Show that the function
$f: R \rightarrow R: f(x)=\left\{\begin{array}{l}1, \text { if } x \text { is rational } \\ -1, \text { if } x \text { is irrational }\end{array}\right.$
is many - one into.
Find
(i) $f\left(\frac{1}{2}\right)$
(ii) $f(\sqrt{2})$
(iii) $f(\pi)$
(iv) $f(2+\sqrt{3})$.
(i) $\mathrm{f}\left(\frac{1}{2}\right)$
Here, $x=1 / 2$, which is rational
$\therefore f(1 / 2)=1$
(ii) $\mathrm{f}(\sqrt{2})$
Here, $x=\sqrt{2}$, which is irrational
$\therefore f(\sqrt{2})=-1$
(iii) $f(\pi)$
Here, $x=\Pi$, which is irrational
$f(\pi)=-1$
(iv) $\mathrm{f}(2+\sqrt{3})$.
Here, $x=2+\sqrt{3}$, which is irrational
$\therefore f(2+\sqrt{3})=-1$
Ans. (i) 1 (ii) $-1$ (iii) $-1$ (iv) $-1$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.