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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Sets and Relations

Let A be the set of all functions f: Z arrow Z and R be a relation...

Let A be the set of all functions f:Z→Zf: \mathbf{Z} \rightarrow \mathbf{Z} and R be a relation on A such that R={(f,g):f(0)=g(1) and f(1)=g(0)}\mathrm{R}=\{(\mathrm{f}, \mathrm{g}): f(0)=\mathrm{g}(1)\text{ and } f(1)=\mathrm{g}(0)\}. Then R is:

Options

  1. ASymmetric and transitive but not reflective
  2. BSymmetric but neither reflective nor transitive
  3. CReflexive but neither symmetric nor transitive
  4. DTransitive but neither reflexive nor symmetric

Correct answer

B. Symmetric but neither reflective nor transitive

Step-by-step solution

R={(f,g):f(0)=g(1) and  f(1)=g(0)} Reflexive: (f,f)∈R=f(0)=f(1) and f(1)=f(0)→ must hold ⇒ but this is not true for all function \begin{aligned} & \mathrm{R}=\{(\mathrm{f}, \mathrm{g}): \mathrm{f}(0)=\mathrm{g}(1) \text{\text{ and } } \mathrm{f}(1)=\mathrm{g}(0)\} \\ & \text{ Reflexive: }(\mathrm{f}, \mathrm{f}) \in \mathrm{R} \\ & =\mathrm{f}(0)=\mathrm{f}(1) \text{ and } \mathrm{f}(1)=\mathrm{f}(0) \rightarrow \text{ must hold } \\ & \Rightarrow \text{ but this is not true for all function }\end{aligned}
so not reflexive
Symmetric: If (f,g)∈R⇒(g,f)∈R(\mathrm{f}, \mathrm{g}) \in \mathrm{R} \Rightarrow(\mathrm{g}, \mathrm{f}) \in \mathrm{R}
Now, g(0)=f(1)andg(1)=f(0)→and g(1)=f(0) \rightarrow true
∴\therefore symmetric
Transitive : If⁡(f,g)∈R\operatorname{If}(\mathrm{f}, \mathrm{g}) \in \mathrm{R} and (g,h)∈R(\mathrm{g}, \mathrm{h}) \in \mathrm{R}
⇒(f,h)∈R\Rightarrow(\mathrm{f}, \mathrm{h}) \in \mathrm{R}
Now (f,g)∈R⇒f(0)=g(1)(f, g) \in R \Rightarrow f(0)=g(1) and f(1)=g(0)
(g,h)∈R⇒g(0)=h(1)andg(1)=h(0)(\mathrm{g}, \mathrm{h}) \in \mathrm{R} \Rightarrow \mathrm{g}(0)=\mathrm{h}(1) and \mathrm{g}(1)=\mathrm{h}(0)
For (f,h)∈R(f, h) \in R we need f(0)=h(1)andf(1)=h(0)f(0)=h(1) and f(1)=h(0)
Now f(0)=g(1)=h(0)andf(1)=g(0)=h(1)f(0)=g(1)=h(0) and f(1)=g(0)=h(1)
Hence not transitive
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