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Question #41899

Difficulty: 8/10 | 🟢 Free

The Question

Let a binary operation $*$ on $\mathbb{R}$ be defined as $a * b = \left( \sum_{r=1}^{\infty} \frac{r}{(r+1)!} \right) (a + b) - \left( \sum_{r=1}^{\infty} \frac{1}{2^r} \right) \{a\}\{b\}$ where $\{x\}$ denotes the fractional part of $x$ . Let $S$ be the set of all invertible elements in $\mathbb{R}$ under the operation $*$ . Which of the following represents the set $S$ ?

Options

A.
$\mathbb{Z}$
B.
$\mathbb{R} \setminus \mathbb{Z}$
C.
$\mathbb{R}$
D.
{0}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Invertibility in Binary Operations|Sigma Notation Properties|Integral Part and Fractional Part

Why This Question Is Tricky

Common Mistakes

This question exploits the common student oversight of assuming that algebraic invertibility holds uniformly across a continuous domain.

The problem uses infinite series involving sigma notation as a 'fog' to consume time, simplifying to a binary operation $a * b = a + b - \{a\}\{b\}$ .

The core trap lies in the fractional part function $\{x\}$ , which introduces a discontinuity.

Students often correctly identify the identity element $e=0$ but fail to rigorously apply the bounds of the fractional part $0 \le \{b\} < 1$ when finding the inverse.

By setting up the inverse equation, the integer constraint forces a contradiction for all non-integer values, revealing that only integers are invertible.

This tests the rigorous application of domain constraints in abstract algebraic structures.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Invertibility in Binary Operations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free