Sigma Percentile
JEE Advanced 2013
LEVELJEE Main

Animated Solution for Mathematics - Functions: Let be defined by , where is a constant such that . Then

Select Answer:

* Multiple Correct

Visualized Solution

Visualizing the Setup

  • Given function: for
  • Parameter constraint:
  • We want to analyze its invertibility and behavior on the domain .

Finding the Derivative

  • To check if the function is one-one, we find its derivative .
  • Using the Quotient Rule:
  • Here, and .

Computing

  • Applying the rule:
  • Simplifying the numerator:
  • Final derivative:

Analyzing the Sign of

  • We have:
  • Since
  • The denominator is always positive for .
  • Therefore, for all .
  • Conclusion: is strictly decreasing, hence one-one.

Finding the Range of

  • Since is strictly decreasing on :
  • Upper bound:
  • Lower bound:
  • Therefore, the Range of is .

Checking Invertibility

  • Given Codomain:
  • Calculated Range:
  • Since Range Codomain, the function is not onto (not surjective).
  • Therefore, is not invertible on .

Finding the Inverse Relation

  • Let
  • Solve for :
  • Thus, on its domain .

Comparing and on

  • Domain of is .
  • Domain of is .
  • Since , the two domains are not identical.
  • Therefore, on the interval .

Evaluating and

  • We found:
  • At :
  • Reciprocal:
  • Since , , so .

The Sigma Insight: Inverse of a Function

Solution Diagram

Analyzing the Machine's Behavior

Imagine you are an engineer designing a machine that takes an input and transforms it into an output defined by the function:
You are told that is a constant, a fixed parameter such that . To determine if this machine is a perfect mirror—that is, if it is invertible—we must analyze its mathematical properties.

Phase 1

The Monotonicity Check
To determine if the function is one-to-one, we examine its derivative. We apply the quotient rule, where and . The derivative formula is:
Substituting our values, we obtain:
Expanding the numerator, we get . The terms cancel out, leaving us with the elegant expression:
Because , we know , which implies that is strictly negative. Since the denominator is a square, it is always positive. Thus, for all in the domain, confirming that the function is strictly decreasing and therefore one-to-one.

Phase 2

The Range Trap
A function is invertible only if it is both one-to-one and onto. We must now determine if the range of covers the entire codomain, .
Given that the function is strictly decreasing, we examine the behavior at the boundaries of the domain :
The range of the function is the open interval . Since $(-1, b) eq \mathbb{R}$, the function is not onto. Consequently, the machine fails the test for global invertibility.

Phase 3

The Inverse Paradox
Although the function is not invertible on , we can derive its inverse relation by solving for :
The algebraic rule for the inverse is identical to the original function. However, we must note that the domain of is , while the domain of is .
Because these domains differ, and are not the same function. You have successfully navigated the traps of this problem by respecting the strict definitions of functions, domains, and ranges.

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