For the relation R defined below, where \( \mathrm{R} \) and N denotes set of real numbers and set of natural numbers respectively, Match the LIST-I with LIST-II
| LIST-I Relation | LIST-II Properties of relation hold by R | ||
| A. | \( R=\{(a,b):a,b\in\mathrm{R};a^{2}+b^{2}=4\} \) | I. | Neither reflexive, nor symmetric, nor transitive |
| B. | \( R=\{(a,b):a,b\in\mathrm{R};1+ab\gt0\} \) | II. | Equivalence relation |
| C. | \( R=\{(a,b):a,b\in N;a+2b=8\} \) | III. | Symmetric but neither reflexive nor transitive |
| D. | On N×N, relation R defined by \( (a,b)R(c,d)\Leftrightarrow a+d=b+c \) | IV. | Reflexive and symmetric, but not transitive |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-IV, C-I, D-II
Each relation is checked for reflexivity, symmetry and transitivity.
(A) \( a^{2}+b^{2}=4 \) matches (III) Symmetric but neither reflexive nor transitive.
Reflexive fails: (1, 1) gives 1 + 1 = 2 ≠ 4.
Symmetric holds: \( a^{2}+b^{2}=b^{2}+a^{2} \).
Transitive fails: (2, 0) and (0, 2) are in R, but (2, 2) is not, since \( 4+4=8\ne4 \).
(B) \( 1+ab\gt0 \) matches (IV) Reflexive and symmetric, but not transitive.
Reflexive: \( 1+a^{2}\gt0 \). Symmetric: \( 1+ab=1+ba \).
Transitive fails: for a = 10, b = 0, c = −10, \( 1+ab=1\gt0 \) and \( 1+bc=1\gt0 \), but \( 1+ac=-99\lt0 \).
(C) \( a+2b=8 \) on N matches (I) Neither reflexive, nor symmetric, nor transitive.
The pairs in R are (6, 1), (4, 2) and (2, 3).
Reflexive fails: (1, 1) ∉ R. Symmetric fails: (6, 1) ∈ R but (1, 6) ∉ R.
Transitive fails: (4, 2) and (2, 3) are in R, but (4, 3) is not, since \( 4+6=10\ne8 \).
(D) \( (a,b)R(c,d)\Leftrightarrow a+d=b+c \) on N×N matches (II) Equivalence relation.
The condition is the same as \( a-b=c-d \). It is reflexive, symmetric and transitive, because equality of the number \( a-b \) has these three properties.
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).