Problem 173 - 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: \(\int_{1 / 3}^{11 / 2}\\{x\\} d x=\frac{185}{72}\) (where \(\\{x\\}\) denotes the fractional part of \(x\\}\) Reason: If \(f(x)\) is a periodic function having period, \(T\), then $$ \int_{a}^{b+n T} f(x) d x=n \int_{0}^{T} f(x) d x+\int_{a}^{b} f(x) d x $$

Step-by-Step Solution

Answer

(D) Assertion (A) is False, Reason (R) is True.

Step 1: Evaluate the Integral of Assertion

To assess the truthfulness of the assertion, we must evaluate \( \int_{1/3}^{11/2} \{x\} \, dx \). The function \( \{x\} \) represents the fractional part of \( x \), which can be expressed as \( x - \lfloor x \rfloor \). We need to consider intervals where the floor function \( \lfloor x \rfloor \) changes—this occurs at whole numbers. Thus, the integral becomes a sum of segments: \( \int_{1/3}^{1} \{x\} \, dx + \sum_{k=1}^{5} \int_{k}^{k+1} \{x\} \, dx \).

Step 2: Calculate Each Integral Segment

1. \( \int_{1/3}^{1} \{x\} \, dx = \int_{1/3}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{1/3}^{1} = \frac{1}{2} - \frac{1}{18} = \frac{8}{18} = \frac{4}{9} \).2. For each \( k \) from 1 to 5, \( \int_{k}^{k+1} \{x\} \, dx = \int_{k}^{k+1} (x-k) \, dx \), giving \( [\frac{(x-k)^2}{2}]_k^{k+1} = \frac{1}{2} \).Sum these segments: \( \frac{4}{9} + 5 \times \frac{1}{2} = \frac{4}{9} + \frac{5}{2} = \frac{4}{9} + \frac{45}{18} = \frac{39}{18} = \frac{13}{6} \)."

Step 3: Compare Calculated Integral to Assertion

The calculated value \( \frac{13}{6} \) doesn't match the given assertion value of \( \frac{185}{72} \). Therefore, the assertion \( \int_{1/3}^{11/2} \{x\} \, dx = \frac{185}{72} \) is False.

Step 4: Verify the Truth of the Reason

The reason states a theorem about the integration of periodic functions over an extended interval. This is a generally valid property and mathematically correct. Therefore, it is independently True.

Step 5: Determine Relationship Between Assertion and Reason

Since the assertion is False and the reason is True, we must choose the option that represents this relationship, which is option (D): Assertion (A) is False, Reason (R) is True.

Key Concepts

Fractional Part Function

Understanding the fractional part function is crucial when dealing with integrals involving non-integer values. This function, denoted as \( \{x\} \), represents the difference between a number \( x \) and its greatest integer less than or equal to \( x \), often called the floor function, \( \lfloor x \rfloor \). In mathematical terms, it can be expressed as:

• \( \{x\} = x - \lfloor x \rfloor \)

The fractional part function, therefore, captures the decimal portion of \( x \) and gives a value from 0 to just below 1. This characteristic makes it particularly interesting in calculus, as it introduces a piecewise quality into integrals that range over integer intervals. Within each whole number interval, the function generates segments of linear behavior: it increases linearly from 0 up to just below 1 before resetting itself.
When integrating the fractional part function, each segment between whole numbers needs individual evaluation. The complexity arises from evaluating these linear segments over their respective intervals, as seen in the original exercise. For example, between any two integers \( k \) and \( k+1 \), the fractional part function behaves as \( x-k \). This analysis allows us to piece together the integral's total value by summing the contributions from each sub-interval.

Definite Integral

A definite integral is a fundamental concept in calculus, representing the accumulation of quantities, such as area under a curve, over a specific interval. In mathematical notation, the definite integral of a function \( f(x) \) from \( a \) to \( b \) is expressed as:

• \( \int_a^b f(x) \, dx \)

This integral computes the net area between the curve \( f(x) \) and the x-axis from \( x = a \) to \( x = b \). The result is influenced by incorporating both the geometry of the function and the defined interval.
When dealing with a fractional part function in the integrand, as in the original problem, each part of the integration must be evaluated separately due to the discontinuous nature at integer points. The process involves determining the integral over each subinterval where the behavior of the function can be clearly defined.
A crucial part of evaluating such integrals is the Fundamental Theorem of Calculus, which connects differentiation and integration, showing that integrating a derivative over a range returns the net change of the original function. This theorem offers an elegant way to evaluate the area under most curves efficiently.

Periodic Functions

Periodic functions are fascinating due to their repetitive behaviors over regular intervals. A function is considered periodic if it behaves the same for any input interval of length equal to its period. The smallest positive value \( T \) for which a function \( f(x) \) satisfies \( f(x + T) = f(x) \) for all \( x \) is called the period of the function.
In our exercise, the reason provided is linked to periodic functions and their properties, especially when calculating integrals over extended intervals that include multiple complete cycles of the function. For any periodic function \( f(x) \) with period \( T \), the integral over a range that extends beyond \( n \) full periods can be simplified by using the formula:

• \( \int_a^{b+nT} f(x) \, dx = n \int_0^T f(x) \, dx + \int_a^b f(x) \, dx \)

This simplifies calculations by breaking down the integral into manageable cyclic quantities.
The central concept here is leveraging the periodic nature to simplify integrals, as repetition implies that the integral over one period can be replicated over multiple periods. This theorem can drastically reduce the computation effort required for analyzing periodic functions over extended sequences.