Problem 171 - A Complete Resource Book in Mathematics for JEE Main

Question

In the following questions an Assertion (A) is given followed by a Reason (R). Mark your responses from the following options: (A) Assertion (A) is True and Reason (R) is True; Reason \((\mathrm{R})\) is a correct explanation for Assertion (A) (B) Assertion (A) is True, Reason \((\mathrm{R})\) is True; Reason (R) is not a correct explanation for Assertion (A) (C) Assertion (A) is True, Reason (R) is False (D) Assertion (A) is False, Reason (R) is True Assertion: If \(a, b(a

Step-by-Step Solution

Answer

(C) Assertion is True, Reason is False.

Step 1: Analyzing the Function

Given the function is iterated three times: \( f \circ f \circ f (x) \), where \( f(x) = \frac{1}{1-x} \). Our goal is to examine when the function is undefined for the calculation.

Step 2: Determining Points of Discontinuity

Evaluate \( f(x) \). It's undefined at \( x = 1 \). If applied iteratively, we need to find where subsequent compositions are undefined. Calculating \( f(f(x)) \) and \( f(f(f(x))) \) confirms each is undefined at \( x = 1 \). No new discontinuities are introduced besides \( x = 1 \). Therefore, \( a eq 0 \) and \( b = 1 \) must be examined further.

Step 3: Integration Setup

For the integral \( \int_{a}^{b} \frac{f(x)}{f(x) + f(1-x)} \, dx \), we consider that it is bounded by unknown \( a \) to \( b = 1 \), focusing on proper evaluation and simplification where \( f(x) = \frac{1}{1-x} \) and \( f(1-x) = \frac{1}{x} \). This results in the integrand becoming \( \frac{\frac{1}{1-x}}{\frac{1}{1-x} + \frac{1}{x}} \).

Step 4: Simplifying the Integrand

Combine and simplify the denominator \( \left(\frac{1}{1-x} + \frac{1}{x}\right) = \frac{2-x}{x(1-x)} \). Substitute into the integrand, simplifying to \( \frac{x}{2-x} \).

Step 5: Integrating Over Interval [a, b]

The resulting integral becomes \( \int_{a}^{1} \frac{x}{2-x} \, dx \). Solving this integral using standard techniques for integration by substitution or partial fractions should yield a value of \( \frac{1}{2} \), as per the problem's assertion result.

Step 6: Evaluating Assertion and Reason

Since the integral evaluates correctly within the given boundary and provided conditions, and we determined the integral's correctness alongside the undefined point \( x = 1 \), the assertion is true. However, \( a eq 0 \), making Reason \( R: a = 0, b = 1 \), incorrect.

Key Concepts

Integration

Integration is a fundamental concept in calculus that involves finding the area under a curve or the accumulation of quantities. It is essentially the reverse operation of differentiation. In mathematical terms, integration allows you to express the sum of infinitely small data points over a continuous interval. This is particularly useful for solving problems related to areas and volumes. Additionally, integrals can help calculate solutions to differential equations.

In this exercise, you are tasked with performing integration over a specific range. The goal is to evaluate the integral \( \int_{a}^{b} \frac{f(x)}{f(x) + f(1-x)} \, dx \). To solve it, one must first simplify the integrand. By substituting \( f(x) = \frac{1}{1-x} \) and \( f(1-x) = \frac{1}{x} \), the integrand reduces to \( \frac{x}{2-x} \). This simplification eases the integration process.

• Understand what the integral represents: Often, it's the area or accumulation between points \(a\) and \(b\) on the x-axis.
• Simplify the integrand whenever possible: This makes the integration more manageable.

When you integrate from \(a\) to \(b\), especially across a region where \(f(x)\) changes dynamically due to its structure, thoroughly prepare by simplifying all expressions involved.

Discontinuity

Discontinuity in a function generally occurs where it is undefined or does not produce finite, predictable results. For a function like \( f(x) = \frac{1}{1-x} \), discontinuities occur where the denominator equals zero. Here, the function is specifically undefined at \( x = 1 \). Discontinuities are critical in calculus because they indicate that a function might not behave well or be integrable over certain points.

For our exercise, we identify discontinuities by composing the function multiple times. This helps ensure we note any new points where the formula fails. After evaluating \( f(f(x)) \) and \( f(f(f(x))) \), it is confirmed that the function remains undefined only at \(x = 1\). It's pivotal to identify all discontinuity points before proceeding with calculations involving integrals, as they determine whether specific limits of integrations are valid.

• Check for undefined values: Find where the denominator of any function becomes zero.
• Evaluate iteratively if needed: When functions compose, further discontinuities may appear.

Discontinuities can potentially change the behavior of a graph or its integral, making it necessary to handle them with caution in calculations.

Function Composition

Function composition involves applying one function to the results of another. Essentially, it is a sequence of function operations, where one outputs to another's input. The notation \( f \circ g(x) \) represents the application of function \( f \) to the result of function \( g \). Function composition allows evaluating complex function dynamics and uncovering deeper behaviors which might not be apparent in simpler forms.

In this task, the function \( f(x) = \frac{1}{1-x} \) is composed repeatedly: \( f(f(f(x))) \). It is necessary to consecutively apply each function to the last equation's output. This process uncovers any additional discontinuities or transformations caused by repetitive application.

• Recognize repeated processes: Applying a function continuously can identify and confirm behaviors.
• Calculate each function step-by-step: In sequential operations, each output informs the subsequent input.

Through composed functions, complex relationships and dependencies are highlighted, enabling a comprehensive understanding of function characteristics.

JEE Mathematics

JEE Mathematics is a significant and challenging component of the Joint Entrance Examination, crucial for accessing prestigious engineering institutes in India. This exam tests students on multiple areas of mathematics, including calculus, algebra, coordinate geometry, and more. It emphasizes not just problem-solving skills but also a deeper understanding of mathematical concepts.

Tackling problems in calculus, like integration and function compositions—requires a solid grasp of underlying theories and practices. As seen in this example, integration and handling discontinuities are common themes. Mastery in these areas can enhance analytical skills crucial for modeling and evaluating mathematical problems in real-world scenarios.

• Practice regularly: Continuous problem-solving builds confidence and skills.
• Focus on core concepts: Understand fundamentals deeply rather than memorizing solutions.

Success in JEE Mathematics relies on a strategic approach towards understanding problems fully, coupled with rigorous practice and application of mathematical principles.