Question

If $f\left(x\right)=\left\{\begin{array}{ll}x,& \mathrm{if}x\mathrm{is}\mathrm{rational}\\ 0,& \mathrm{if}x\mathrm{is}\mathrm{irrational}\end{array}\right.$ and $g\left(x\right)=\left\{\begin{array}{ll}0,& \mathrm{if}x\mathrm{is}\mathrm{rational}\\ x,& \mathrm{if}x\mathrm{is}\mathrm{irrational}\end{array}\right.$then $\left(f-g\right)\left(x\right)$ is:

• A. one-one and into
• B. neither one-one nor onto
• C. many-one and onto
• D. one-one and onto

Answer 1

The correct option is D

one-one and onto

Explanation for the correct option:

Finding $\left(f-g\right)\left(x\right)$:

We have

$f\left(x\right)=\left\{\begin{array}{ll}x,& \mathrm{if}x\mathrm{is}\mathrm{rational}\\ 0,& \mathrm{if}x\mathrm{is}\mathrm{irrational}\end{array}\right.$

$g\left(x\right)=\left\{\begin{array}{ll}0,& \mathrm{if}x\mathrm{is}\mathrm{rational}\\ x,& \mathrm{if}x\mathrm{is}\mathrm{irrational}\end{array}\right.$

So,

$\left(f-g\right)\left(x\right)=\left\{\begin{array}{ll}x,& \mathrm{if}x\mathrm{is}\mathrm{rational}\\ -x,& \mathrm{if}x\mathrm{is}\mathrm{irrational}\end{array}\right.$

Domain of $\left(f-g\right)\left(x\right)$ will be (rational $\cup$ irrational) = Real Numbers and its range is codomain.

So, for any $x$,there is only one value of $\left(f-g\right)\left(x\right)$ and $x\in \mathrm{ℝ}$ and $\left(f-g\right)\left(x\right)\in \mathrm{ℝ}$ so it is onto function.

Hence, option D is correct.