Sigma Percentile
JEE Main 2022 (28 July Shift 2)
LEVELJEE Main

Animated Solution for Mathematics - Basic Mathematics: Let and . Then the number of elements in is

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Visualized Solution

Defining the Sets and

  • We need to find the number of elements in .
  • Set : satisfying .
  • Set : Integers satisfying .

Analyzing Set : The Numerator

  • Consider the inequality:
  • Critical points of the numerator:
  • or
  • or

Analyzing Set : The Denominator

  • Critical points of the denominator:
  • or
  • Note: The denominator cannot be zero, so and .

Solving the Inequality for

  • Critical points: .
  • The expression is positive when numerator and denominator have the same sign.
  • Solving this yields: .

Applying Domain Constraints to

  • Given domain for :
  • Intersecting with our solution:

Analyzing Set : Quadratic Inequality

  • Set condition: for
  • Notice that .
  • Let . The inequality becomes:

Finding Roots for Set

  • Solve using the quadratic formula:

Determining Integer Values for

  • Roots are approximately and
  • So,
  • Since , must be an integer:

Listing the Elements of Set

  • If , then can be positive or negative.

Finding the Intersection

  • Integers in :
  • Elements of :
  • Common elements ():

Final Conclusion

  • The elements in are .
  • The number of elements is 3.

The Sigma Insight: Properties of Modulus Function

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape! Today, we are embarking on a journey to solve a problem that tests your precision, logic, and ability to navigate the traps of absolute values and rational inequalities.
We are tasked with finding the intersection for two sets defined under specific constraints. Let us begin our systematic approach.

Phase 1

The Rational Inequality (Set )
Our first challenge is to define Set using the condition:
The domain is constrained to .
To solve this, we identify the critical points. The numerator vanishes when , which yields or .
The denominator vanishes when , giving or . Since the denominator cannot be zero, and are strictly excluded from the solution set.
Using the Wavy Curve Method across the critical points , we determine the expression is non-negative when .
Intersecting this result with the domain , we obtain:

Phase 2

The Quadratic Elegance (Set )
Now, we define Set by the condition , where . Noting that , we substitute to get the quadratic inequality:
Applying the quadratic formula, the roots are:
Given , the roots are approximately and . Thus, the inequality holds for .
Since must be an integer, can only take the values or . Consequently, the set is:

Phase 3

The Final Convergence
We now find the intersection by checking which elements of reside within the intervals of .
Evaluating the elements of : is in . is in . is in . The values and do not satisfy the conditions of .
Thus, the intersection is:
The number of elements in the intersection is 3. We have successfully conquered the problem through rigorous logical steps.

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