Problem 15 - A Complete Resource Book in Mathematics for JEE Main

Question

The number of solutions of the equation \(2 \sin ^{-1} \sqrt{x^{2}-x+1}+\cos ^{-1}\left(\sqrt{x^{2}-x}\right)=\frac{3 \pi}{2}\) is (A) 0 (B) infinite (C) 2 (D) 4

Step-by-Step Solution

Answer

0 solutions (Option A).

Step 1: Analyze Domain Restrictions

First, note the domain restrictions for the inverse sine and inverse cosine functions. The expression inside \( \sin^{-1} \) requires \( 0 \leq \sqrt{x^2 - x + 1} \leq 1 \), while the expression inside \( \cos^{-1} \) requires \( 0 \leq \sqrt{x^2 - x} \leq 1 \).

Step 2: Simplify Domain Conditions

\( \sqrt{x^2 - x + 1} \) is always greater than \( 1 \) because its minimum value is 1 (by completing the square). Therefore, \( 2 \sin^{-1} \sqrt{x^2 - x + 1} \) is not defined. Similarly, \( \sqrt{x^2 - x} \) implies \( x(x-1) \geq 0 \), leading to \( x \leq 0 \) or \( x \geq 1 \). However, this is irrelevant if the first condition fails.

Step 3: Conclude with Solution Count

Since \( 2 \sin^{-1} \sqrt{x^2 - x + 1} \) is not defined for any real number \( x \), the equation has no real solution.

Key Concepts

Domain Restrictions

When working with inverse trigonometric functions, it's crucial to understand domain restrictions. These restrictions ensure that the functions maintain their properties and remain well-defined. For \( \sin^{-1} \sqrt{x^2 - x + 1} \), the value inside the square root must be within the range \( 0 \leq y \leq 1 \). However, upon analyzing \( \sqrt{x^2 - x + 1} \), you will notice it is always greater than or equal to 1, due to its structure when simplified. This makes it impossible to meet the domain requirement for the inverse sine, rendering the function undefined for real numbers.
On the other hand, the expression \( \cos^{-1}(\sqrt{x^2 - x}) \) requires that the argument inside the square root be in \( 0 \leq y \leq 1 \) and also satisfy \( x(x-1) \geq 0 \). However, none of this matters if the sine part of the function cannot be satisfied to begin with. Understanding these conditions can often dictate the feasibility of finding any solutions.

Equation Solving

Solving equations involving inverse trigonometric functions demands a precise approach to simplify and examine each component for real and valid solutions. It's important to substitute values only within their allowed domains. In this scenario, analyzing both \( 2 \sin^{-1} \sqrt{x^2 - x + 1} \) and \( \cos^{-1}(\sqrt{x^2 - x}) \) is necessary.
The entire process involves verifying whether valid solutions can emerge for any \( x \) that will satisfy the entire equation \( 2 \sin^{-1} \sqrt{x^{2}-x+1}+\cos^{-1}\left(\sqrt{x^{2}-x}\right)=\frac{3 \pi}{2} \). If any part of the equation is not well-defined due to the domain issues as discussed previously, it's impossible to solve the equation. Hence, the original step in this solution process hinges on identifying the non-existence of valid \( x \) values due to the inverse functions being out of their range.

Trigonometric Equations

Trigonometric equations often provide insightful solutions linked to periodic properties of trigonometric functions and their inverses. When you involve inverse functions like \( \sin^{-1} \) and \( \cos^{-1} \), ensuring correctness of domains becomes even more crucial.
In the given equation, it combines both inverse sine and cosine in a complex manner. You should always check the feasibility of the expressions within the inverse functions before proceeding, as many trigonometric equations come with such constraints. Also, recognize that mistakes in evaluating such equations often stem from not observing these rules.

• Inverse sine requires inputs between -1 and 1 only.
• Inverse cosine similarly requires inputs within this range.

These rules allow functions to remain defined and meaningful, ensuring we end up with valid solutions if possible. Unfortunately, here due to the failings in domain restrictions, no meaningful trigonometric solution can emerge.