Question

The domain and range of the relation $R$ given by $R=\{(x,y):y=x+\frac{6}{x};$ where $x,y\in N$ and $x<6\}$ 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, $R=\{(x,y):y=x+\frac{6}{x}$,
where $x,y\in N$ and $x<6\}$
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 $R=\{7,5\}$