Problem 222 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(g(x)=\int_{0}^{x} \cos 4 t d t\), then \(g(x+\pi)\) equals \([2012]\) (A) \(\frac{g(x)}{g(\pi)}\) (B) \(g(x)+g(\pi)\) (C) \(g(x)-g(\pi)\) (D) \(g(x) \cdot g(\pi)\)

Step-by-Step Solution

Answer

(B) g(x) + g(π)

Step 1: Understanding the Function

We are given the function \( g(x) = \int_{0}^{x} \cos 4t \, dt \). This integral represents the accumulation of the cosine function with respect to \( t \) from 0 to \( x \).

Step 2: Differentiating to Find g'

To understand the behavior of the function \( g(x) \), we differentiate it: \( g'(x) = \cos(4x) \). This shows that \( g(x) \) is the antiderivative of \( \cos(4x) \).

Step 3: Substituting x with (x + π)

Now we want to find \( g(x+\pi) \). Substitute \( x + \pi \) into the function: \( g(x+\pi) = \int_{0}^{x+\pi} \cos 4t \, dt \). This can be split into two integrals: \( g(x+\pi) = \int_{0}^{x} \cos 4t \, dt + \int_{x}^{x+\pi} \cos 4t \, dt \), which simplifies to \( g(x) + \int_{x}^{x+\pi} \cos 4t \, dt \).

Step 4: Computing Second Integral

Calculate \( \int_{x}^{x+\pi} \cos 4t \, dt \). This can be rewritten using the antiderivative of \( \cos 4t \), which is \( \frac{1}{4}\sin(4t) \). Evaluating, we get \( \int_{x}^{x+\pi} \cos 4t \, dt = \left. \frac{1}{4}\sin(4t) \right|_{x}^{x+\pi} \). Substitute values to find \( \frac{1}{4}(\sin(4(x+\pi)) - \sin(4x)) = \frac{1}{4}(\sin(4x) - \sin(4x)) = 0 \).

Step 5: Summing the Integrals

Now substitute back to find \( g(x+\pi) \): \( g(x+\pi) = g(x) + 0 = g(x) + g(\pi) \), where \( g(\pi) \) accounts for a full period of the cosine function contributing zero. Thus, the solution is \( g(x+\pi) = g(x) + g(\pi) \).

Step 6: Analyzing Answer Choices

From the calculation, we find that \( g(x+\pi) = g(x) + g(\pi) \) matches choice (B).

Key Concepts

Definite Integrals

One of the key components of this exercise is the concept of definite integrals. A definite integral computes the accumulation of a function over a specific interval. In this case, we look at the behavior of the function \( \cos 4t \) from 0 to \( x \).
This definite integral is represented by \( \int_{0}^{x} \cos 4t \, dt \).

• Limits of Integration: In the expression \( \int_{0}^{x} \), 0 and \( x \) are the limits. The lower limit is the starting point of the integral, and the upper limit is where the accumulation ends.
• Area Under the Curve: The integral represents the net area under the curve of \( \cos 4t \) within the interval \([0, x]\). If the curve is entirely above the x-axis, this is a straightforward area. If the curve dips below, it contributes negatively to the total area.

Understanding definite integrals helps you appreciate how the function accumulates value over an interval, which is crucial for solving problems involving functions like \( g(x) = \int_{0}^{x} \cos 4t \, dt \).

Cosine Function

The cosine function is periodic and plays a central role in this exercise. Specifically, our integral involves \( \cos 4t \), which is a variation of the base cosine function.

• Periodicity: The standard cosine function, \( \cos(t) \), has a period of \( 2\pi \). This means it repeats every \( 2\pi \) units. The integral over a full period will sum to zero due to symmetry.
• Frequency Adjustment: With \( \cos 4t \), the frequency is adjusted to repeat every \( \pi / 2 \) because \( 4 \times \frac{\pi}{2} = 2\pi \). This increases the oscillations per unit interval, impacting the integral's value over different intervals.

The knowledge of how the cosine function behaves helps to break down the integral \( \int_{0}^{x+\pi} \cos 4t \, dt \) into recognizable patterns and behavior, which is integral to the solution.

Antiderivative

The antiderivative, or indefinite integral, is what you get when you reverse the process of differentiation.
For this exercise, if you differentiate \( g(x) = \int_{0}^{x} \cos 4t \, dt \), you get \( g'(x) = \cos(4x) \).

• Chain Rule and Antiderivatives: To find the antiderivative of \( \cos 4t \), we use the knowledge that the derivative of \( \sin(4t) \) is \( 4\cos(4t) \). So, we adjust by multiplying by \( \frac{1}{4} \), resulting in \( \frac{1}{4}\sin(4t) \).
• Checking Your Work: Differentiating \( \frac{1}{4}\sin(4t) \) gives back \( \cos(4t) \), confirming the antiderivative is correct.

The concept of antiderivatives is critical when spliting definite integrals into recognizable parts, as done in this exercise to evaluate \( \int_{x}^{x+\pi} \), eventually simplifying the solution to \( g(x+\pi) = g(x) + g(\pi) \).