Problem 12 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(f(x)=\log _{e}\left(\frac{1-x}{1+x}\right),|x|<1\), then \(f\left(\frac{2 x}{1+x^{2}}\right)\) is equal to: [April 8, 2019 (I)] (a) \(2 f(x)\) (b) \(2 f\left(x^{2}\right)\) (c) \((f(x))^{2}\) (d) \(-2 f(x)\)

Step-by-Step Solution

Answer

The expression \( f\left(\frac{2x}{1+x^2}\right) \) is equal to \(-2f(x)\).

Step 1: Understanding the Problem

We need to find the expression for \( f\left(\frac{2x}{1+x^2}\right) \) when \( f(x) = \log_e\left(\frac{1-x}{1+x}\right) \). We will substitute \( x \) in \( f(x) \) with \( \frac{2x}{1+x^2} \).

Step 2: Substitution in the Function

Substitute \( x = \frac{2x}{1+x^2} \) into the expression for \( f(x) \). We get:\[ f\left(\frac{2x}{1+x^2}\right) = \log_e\left(\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right) \]

Step 3: Simplifying the Expression

Simplify the expression:\[ \frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}} = \frac{\frac{(1+x^2) - 2x}{1+x^2}}{\frac{(1+x^2) + 2x}{1+x^2}} = \frac{(1 - 2x + x^2)}{(1 + 2x + x^2)} \]

Step 4: Identifying Symmetry or Patterns

Recognize that the simplified form, \( \frac{1 - 2x + x^2}{1 + 2x + x^2} \), is equivalent to the substitution of \( \frac{1-x}{1+x} \) for \( (-x) \). This pattern leads to:\[ f\left(\frac{2x}{1+x^2}\right) = \log_e\left(\frac{1-x^2}{1+x^2}\right) \]

Step 5: Expanding the Logarithm

Notice that \( \frac{1-x^2}{1+x^2} \) can be thought of as two fractions multiplied under a logarithm, leading to:\[ \log_e\left(\frac{(1-x)(1+x)}{(1+x)(1-x)}\right) = 2 \cdot \log_e\left(\frac{1-x}{1+x}\right) \], since the expression squares the original argument inside the log.

Step 6: Final Expression

Hence, we arrive at the conclusion that:\[ f\left(\frac{2x}{1+x^2}\right) = -2\log_e\left(\frac{1-x}{1+x}\right) = -2f(x) \].

Key Concepts

Properties of Logarithms

Logarithms have several useful properties that can simplify complex expressions. When dealing with logarithmic functions, these properties are crucial for making substitutions and simplifications. One key property is the product rule, which states:

• \( \log_b(MN) = \log_b(M) + \log_b(N) \)

This allows us to split the logarithm of a product into a sum of logarithms.
Another important rule is the power rule:

• \( \log_b(M^n) = n \times \log_b(M) \)

This property shows that an exponent can be brought in front of the logarithm as a multiplier.
Lastly, the quotient rule is often used:

• \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)

In our exercise, these log rules help simplify \( \log_e\left(\frac{1-x^2}{1+x^2}\right) \) to a more manageable form by recognizing that it represents the square of the original expression \( \log_e\left(\frac{1-x}{1+x}\right) \), thus allowing us to use the power rule to bring out the factor of 2.

Function Substitution

Function substitution is a process where we replace the variable in a function with another expression or value. This technique is useful for analyzing how the function behavior changes under transformations. For instance, in our exercise, we substitute the variable \( x \) in the function \( f(x) = \log_e\left(\frac{1-x}{1+x}\right) \) with the expression \( \frac{2x}{1+x^2} \).
The substitution looks like this:

• \( f\left(\frac{2x}{1+x^2}\right) = \log_e\left(\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right) \)

After the substitution, simplifying the expression is crucial to gain insight into how the function's output changes. This step is similar to adjusting the lens on a camera to see a sharper picture. Ultimately, recognizing patterns like symmetry can often make complex substitutions simpler to handle.

Algebraic Simplification

Algebraic simplification involves reducing expressions to simpler forms by performing algebraic operations. This process is vital for solving problems that involve complex fractions or elaborate functions. In the context of the given exercise, we needed to simplify the fraction inside the logarithm:

• \( \frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}} \)

We achieve this by combining the fractions:

• First, clear the complex fraction by multiplying both the numerator and denominator by \( 1 + x^2 \).
• This results in: \( \frac{1 - 2x + x^2}{1 + 2x + x^2} \)

Next, we identify that the pattern matches a known form, where substituting one part with \(-x\) aids in simplification.
Finally, the equivalent expression \( \log_e\left(\frac{1-x^2}{1+x^2}\right) \) simplifies further using the properties of logarithms, arriving at a manageable expression that reveals the relationship of the original function's transformation. These steps illustrate how breaking down and manipulating expressions can uncover intricate relationships and insights within mathematical functions.