Sets & Relations
Equivalence relation on functions
nta_pyq_2025_apr
Grade 12
Question:
Let A be the set of all functions f : Z$\to$Z and R be a relation$o_n$A such that$R = {(f$, g) : f$(0) = g(1)$and f$(1) = g(0)}$. Then R is:
Symmetric and transitive but not reflective
Symmetric but neither reflective nor transitive
Reflexive but neither symmetric nor transitive
Transitive but neither reflexive nor symmetric
Step-by-Step Solution
Key Concept: Translate the finite relation rule into explicit admissible ordered pairs and count the required set.
$R = {(f$, g) : f$(0) = g(1)$and f$(1) = g(0)}$Reflexive: (f , f )$\ in $R$(2) = f$$(0) = f$(1) and f$(1) = f$(0)$\to$must hold $\Rightarrow$ but this is not true for all function so not reflexive Symmetric: If (f , g)$\ in $R $\Rightarrow$ (g, f )$\ in $R Now,$g(0) = f$(1) and$g(1) = f$(0)$\to$true ∴ symmetric Transitive : If(f , g)$\ in $R and (g, h)$\ in $R $\Rightarrow$ (f , h)$\ in $R Now (f , g)$\ in $R $\Rightarrow$ f$(0) = g(1)$and f$(1) = g(0)$(g, h)$\ in $R $\Rightarrow$$g(0) = h(1)$and$g(1) = h(0)$For (f , h)$\ in $R we need f$(0) = h(1)$and f$(1) = h(0)$Now f$(0) = g(1) = h(0)$and f$(1) = g(0) = h(1)$Hence not transitive
Correct Answer: 2