If $f(x)=\left\{\begin{matrix}x,&if\,x\,is\,rational\\0,&if\,x\,is\,irrational\end{matrix}\right.$ and, $g(x)=\left\{\begin{matrix}0,&if\,x\,is\,rational\\x,&if\,x\,is\,irrational\end{matrix}\right.$ Then, $f-g$ is

one-one and onto

The correct answer is Option (4) → one-one and onto

We have,

$(f-g) (x) = f(x) - g(x)$

$⇒(f-g) (x) =\left\{\begin{matrix}x-0=x,&if\,x\,is\,rational\\0-x=-x,&if\,x\,is\,irrational\end{matrix}\right.$

Let $h=f-g$.

Let x, y be any two distinct real numbers. Then,

$x ≠ y ⇒ -x≠-y$

$∴x≠y⇒ h(x)≠ h (y) ⇒ (f−g) (x) ≠ (f − g) (y)$

$⇒ f -g$ is one-one.

Let y be any real number.

If y is a rational number, then $h (y) = y$ i.e. $(f-g) (y) = y$.

If y is an irrational number, then $h (- y) = y$

i.e.$ (f-g) (- y) = y$

Thus, every $y∈R$ (Co-domain) has its pre-image in R (domain). So, $f-g: R→R$ is onto.

Hence, $f-g: R→ R$ is both one-one and onto.