Problem 25 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(x \in(0,1) .\) The set of all \(x\) such that \(\sin ^{-1} x>\cos ^{-1} x\), is the interval: \([\) Online April \(25,2013 \mid\) (a) \(\left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)\) (b) \(\left(\frac{1}{\sqrt{2}}, 1\right)\) (c) \((0,1)\) (d) \(\left(0, \frac{\sqrt{3}}{2}\right)\)

Step-by-Step Solution

Answer

The interval is \( \left( \frac{1}{\sqrt{2}}, 1 \right) \).

Step 1: Understand the Inverse Functions

Recall that the inverse sine function, \( \sin^{-1} x \), and the inverse cosine function, \( \cos^{-1} x \), are defined such that \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \). This implies that if \( \sin^{-1} x > \cos^{-1} x \), then \( \sin^{-1} x \) must be greater than \( \frac{\pi}{4} \).

Step 2: Set the Inequality

To find when \( \sin^{-1} x > \cos^{-1} x \), set \( \sin^{-1} x > \frac{\pi}{4} \). This is because the range for the sum to reach \( \frac{\pi}{2} \) is valid when \( \cos^{-1} x < \frac{\pi}{4} \). Thus, solve for \( x \) when \( \sin^{-1} x > \frac{\pi}{4} \).

Step 3: Solve the Inequality for \( x \)

Since \( \sin \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \), the inequality \( \sin^{-1} x > \frac{\pi}{4} \) implies that \( x > \frac{1}{\sqrt{2}} \). Therefore, the set of all \( x \) in the interval \( (0,1) \) that satisfies this is \( \left( \frac{1}{\sqrt{2}}, 1 \right) \).

Step 4: Check the Interval

Verify by substituting values above and below \( \frac{1}{\sqrt{2}} \). For instance, at \( x = \frac{1}{\sqrt{2}} \), \( \sin^{-1} x = \frac{\pi}{4} \) which equals \( \cos^{-1} x \). For \( x > \frac{1}{\sqrt{2}} \), \( \sin^{-1} x > \frac{\pi}{4} \) holds. Therefore, the correct interval is \( \left( \frac{1}{\sqrt{2}}, 1 \right) \).

Key Concepts

Inequalities

When working with inequalities in mathematics, it is essential to understand their fundamental role in comparing two quantities or expressions. An inequality shows that one side of a mathematical statement is larger or smaller than the other. In this exercise, the inequality \( \sin^{-1} x > \cos^{-1} x \) portrays where the inverse sine of \( x \) is greater than the inverse cosine of \( x \).

To tackle such inequalities involving trigonometric functions, we often translate them into conditions that \( x \) must satisfy. For example, using the identity \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \), we reassess the inequality as \( \sin^{-1} x > \frac{\pi}{4} \). This conversion allows us to work directly with known trigonometric values. Subsequently, the inequality \( \sin^{-1} x > \frac{\pi}{4} \) tells us that:

• \( x \) must be such that its sine value is greater than \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).

Recognizing and transforming the initial inequality into a format that utilizes known trigonometric values makes it easier to solve for \( x \) over its domain.

Interval Notation

Interval notation is a concise way to represent a set of numbers on the real number line. It describes the start and end points of the interval and whether those points are included or excluded, simplifying the expression of solutions to inequalities. In this scenario, we are interested in determining when \( \sin^{-1} x > \cos^{-1} x \) holds within the given interval of \( x \in (0,1) \).

Intervals use brackets \( ( ) \) or \( [ ] \) to indicate whether endpoints are excluded or included:

• Round brackets \(( )\) mean the endpoint is not included in the interval, known as "open."
• Square brackets \([ ]\) include the endpoint, considered "closed."

In the solution to this problem, we find that the values of \( x \) satisfying the inequality belong to the interval \( \left( \frac{1}{\sqrt{2}}, 1 \right) \). This open interval tells us that neither endpoint \( \frac{1}{\sqrt{2}} \) nor 1 is included in the set of valid \( x \)-values.

Trigonometric Identities

Trigonometric identities are vital tools that simplify the relationships between different trigonometric functions. In solving this problem, the identity \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) is key. It tells us that for any \( x \), the sum of its inverse sine and inverse cosine equals \( \frac{\pi}{2} \), ensuring that each value of these inverse functions is consistent within their respective ranges.

When we apply this identity, it becomes apparent:

• If \( \sin^{-1} x > \cos^{-1} x \), then \( \sin^{-1} x \) has exceeded \( \frac{\pi}{4} \).
• This implies the sine of \( x \) exceeds the value of \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).

Understanding and using these types of trigonometric identities helps not only in solving inequalities but also in exploring deeper facets of trigonometry, such as transformations and angles. Knowing these identities gives students tools to approach more complex problems with confidence.