First slide

Functions (XII)

Question

$If the functions f (x) and g (x) are defined on R \to R such that f (x) = \{\begin{cases} 0, x \in rational \\ x, x \in irrational \end{cases}$
$and g (x) = \{\begin{matrix} 0, x \in irrational \\ x, x \in rational \end{matrix} then (f - g) (x) is$

Moderate

A

one-one and onto
B

neither one-one nor onto
C

one-one but not onto
D

onto but not one-one

Solution

We are given that
$\begin{matrix} f : R \to R, f (x) = \{\begin{matrix} 0, x \in rational \\ x, x \in irrational \end{matrix} \\ g : R \to R, g (x) = \{\begin{matrix} 0, x \in irrational \\ x, x \in rational \end{matrix} \\ ∴ (f - g) : R \to R such that \\ (f - g) (x) = \{\begin{cases} - x, if x \in rational \\ x, if x \in irrational \end{cases} \end{matrix}$
$Since (f - g) : R \to R for any x, there is only one value of (f (x) - g (x)) whether x is rational or irrational. Moreover,$
$as x \in R, f (x) - g (x) also belongs to R . Therefore, (f - g) is one-one and onto.$

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