Problem 14 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If \(f(x)=\log \left(\frac{1+x}{1-x}\right)\) when \(-1

Step-by-Step Solution

Answer

The answer is (a) \( f\left(\frac{x_1 + x_2}{1 + x_1 x_2}\right) \).

Step 1: Express the Function Using Properties of Logarithms

The function given is \( f(x) = \log \left(\frac{1+x}{1-x}\right) \). We'll use properties of logarithms to simplify the expression. The sum \( f(x_1) + f(x_2) \) can be rewritten using the property \( \log(a) + \log(b) = \log(ab) \): \[f(x_1) + f(x_2) = \log \left(\frac{1+x_1}{1-x_1}\right) + \log \left(\frac{1+x_2}{1-x_2}\right)\] \[= \log \left( \frac{1+x_1}{1-x_1} \times \frac{1+x_2}{1-x_2} \right)\] \[= \log \left( \frac{(1+x_1)(1+x_2)}{(1-x_1)(1-x_2)} \right)\]

Step 2: Simplify the Fraction

We now simplify the expression: 1. Expand the numerators and denominators: - Numerator: \((1 + x_1)(1 + x_2) = 1 + x_1 + x_2 + x_1x_2\). - Denominator: \((1 - x_1)(1 - x_2) = 1 - x_1 - x_2 + x_1x_2\).2. Rewrite the logarithmic fraction using this expansion: \[= \log \left( \frac{1 + x_1 + x_2 + x_1x_2}{1 - x_1 - x_2 + x_1x_2}\right)\]

Step 3: Relate to Function Arguments

Notice the similarity to the argument structure of \( f(x) \), as \( f(x) = \log \left(\frac{1+x}{1-x}\right) \). The numerator and denominator can be matched to possible forms from the answer choices: \[= f\left(\frac{(x_1 + x_2)}{1 + x_1x_2}\right)\] This directly corresponds to choice (a).

Step 4: Conclusion: Verify Answer Choice

The simplified expression we derived, \( f\left(\frac{x_1 + x_2}{1 + x_1 x_2}\right) \), matches answer (a). Therefore, the solution to the problem is that \( f(x_1) + f(x_2) \) is equal to choice (a): \[f\left(\frac{x_1+x_2}{1+x_1 x_2}\right)\]

Key Concepts

Properties of Logarithms

Logarithms have several important properties that are key in simplifying expressions. For example, the property \( \log(a) + \log(b) = \log(ab) \) allows us to combine the log of two numbers into a single expression. This property is particularly useful in algebraic manipulation. When dealing with rational functions like \( \frac{1+x}{1-x} \), applying these properties can reveal simpler or equivalent forms, aiding in finding solutions to complex logarithmic problems. Understanding how to break down and recombine log expressions is crucial for solving such exercises efficiently.

Function Transformation

In mathematical terms, a transformation can often help us to interpret functions in different ways. In this case, the transformation of \( f(x) = \log\left(\frac{1+x}{1-x}\right) \) is key. By considering \( f(x_1) + f(x_2) \), we are encouraged to transform and combine these expressions into a single transformed function using the logarithm properties. Transformations can demonstrate equivalency and help to better visualize complex relationships between different functions. This is particularly helpful when testing against multiple-choice answer options.

Algebraic Manipulation

Algebraic manipulation involves the strategic rearrangement or simplification of equations to make them easier to solve or understand. For the given problem, after using properties of logarithms, we expanded and simplified the expression. This involved multiplying terms to simplify a rational function: \( \frac{(1+x_1)(1+x_2)}{(1-x_1)(1-x_2)} \). Through expanding, where each term is multiplied out, algebraic manipulation allows for significant simplification. This method is vital for correctly identifying which function form matches the provided answer choices. Skills in algebraic manipulation are foundational for breaking down complex logarithmic functions.