Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Sets and Relations question

2021 · 26 Feb · Shift 1 · Q28
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Sets and Relations
  5. /2021 · 26 Feb · Shift 1 · Q28

Sets and Relations question

2021 · 26 Feb · Shift 1 · Q28

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let R = {(P, Q) | P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1, −-− 1) is the set :
  1. A
    S={(x,y)∣x2+y2=2}S = \{ (x,y)|{x^2} + {y^2} = \sqrt 2 \}S={(x,y)∣x2+y2=2​}
  2. B
    S={(x,y)∣x2+y2=2}S = \{ (x,y)|{x^2} + {y^2} = 2\}S={(x,y)∣x2+y2=2}
  3. C
    S={(x,y)∣x2+y2=1}S = \{ (x,y)|{x^2} + {y^2} = 1\}S={(x,y)∣x2+y2=1}
  4. D
    S={(x,y)∣x2+y2=4}S = \{ (x,y)|{x^2} + {y^2} = 4\}S={(x,y)∣x2+y2=4}
View written solutionFree

Correct answer: B

  1. Understand the relation

    The relation is R={(P,Q)∣P and Q are at the same distance from the origin}.R = \{(P,Q) \mid P \text{ and } Q \text{ are at the same distance from the origin}\}.R={(P,Q)∣P and Q are at the same distance from the origin}.

    So two points are related if their distances from the origin are equal.

  2. Find the distance of the point (1,−1)(1,-1)(1,−1) from the origin

    For a point (x,y)(x,y)(x,y), distance from origin is x2+y2.\sqrt{x^2+y^2}.x2+y2​.

    Therefore, for (1,−1)(1,-1)(1,−1): 12+(−1)2=1+1=2.\sqrt{1^2+(-1)^2} = \sqrt{1+1} = \sqrt{2}.12+(−1)2​=1+1​=2​.

  3. Find its equivalence class

    The equivalence class of (1,−1)(1,-1)(1,−1) consists of all points (x,y)(x,y)(x,y) whose distance from the origin is also 2\sqrt{2}2​.

    So, x2+y2=2.\sqrt{x^2+y^2} = \sqrt{2}.x2+y2​=2​.

    Squaring both sides, x2+y2=2.x^2+y^2=2.x2+y2=2.

  4. Match with the options

    • A: x2+y2=2x^2+y^2=\sqrt{2}x2+y2=2​ ❌
    • B: x2+y2=2x^2+y^2=2x2+y2=2 ✅
    • C: x2+y2=1x^2+y^2=1x2+y2=1 ❌
    • D: x2+y2=4x^2+y^2=4x2+y2=4 ❌

  5. Conclusion

    The equivalence class of (1,−1)(1,-1)(1,−1) is {(x,y)∣x2+y2=2}.\{(x,y) \mid x^2+y^2=2\}.{(x,y)∣x2+y2=2}.

    Hence, the correct option is B.

PreviousNext

More from Sets and Relations

  • If A = {x ∈ R : |x − 2| > 1}, B = {x ∈ R : x2−3​> 1}, C = {x ∈ R : |x − 4|≥ 2} and Z is the set of all integers, then the number of subsets of the set (A ∩ B ∩ C)c ∩ Z is ​…2021 · Numerical
  • Let N be the set of natural numbers and a relation R on N be defined by R={(x,y)∈N×N:x3−3x2y−xy2+3y3=0}. Then the relation R is :2021 · MCQ
  • Let A = {n ∈ N | n2 ≤ n + 10,000}, B = {3k + 1 | k ∈ N} an dC = {2k | k ∈ N}, then the sum of all the elements of the set A ∩(B − C) is equal to ​.2021 · Numerical
  • Which of the following is not correct for relation R on the set of real numbers ?2021 · MCQ
  • If R = {(x, y) : x, y ∈ Z, x2 + 3y2 ≤ 8} is a relation on the set of integers Z, then the domain of R–1 is :2020 · MCQ
  • Consider the two sets : A = {m ∈ R : both the roots of x2 – (m + 1)x + m + 4 = 0 are real} and B = [–3, 5). Which of the following is not true?2020 · MCQ
  • Let R1 and R2 be two relation defined as follows : R1 = {(a, b) ∈ R2 : a2 + b2 ∈ Q} and R2 = {(a, b) ∈ R2 : a2 + b2 otin Q}, where Q is the set of all rational numbers. Then :2020 · MCQ
  • A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be:2020 · MCQ