Problem 95 - 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 \(f^{\prime}(0)=2\). \(f(x)\) is equal to (A) \(\tan ^{-1} x\) (B) \(2 \tan ^{-1} x\) (C) \(4 \tan ^{-1} x\) (D) None of these

Step-by-Step Solution

Answer

The function is \( 2 \tan^{-1}(x) \), so the answer is (B).

Step 1: Identify the functional equation

The given functional equation is \( f(x) + f(y) = f\left(\frac{x+y}{1-xy}\right) \). This equation is reminiscent of the angle addition formula for tangent.

Step 2: Inspect angle addition formula

The angle addition formula for arctangent is \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) for values where the right-hand side is defined. Comparing this equation to our functional equation suggests a connection with the \( \tan^{-1} \) function.

Step 3: Hypothesize potential solutions

Given the structure of the equation, hypothesize that \( f(x) = C \tan^{-1}(x) \) for some constant \( C \). We choose this form to verify if it satisfies the original equation.

Step 4: Differentiate the hypothesized function

Assume \( f(x) = C \tan^{-1}(x) \). Then the derivative \( f'(x) = \frac{C}{1+x^2} \). We know from the problem that \( f'(0) = 2 \), so we find \( C \) by setting \( f'(0) = \frac{C}{1+0^2} = C = 2 \).

Step 5: Verify the hypothesis

Substitute \( f(x) = 2 \tan^{-1}(x) \) back into the original functional equation: - Left-hand side: \( 2\tan^{-1}(x) + 2\tan^{-1}(y) = 2(\tan^{-1}(x) + \tan^{-1}(y)) \).- Right-hand side: \( 2\tan^{-1}\left(\frac{x+y}{1-xy}\right) \).- Since \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \), both sides match confirming \( f(x) = 2\tan^{-1}(x) \) is correct.

Key Concepts

Angle Addition Formula

The angle addition formula is an essential concept in trigonometry. It helps in finding the trigonometric functions of the sum of two angles using the functions of those angles individually. For the tangent function, the formula is:

• \( \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \)

This formula is crucial when dealing with functional equations similar to the one in our exercise. The exercise uses the angle addition formula for the inverse tangent, also known as the arctangent.

When equating the function to \(\tan^{-1}(x)\), we use the arctangent form of the angle addition formula, which is:

• \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \)

Provided the condition \(1-xy eq 0\). This comparison is crucial in solving functional equations like the one provided, as it matches perfectly with the given equation explaining the structure of function \(f\) involved.

Differentiable Functions

A differentiable function is one where its derivative exists at each point in its domain. Differentiability implies continuity, which signals that the function changes smoothly, without any jumps or abrupt angles.

In this exercise, the function \(f\) is known to be differentiable over all real numbers, which means we can find its derivative at any point.Differentiation plays a key role in verifying the solution since we know from the problem that \(f'(0) = 2\).

• This condition helps determine the constant \(C\) when hypothesizing that \(f(x) = C \tan^{-1}(x)\).
• By differentiating this hypothesis, we get \(f'(x) = \frac{C}{1+x^2}\).

Setting \(x = 0\), we find \(f'(0) = C\), thus \(C = 2\). Differentiability ensures that our function behaves nicely enough for such manipulations to be valid.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the sine, cosine, and tangent functions, often denoted as \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and \(\tan^{-1}(x)\). They are integral in trigonometry as they allow us to work backward from the value of a trigonometric function to the corresponding angle.

For the tangent function, \(\tan^{-1}(x)\) represents the angle whose tangent is \(x\).

• Specifically, it maps input values from any real number to angles between \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• This range corresponds to the function \(f\) in our exercise, as it is given as \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

These functions are not just algebraic inverses but also exhibit important properties and restrictions, essential in calculus and solving equations. By understanding these properties, we can more effectively find solutions to problems involving inverse trigonometric functions, like the original one from this exercise, confirming that \(f(x) = 2 \tan^{-1}(x)\) fits within these parameters.