Problem 55 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}\), then for \(x \geq 1, f(x)\) is equal to (A) \(\pi\) (B) \(2 \pi\) (C) \(\frac{\pi}{2}\) (D) none of these

Step-by-Step Solution

Answer

The value of \( f(x) \) for \( x \geq 1 \) is \( \pi \).

Step 1: Identify the Function Components

The function given is \( f(x) = 2 \tan^{-1} x + \sin^{-1} \frac{2x}{1+x^2} \). We need to evaluate this function for \( x \geq 1 \). First, recognize the components: two inverse trigonometric functions, \( \tan^{-1}\) and \( \sin^{-1}\).

Step 2: Simplify the Inverse Sine Expression

The term \( \sin^{-1} \frac{2x}{1+x^2} \) can be rewritten using the identity \( \sin^{-1} y = 2 \tan^{-1} \left( \frac{y}{1 + \sqrt{1-y^2}} \right) \) for \( y = \frac{2x}{1+x^2} \). By identity, \( \tan t = \frac{2x}{1+x^2} \), where \( t = \tan^{-1} x \).

Step 3: Reformulate the Expression Using Identities

Reformulate using the fact \( \sin \theta = \frac{2 \tan \left(\theta/2\right)}{1 + \tan^2 \left(\theta/2\right)} \) to express \( \sin^{-1} \) in terms of \( \tan^{-1} \). Since \( \theta = \tan^{-1} x \), the expression simplifies as \( \tan^{-1}(x) = t \), hence \( \frac{2x}{1+x^2} = \sin(t) \), where \( t = \tan^{-1} x \).

Step 4: Calculate \(f(x)\)

Now substitute \( \tan t = x \) into \( f(x) = 2 \tan^{-1} x + \sin^{-1} \frac{2x}{1+x^2} = 2t + t = 3t \). As \( x \to \infty \), \( t \to \pi/2 \). For \( x = 1 \), \( t = \pi/4 \), hence \( 2t = \pi/2 \) and \( \sin^{-1}(1) = \pi/2 \). Thus, for \( x \geq 1 \), \( f(x) = \pi \).

Step 5: Result Verification

After simplification and considering the identity, for all \( x \geq 1 \), the calculated \( f(x) \) is constantly \( \pi \). Therefore, the entire function evaluates to a constant value \( \pi \) for the specified domain.

Key Concepts

Tangent Inverse

The tangent inverse function, denoted as \( \tan^{-1}(x) \), is used to find the angle whose tangent is \( x \). This function holds immense importance in solving trigonometric equations where the tangent values are known but the angles aren’t. By understanding \( \tan^{-1}(x) \), we can transform complex trigonometric expressions into simpler ones using inverse trigonometric identities. In this exercise, \( \tan^{-1}(x) \) is a key element, as it allows us to handle the given expressions efficiently. By realizing \( t = \tan^{-1}(x) \), you can see how seamlessly the problem was transformed to find \( f(x) = \pi \).
The range of \( \tan^{-1} x \) is \(-\frac{\pi}{2} \leq \tan^{-1} x \leq \frac{\pi}{2} \), which means it maps to all real values of \( x \), making it a particularly flexible tool in trigonometric problem-solving.

Sine Inverse

The sine inverse, written as \( \sin^{-1}(x) \) or arcsine, is used to find the angle whose sine is \( x \). Ranging from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), this function is crucial for adjusting between trigonometric and inverse trigonometric forms. It allows us to interpret trigonometric expressions with a real-world angle perspective.
In the problem, transforming \( \sin^{-1} \frac{2x}{1+x^2} \) involved substituting conscious identities that express it in terms of a tangent, specifically \( y = \frac{2x}{1+x^2} \). Understanding these transformations is key, and here it's neatly demonstrated how inverse functions help in simplifying intricate expressions. When \( y = 1 \), we know \( \sin^{-1}(1) = \frac{\pi}{2} \), guiding us to discern the function's behavior for larger values of \( x \).

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the involved variables. They are tools that simplify trigonometric calculations and transformations. For this topic, the identity \( \tan^{-1}(x) + \sin^{-1}(y) = \text{constant} \), especially when it equals \( \pi \), played a crucial role in the given problem.
Understanding these identities, such as \( \sin \theta = \frac{2 \tan(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} \), and their application in expressing \( \sin^{-1} \) in terms of \( \tan^{-1} \), allows for transformation of complex expressions into varied forms. Thus, making calculations more manageable. The application of these identities in simplifying \( f(x) \) was essential in reaching the conclusion that \( f(x) = \pi \) for \( x \geq 1 \).

Function Evaluation

Function evaluation is the process of determining the output of a function for a specific input. In our exercise, evaluating \( f(x) \) for \( x \geq 1 \) means substituting values into the function \( f(x) = 2 \tan^{-1} x + \sin^{-1} \frac{2x}{1+x^2} \) and simplifying.
During the evaluation, we leveraged trigonometric identities to transform and substitute inverse trigonometric expressions accurately. This exercise highlights how, with the right identities and substitutions, a seemingly complex function can reveal a simple constant solution. This offers an elegant demonstration of the harmony between different branches of trigonometry, leading to the realization that \( f(x) \) constantly equals \( \pi \) for \( x \geq 1 \).