Problem 148 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\int_{a}^{b}|\sin x| d x=8\) and \(\int_{a}^{a+b}|\cos x| d x=\frac{9}{2}\), then (A) \(a=\frac{\pi}{2}\) (B) \(b=\frac{11 \pi}{4}\) (C) \(a=\frac{\pi}{4}\) (D) \(b=\frac{17 \pi}{4}\)

Step-by-Step Solution

Answer

(C) \(a=\frac{\pi}{4}\)

Step 1: Understanding the Problem

We are given two integrals: \(\int_{a}^{b}|\sin x| dx=8\) and \(\int_{a}^{a+b}|\cos x| dx=\frac{9}{2}\). We are tasked to find the values of \(a\) and \(b\) from the given options.

Step 2: Analyzing the Sine Integral

To solve \(\int_{a}^{b}|\sin x| dx=8\), recognize that the absolute value of sine can be broken into period intervals. Since \(|\sin x|\) has a period of \(\pi\), set \(b-a\) as a multiple of \(\pi\), say \(n \pi\). The integral over one period is 2, so \(n \times 2 = 8\), implying \(n = 4\). Thus, \(b-a = 4\pi\).

Step 3: Analyzing the Cosine Integral

Now consider \(\int_{a}^{a+b}|\cos x| dx=\frac{9}{2}\). The integral of \(|\cos x|\) over one full period \([0,\pi]\) is 2. Set the interval length \(a+b-a = b\) as a multiple of half-periods, \(m \pi/2\), where the half-period integral of \(|\cos x|\) is 1. Therefore, \((m/2) \times 1 = \frac{9}{2}\), giving \(m = 9\), or \(b = \frac{9\pi}{4}\).

Step 4: Finding Values of a and b

Given \(b-a = 4\pi\) from the sine integral, substitute \(b = \frac{9\pi}{4}\) into the equation: \(\frac{9\pi}{4} - a = 4\pi\). Solve for \(a\) to find: \(a = \frac{\pi}{4}\).

Step 5: Selecting the Correct Options

The derived values are \(a = \frac{\pi}{4}\) and \(b = \frac{9\pi}{4}\). Compare these with the given options: Option (C) matches \(a = \frac{\pi}{4}\); Option (B) does not match \(b\) as derived. Thus, select (C) for \(a\).

Key Concepts

Sine and Cosine Functions

The sine and cosine functions are fundamental in trigonometry, particularly when studying oscillations and periodic motions.

• The sine function, represented as \( \sin x \), oscillates between -1 and 1 and has a period of \(2\pi\).
• The cosine function, or \( \cos x \), similarly oscillates between -1 and 1 and shares the period of \(2\pi\). However, it is shifted to the left by \(\frac{\pi}{2}\).

Both of these functions are key when dealing with integrals that require consideration of their absolute values.The absolute value functions \(|\sin x|\) and \(|\cos x|\) influence the periodic nature:

• The absolute value \(|\sin x|\) makes all negative sine values positive, effectively doubling the frequency within one period from \(2\pi\) to \(\pi\).
• Similarly, \(|\cos x|\) modifies the cosine function, resulting in effective periods that must be considered correctly in integral calculations.

Understanding these properties allows for accurate solutions of integrals involving sine and cosine functions.

Calculus Problem Solving

The approach to solving calculus problems often starts with analyzing the given information and then applying suitable mathematical strategies.

• First, translate the problem's requirements to understand what is being asked. In our example, two integrals are given, and we need to find specific variables \(a\) and \(b\).
• Understanding the behavior of functions and their integrals over certain intervals is crucial. This means knowing the periodicity and symmetry of the functions involved.

Once this understanding is in place, breaking down the problem into manageable parts, such as identifying relevant function periods and simplifying expressions, helps in finding the solution. By systematically following steps like determining the correct interval size from known integral values, we can methodically and accurately solve these types of calculus problems.

Integral Calculus

Integral calculus is a vast area dealing with integrals' computation, especially definite integrals, which are used to find areas under curves.

• Definite integrals, such as \( \int_{a}^{b} f(x) \, dx \), calculate the signed area between the function \( f(x) \) and the x-axis over the interval \([a, b]\).
• When dealing with integrals of absolute values like \(|\sin x|\) or \(|\cos x|\), consider their periodic properties. We know integrals over one period, like from 0 to \(\pi\), will repeat themselves.

In practice, understanding how these principles apply allows you to break down complex integral expressions into more manageable parts. For example, determining the length of intervals or translating repeating patterns into summed areas. This knowledge is then applied to solve specific values from given integrals, just like we found \(a\) and \(b\) by setting up equations based on these integral results.