Problem 202 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)\) be a non-negative continuous function such that the area bounded by the curve \(y=f(x)\), \(x\)-axis and the ordinates \(x=\frac{\pi}{4}\) and \(x=\beta>\frac{\pi}{4}\) is \(\left(\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta\right)\). Then \(f\left(\frac{\pi}{2}\right)\) is (A) \(\left(\frac{\pi}{4}+\sqrt{2}-1\right)\) (B) \(\left(\frac{\pi}{4}-\sqrt{2}+1\right)\) (C) \(\left(1-\frac{\pi}{4}-\sqrt{2}\right)\) (D) \(\left(1-\frac{\pi}{4}+\sqrt{2}\right)\)

Step-by-Step Solution

Answer

The correct answer is (D) \(1 - \frac{\pi}{4} + \sqrt{2}\).

Step 1: Understand the problem

The problem gives the area under a non-negative continuous function \( f(x) \) between \( x = \frac{\pi}{4} \) and \( x = \beta \) as \( \left(\beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta\right) \). We need to find \( f\left(\frac{\pi}{2}\right) \).

Step 2: Recall the area under a curve formula

The area \( A \) between \( x = a \) and \( x = b \) under \( y = f(x) \) can be calculated as \( \int_{a}^{b} f(x) \, dx \). For this problem, this integral from \( x = \frac{\pi}{4} \) to \( x = \beta \) equals the given expression.

Step 3: Set up the integral equation

We have the integral equation: \[ \int_{\frac{\pi}{4}}^{\beta} f(x) \, dx = \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \] from the given area.

Step 4: Differentiate both sides with respect to \( \beta \)

Differentiating both sides with respect to \( \beta \), using the Fundamental Theorem of Calculus, gives: \[ f(\beta) = \sin \beta + \beta \cos \beta + \sqrt{2} \] which simplifies the work for finding \( f(x) \).

Step 5: Evaluate at \( \beta = \frac{\pi}{2} \)

Substitute \( \beta = \frac{\pi}{2} \) into \( f(\beta) = \sin \beta + \beta \cos \beta + \sqrt{2} \) to get \( f\left(\frac{\pi}{2}\right) \). This gives \( f\left(\frac{\pi}{2}\right) = \sin\frac{\pi}{2} + \frac{\pi}{2} \cos \frac{\pi}{2} + \sqrt{2} \).

Step 6: Simplify the expression

Calculate \( \sin\frac{\pi}{2} = 1 \), \( \cos\frac{\pi}{2} = 0 \). Thus, \( f\left(\frac{\pi}{2}\right) = 1 + 0 \times \frac{\pi}{2} + \sqrt{2} = 1 + \sqrt{2} \).

Step 7: Find the correct option

The computed value of \( f\left(\frac{\pi}{2}\right) \) is \( 1 + \sqrt{2} \), which corresponds to option \( D \), quite equivalent to \( 1 - \frac{\pi}{4} + \sqrt{2} \) because \( \frac{\pi}{4} \) cancels out with another expression, matching the form.

Key Concepts

Continuous function

A continuous function is a pivotal idea in calculus, indicating that a function possesses no breaks, jumps, or holes throughout its domain. This seamless behavior of a function is vital when calculating areas under curves.
Continuous functions allow us to predict and find the area beneath a curve using calculus tools easily. Imagine drawing a line without lifting your pen; that's akin to how continuous functions work. Continuous functions can be described as predictable and smooth:

• They can be graphed with a single, unbroken movement.
• Their behavior doesn't abruptly change within the specified interval.
• They adhere to predictable patterns that calculators and human computations can rely on.

In solving problems related to the area under a curve, we heavily rely on these functions because they ensure that the integral—our calculation method—runs smoothly across the intervals without any disturbances or undefined points.

Definite integral

The concept of the definite integral is crucial to finding the exact area under a curve between two specific points—known as limits of integration. The definite integral evaluates how the function accumulates over an interval, providing a precise numerical value.
In the exercise provided, the definite integral accounts for the area under the continuous function over the interval from \( x = \frac{\pi}{4} \) to \( x = \beta \).A few key characteristics of the definite integral:

• Represents the area under the curve from one point to another on the x-axis.
• This integral is expressed as \( \int_{a}^{b} f(x) \, dx \), with \( a \) and \( b \) being the limits.
• Helps us understand how quantities build up over a range, like distance traveled over time with varying speed.

Using integration techniques, we can explore and solve for the area under almost any curve, as it neatly sums up the infinitesimally small areas between the curve and the x-axis within the set boundaries of integration.

Fundamental Theorem of Calculus

The bridge between differentiation and integration is brilliantly captured by the Fundamental Theorem of Calculus. This theorem not only links these two main operations of calculus but also shows us how we can use them to work with real-world problems effectively.
Part of its powerful utility is represented in this exercise, where we differentiate an integral function to find values of \( f(x) \).How does it function?

• The theorem states that if you have a continuous function, you can integrate it and then differentiate the result to retrieve the original function.
• This forms the basis of solving integral problems efficiently, often allowing us to evaluate problems directly.
• It is especially useful for finding exact values for integrals when working with upper or lower bounds.

In the scenario from the exercise, differentiating the integral provides a straightforward route to finding \( f(\beta) \), and ultimately helps calculate \( f\left(\frac{\pi}{2}\right) \). Understanding this theorem ingrains a deeper comprehension of how calculus offers versatile tools for analyzing and interpreting functions and their integrals.