Problem 174 - 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 \(f(x)\) is a non-negative continuous function such that \(f(x)+f\left(x+\frac{1}{2}\right)=1\), then \(\int_{0}^{2} f(x) d x=1\) Reason: \(f(x)\) is a periodic function having period \(1 .\)

Step-by-Step Solution

Answer

Correct response is (B): Assertion is true, reason is true but not the correct explanation.

Step 1: Analyze the Assertion

The assertion states that if \( f(x) \) is a non-negative continuous function such that \( f(x) + f(x + \frac{1}{2}) = 1 \), then \( \int_{0}^{2} f(x) \, dx = 1 \). Let's verify if the assertion is true by evaluating the integral. Notice that the function value and its value at half a unit away sum to 1.

Step 2: Evaluate the Integral

Split the integral \( \int_{0}^{2} f(x) \, dx \) into two parts: \( \int_{0}^{1} f(x) \, dx \) and \( \int_{1}^{2} f(x) \, dx \). Use the property \( f(x) + f\left(x+\frac{1}{2}\right) = 1 \) to see how this affects the integral over the range.

Step 3: Use Continuity and the Sum Property

Given \( f(x) + f(x+\frac{1}{2}) = 1 \), observe over \([0, 1)\) and \([1, 2)\), for any \( x \), \( f(x) \) and \( f(x + \frac{1}{2}) \) cover the range such that their combined value over these regions remains constant. Hence, the relationship creates a condition that completes the integral cycle over \([0,2]\).

Step 4: Verify the Reason

The reason given is that \( f(x) \) is a periodic function with a period of 1. However, for a function to be periodic with period 1, \( f(x) = f(x+1) \) must hold, which is not explicitly derived from the given condition \( f(x) + f(x+\frac{1}{2}) = 1 \). Instead, this condition suggests a different symmetry rather than periodicity. Therefore, the reason doesn't directly corroborate the structure necessary for the integral assertion.

Key Concepts

Assertion and Reasoning Questions

Understanding assertion and reasoning questions is crucial for developing strong mathematical reasoning skills. These questions typically involve an assertion followed by a reason. Your task is to evaluate if the assertion is true, if the reason is true, and most importantly, if the reason correctly explains the assertion. In our example, the assertion stated a property regarding a continuous function and an integral, while the reason provided an explanation using periodic functions. Assessing both statements and their relationship requires critical thinking and a good grasp of related mathematical concepts.

Continuity in Functions

A function is said to be continuous if there are no abrupt changes or holes in its graph. This means that for any point within the domain, the function does not break, which implies that you can draw it without lifting your pen from the paper. In the given exercise, the function \( f(x) \) is continuous. This characteristic is crucial because it ensures that the integral of the function over a specified interval will be well-defined and finite. When analyzing integrals of continuous functions, it simplifies the process as there are no discontinuities to complicate the calculations.

Integral Calculus

Integral calculus involves the concept of integrals, which are used to find areas under curves, among other things. The integral given in the example, \( \int_{0}^{2} f(x) \, dx \), is examined by breaking it into smaller, more manageable sections: \( \int_{0}^{1} f(x) \, dx \) and \( \int_{1}^{2} f(x) \, dx \). This approach uses the function's properties, such as \( f(x) + f(x + \frac{1}{2}) = 1 \), to explore and leverage symmetrical attributes of the function. This method helps simplify complex integrals and make calculations more approachable by utilizing known function behaviors.

Periodic Functions

A periodic function is one that repeats at regular intervals, known as its period. For instance, a function has a period \(T\) if \( f(x + T) = f(x) \) for all \(x\). However, in the problem, the function \(f(x) + f(x + \frac{1}{2}) = 1\) does not directly exhibit standard periodic behavior with period 1. Instead, it demonstrates a unique property resembling symmetry over intervals of length 1, rather than repeating values. Thus, while periodic functions typically help identify the repeating nature of a function, in this context the reasoning explanation wasn't enough to validate the assertion precisely based on the periodicity argument.