Problem 220 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=\int_{0}^{x} \sqrt{t} \sin t d t\), for \(x \in\left(0, \frac{5 \pi}{2}\right)\). Then, \(f\) has (A) local minimum at \(\pi\) and \(2 \pi\) (B) local minimum at \(\pi\) and local maximum at \(2 \pi\) (C) local maximum at \(\pi\) and local minimum at \(2 \pi\) (D) local maximum at \(\pi\) and \(2 \pi\)

Step-by-Step Solution

Answer

The function \( f(x) \) has a local maximum at \( \pi \) and a local minimum at \( 2\pi \), so the answer is (C).

Step 1: Understand the Problem

The function given is \( f(x) = \int_{0}^{x} \sqrt{t} \sin t \, dt \). We need to determine where this function has local maxima and minima within the given interval \( x \in \left(0, \frac{5\pi}{2}\right) \).

Step 2: Find the Derivative

To analyze maxima and minima, we find the derivative of \( f(x) \). By the Fundamental Theorem of Calculus, \( f'(x) = \sqrt{x} \sin x \).

Step 3: Critical Points Calculation

We set \( f'(x) = \sqrt{x} \sin x = 0 \) to find critical points. This yields \( \sin x = 0 \). Thus, critical points in the interval are \( x = \pi \) and \( x = 2\pi \).

Step 4: Second Derivative Test

Calculate the second derivative: \( f''(x) = \frac{1}{2\sqrt{x}} \sin x + \sqrt{x} \cos x \). For \( x = \pi \), \( f''(\pi) = \frac{1}{2\sqrt{\pi}} \cdot 0 + \sqrt{\pi} \cdot (-1) < 0 \) indicates a local maximum. For \( x = 2\pi \), \( f''(2\pi) = \frac{1}{2\sqrt{2\pi}} \cdot 0 + \sqrt{2\pi} \cdot 1 > 0 \) indicates a local minimum.

Key Concepts

Fundamental Theorem of Calculus

The **Fundamental Theorem of Calculus** is a cornerstone of integral calculus.This theorem provides a connection between differentiation and integration in mathematics.It consists of two parts that establish this profound link:

• The first part states that if a function is continuous on [a, b], then it has an antiderivative over this interval. This means that you can integrate the continuous function to get another function whose derivative is the original function.
• The second part states that if you have a function that is the integral of another function, then by differentiating it, you can obtain the original function.

In our exercise, we use the second part of this theorem.The function provided is an integral from 0 to x of \( \sqrt{t} \sin t \, dt \). Using the Fundamental Theorem of Calculus, the derivative of this integral is \( \sqrt{x} \sin x \).This crucial step helps us move forward in identifying critical points.

Critical Points

**Critical points** of a function are where its derivative is either zero or undefined.These points are significant because they can indicate potential locations of local maxima or minima. To find critical points in the function \(f(x) = \int_{0}^{x} \sqrt{t} \sin t \, dt\), we first find its derivative: \(f'(x) = \sqrt{x} \sin x\).

To identify critical points:

• Set the derivative equal to zero: \(\sqrt{x} \sin x = 0\).
• Since \(\sqrt{x} eq 0\) for any \(x > 0\), it must be \(\sin x = 0\).

This occurs at integer multiples of \(\pi\), which means critical points in our interval are at \(x = \pi\) and \(x = 2\pi\).Finding these critical points is essential, as they are candidates for local extrema.

Second Derivative Test

The **Second Derivative Test** is a powerful tool for determining whether critical points are local maxima, minima, or points of inflection.By taking the second derivative and evaluating it at critical points, we can understand the curvature of the graph:

• If \(f''(x) > 0\), the function is concave up, indicating a local minimum at that point.
• If \(f''(x) < 0\), the function is concave down, indicating a local maximum.

For the exercise function, we compute:\(f''(x) = \frac{1}{2\sqrt{x}} \sin x + \sqrt{x} \cos x\).
At \(x = \pi\):

• \(f''(\pi) = \sqrt{\pi} \cdot (-1) < 0\), indicates a local maximum.

At \(x = 2\pi\):

• \(f''(2\pi) = \sqrt{2\pi} \cdot 1 > 0\), indicates a local minimum.

This analysis helps pinpoint where the function has peaks and valleys, completing our understanding of its behavior.