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

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 \(2\pi\) and local maximum at \(\pi\).

Step 1: Understand the Function

The given function is \( f(x) = \int_0^x \sqrt{t} \sin t \, dt \). We need to analyze it to find local maxima or minima. The interval of interest is \( (0, \frac{5\pi}{2}) \).

Step 2: Find the First Derivative

To determine the critical points where maxima or minima can occur, find the first derivative of \( f(x) \) using the Fundamental Theorem of Calculus. The derivative \( f'(x) \) is \( \sqrt{x} \sin x \).

Step 3: Determine Critical Points

Critical points occur when \( f'(x) = 0 \). Therefore, set \( \sqrt{x} \sin x = 0 \). This means either \( \sqrt{x} = 0 \) or \( \sin x = 0 \). Since \( x > 0 \), only \( \sin x = 0 \) is relevant. Thus, \( x = n\pi \) where \( n \) is an integer.

Step 4: Check Points in the Interval

Within \( (0, \frac{5\pi}{2}) \), the critical points based on \( \sin x = 0 \) are \( x = \pi, 2\pi \).

Step 5: Determine Nature of Critical Points

To determine if these critical points are maxima or minima, examine the second derivative using the second derivative test. Compute \( f''(x) = \frac{1}{2\sqrt{x}} \sin x + \sqrt{x} \cos x \).

Step 6: Evaluate the Second Derivative

Evaluate \( f''(x) \) at \( x = \pi \) and \( x = 2\pi \):\[ f''(\pi) = \frac{1}{2\sqrt{\pi}} \sin \pi + \sqrt{\pi} \cos \pi = -\sqrt{\pi} < 0 \], which indicates a local maximum at \( \pi \). \[ f''(2\pi) = \frac{1}{2\sqrt{2\pi}} \sin 2\pi + \sqrt{2\pi} \cos 2\pi = \sqrt{2\pi} > 0 \], which indicates a local minimum at \( 2\pi \).

Step 7: Conclusion

Based on the analysis, \( f \) has a local maximum at \( x = \pi \) and a local minimum at \( x = 2\pi \). The correct answer is (D).

Key Concepts

Integration

Integration is a central concept in calculus that involves finding the accumulation of quantities, often areas under curves. In this context, we are given a function defined as an integral: \[ f(x) = \int_0^x \sqrt{t} \sin t \, dt \] This means the function \(f(x)\) represents the area under the curve of \( \sqrt{t} \sin t \) from 0 to \(x\). Understanding integration well includes knowing several rules and theorems:

• Definite integrals: This represents the total accumulation from one point to another, which could be area, volume, or other physical quantities.
• Fundamental Theorem of Calculus: It connects differentiation with integration and provides a method to evaluate definite integrals.

In solving the exercise, this fundamental theorem helps us differentiate and understand the behavior of the function \(f(x)\) by finding \(f'(x)\). This knowledge of integration helps tackle complex problems by breaking them into smaller, solvable parts.

Critical Points

Critical points are locations on a curve where the first derivative, \(f'(x)\), is zero or undefined. Identifying these points is crucial because they provide potential locations of local maxima or minima.For the given problem, after finding the first derivative \( f'(x) = \sqrt{x} \sin x \), we position ourselves to determine critical points:

• Set \( f'(x) = 0 \), resulting in either \(\sqrt{x}=0\) or \(\sin x = 0\).
• Since \( x > 0 \), only \(\sin x = 0\) remains relevant.
• This translates into \( x = n\pi \) (where \(n\) is an integer) within the interval \((0, \frac{5\pi}{2})\).

This approach finds us critical points at \( x = \pi \) and \( x = 2\pi \). Both points need further inspection to understand whether they indeed are local maxima or minima.To effectively use critical points, always check if they lie within the specific limits of your problem context. Identifying them accurately helps focus on potential solutions.

Second Derivative Test

The second derivative test is a useful tool for determining the nature of critical points. It involves computing the second derivative, \(f''(x)\), and analyzing its sign at these points.Once we have critical points from \(f'(x)\), compute the second derivative:\[ f''(x) = \frac{1}{2\sqrt{x}} \sin x + \sqrt{x} \cos x \]Evaluate this at each critical point:

• At \(x = \pi\): \[ f''(\pi) = -\sqrt{\pi} < 0 \] implying a local maximum, because \(f''(x)\) is negative.
• At \(x = 2\pi\): \[ f''(2\pi) = \sqrt{2\pi} > 0 \] indicating a local minimum, since \(f''(x)\) is positive.

This test provides a straightforward method to identify if the curve is

• Concave up: \(f''(x) > 0\) indicating local minima
• Concave down: \(f''(x) < 0\) suggesting local maxima

In essence, understanding and applying the second derivative test converts critical points into definitive maxima or minima based on concavity, making it an essential analysis step in calculus.