Problem 67 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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

Step-by-Step Solution

Answer

(D) Local minimum at \(\pi\) and local maximum at \(2\pi\).

Step 1: Differentiate the given function

First, find the derivative of the function \( f(x) \) using the Fundamental Theorem of Calculus. The fundamental theorem states that if \( F(x) = \int_{a}^{x} g(t) \, dt \), then \( F'(x) = g(x) \). So, \( f'(x) = \sqrt{x} \sin x \).

Step 2: Find critical points

To find critical points, set the derivative equal to zero: \( f'(x) = \sqrt{x} \sin x = 0 \). This results in two cases: \( \sqrt{x} = 0 \) or \( \sin x = 0 \). Since \( \sqrt{x} = 0 \) only at \( x = 0 \), focus on \( \sin x = 0 \) for critical points within the domain. \( \sin x = 0 \) at \( x = n\pi \) where \( n \) is an integer. Within \( (0, \frac{5\pi}{2}) \), this gives \( x = \pi \) and \( x = 2\pi \) as critical points.

Step 3: Determine nature of critical points using second derivative

Use the second derivative test to determine whether these critical points are maxima or minima. First, differentiate \( f'(x) = \sqrt{x} \sin x \) to find \( f''(x) = \frac{1}{2\sqrt{x}} \sin x + \sqrt{x} \cos x \). Evaluate \( f''(x) \) at \( x = \pi \) and \( x = 2\pi \):- At \( x = \pi \), \( f''(\pi) \) yields a negative result.- At \( x = 2\pi \), \( f''(2\pi) \) yields a positive result.

Step 4: Conclusion on local extremas

The second derivative test results indicate:- At \( x = \pi \), \( f''(\pi) < 0 \), indicating a local maximum.- At \( x = 2\pi \), \( f''(2\pi) > 0 \), indicating a local minimum.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone in the field of Calculus. It connects differentiation and integration, two of the primary concepts in Calculus. Here is how it works:

• If you have a function defined as an integral with a variable as an upper limit, like \( F(x) = \int_{a}^{x} g(t) \, dt \), the derivative of this function is simply the integrand evaluated at \( x \). So, \( F'(x) = g(x) \).
• This means if you differentiate the integral \( F(x) \), you get back the original function \( g(t) \) at \( x \).

This theorem allows us to quickly find the derivative of a function that is defined as an integral. In the exercise, the given function \( f(x) = \int_{0}^{x} \sqrt{t}\sin t \, dt \) uses the Fundamental Theorem to yield \( f'(x) = \sqrt{x}\sin x \). This process involves replacing \( t \) with \( x \) in the integrand and differentiating.

Critical Points

Critical points occur in a function where its derivative is either zero or undefined. These points are crucial because they can indicate where a function has a maximum, minimum, or inflection point.

• To find critical points, set the derivative equal to zero and solve for \( x \).
• In this exercise, \( f'(x) = \sqrt{x}\sin x = 0 \) gives the conditions \( \sqrt{x} = 0 \) or \( \sin x = 0 \).
• We disregard \( \sqrt{x} = 0 \) since it only holds when \( x = 0 \), which is outside of our domain of interest \( (0, \frac{5\pi}{2}) \).
• The condition \( \sin x = 0 \) leads to \( x = n\pi \), where \( n \) is an integer within our interval. This identifies the critical points at \( x = \pi \) and \( x = 2\pi \).

These are the specifics of critical points calculation, which help us further explore the nature of the function's behavior at these points.

Second Derivative Test

The Second Derivative Test helps us determine whether a critical point is a local maximum or a minimum. Here's how it can be used:

• First, compute the second derivative of the function. For \( f(x) = \int_{0}^{x} \sqrt{t}\sin t \, dt \), after finding \( f'(x) = \sqrt{x}\sin x \), we find the second derivative \( f''(x) = \frac{1}{2\sqrt{x}} \sin x + \sqrt{x} \cos x \).
• Evaluate the second derivative at each critical point.
• If \( f''(x) > 0 \) at a critical point, the function has a local minimum there. If \( f''(x) < 0 \), the function has a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and other methods might be needed.

For the function in this exercise:

• At \( x = \pi \), \( f''(\pi) < 0 \), indicating a local maximum.
• At \( x = 2\pi \), \( f''(2\pi) > 0 \), indicating a local minimum.

Understanding this test makes it simpler to determine the nature of extremas in a function, fully leveraging the properties of calculus.