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

& 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{aligned}$$ . Also, as the relation is given to be on the set of integers, we should find integer values of x and y such that equation (1.1) is satisfied………………………………..(1.2) We know from (1.2) that as y is an integer, it should be real. Thus, the value inside the square root in equation (1.1) should be positive. Therefore, $\begin{aligned} & {{5}^{2}}-{{x}^{2}}\ge 0 \\\ & \Rightarrow {{x}^{2}}\le {{5}^{2}}\Rightarrow \left| x \right|\le \left| 5 \right| \\\ & \Rightarrow \left| x \right|\le 5............................(1.3) \\\ \end{aligned}$ Therefore, from equation (1.2) and (1.3), we find that the values of y can be -5, -4, -3, -2, -1, 0, 1, 2, 3, 4 and 5. Also, we can find the corresponding values of y as If x=-5, $y=\pm \sqrt{{{5}^{2}}-{{x}^{2}}}=\pm \sqrt{{{5}^{2}}-{{\left( -5 \right)}^{2}}}=\pm \sqrt{25-25}=0$ If x=-4, $$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{aligned} & {{x}^{2}}+{{y}^{2}}=25 \\\ & \Rightarrow x=\pm \sqrt{25-{{y}^{2}}}=\pm \sqrt{{{5}^{2}}-{{y}^{2}}}..............................(1.6) \\\ \end{aligned}$$ 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.