Functions Ques 19
- $f(x)=\left\{\begin{array}{ll}x, & \text { if } x \text { is rational } \\ 0, & \text { if } x \text { is irrational }\end{array}, g(x)= \begin{cases}0, & \text { if } x \text { is rational } \\ x, & \text { if } x \text { is irrational }\end{cases}\right.$ Then, $f-g$ is
(2005, 1M)
(a) one-one and into
(b) neither one-one nor onto
(c) many one and onto
(d) one-one and onto
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Answer:
Correct Answer: 19.(d)
Solution: (d) Let $\phi(x)=f(x)-g(x)= \begin{cases}x, & x \in Q \\ -x, & x \notin Q\end{cases}$
Now, to check one-one.
Take any straight line parallel to $X$-axis which will intersect $\phi(x)$ only at one point.
$\Rightarrow \phi(x)$ is one-one.
To check onto
As $f(x)=\left\{\begin{array}{ll}x, & x \in Q \\ -x, & x \notin Q\end{array}\right.$, which shows
$y=x$ and $y=-x$ for rational and irrational values
$\Rightarrow y \in$ real numbers.
$\therefore \quad$ Range $=$ Codomain $\Rightarrow$ onto
Thus, $f-g$ is one-one and onto.
BETA
