Question

If the functions $f$ and $g$ are defined on $R\to R$ such that
$f\left(x\right)=\{\begin{array}{c}0,\text{if}x\text{is rational}\\ x,\text{if}x\text{is irrational}\end{array}$ and
$g\left(x\right)=\{\begin{array}{c}0,\text{if}x\text{is irrational}\\ x,\text{if}x\text{is rational}\end{array},$
then $(f-g)\left(x\right)$ is

• A. a bijective function
• B. neither a one-one nor an onto function
• C. a one-one but not an onto function
• D. an onto but not a one-one function

Answer 1

The correct option is A a bijective function

$(f-g)\left(x\right)=\{\begin{array}{c}-x,\text{if}x\text{is rational}\\ x,\text{if}x\text{is irrational}\end{array}$

Let $x_1,x_2\in$ rational $(f-g)\left(x_1\right)=(f-g)\left(x_2\right)\Rightarrow -x_1=-x_2\Rightarrow x_1=x_2\cdots \left(i\right)$

Let $x_1,x_2\in$ irrational
$(f-g)\left(x_1\right)=(f-g)\left(x_2\right)\Rightarrow x_1=x_2\cdots \left(ii\right)$

Let $x_1\in$ irrational, $x_2\in$ rational
$(f-g)\left(x_1\right)=(f-g)\left(x_2\right)\Rightarrow x_1=-x_2\Rightarrow x+x_2=0\cdots \left(iii\right)$
This is not possible as sum of rational and irrational cannot be $0.$
So, this is a contradiction to assumption $x_1\in$ irrational, $x_2\in$ rational
$\therefore$ From equations $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we get
$(f-g)\left(x\right)$ is one-one function.

Let $y\in$ rational
$\Rightarrow y=(f-g)\left(x\right)\Rightarrow y=-x$
$\Rightarrow x=-y\in$ rational $\cdots \left(iv\right)$

$y\in$ Irrational
$\Rightarrow y=(f-g)\left(x\right)\Rightarrow y=x$
$\Rightarrow x\in$ irrational $\cdots \left(v\right)$

$\therefore$ from $\left(iv\right)$ and $\left(v\right)$ $(f-g)\left(x\right)$ is onto function.
Hence $(f-g)\left(x\right)$ is a bijective function.