Problem 74 - Calculus for JEE ADVANCED

Question

$$ \begin{aligned} f(-x)=\frac{\cos (-x)}{\left[-\frac{2 x}{\pi}\right]+\frac{1}{2}} &=\frac{\cos x}{-1-\left[\frac{2 x}{\pi}\right]+\frac{1}{2}} \quad \begin{array}{r} \text { (As } x \text { is not an integral } \\ \text { multiple of } \pi) \end{array} \\ &=-\frac{\cos x}{\left[\frac{2 x}{\pi}\right]+\frac{1}{2}}=-f(x) \end{aligned} $$ Therefore, \(f(x)\) is an odd function.

Step-by-Step Solution

Answer

The function \( f(x) \) is odd: \( f(-x) = -f(x) \).

Step 1: Identify the Transformation for Odd Functions

Recognize that for a function to be odd, the following must be true: \( f(-x) = -f(x) \). We need to verify if this condition holds for the given function.

Step 2: Simplify the Expression for \( f(-x) \)

Start with the given expression for \( f(-x) \): \[f(-x) = \frac{\cos(-x)}{\left[-\frac{2x}{\pi}\right] + \frac{1}{2}}.\]Use the trigonometric identity \( \cos(-x) = \cos(x) \) to simplify: \[f(-x) = \frac{\cos(x)}{\left[-\frac{2x}{\pi}\right] + \frac{1}{2}}.\]

Step 3: Adjust the Denominator

Observe the denominator transformation:\[\left[-\frac{2x}{\pi}\right] + \frac{1}{2} = -1 - \left[\frac{2x}{\pi}\right] + \frac{1}{2}.\]This further simplifies to: \[-1 + \frac{1}{2} = -\frac{1}{2} - \left[\frac{2x}{\pi}\right].\]

Step 4: Write the Function \( f(-x) \) and Compare

We can now express \( f(-x) \) in simplified form:\[f(-x) = \frac{\cos x}{-\left[\frac{2x}{\pi}\right] - \frac{1}{2}}.\]Notice the similarity in structure with \( f(x) \), suggesting \( f(-x) = -f(x) \).

Step 5: Verification of Odd Function Property

Compare with \( f(x) \), which is expressed as:\[f(x) = \frac{\cos x}{\left[\frac{2x}{\pi}\right] + \frac{1}{2}}.\]Therefore, we indeed have:\[f(-x) = -f(x).\]This confirms that \( f(x) \) is an odd function because it satisfies the property \( f(-x) = -f(x) \).

Key Concepts

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for every value of the occurring variables. A key identity used in the exercise is \( \cos(-x) = \cos(x) \). This identity is derived from the cosine function being an even function, meaning it reflects across the y-axis.

Understanding and using these identities is crucial for simplifying trigonometric expressions or solving trigonometric equations. Other basic identities include \( \sin(-x) = -\sin(x) \), \( \tan(-x) = -\tan(x) \), and the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \). In our problem, utilizing the identity of the even nature of cosine helps in transforming \( f(-x) \) to simpler terms, making the verification of whether the function is odd much more straightforward.

Simplifying Expressions

Simplifying expressions is about rewriting them in an easier or reduced form. It often involves using algebraic operations or applying known identities to make the expression more manageable.

In the exercise, the expression for \( f(-x) \) initially appears complicated. By breaking it down and applying trigonometric identities, we can simplify the expression significantly. Another tool used here is algebra, specifically the manipulation of the denominator. The problem demonstrates splitting and rearranging terms like \(-\frac{2x}{\pi} + \frac{1}{2}\).

Simplifying expressions is essential in solving mathematical problems as it often clears the way for a clearer understanding or a direct solution.

Verification of Function Properties

The verification of function properties is the process of proving or validating the characteristics of a function. In mathematical problems, this often involves checking if a function is odd, even, or periodic.

In this exercise, we aim to verify that \( f(x) \) is an odd function. To do so, we need to demonstrate that \( f(-x) = -f(x) \). This is done by substituting \(-x\) into the function and comparing the new expression with \(-f(x)\).

This verification process helps us understand the symmetries and behaviors of the function, forming an essential step in further problem-solving, such as integrating or differentiating the function.

Calculus Techniques

Calculus techniques involve methods and strategies used within calculus to analyze and solve problems. These may include differentiation, integration, and understanding properties of functions like being odd or even.

While the exercise primarily focuses on algebraic manipulation and identities, calculus techniques would come into play when these properties are used for further analysis. For example:

• Integrating an odd function over a symmetric interval yields zero, simplifying calculations in integral problems.
• If we differentiated \( f(x) \) and determined it was odd, it could indicate more complex behaviors over its domain.

Even though this particular exercise doesn't directly require differentiation or integration, understanding the odd property of \( f(x) \) sets the foundation for more advanced calculus applications. Recognizing these properties helps streamline the approach towards solving larger calculus-based problems.