Let A be the set of all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ and R be a relation on A such that $R = \{(f, g): f(0) = g(1)$ and $f(1) = g(0)\}$. Then R is :
(A)
Symmetric and transitive but not reflective
(B)
Symmetric but neither reflective nor transitive
(C)
Transitive but neither reflexive nor symmetric
(D)
Reflexive but neither symmetric nor transitive
Reflexivity: For R to be reflexive, we need $(f, f) \in R$ for all $f \in A$. This means $f(0) = f(1)$ and $f(1) = f(0)$. Since $f(1) = f(0)$ is always true if $f(0) = f(1)$, R is reflexive.
Symmetry: For R to be symmetric, if $(f, g) \in R$, then $(g, f) \in R$. Given $(f, g) \in R$, we have $f(0) = g(1)$ and $f(1) = g(0)$. For $(g, f) \in R$, we need $g(0) = f(1)$ and $g(1) = f(0)$. Since $f(0) = g(1)$ implies $g(1) = f(0)$ and $f(1) = g(0)$ implies $g(0) = f(1)$, R is symmetric.
Transitivity: For R to be transitive, if $(f, g) \in R$ and $(g, h) \in R$, then $(f, h) \in R$. Given $(f, g) \in R$, we have $f(0) = g(1)$ and $f(1) = g(0)$. Given $(g, h) \in R$, we have $g(0) = h(1)$ and $g(1) = h(0)$. For $(f, h) \in R$, we need $f(0) = h(1)$ and $f(1) = h(0)$.
We have $f(0) = g(1) = h(0)$ and $f(1) = g(0) = h(1)$. Thus, $f(0) = h(0)$ and $f(1) = h(1)$.
However, for transitivity, we need $f(0) = h(1)$ and $f(1) = h(0)$.
Let's consider a counterexample:
Let $f(x) = x$, $g(x) = 1-x$, and $h(x) = x$.
Then $f(0) = 0$, $f(1) = 1$, $g(0) = 1$, $g(1) = 0$, $h(0) = 0$, $h(1) = 1$.
$(f, g) \in R$ because $f(0) = g(1) = 0$ and $f(1) = g(0) = 1$.
$(g, h) \in R$ because $g(0) = h(1) = 1$ and $g(1) = h(0) = 0$.
For $(f, h) \in R$, we need $f(0) = h(1)$ and $f(1) = h(0)$.
$f(0) = 0$ and $h(1) = 1$, so $f(0) \neq h(1)$.
$f(1) = 1$ and $h(0) = 0$, so $f(1) \neq h(0)$.
Therefore, $(f, h) \notin R$, and R is not transitive.
Correct Answer: Symmetric but neither reflexive nor transitive<\/strong>
AI Suggestion: Option B
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Pedagogical Audit
Bloom's Analysis:
This is an APPLY question because the student needs to apply the definitions of reflexive, symmetric, and transitive relations to the given relation R defined on functions.
Knowledge Dimension:CONCEPTUAL
Justification:The question requires understanding the concepts of relations, functions, reflexivity, symmetry, and transitivity. It involves applying these concepts to a specific relation defined on a set of functions.
Syllabus Audit:
In the context of JEE, this is classified as COMPETENCY. The question assesses the student's ability to apply the definitions of relations and their properties (reflexivity, symmetry, transitivity) to a specific context involving functions, which goes beyond rote memorization and requires problem-solving skills.