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Compute R∘R⁻¹ for the "less than" Relation A to B | JEE

JEE Maths question with a full step-by-step solution.

Question
Let RR be the relation "less than" from A={1,2,3,4}A = \{1, 2, 3, 4\} to B={1,3,5}B = \{1, 3, 5\}, i.e. (a,b)∈R⇔a<b(a, b) \in R \Leftrightarrow a < b. Then R∘R−1R \circ R^{-1} is:
A{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)}\{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}
B{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}\{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}
C{(3,3),(3,5),(5,3),(5,5)}\{(3, 3), (3, 5), (5, 3), (5, 5)\}correct
D{(3,3),(3,4),(4,5)}\{(3, 3), (3, 4), (4, 5)\}
Solution
Step 1: Write RR (pairs with a<ba < b) and its inverse:
R={(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)}R = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}
R−1={(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}.R^{-1} = \{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}.
 Step 2: (x,z)∈R∘R−1⇔∃ y:(x,y)∈R−1 and (y,z)∈R.(x, z) \in R \circ R^{-1} \Leftrightarrow \exists\, y : (x, y) \in R^{-1} \text{ and } (y, z) \in R.
x=3: y∈{1,2}⇒z∈{3,5}⇒(3,3),(3,5)x = 3: \ y \in \{1, 2\} \Rightarrow z \in \{3, 5\} \Rightarrow (3, 3), (3, 5)
x=5: y∈{1,2,3,4}⇒z∈{3,5}⇒(5,3),(5,5)x = 5: \ y \in \{1, 2, 3, 4\} \Rightarrow z \in \{3, 5\} \Rightarrow (5, 3), (5, 5)
∴R∘R−1={(3,3),(3,5),(5,3),(5,5)}.\therefore R \circ R^{-1} = \{(3, 3), (3, 5), (5, 3), (5, 5)\}.
 Correct answer: (3)
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