Question

Sum of the number of elements in the domain and the range of the relation R is given by $R = \{ (x,y):y = x + \dfrac{6}{x};|\,where\,x,y \in N|\,and\,x < 6\} $ is

Answer 1

Hint: First, we will write the possible values of $x$ as it is given $x < 6$. Then, we will put every possible value of x in a given function to get the value of y. And will select only those values which are natural numbers. Then will sum the elements in the domain and the range of the relation to get the required answer.
Complete step by step solution: Given, relation is $R = \{ (x,y):y = x + \dfrac{6}{x};|\,where\,x,y \in N|\,and\,x < 6\} $
$y = x + \dfrac{6}{x}$, $x < 6$ --- (1)
So, $x = 1,2,3,4,5$ as $x < 6$
When $x = 1$
Putting in the equation (1)
$y = 1 + \dfrac{6}{1}$
After solving the above expression
$y = 1 + 6$
After adding, we will get
$y = 7$
When $x = 2$
Putting in the equation (1)
$y = 2 + \dfrac{6}{2}$
After solving the above expression
$y = 2 + 3$
After adding, we will get
$y = 5$
When $x = 3$
Putting in the equation (1)
$y = 3 + \dfrac{6}{3}$
After solving the above expression
$y = 3 + 2$
After adding, we will get
$y = 5$
When $x = 4$
Putting in the equation (1)
$y = 4 + \dfrac{6}{4}$
After taking LCM, we will get
$y = \dfrac{{16 + 6}}{4}$
After adding, we will get
$y = \dfrac{{22}}{4}$
After solving the above expression
$y = 5.5 \notin N$
When $x = 5$
Putting in the equation (1)
$y = 5 + \dfrac{6}{5}$
After taking LCM, we will get
$y = \dfrac{{25 + 6}}{5}$
After adding, we will get
$y = \dfrac{{31}}{5}$
After solving the above expression
$y = 6.2 \notin N$
Thus, $R = \{ (1,7),\,(2,5),\,(3,5)\} $
Hence, Domain of relation is $\{ 1,2,3\} $
Range of relation is $\{ 5,7\} $
Sum of elements of domain and range of relation $ = 3 + 2$
$ = 5$
Hence, the sum of elements of domain and range of relation is $5$
Note: While writing the values of x students should not include 6 as it is given that $x < 6$. After getting the values of y should select only natural numbers not integers.