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Sets and Relations question

2025 · 3 Apr · Shift 1 · Q30
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Sets and Relations question

2025 · 3 Apr · Shift 1 · Q30

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let A={−3,−2,−1,0,1,2,3}\mathrm{A}=\{-3,-2,-1,0,1,2,3\}A={−3,−2,−1,0,1,2,3}. Let R be a relation on A defined by xRyx \mathrm{R} yxRy if and only if 0≤x2+2y≤40 \leq x^2+2 y \leq 40≤x2+2y≤4. Let lll be the number of elements in R and mmm be the minimum number of elements required to be added in R to make it a reflexive relation. Then l+ml+ml+m is equal to
  1. A
    18
  2. B
    20
  3. C
    17
  4. D
    19
View written solutionFree

Correct answer: A

  1. Given set and relation

We have A={−3,−2,−1,0,1,2,3}A=\{-3,-2,-1,0,1,2,3\}A={−3,−2,−1,0,1,2,3} and relation RRR on AAA defined by xRy  ⟺  0≤x2+2y≤4.xRy \iff 0\le x^2+2y\le 4.xRy⟺0≤x2+2y≤4.

We need:

  • l=l =l= number of ordered pairs in RRR
  • m=m =m= minimum number of ordered pairs to be added to make RRR reflexive
  • then compute l+ml+ml+m
  1. Find all ordered pairs in RRR

For each x∈Ax\in Ax∈A, solve 0≤x2+2y≤4.0\le x^2+2y\le 4.0≤x2+2y≤4. This gives −x2≤2y≤4−x2-x^2\le 2y\le 4-x^2−x2≤2y≤4−x2 so −x22≤y≤4−x22.\frac{-x^2}{2}\le y\le \frac{4-x^2}{2}.2−x2​≤y≤24−x2​. Since y∈Ay\in Ay∈A, we check each possible xxx.

  1. Casewise counting

Case 1: x=±3x=\pm 3x=±3

Then x2=9x^2=9x2=9. Condition becomes 0≤9+2y≤4.0\le 9+2y\le 4.0≤9+2y≤4. From the right inequality, 9+2y≤4  ⟹  2y≤−5  ⟹  y≤−2.5.9+2y\le 4 \implies 2y\le -5 \implies y\le -2.5.9+2y≤4⟹2y≤−5⟹y≤−2.5. From the left inequality, 9+2y≥0  ⟹  2y≥−9  ⟹  y≥−4.5.9+2y\ge 0 \implies 2y\ge -9 \implies y\ge -4.5.9+2y≥0⟹2y≥−9⟹y≥−4.5. Thus −4.5≤y≤−2.5.-4.5\le y\le -2.5.−4.5≤y≤−2.5. In AAA, only y=−3y=-3y=−3 satisfies this.

So for each of x=3x=3x=3 and x=−3x=-3x=−3, there is 1 value of yyy.

Contribution: 222 pairs.

Case 2: x=±2x=\pm 2x=±2

Then x2=4x^2=4x2=4. Condition becomes 0≤4+2y≤4.0\le 4+2y\le 4.0≤4+2y≤4. From 0≤4+2y  ⟹  y≥−2,0\le 4+2y \implies y\ge -2,0≤4+2y⟹y≥−2, and 4+2y≤4  ⟹  y≤0.4+2y\le 4 \implies y\le 0.4+2y≤4⟹y≤0. So −2≤y≤0.-2\le y\le 0.−2≤y≤0. Hence y∈{−2,−1,0}.y\in\{-2,-1,0\}.y∈{−2,−1,0}.

So for each of x=2x=2x=2 and x=−2x=-2x=−2, there are 3 values of yyy.

Contribution: 666 pairs.

Case 3: x=±1x=\pm 1x=±1

Then x2=1x^2=1x2=1. Condition becomes 0≤1+2y≤4.0\le 1+2y\le 4.0≤1+2y≤4. From 1+2y≥0  ⟹  y≥−12,1+2y\ge 0 \implies y\ge -\tfrac12,1+2y≥0⟹y≥−21​, and 1+2y≤4  ⟹  y≤32.1+2y\le 4 \implies y\le \tfrac32.1+2y≤4⟹y≤23​. Thus −12≤y≤32.-\tfrac12\le y\le \tfrac32.−21​≤y≤23​. In AAA, this gives y∈{0,1}.y\in\{0,1\}.y∈{0,1}.

So for each of x=1x=1x=1 and x=−1x=-1x=−1, there are 2 values of yyy.

Contribution: 444 pairs.

Case 4: x=0x=0x=0

Then x2=0x^2=0x2=0. Condition becomes 0≤2y≤4,0\le 2y\le 4,0≤2y≤4, so 0≤y≤2.0\le y\le 2.0≤y≤2. Hence y∈{0,1,2}.y\in\{0,1,2\}.y∈{0,1,2}.

Contribution: 333 pairs.

  1. Total number of elements in RRR

Therefore, l=2+6+4+3=15.l=2+6+4+3=15.l=2+6+4+3=15.

  1. Find mmm to make RRR reflexive

A relation on AAA is reflexive if every diagonal pair (a,a)(a,a)(a,a) belongs to RRR for all a∈Aa\in Aa∈A.

So check when (x,x)∈R(x,x)\in R(x,x)∈R: 0≤x2+2x≤4.0\le x^2+2x\le 4.0≤x2+2x≤4. Evaluate for all x∈Ax\in Ax∈A:

  • x=−3x=-3x=−3: x2+2x=9−6=3x^2+2x=9-6=3x2+2x=9−6=3 ✓
  • x=−2x=-2x=−2: 4−4=04-4=04−4=0 ✓
  • x=−1x=-1x=−1: 1−2=−11-2=-11−2=−1 ✗
  • x=0x=0x=0: 000 ✓
  • x=1x=1x=1: 1+2=31+2=31+2=3 ✓
  • x=2x=2x=2: 4+4=84+4=84+4=8 ✗
  • x=3x=3x=3: 9+6=159+6=159+6=15 ✗

So the missing diagonal pairs are (−1,−1), (2,2), (3,3).(-1,-1),\, (2,2),\, (3,3).(−1,−1),(2,2),(3,3). Hence the minimum number of pairs to add is m=3.m=3.m=3.

  1. Compute l+ml+ml+m

l+m=15+3=18.l+m=15+3=18.l+m=15+3=18.

  1. Compare with stored answer

Derived answer is 18, which matches option A.

Therefore, the stored correct answer is correct.

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