Problem 94 - A Complete Resource Book in Mathematics for JEE Main

Question

A function \(f: R \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) satisfies the equation \(f(x)+f(y)=f\left(\frac{x+y}{1-x y}\right)\). The function \(f\) is differentiable on \(R\) and \(f^{\prime}(0)=2\). \(f^{\prime}(x)\) is equal to (A) \(\frac{1}{1+x^{2}}\) (B) \(\frac{1}{1-x^{2}}\) (C) \(\frac{2}{1+x^{2}}\) (D) \(\frac{2}{x^{2}-1}\)

Step-by-Step Solution

Answer

The derivative is \(f'(x) = \frac{2}{1+x^2}\). The correct answer is (C).

Step 1: Set Up Given Equation

We are given the functional equation \( f(x) + f(y) = f\left(\frac{x+y}{1-xy}\right) \). This resembles the angle addition formula for the tangent function.

Step 2: Consider Possible Function Form

The functional form \( f(x) = \tan^{-1}(a x) \) (inverse tangent) is a natural candidate for this problem because it maps to \((-\frac{\pi}{2}, \frac{\pi}{2})\) and satisfies a similar addition formula: \( \tan^{-1}(a x) + \tan^{-1}(a y) = \tan^{-1}\left(\frac{a(x+y)}{1-a^2 xy}\right) \).

Step 3: Compare To Identity

Given the structure, we compare: \( f(x) = \tan^{-1}(x) \) satisfies the same functional equation: \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \).

Step 4: Differentiate the Function

Assuming \( f(x) = \tan^{-1}(x) \), differentiate to find \( f'(x) = \frac{1}{1+x^2} \).

Step 5: Validate Given Condition

The problem states \( f'(0) = 2 \). For \( f(x) = a \tan^{-1}(x) \), differentiating gives \( f'(x) = \frac{a}{1+x^2} \). Given \( f'(0) = 2 \), we have: \( a \cdot 1 = 2 \), thus \( a = 2 \).

Step 6: Finalize the Derivative

With \( a = 2 \), the derivative becomes \( f'(x) = \frac{2}{1+x^2} \). Compare this result with the options to find the match.

Key Concepts

Functional Equations

Functional equations are intriguing mathematical expressions that are determined entirely by their operations, without the explicit need for numeric values. In this case, we have \( f(x) + f(y) = f\left(\frac{x+y}{1-xy}\right) \), a classical form often related to trigonometric identities. When dealing with such functional equations, especially those that remind us of known formulas, it's beneficial to consider similar established functions. These equations often hold a geometric meaning or a relation to trigonometric functions, which can give us hints on their solutions.

For instance, the given equation strongly resembles the addition formula for the tangent function, \( \tan(x+y) = \frac{\tan(x) + \tan(y)}{1- \tan(x) \tan(y)} \). Recognizing this similarity helps us deduce that the function \( f \) is likely related to the inverse tangent function, streamlining our search for a solution.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \tan^{-1} \), serve as the cornerstone for solving trigonometric expressions when the output angle is needed. In our problem, the potential solution \( f(x) = \tan^{-1}(a x) \) is proposed because this function naturally fits the given range \((-\frac{\pi}{2}, \frac{\pi}{2})\).

The hallmark of \( \tan^{-1}(x) \) is its ability to convert additions and combinations of angles using familiar forms, like \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-x y}\right) \). This identity simplifies the search for the right function form of \( f(x) \).

Using inverse trigonometric functions offers us a strategic advantage by allowing us to easily work within certain numeric ranges and equational constraints, thereby validating our proposed function form by mapping it directly to the given functional equation.

Derivative Calculations

Understanding derivatives in the context of a solved function confirms the differentiability and the rate of change properties of that function. Upon assuming \( f(x) = a \tan^{-1}(x) \), we proceed to find its derivative to confirm whether it aligns with the provided condition that \( f'(0) = 2 \).

This process involves the basic differentiation rule for the inverse tangent. We know that the derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \). Consequently, for \( f(x) = a \tan^{-1}(x) \), differentiation results in \( f'(x) = \frac{a}{1+x^2} \).

Given that \( f'(0) = 2 \), we substitute \( x = 0 \) into the derivative \( \frac{a}{1+0^2} = a \) to determine \( a \). This yields \( a = 2 \). Substituting back, the definitive expression for \( f'(x) \) becomes \( \frac{2}{1+x^2} \), allowing us to match our calculation with the given options and confirm the answer.