Q29 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
Let $\mathrm{A}=\{-2,-1,0,1,2,3,4\}$. Let R be a relation on A defined by $x \mathrm{Ry}$ if and only if $2 x+y \leqslant 2$. Let $l$ be the number of elements in R. Let m and n be the minimum number of elements required to be added in $R$ to make it reflexive and symmetric relations respectively. Then $l+m+n$ is equal to :
- A35
- B34
- Correct answerC33
- D32
Solution
Given $A = \{-2, -1, 0, 1, 2, 3, 4\}$ and $xRy$ iff $2x + y \leqslant 2$.
For $x = -2$: $y \leqslant 6$, all 7 elements valid (7 pairs).
For $x = -1$: $y \leqslant 4$, all 7 elements valid (7 pairs).
For $x = 0$: $y \leqslant 2 \Rightarrow y \in \{-2,-1,0,1,2\}$ (5 pairs).
For $x = 1$: $y \leqslant 0 \Rightarrow y \in \{-2,-1,0\}$ (3 pairs).
For $x = 2$: $y \leqslant -2 \Rightarrow y \in \{-2\}$ (1 pair).
For $x = 3, 4$: no valid $y$.
So $l = 7+7+5+3+1 = 23$.
For reflexive: $(x,x) \in R \Rightarrow 3x \leqslant 2 \Rightarrow x \leqslant 2/3$. Already present for $x = -2, -1, 0$. Missing for $x = 1, 2, 3, 4$, so $m = 4$.
For symmetric: need $(y,x) \in R$ for every $(x,y) \in R$. Missing pairs: $(3,-2), (4,-2), (2,-1), (3,-1), (4,-1), (2,0)$, so $n = 6$.
$l + m + n = 23 + 4 + 6 = 33$.
For $x = -2$: $y \leqslant 6$, all 7 elements valid (7 pairs).
For $x = -1$: $y \leqslant 4$, all 7 elements valid (7 pairs).
For $x = 0$: $y \leqslant 2 \Rightarrow y \in \{-2,-1,0,1,2\}$ (5 pairs).
For $x = 1$: $y \leqslant 0 \Rightarrow y \in \{-2,-1,0\}$ (3 pairs).
For $x = 2$: $y \leqslant -2 \Rightarrow y \in \{-2\}$ (1 pair).
For $x = 3, 4$: no valid $y$.
So $l = 7+7+5+3+1 = 23$.
For reflexive: $(x,x) \in R \Rightarrow 3x \leqslant 2 \Rightarrow x \leqslant 2/3$. Already present for $x = -2, -1, 0$. Missing for $x = 1, 2, 3, 4$, so $m = 4$.
For symmetric: need $(y,x) \in R$ for every $(x,y) \in R$. Missing pairs: $(3,-2), (4,-2), (2,-1), (3,-1), (4,-1), (2,0)$, so $n = 6$.
$l + m + n = 23 + 4 + 6 = 33$.
Hence, the correct answer is option: 'C'.