Problem 162 - A Complete Resource Book in Mathematics for JEE Main

Question

A function \(f\) defined on \([a, b]\) is bounded above if there exists a real number \(K\) such that \(|f(x)| \leq K, \forall x \in[a, b]\). The real number \(K\) is called an upper bound of \(f . f\) is called bounded below if there exists a real number \(k\) such that \(|f(x)| \geq k, \forall x \in[a, b]\). The real number \(k\) is called a lower bound of \(f\). It may be observed that if \(K\) is an upper bound of \(f\), then every real number greater than or equal to \(K\) is also an upper bound of \(f\). The smallest of all the upper bounds of \(f\) is called the least upper bound or the supremum (sup.) of \(f\). Similarly, the greatest of all the lower bounds of \(f\) is called the greatest lower bound or the infimum (Inf.) of \(f\). If the functions \(f\) and \(g\) are bounded and continuous on \([a, b]\) and \(g\) keeps the same sign on \([a, b]\), then there exists a number \(\mu\) between the infimum and supremum of \(f\) on \([a, b]\) such that $$ \int_{a}^{b} f(x) g(x) d x=\mu \int_{a}^{b} g(x) d x $$. \(\int_{\pi / 6}^{\pi / 2} \frac{x}{\sin x} d x\) lies between (A) \(\frac{\pi^{2}}{3}\) and \(\frac{2 \pi^{2}}{3}\) (B) \(\frac{\pi^{2}}{9}\) and \(\frac{2 \pi^{2}}{9}\) (C) \(\frac{2 \pi^{2}}{9}\) and \(\frac{4 \pi^{2}}{9}\) (D) None of these

Step-by-Step Solution

Answer

Option (A) is correct.

Step 1: Identify the given function

The given integral to evaluate is \( \int_{\pi/6}^{\pi/2} \frac{x}{\sin x} \, dx \). Here, \( f(x) = \frac{x}{\sin x} \) and \( g(x) = 1 \). The function \( g(x) \) is positive on \([\pi/6, \pi/2]\).

Step 2: Determine bounds for \(f(x)\)

For \( f(x) = \frac{x}{\sin x} \), observe the identity of \( \sin x \leq x \) for \(x\) in the interval \((0, \pi/2)\), hence \( 1 \leq \frac{x}{\sin x} \). This gives a lower bound.

Step 3: Compute the upper bound for \( f(x) \)

In the interval \([\pi/6, \pi/2]\), \( \sin x \) is minimized at \( x = \pi/6 \) where \( \sin(\pi/6) = 1/2 \). Therefore, \( \frac{x}{\sin x} \leq \frac{\pi/2}{1/2} = \pi \). Thus, the upper bound of \( f(x) \) is \( \pi \).

Step 4: Calculate the supremum and infimum

The infimum of \( f(x) = \frac{x}{\sin x} \) on \([\pi/6, \pi/2]\) is 1, and the supremum is \( \pi \).

Step 5: Apply the Mean Value Theorem for integrals

According to the theorem, since \(f(x)\) is bounded, there exists \( \mu \) such that \( 1 \leq \mu \leq \pi \) and \( \int_{\pi/6}^{\pi/2} f(x) \, dx = \mu \int_{\pi/6}^{\pi/2} 1 \, dx \).

Step 6: Evaluate \(\int_{\pi/6}^{\pi/2} 1 \, dx\)

Calculate \( \int_{\pi/6}^{\pi/2} 1 \, dx = \left[ x \right]_{\pi/6}^{\pi/2} = \pi/2 - \pi/6 = \pi/3 \).

Step 7: Evaluate the integral using the theorem

The integral becomes \( \mu \times \pi/3 \), hence the value of \( \int_{\pi/6}^{\pi/2} \frac{x}{\sin x} \, dx \) lies between \(1 \times \pi/3 = \pi/3\) and \(\pi \times \pi/3 = \pi^2/3\).

Step 8: Determine which interval the solution falls into

Since \( \pi^2/9 \approx 1.097 \), \( 2\pi^2/9 \approx 2.194 \), \( 4\pi^2/9 \approx 4.387 \), and \( \pi^2/3 \approx 3.289 \), the integral value \( \pi^2/3 \approx 3.289 \) fits within the interval \([\pi^2/3, 2\pi^2/3]\). Therefore, option (A) is correct.

Key Concepts

Bounded Functions

In mathematics, a bounded function is one where the values of the function remain within a fixed range. More formally, a function \( f(x) \) is called bounded on an interval \([a, b]\) if there exists a real number \( K \) such that \( |f(x)| \leq K \) for all \( x \) in \([a, b]\). This real number \( K \) is termed as the upper bound of the function. Conversely, if there exists a real number \( k \) such that \( |f(x)| \geq k \), then \( k \) is called a lower bound.
A key aspect of bounded functions is understanding that any number greater than \( K \) is also an upper bound, although \( K \) is usually taken as the smallest such bound. Controlling these bounds is essential for computing integrals over specific intervals, particularly in definite integrals where the function is evaluated within limits \( a \) and \( b \). This ensures that the function does not "blow up" to infinity, making calculations manageable and the application of integration theorems feasible.

Supremum and Infimum

The concepts of supremum and infimum are fundamental when dealing with bounded functions. The supremum, or least upper bound, of a function is the smallest real number that is greater than or equal to every value of the function on the interval. For a function \( f(x) \) defined on \([a, b]\), the supremum is denoted as \( \text{sup} f \).
The infimum, on the other hand, is the greatest lower bound. It is the largest real number that is less than or equal to every value of the function in the interval, denoted as \( \text{inf} f \).
These concepts are particularly useful as they allow for precise determination of the range within which function values actually lie. For instance, in integration problems, knowing the supremum and infimum helps in establishing bounds for further calculations like integrals, as shown in the original exercise.

Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals is a pivotal result in calculus, extending the Mean Value Theorem from differentiation to integration. It states that if \( f \) is continuous on \([a, b]\) and \( g \) is integrable with consistent sign on this interval, then there exists a \( \mu \) such that:

• \( f(x) \) is bounded and continuous on \([a, b]\), as is \(g(x)\)
• \( \text{inf} \, f \leq \mu \leq \text{sup} \, f \) on \([a, b]\).

The theorem provides the result:\[\int_{a}^{b} f(x) g(x) \, dx = \mu \int_{a}^{b} g(x) \, dx\]This theorem is crucial for calculating the "average" value that the function \( f(x) \) takes on over \([a, b]\). By controlling \( \mu \)'s bounds using the infimum and supremum, it ensures the solution is well within the calculated upper and lower limits. Application of this theorem simplifies the process of integrating potentially complex functions by transforming the problem into evaluating integrals of simpler bounding functions.