Question

The domain and range of the relation $R$ given by $R=\left\{(x, y): y=x+\frac{6}{x} ;\right.$ where $x, y \in N$ and $\left.x<6\right\}$ are

• A. $\{1,2,3\}$ and $\{3,5\}$
• B. $\{1,2,3\}$ and $\{5,7\}$
• C. $\{5,7\}$ and $\{1,2,3\}$
• D. $\{3,5\}$ and $\{1,2,3\}$

Answer 1

Answer: (B)

Given, $R=\left\{(x, y): y=x+\frac{6}{x}\right.$,
where $x, y \in N$ and $\left.x< 6\right\}$
For, $ x=1, y=1+\frac{6}{1}=7 $
$ x=2, y=2+\frac{6}{2}=5$
$x=3, y=3+\frac{6}{3}=5$
$x=4, y=4+\frac{6}{4}=\frac{11}{2} \notin N$
$x=5, y=5+\frac{6}{5}=\frac{31}{5} \notin N$
$\therefore$ The given relation in Roster form is
$R=\{(1,7),(2,5),(3,5)\}$
Hence, domain of $R=\{1,2,3\}$
Range of $R=\{7,5\}$