
If $R = \{ (x,y):x,y \in I\,and\,{x^2} + {y^2} \leqslant 4\} $ is a relation in I, then the number of elements in domain of R is
Answer
162.6k+ views
Hint: First we will put $x = 0$ in the given equation ${x^2} + {y^2} \leqslant 4$ to get the value of y. And in the same way will put $x = \pm 1$ and $x = \pm 2$ in the given equation and find the value of y. As we will see those values will satisfy the equation so that that will be the domain of the relation.
Complete step by step solution: Given, relation is $R = \{ (x,y):x,y \in I\,and\,{x^2} + {y^2} \leqslant 4\} $
${x^2} + {y^2} \leqslant 4$ (1)
$x,y \in I\,$
Put $x = 0$ in equation (1)
$0 + {y^2} \leqslant 4$
${y^2} \leqslant 4$
After solving, we will get
$ \Rightarrow y = 0, \pm 1, \pm 2$
Put $x = \pm 1$ in equation (1)
${( \pm 1)^2} + {y^2} \leqslant 4$
$1 + {y^2} \leqslant 4$
Subtracting 1 from both the sides
${y^2} \leqslant 3$
After solving, we will get
$ \Rightarrow y = 0, \pm 1$
Put $x = \pm 2$ in equation (2)
${( \pm 2)^2} + {y^2} \leqslant 4$
$4 + {y^2} \leqslant 4$
Subtracting 4 from both the sides
${y^2} \leqslant 0$
After solving, we will get
$ \Rightarrow y = 0$
Now if we put $x > 2$ then $y^2 < 0$ which is not possible.
Hence, Domain of R is $\{ 0, \pm 1, \pm 2\} $
That is domain of R is $\{ - 1, - 2,0,1,2\} $
Therefore, the number of elements in the domain of R is $5$
Note: While solving the question Students should use the accurate value of x. They should choose only integers as it is given in the question that $x,y \in I\,$, and should avoid any other types of numbers.
Complete step by step solution: Given, relation is $R = \{ (x,y):x,y \in I\,and\,{x^2} + {y^2} \leqslant 4\} $
${x^2} + {y^2} \leqslant 4$ (1)
$x,y \in I\,$
Put $x = 0$ in equation (1)
$0 + {y^2} \leqslant 4$
${y^2} \leqslant 4$
After solving, we will get
$ \Rightarrow y = 0, \pm 1, \pm 2$
Put $x = \pm 1$ in equation (1)
${( \pm 1)^2} + {y^2} \leqslant 4$
$1 + {y^2} \leqslant 4$
Subtracting 1 from both the sides
${y^2} \leqslant 3$
After solving, we will get
$ \Rightarrow y = 0, \pm 1$
Put $x = \pm 2$ in equation (2)
${( \pm 2)^2} + {y^2} \leqslant 4$
$4 + {y^2} \leqslant 4$
Subtracting 4 from both the sides
${y^2} \leqslant 0$
After solving, we will get
$ \Rightarrow y = 0$
Now if we put $x > 2$ then $y^2 < 0$ which is not possible.
Hence, Domain of R is $\{ 0, \pm 1, \pm 2\} $
That is domain of R is $\{ - 1, - 2,0,1,2\} $
Therefore, the number of elements in the domain of R is $5$
Note: While solving the question Students should use the accurate value of x. They should choose only integers as it is given in the question that $x,y \in I\,$, and should avoid any other types of numbers.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
