Problem 102 - A Complete Resource Book in Mathematics for JEE Main

Question

Domain of definition of the function \(f(x)=\sqrt{\sin x}+\sin ^{-1}\left(\frac{2|x|}{1+x^{2}}\right)\) is (A) \((2 n \pi,(2 n+1) \pi), n \in I\) (B) \([2 n \pi,(2 n+1) \pi], n \in I\) (C) \(R\) (D) None of these

Step-by-Step Solution

Answer

The domain is \([2n\pi, (2n+1)\pi], n \in \mathbb{Z}\).

Step 1: Analyze the Function Inside the Square Root

For the function \(f(x) = \sqrt{\sin x} + \sin^{-1}\left(\frac{2|x|}{1+x^{2}}\right)\), we need to first determine the domain of \(\sqrt{\sin x}\). The square root function requires that \(\sin x \geq 0\), which occurs for \(x \in [2n\pi, (2n+1)\pi]\) where \(n\in \mathbb{Z}\).

Step 2: Analyze the Function Inside the Inverse Sine

Next, consider the domain for \(\sin^{-1}\left(\frac{2|x|}{1+x^{2}}\right)\). The domain of \(\sin^{-1}(u)\) is \(-1 \leq u \leq 1\). Thus, \(-1 \leq \frac{2|x|}{1+x^{2}} \leq 1\).

Step 3: Solve Inequalities for Second Term

We solv for \(\frac{2|x|}{1+x^{2}} \leq 1\). This inequality simplifies to \(2|x| \leq 1+x^{2}\), and further to \(x^{2} - 2|x| + 1 \geq 0\), which is always true for all real \(x\). Therefore, this part does not restrict the domain.

Step 4: Combine the Domains

Since the restriction from the \(\sin^{-1}\) term is satisfied for all real \(x\), and the restriction from the \(\sqrt{\sin x}\) term limits it to \(x \in [2n\pi, (2n+1)\pi]\), the combined domain where both conditions are satisfied is \(x \in [2n\pi, (2n+1)\pi]\), taking into account only closed intervals where \(\sin x \geq 0\). Therefore, the domain is option \(B\).

Key Concepts

Square Root Function

The square root function is a basic but essential mathematical concept. When dealing with square roots, it's crucial to understand that you can only take the square root of non-negative numbers. That's because you cannot have a real number as a square root of a negative number.
In our exercise, the function includes a square root of \( \sin x \). This means we need \( \sin x \geq 0 \).
This condition identifies intervals on the x-axis where the sine function is non-negative.

• Remember that \( \sin x \) changes its sign periodically.
• It is non-negative from 0 to \( \pi \), then negative from \( \pi \) to \( 2\pi \).

The domain for taking the square root requires selecting the correct intervals:

• This means we look at intervals formatted as \( [2n\pi, (2n+1)\pi] \) where \( n \) is an integer.

Understanding these intervals helps us establish where the square root term is valid.

Inverse Trigonometric Functions

Inverse trigonometric functions are extensions of trigonometric functions that enable us to find angles given trigonometric values.
They include functions like \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \).
In our exercise, the function contains an inverse sine, expressed as \( \sin^{-1}\left(\frac{2|x|}{1+x^{2}}\right) \).
For inverse sine, the argument must lie between -1 and 1, inclusive. This property establishes:

• \(-1 \leq \frac{2|x|}{1+x^2} \leq 1\)

Ensuring this inequality holds is crucial for determining the domain of the function. Interestingly, our problem shows that \( \frac{2|x|}{1+x^{2}} \leq 1\) simplifies to always being true for all real \( x \).
This characteristic means the inverse sine term does not further restrict the domain.
Connecting this to our square root considerations reveals the intervals where the function behaves properly.

Trigonometric Inequalities

Trigonometric inequalities often determine the limits of where a function is defined or where it takes certain values. In this scenario, trigonometric inequalities come into play when combining domain conditions.
Solve the inequality for \( \sin x \geq 0 \) to find the proper domain parts from previous sections:

• This is where sine is non-negative.
• Focus on intervals like \( [2n\pi, (2n+1)\pi] \).

These are intervals where both the square root and inverse trigonometric conditions coexist properly.
Once we ensure both inequalities are satisfied, the combined domain defines where the entire function is valid. This approach showcases how solving inequalities accurately assists in understanding trigonometric functions' domains and combined terms in more complex equations.