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Question #41896

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = [0, 1)$ . A binary operation $*$ is defined on $S$ by $x * y = \{x+y\}$ , where $\{ \cdot \}$ denotes the fractional part function. Let $x^{-1}$ denote the inverse of $x$ in the group $(S, *)$ . A relation $R$ is defined on $S$ by: $x R y \iff x^{-1} * y = x * y^{-1}$ Which of the following statements is TRUE?

Options

A.
$R$ is an antisymmetric relation on $S$ .
B.
$R$ is an equivalence relation on $S$ and every equivalence class has exactly $2$ elements.
C.
$R$ is an equivalence relation on $S$ and exactly one equivalence class has $1$ element.
D.
$R$ is reflexive and symmetric, but not transitive.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Symmetry vs. Antisymmetry|Invertibility in Binary Operations|Integral Part and Fractional Part

Why This Question Is Tricky

Common Mistakes

This question tests the properties of binary operations and relations involving the fractional part function.

The core trap lies in the invertibility condition within the group $([0, 1), *)$ .

Students often assume that the relation $x^{-1} * y = x * y^{-1}$ implies $x = y$ , which would make the relation antisymmetric.

However, the periodic nature of the fractional part function creates 'ghost symmetry' points.

Specifically, the equation $\{y-x\} = \{x-y\}$ has non-trivial solutions $|x-y| = 1/2$ due to the mapping onto $[0, 1)$ .

This invalidates antisymmetry and instead forms an equivalence relation where every element is paired with exactly one 'ghost' element, making all equivalence classes of size 2.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Symmetry vs. Antisymmetry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free