Problem 27 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\cos ^{-1} \sqrt{\frac{a-x}{a-b}}=\sin ^{-1} \sqrt{\frac{x-b}{a-b}}\) is possible if (A) \(a>x>b\) or \(ab\) and \(x\) takes any value (D) \(a

Step-by-Step Solution

Answer

The correct answer is (A) \(a > x > b\) or \(a < x < b\).

Step 1: Understand the Functions

The given equation is \(\cos^{-1} \sqrt{\frac{a-x}{a-b}} = \sin^{-1} \sqrt{\frac{x-b}{a-b}}\). This implies that both the cosine inverse and sine inverse functions are used. It is important to remember that \(\cos^{-1}(y)\) is defined for \(-1 \leq y \leq 1\) and \(\sin^{-1}(y)\) is defined similarly.

Step 2: Set Up the Range Conditions

Since \(\cos^{-1}(y)\) and \(\sin^{-1}(y)\) are defined over the same range, both expressions under the radicals must independently fall between 0 and 1. This gives us: 1. \(0 \leq \frac{a-x}{a-b} \leq 1\)2. \(0 \leq \frac{x-b}{a-b} \leq 1\)

Step 3: Analyze the First Inequality

For the first inequality \(\frac{a-x}{a-b}\), consider: - To satisfy \(0 \leq \frac{a-x}{a-b} \leq 1\), both \(a-x\) and \(a-b\) must be positive, meaning \(a > x\) and \(a > b\).

Step 4: Analyze the Second Inequality

For the second inequality \(\frac{x-b}{a-b}\), consider: - To satisfy \(0 \leq \frac{x-b}{a-b} \leq 1\), both \(x-b\) and \(a-b\) must be positive, meaning \(x > b\) and \(a > b\).

Step 5: Combine Results

From analyzing both inequalities, the conditions \(a > x > b\) arise when both are satisfied together. This matches the condition from the given options when \(a > x > b\).

Step 6: Conclusion

The solution that satisfies both inequalities and the condition is \(a > x > b\). Thus, the correct option is the one that includes the condition \(a > x > b\), which corresponds to option (A) \(a > x > b\) or \(a < x < b\).

Key Concepts

Range of Trigonometric Functions

Understanding the range of trigonometric functions is crucial when solving equations involving inverse trigonometric functions. In the given exercise, we deal with

• the inverse cosine function, \(\cos^{-1}(y)\), which is defined for -1 ≤ y ≤ 1,
• and the inverse sine function, \(\sin^{-1}(y)\), also defined for -1 ≤ y ≤ 1.

This congruity in their domains allows us to equate functions involving these inverses, under the constraint that their arguments lie within this range. Specifically, when manipulating an expression that involves radicals inside these inverse functions, like \sqrt{\frac{a-x}{a-b}}\, the expression must result in a value between 0 and 1. Essentially, this process involves ensuring that both the numerator and the denominator in the fraction represent positive values.
Fulfilling this condition ensures that the radicals involved remain valid under the functions' definitions. This understanding of ranges is fundamental to comprehending how to transition between different mathematical representations of inverse trigonometric equations.

Inequalities in Trigonometry

Inequalities play a pivotal role in steering solutions toward the right domain and range in trigonometric problems. In the original exercise, we focus on ensuring that the expressions inside the inverse functions meet certain inequalities.
Given the inequalities:

• \(0 \leq \frac{a-x}{a-b} \leq 1\)
• \(0 \leq \frac{x-b}{a-b} \leq 1\)

the task is to solve them under the precondition that \(a > b\). Examining the expressions, \frac{a-x}{a-b}\ essentially requires that both \(a-x\) and \(a-b\) are positive, leading to \(a > x\) and \(a > b\). Similarly, for the second inequality, \(x-b\) and \(a-b\) being positive implies \(x > b\) and \(a > b\).
Combining these conditions helps establish that \(a > x > b\), which aligns perfectly with Option (A) in the provided choices. Understanding how inequalities restrict and shape the solution space is an invaluable tool in solving such equations.

Trigonometric Identities

Trigonometric identities form the backbone of many calculations in trigonometry by allowing simplification and transformation of expressions. In the context of inverse trigonometric functions, such identities can help double-check the validity of expressions or aid in simplifying complex irrational components.
For instance, when encountering functions like \cos^{-1}\ and \sin^{-1}\ within the same problem, identities such as \(\cos(\theta) = \sin(\frac{\pi}{2} - \theta)\) can offer insights into solutions by characterizing the change from one trigonometric form to another. Though the exercise provided here is more about ensuring correct ranges and inequalities, introducing identities can enhance overall understanding.
By understanding and thoroughly applying trigonometric identities, simplifying equations even amidst complex transformations becomes more intuitive, effectively making learning more comprehensive and less daunting.