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

Question

If \(g(x)=\int_{0}^{x} \cos 4 t d t\), then \(g(x+\pi)\) equals (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) \(g(x) + g(\pi)\)

Step 1: Understand the Definition of g(x)

The function \(g(x)\) is defined as an integral from 0 to \(x\) of the function \(\cos 4t\). We can express it as \(g(x) = \int_{0}^{x} \cos 4t \, dt\). This represents the area under the curve \(\cos 4t\) from 0 to \(x\).

Step 2: Express g(x+π) as an Integral

We need to find \(g(x+\pi)\). By substitution in the integral, we have: \(g(x+\pi) = \int_{0}^{x+\pi} \cos 4t \, dt\).

Step 3: Use the Property of Definite Integrals

Using the property of definite integrals, we can split the integral: \(g(x+\pi) = \int_{0}^{x} \cos 4t \, dt + \int_{x}^{x+\pi} \cos 4t \, dt\). Here, \(\int_{0}^{x} \cos 4t \, dt = g(x)\).

Step 4: Evaluate the Integral from x to x+π

Now, evaluate \(\int_{x}^{x+\pi} \cos 4t \, dt\). This is the integral of one complete period of the function \(\cos 4t\), since the period is \(\pi\) for \(\cos 4t\). The integral over a complete period of \(\cos\) is zero. Thus, \(\int_{x}^{x+\pi} \cos 4t \, dt = 0\).

Step 5: Combine the Results to Find g(x+π)

Substituting back, we have: \(g(x+\pi) = g(x) + 0 = g(x)\). Therefore, \(g(x+\pi) = g(x) + g(\pi)\).

Step 6: Interpret the Solution

The expression \(g(x+\pi) = g(x) + g(\pi)\) corresponds to option (B) in the original problem, meaning that \(g(x+\pi)\) is the sum of \(g(x)\) and \(g(\pi)\).

Key Concepts

Definite Integrals

When we talk about definite integrals, we're essentially talking about calculating the area under a curve, specifically from a starting point to an endpoint. In the exercise, we have the integral from 0 to \(x\) of \(\cos 4t\), which gives us the function \(g(x) = \int_{0}^{x} \cos 4t \, dt\). This helps in understanding how functions change over the interval from 0 to \(x\).

Let's break down what this means:

• The integral \(\int_{0}^{x} \cos 4t \, dt\) represents a summation of infinite tiny slices of areas under the curve \(\cos 4t\) from 0 to \(x\).
• Defining \(g(x)\) this way, we effectively 'measure' how much 'area' has accumulated as \(x\) increases.
• This definite integral results in a real number representing that area, unlike indefinite integrals that yield functions.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that contain trigonometric identities, such as \(\cos 4t\) in our example. Integrating trigonometric functions often involves understanding their periodic nature and symmetry. This knowledge is crucial.

For \(g(x) = \int_{0}^{x} \cos 4t \, dt\):

• The function \(\cos 4t\) is periodic, with a period of \(\frac{2\pi}{4} = \frac{\pi}{2}\); however, its integral over one period (\(0\) to \(\pi\)) results in zero. This indicates a balance of positive and negative areas.
• This property helps when splitting intervals as seen in the integral \(\int_{0}^{x+\pi} \cos 4t \, dt = \int_{0}^{x} \cos 4t \, dt + \int_{x}^{x+\pi} \cos 4t \, dt\).
• Recognizing the zero sum over a complete period, the trigonometric integral property simplifies the calculation significantly, helping articulate \(g(x+\pi)\) to be \(g(x) + 0\).

Properties of Integrals

One key thing to understand about integrals is their properties, which make them easier to handle. In the case of our exercise, we're using the linearity and additivity of integrals.

Here are some integral properties that apply:

• Linearity Property: The integral of a sum is the sum of the integrals, \(\int (f(t) + g(t)) \, dt = \int f(t) \, dt + \int g(t) \, dt\). This property allows splitting \(\int_{0}^{x+\pi} \cos 4t \, dt\) into two parts.
• Additivity: We can safely break an interval: \(\int_{a}^{b+c} f(t) \, dt = \int_{a}^{b} f(t) \, dt + \int_{b}^{b+c} f(t) \, dt\). This is used when determining \(g(x+\pi) = g(x) + \int_{x}^{x+\pi} \cos 4t \, dt\).

Understanding these properties is crucial; they not only simplify solving the integral but also clarify why certain terms become zero or transform into something easier to compute. This can often be the deciding factor in quickly recognizing the correct solution for integral-related problems.