Problem 220 - A Complete Resource Book in Mathematics for JEE Main 2017

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

(C) local maximum at \(\pi\) and local minimum at \(2 \pi\)

Step 1: Understand the Fundamental Theorem of Calculus

Given that \( f(x) = \int_{0}^{x} \sqrt{t} \sin t \, dt \), this can be connected to the derivative \( f'(x) \) using the Fundamental Theorem of Calculus. Specifically, \( f'(x) = \sqrt{x} \sin x \).

Step 2: Find Critical Points

To find critical points, we need to solve \( f'(x) = 0 \). This gives \( \sqrt{x} \sin x = 0 \). Thus, \( \sin x = 0 \), which occurs at \( x = n \pi \), where \( n \in \mathbb{Z} \). Considering \( x \in \left(0, \frac{5\pi}{2}\right) \), the relevant critical points are \( x = \pi \) and \( x = 2\pi \).

Step 3: Apply the Second Derivative Test

To apply the second derivative test, compute \( f''(x) \). First, find \( f'(x) = \sqrt{x} \sin x \). Use the product rule to find \( f''(x) = \frac{1}{2\sqrt{x}} \sin x + \sqrt{x} \cos x \). Evaluate \( f''(x) \) at the critical points: - \( f''(\pi) = \frac{1}{2\sqrt{\pi}} \cdot 0 + \sqrt{\pi} \cdot (-1) = -\sqrt{\pi} < 0 \), so \( x = \pi \) is a local maximum.- \( f''(2\pi) = \frac{1}{2\sqrt{2\pi}} \cdot 0 + \sqrt{2\pi} \cdot 1 = \sqrt{2\pi} > 0 \), so \( x = 2\pi \) is a local minimum.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key concept linking differentiation with integration, allowing us to evaluate definite integrals through antiderivatives. When we have a function defined as an integral, like \( f(x) = \int_{0}^{x} \sqrt{t} \sin t \, dt \), this theorem helps us find its derivative easily.

According to the theorem, if \( F(t) \) is an antiderivative of \( \sqrt{t} \sin t \), then the derivative of the integral function, \( f(x) \), with respect to \( x \) is simply the original integrand evaluated at \( x \). The formula can be expressed as:

• \( f'(x) = \sqrt{x} \sin x \)

This theorem simplifies finding the rate of change of a function when it is expressed as an integral. Thus, applying this principle, we transform complex integral expressions into manageable derivative evaluations. This is the first step in understanding the behavior of \( f(x) \) in this context.

Critical Points

Critical points of a function occur where the derivative is zero or undefined. These points are significant because they often correspond to local maxima, minima, or points of inflection.

For the function \( f(x) = \int_{0}^{x} \sqrt{t} \sin t \, dt \), the derivative \( f'(x) = \sqrt{x} \sin x \) is zero when any of the factors in the expression is zero.

• \( \sqrt{x} = 0 \) is only true when \( x = 0 \), not in our interval of interest.
• \( \sin x = 0 \) when \( x = n\pi \), \( n \in \mathbb{Z} \), which gives us our critical points within the interval \( (0, \frac{5\pi}{2}) \) as \( x = \pi \) and \( x = 2\pi \).

Understanding where the derivative equals zero within our interval focuses our attention on these critical points, where the function \( f(x) \) might have local maxima or minima.

Second Derivative Test

The Second Derivative Test provides a way to determine the nature of a critical point—whether it is a local maximum, minimum, or neither.

To apply this test, we take the second derivative of the function and evaluate it at the critical points. For our function, we begin with the first derivative:

• \( f'(x) = \sqrt{x} \sin x \)
• Then, using the product rule, we find the second derivative as: \( f''(x) = \frac{1}{2\sqrt{x}} \sin x + \sqrt{x} \cos x \)

Next, we evaluate \( f''(x) \) at the critical points:

• At \( x = \pi \), \( f''(\pi) = -\sqrt{\pi} \), which is less than zero, indicating a local maximum.
• At \( x = 2\pi \), \( f''(2\pi) = \sqrt{2\pi} \), which is greater than zero, indicating a local minimum.

This test is valuable because it allows us to quickly categorize the nature of critical points using simple calculations of the second derivative, providing insights into the function's curvature at these points.