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JEE Main 20253 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Sets and Relations

Let A=\ -3,-2,-1,0,1,2,3 , . Let R be a relation on A defined by x...

Let A={−3,−2,−1,0,1,2,3A=\{-3,-2,-1,0,1,2,3, . Let R be a relation on A defined by xRyx R y if and only if 0≤x2+2y≤40 \leq x^2+2 y \leq 4.
Let ll be the number of elements in R and mm be the minimum number of elements required to be added in R to make it a reflexive relation. then l+ml+m is equal to

Options

  1. A1919
  2. B2020
  3. C1717
  4. D1818

Correct answer

D. 1818

Step-by-step solution

A={−3,−2,−1,0,1,2,3}−2y≤x2≤4−2y\begin{aligned} & \mathrm{A}=\{-3,-2,-1,0,1,2,3\} \\ & -2 \mathrm{y} \leq \mathrm{x}^2 \leq 4-2 \mathrm{y}\end{aligned}
y=−36≤x2≤10⇒x∈{−3,3}y=−24≤x2≤8⇒x∈{−2,2}y=−12≤x2≤6⇒x∈{−2,2}y=00≤x2≤4⇒x∈{−2,−1,0,1,2}y=1−2≤x2≤2⇒x∈{−1,0,1}y=2−4≤x2≤0⇒x∈{0}y=3−6≤x2≤−2⇒ No x-Exist \begin{array}{lll}y=-3 & 6 \leq x^2 \leq 10 & \Rightarrow x \in\{-3,3\} \\ y=-2 & 4 \leq x^2 \leq 8 & \Rightarrow x \in\{-2,2\} \\ y=-1 & 2 \leq x^2 \leq 6 & \Rightarrow x \in\{-2,2\} \\ y=0 & 0 \leq x^2 \leq 4 & \Rightarrow x \in\{-2,-1,0,1,2\} \\ y=1 & -2 \leq x^2 \leq 2 & \Rightarrow x \in\{-1,0,1\} \\ y=2 & -4 \leq x^2 \leq 0 & \Rightarrow x \in\{0\} \\ y=3 & -6 \leq x^2 \leq-2 & \Rightarrow \text{ No } x \text{-Exist }\end{array}
R={(−3,−3)(−3,3),(−2,−2)(−2,2)(−1,−2)(−1,2)(0,−2)(0,−1)(0,0)(0,1)(0,2)(1,−1)(1,0)(1,1)(2,0)}∴ℓ=15\begin{aligned} & \mathrm{R}=\{(-3,-3)(-3,3),(-2,-2)(-2,2)(-1,-2)(-1,2) \\ & (0,-2)(0,-1)(0,0)(0,1)(0,2)(1,-1)(1,0)(1,1) \\ & (2,0)\} \\ & \therefore \ell=15\end{aligned}
To make it reflexive we will add
{(−1,−1),(2,2),(3,3)}∴m=3∴ℓ+m=15+3=18\begin{aligned} & \{(-1,-1),(2,2),(3,3)\} \quad \therefore \mathrm{m}=3 \\ & \therefore \ell+\mathrm{m}=15+3=18\end{aligned}
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