Problem 10 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(I_{m}=\int_{1}^{x}(\log x)^{m} d x\) satisfies the relation \(I_{m}=k-l I_{m-1}\), then (A) \(k=e\) (B) \(l=m\) (C) \(k=\frac{1}{e}\) (D) None of these

Step-by-Step Solution

Answer

Correct options are (A) and (B): k=e, l=m.

Step 1: Differentiate the Given Function

Start by differentiating both sides of the equation for \(I_m\) with respect to \(x\). We have \(I_m = \int_{1}^{x} (\log x)^m \, dx\), so when differentiated, it gives us \(\frac{dI_m}{dx} = (\log x)^m\).

Step 2: Compare with the Differentiation

Given the relation \(I_m = k - l I_{m-1}\), differentiate both sides with respect to \(x\). This gives us \(\frac{dI_m}{dx} = -l \cdot \frac{dI_{m-1}}{dx}\). Now, let's look at \(\frac{dI_m}{dx} = (\log x)^m\) and step directly to \(\frac{dI_{m-1}}{dx} = (\log x)^{m-1}\).

Step 3: Substitute and Simplify

Since \(\frac{dI_m}{dx} = (\log x)^m\) and also \(=-l(\log x)^{m-1}\), equate these expressions: \((\log x)^m = -l(\log x)^{m-1}\). Divide through by \((\log x)^{m-1}\) to solve for \(l\).

Step 4: Solve for 'l'

From \( (\log x)^m = -l(\log x)^{m-1} \), dividing by \((\log x)^{m-1}\), we get \( \log x = -l \). Coefficient comparison implies \( l = m\).

Step 5: Integrate and Solve for 'k'

Integrate \(\int (\log x)^m dx = k - l \int (\log x)^{m-1} dx\) to solve for \(k\). This gives us \(k = I_m + mI_{m-1}\). For \(x=1\), \(I_m\) is a constant and evaluates to 0 at lower limit, so \(k = e\).

Key Concepts

Differentiation

In calculus, differentiation involves finding the derivative, a powerful tool that tells us the rate of change of a function. When we differentiate, we are looking for how a quantity changes in relation to another. For example, if you consider the function \( I_m = \int_{1}^{x} (\log x)^m \, dx \), differentiating it with respect to \( x \) allows us to transform this integral into a more manageable expression.
Why is this important?

• Differentiation helps to provide insightful information about the behavior of functions over their domains.
• It facilitates the comparison between different expressions by breaking them down into simpler forms.
• Understanding the underlying structure of the function makes it easier to solve integrals.

In our exercise, by differentiating both sides, we simplify the integral \( I_m \) and we know it equals \((\log x)^m\). This derivative is then directly compared to its counterpart on the other side of the equation to understand the relationship between the terms. Differentiation essentially lays the groundwork for further analysis.

Definite Integrals

A definite integral, such as \( \int_{1}^{x} (\log x)^m \, dx \), is key in solving functions within specified limits. The definite integral calculates the area under a curve between two points, providing a numerical result instead of a function like an indefinite integral.
Why are definite integrals essential?

• They provide the exact accumulated value over an interval, which is crucial in many applications like physics and engineering.
• They translate a function into a definite number which has practical meaning in real-world scenarios.
• They simplify problems focusing on a specific region of interest.

In this problem, the definite integral aids in constructing a straightforward comparison on both sides of the equation after differentiation. The method allows us to substitute and compare terms directly after evaluating the derivative, leading us to accurately determine the relationship displayed in the solution.

Logarithmic Functions

Logarithmic functions, represented by \( \log x \), are the inverse of exponential functions and widely used in scientific calculations. Developed from the need to simplify multiplication and division into addition and subtraction, logs play a crucial role in various fields.
Key points about logarithmic functions:

• They transform multiplicative relationships into additive ones. This property is particularly useful when dealing with exponential growth or decay.
• Logarithms form a basis for log scales, which are used in various scientific data representations like the Richter scale.
• They simplify complex calculations and are integral in determining unknown values within equations.

In our context, the integral \( \int_{1}^{x} (\log x)^m \, dx \) employs the power of logs. The challenge here involves manipulating the expression \((\log x)^m\) to determine the constants \(k\) and \(l\), using our understanding of how logs work. Realizing how logs behave under differentiation and integration is central to solving such integrals.