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
Answer & explanation
Correct answer: option 4
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.