Problem 55 - A Complete Resource Book in Mathematics for JEE Main 2017

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

None of these.

Step 1: Understanding the Functions

Identify the two functions involved in the expression: the inverse tangent function \( \tan^{-1} x \) and the inverse sine function \( \sin^{-1} \frac{2x}{1+x^2} \).

Step 2: Recognizing the Identity

Recall the identity \( \tan(\sin^{-1} t) = \frac{t}{\sqrt{1-t^2}} \). Notice that \( \frac{2x}{1+x^2} \) can be rewritten in terms of the identity of the tangent of a double angle.

Step 3: Simplifying the Expression

Recognize that \( \frac{2x}{1+x^2} \) is the tangent of the angle \( \tan^{-1}(x) \). Therefore, the expression \( \sin^{-1} \frac{2x}{1+x^2} \) is equivalent to \( \tan^{-1} x \).

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

Calculate \( f(x) = 2 \tan^{-1} x + \tan^{-1} x = 3 \tan^{-1}(x) \).

Step 5: Evaluate for \(x \geq 1 \)

When \( x \geq 1 \), \( \tan^{-1} x \) will approach \( \frac{\pi}{4} \) as \( x \) approaches 1. Thus, \( 3 \tan^{-1}(x) \) approaches \( 3 \times \frac{\pi}{4} = \frac{3\pi}{4} \), which does not match any of the given options.

Key Concepts

Tangent Inverse

The tangent inverse function, often denoted as \( \tan^{-1} x \), is the inverse of the tangent function. This function helps us determine the angle whose tangent value is \( x \). It's crucial in trigonometry because it allows us to work backward from a given tangent value to find an angle. For example, if \( \tan \theta = x \), then \( \theta = \tan^{-1} x \).Understanding the behavior of the tangent inverse function is important:

• The range of \( \tan^{-1} x \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
• It is an increasing function, which means as \( x \) increases, \( \tan^{-1} x \) also increases.
• Its graph is symmetric about the origin, reflecting its property as an odd function: \( \tan^{-1} (-x) = -\tan^{-1} x \).

In the context of this exercise, the expression \( 2\tan^{-1} x \) means the angle corresponding to \( \tan^{-1} x \) is doubled, crucial for solving the problem by simplifying the trigonometric identities.

Sine Inverse

The sine inverse function, represented as \( \sin^{-1} y \), is used to find the angle whose sine is a given number \( y \). This function is particularly helpful when solving trigonometric equations or finding angles in right triangles.Key properties of the sine inverse function include:

• The range of \( \sin^{-1} y \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
• It's a smooth, continuous function that increases on its interval.
• It is an odd function, meaning \( \sin^{-1}(-y) = -\sin^{-1}(y) \).

In this exercise, \( \sin^{-1} \frac{2x}{1+x^2} \) is simplified using the identity: \( \tan(\sin^{-1} t) = \frac{t}{\sqrt{1-t^2}} \). This identity is important because it allows us to express the sine inverse term in terms of tangent, eventually simplifying to \( \tan^{-1} x \). Understanding how to maneuver between these inverse trigonometric functions is key to solving the problem.

Trigonometric Identities

Trigonometric identities are equations that hold true for all values of the variables involved. These identities are fundamental tools for simplifying expressions and solving trigonometric equations. Familiarity with them can dramatically simplify seemingly complex problems.Key identities related to inverse functions include:

• \( \tan(\sin^{-1} t) = \frac{t}{\sqrt{1-t^2}} \)
• \( \sin(\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}} \)
• \( \cos(\tan^{-1} x) = \frac{1}{\sqrt{1+x^2}} \)

These identities allow conversions between sine, cosine, and tangent, granting a new perspective to tackle trigonometric functions.In the exercise, the transformation of \( \sin^{-1} \frac{2x}{1+x^2} \) to \( \tan^{-1} x \) is based on recognizing \( \frac{2x}{1+x^2} \) as the tangent of a double angle. Utilizing this key identity allowed the expression \( f(x) \) to be re-written and calculated, resulting in an expression with tangents solely. This leads to a more straightforward evaluation of \( f(x) \) as it approaches specific values.