begin{matrix} 0, & x \in rational \\ x, & x \in irrational \end{matrix}$

$g (x) =$ ${\begin{matrix} 0, & x \in irrational \\ x, & x \in rational \end{matrix}$

$∴$ $(f - g) : ℝ \to ℝ such that$

$(f - g) (x) =$ ${\begin{matrix} - x, & x \in rational \\ x, & x \in irrational \end{matrix}$

Since $(f - g) : ℝ \to ℝ$ , for any $x$ , there is only one value of $(f (x) - g (x))$ whether $x$ is rational or irrational. Moreover, as $x \in ℝ$ , $f (x) - g (x)$ also belongs to $R$ . Therefore $(f - g)$ is one-one onto.

Solution

Q.
If the functions $f (x)$ and $g (x)$ are defined on $ℝ \to ℝ$ such that

$f (x) =$ ${\begin{matrix} 0, & x \in rational \\ x, & x \in irrational \end{matrix};$ $g (x) =$ ${\begin{matrix} 0, & x \in irrational \\ x, & x \in rational \end{matrix}$

then $(f - g) (x)$ is [2005]

1 one-one & onto

2 neither one-one nor onto

3 one-one but not onto

4 onto but not one-one

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