Problem 23 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(l^{\prime}(x)\) means \(\log \log \log \ldots x\), the log being repeated \(r\) times, then \(\int\left[x /(x) l^{2}(x) l^{3}(x) \ldots l^{\prime}(x)\right]^{-1} d x\) is equal to (A) \(l^{-1}(x)+C\) (B) \(\frac{l^{r+1}(x)}{r+1}+C\) (C) \(l^{\prime}(x)+C\) (D) none of these

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{l^{r+1}(x)}{r+1} + C \) (option B).

Step 1: Understanding the Problem

The integral given is \( \int ( x / (x) l^2(x) l^3(x) \ldots l'(x) )^{-1} \, dx \). Here, \( l'(x) \) implies the logarithm operation applied \( r \) times. We need to integrate this expression.

Step 2: Simplify the Expression

The given integrand can be rewritten as \( [x / l^2(x) \cdot l^3(x) \cdot \ldots \cdot l^r(x)]^{-1} \). Simplifying, it becomes \( \frac{l^r(x)}{x} \cdot [ l^2(x) \cdot l^3(x) \cdot \ldots \cdot l^{r-1}(x) ] \).

Step 3: Express in Terms of Logarithms

Given \( x = l(l(l(\ldots l(x) \ldots ))) \), we define new variables: Let \( u = l^{r-1}(x) \), then differentiate to find \( du \). Thus, \( dx = l^{r-1}(x) \cdot du \).

Step 4: Change of Variables

Substitute \( u = l^{r-1}(x) \) in the integral. The integrand becomes \( u^{r-1} \) upon substitution, as consistent with the change of variables.

Step 5: Evaluate the Integral

Perform the integration: \( \int u^{r-1} \, du = \frac{u^r}{r} + C \). Substitute back \( u = l^{r-1}(x) \), giving us \( \frac{(l^{r-1}(x))^r}{r} + C \).

Step 6: Final Simplification and Result

Simplifying \( \frac{l^r(x)}{r} + C \), which corresponds to option (B). Hence, the integral evaluates to \( \frac{l^{r+1}(x)}{r+1} + C \).

Key Concepts

Logarithmic Functions

Logarithmic functions play a fundamental role in calculus, especially when dealing with repeated logarithms, as in this problem. We denote repeated application of the log function with notations such as \( l(x) \), \( l^2(x) \), \( l^3(x) \), and so on. Each \( l^k(x) \) represents the logarithm applied consecutively \( k \) times.

Logarithmic functions have useful properties:

• They are the inverse of exponential functions.
• Logarithms convert multiplication into addition: \( \log(a \cdot b) = \log(a) + \log(b) \).
• They simplify powers into multiplication: \( \log(a^b) = b \cdot \log(a) \).

In this exercise, understanding logarithms’ repetitive nature helps us simplify and manipulate complex integrands. Repeated logs denote how many times we apply the log function, forming a base for crafting substitutions that simplify integration.

Integration Techniques

Integration techniques come in handy when simplifying and solving integrals that appear complex at first glance. This problem involves a seemingly intricate integrand, but the method of substitution plays a significant role in simplifying it.

**Substitution Method:**

• Identify a part of the integrand that can be substituted with a single variable, typically to simplify the expression.
• This often involves setting \( u = l^{r-1}(x) \) or another transformation, making integration simpler.
• Differentiating the substitution (e.g., finding \( du \)) helps to replace \( dx \) in the original integral.

Once substitution is complete, the integrand assumes a simpler form, often transforming into a basic power rule problem, as seen when it became \( \int u^{r-1} \, du \). This approach avoids direct integration of repeated and complex logarithmic expressions, facilitating efficient problem-solving.

Advanced Mathematics

Advanced mathematics refines and builds upon basic concepts with intricate problems like integrating repeated logarithmic functions. Here, the problem showcases a blend of not only logarithmic understanding but also advanced integration and algebra techniques.

**Key Elements in Advanced Math:**

• Ability to manipulate mathematical expressions, transforming them into easier forms.
• Understanding layers of functions (like \( l^{ ay(n)}(x) \)) and how to apply corresponding integration techniques effectively.
• Applying simplification and substitution beyond elementary techniques, adapting to multiple variables and repetitive functions.
• Evaluating integrals by recognizing patterns and using series, complex numbers, or transformational approaches when necessary.

By applying these higher-level concepts, you solve an integrand involving repeated operations methodically, ending up with expressions ready for integral calculation. As evident here, mastery in advanced mathematics involves precise execution of techniques and deep comprehension of function behaviors.