Question

A relation R is defined on the set Z of integers as follows: \[R=\left( x,y \right)\in R:{{x}^{2}}+{{y}^{2}}=25\]. Express \[R\] and \[{{R}^{-1}}\] as the set of ordered pairs and hence find their respective domains
a) 0
b) Domain of $R=\{0,\pm 3\}$ = Domain of ${{R}^{-1}}$
c) Domain of $R=\{0,\pm 3,\pm 4\}$ = Domain of ${{R}^{-1}}$
d) Domain of $R=\{0,\pm 3,\pm 4,\pm 5\}$ = Domain of ${{R}^{-1}}$

Answer 1

Hint: In this question, we are given a relation with a given property of the elements x and y. Therefore, we should try to find the values of x and y for which the given relation is satisfied. Then, we can form these ordered pairs and the values of x in these ordered pairs should be the domain of R. Also, as ${{R}^{-1}}$ is the inverse relation, the ordered pairs for ${{R}^{-1}}$ can be found by just interchanging the values of x and y in the ordered pairs for $R$ and thus the first elements of these ordered pairs will give us the domain of ${{R}^{-1}}$ .

Complete step by step solution:
The given relation is
 \[\begin{align}
  & R=\left( x,y \right)\in R:{{x}^{2}}+{{y}^{2}}=25 \\
 & \Rightarrow y=\pm \sqrt{25-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}..............................(1.1) \\
\end{align}\] .
Also, as the relation is given to be on the set of integers, we should find integer values of x and y such that equation (1.1) is satisfied………………………………..(1.2)
We know from (1.2) that as y is an integer, it should be real. Thus, the value inside the square root in equation (1.1) should be positive. Therefore,
$\begin{align}
  & {{5}^{2}}-{{x}^{2}}\ge 0 \\
 & \Rightarrow {{x}^{2}}\le {{5}^{2}}\Rightarrow \left| x \right|\le \left| 5 \right| \\
 & \Rightarrow \left| x \right|\le 5............................(1.3) \\
\end{align}$
Therefore, from equation (1.2) and (1.3), we find that the values of y can be -5, -4, -3, -2, -1, 0, 1, 2, 3, 4 and 5. Also, we can find the corresponding values of y as
If x=-5, $y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( -5 \right)}^{2}}}=\pm \sqrt{25-25}=0$
If x=-4, \[y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( -4 \right)}^{2}}}=\pm \sqrt{25-16}=\pm \sqrt{9}=\pm 3\]
If x=-3, $y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( -3 \right)}^{2}}}=\pm \sqrt{25-9}=\pm \sqrt{16}=\pm 4$
If x=-2, $y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( -2 \right)}^{2}}}=\pm \sqrt{25-4}=\pm \sqrt{21}=\pm 4.5825$
If x=-1, $y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( -1 \right)}^{2}}}=\pm \sqrt{25-1}=\pm \sqrt{24}=\pm 4.89$
If x=0, $y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{0}^{2}}}=\pm \sqrt{25}=\pm 5$
If x=1, $y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( 1 \right)}^{2}}}=\pm \sqrt{25-1}=\pm \sqrt{24}=\pm 4.89$
If x=2, \[y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( 2 \right)}^{2}}}=\pm \sqrt{25-4}=\pm \sqrt{21}=\pm 4.5825\]
If x=3, $y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( 3 \right)}^{2}}}=\pm \sqrt{25-9}=\pm \sqrt{16}=\pm 4$
If x=4, \[y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( 4 \right)}^{2}}}=\pm \sqrt{25-16}=\pm \sqrt{9}=\pm 3\]
If x=5, \[y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( 5 \right)}^{2}}}=\pm \sqrt{25-25}=0\] …………………………………………….(1.4)
Thus, the values of x for which x and y are integers and satisfy the condition are x=-5,-4,-3, 0, 3, 4, 5. Thus R can be represented in terms of ordered pairs as $\left\{ \left( -5,0 \right),\left( -4,\pm 3 \right),\left( -3,\pm 4 \right),\left( 0,\pm 5 \right),\left( 3,\pm 4 \right),\left( 4,\pm 3 \right),\left( 5,0 \right) \right\}$ domain of R should be the set of the first elements of these ordered pairs=$\left\{ 0,\pm 3,\pm 4,\pm 5 \right\}$ ……………………(1.5)
Similarly, to find the domain of the inverse relation, we can rewrite equation (1.1) to find the value of x from y as
\[\begin{align}
  & {{x}^{2}}+{{y}^{2}}=25 \\
 & \Rightarrow x=\pm \sqrt{25-{{y}^{2}}}=\pm \sqrt{{{5}^{2}}-{{y}^{2}}}..............................(1.6) \\
\end{align}\]
Now, as this exactly the same formula as (1.1) with x and y interchanged, we can follow the same procedures and thus obtain the same domain so the correct answer should be
Domain of $R=\{0,\pm 3,\pm 4,\pm 5\}$ = Domain of ${{R}^{-1}}$
Which matches option (d) given in the question.

Note: We should be careful to include the $\pm $ sign in equations (1.1) because the given relation is quadratic and hence for a given value of x, there should be two values of y satisfying the given relation. Similarly, while finding the values of y in equation (1.4), we should be careful to include both the positive and negative values to find the ordered pairs satisfying the given relation.