Problem 43 - A Complete Resource Book in Mathematics for JEE Main

Question

The number of real solutions of the equations \(\tan ^{-1} \sqrt{x^{2}-3 x+2}+\cos ^{-1} \sqrt{4 x-x^{2}-3}=\pi\) is (A) one (B) two (C) zero (D) infinite

Step-by-Step Solution

Answer

The number of real solutions is one.

Step 1: Understanding the Equation

The given equation is \( \tan^{-1}(\sqrt{x^2-3x+2}) + \cos^{-1}(\sqrt{4x-x^2-3}) = \pi \). We need to find the number of real solutions for this equation.

Step 2: Analyze Range of Functions

The range of \( \tan^{-1}(y) \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). The domain for \( \tan^{-1}(\sqrt{x^2-3x+2}) \) implies \( x^2-3x+2 \geq 0 \), which factors as \((x-1)(x-2) \geq 0 \). This means \( x \leq 1 \) or \( x \geq 2 \).The range of \( \cos^{-1}(y) \) is \([0, \pi]\), and the domain for \( \cos^{-1}(\sqrt{4x-x^2-3}) \) is \( 4x-x^2-3 \geq 0 \), which simplifies to \( -(x^2-4x+3) \geq 0 \) or \( (x-1)(x-3) \leq 0 \). This means \( 1 \leq x \leq 3 \).

Step 3: Determine Intersection of Domains

To solve the equation, intersect the domains: - From \( \tan^{-1}(\sqrt{x^2-3x+2}) \), allowed \( x \) values are \( x \leq 1 \) or \( x \geq 2 \).- From \( \cos^{-1}(\sqrt{4x-x^2-3}) \), allowed \( x \) values are \( 1 \leq x \leq 3 \).The intersection gives \( x \in [2, 3] \).

Step 4: Assess Compatibility with the Equation

For \( \tan^{-1}(a) + \cos^{-1}(b) = \pi \), either \( 1 + b = 0 \) or \( a = 0 \) (as both angles should sum up to \(\pi\)).- For \( a = \tan^{-1}(\sqrt{x^2-3x+2}) = 0 \Rightarrow \sqrt{x^2-3x+2} = 0 \Rightarrow x^2-3x+2=0 \Rightarrow (x-1)(x-2)=0 \Rightarrow x=1 \text{ or } x=2 \). Only \( x = 2 \) is in the intersection.- For \( \cos^{-1}(b) = \cos^{-1}(-1) \Rightarrow b = -1 \), impossible since \( b = \sqrt{4x-x^2-3} \) must be non-negative.

Step 5: Solution Validation

Check \( x = 2 \): The expression for \( \tan^{-1}(\sqrt{x^2-3x+2}) \) at \( x = 2 \) is \( 0 \) and \( \cos^{-1}(\sqrt{4x-x^2-3}) \) at \( x = 2 \) is \( \pi \). The equation holds true.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are extensions of regular trigonometric functions but work in reverse. They allow us to determine angles based on given trigonometric values. Here are some crucial points about them:

• **Notation**: They are usually denoted as \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), \( \tan^{-1}(x) \), corresponding to arcsin, arccos, and arctan, respectively.
• **Range**: Each inverse function has a specific range within which it outputs values for angles. For example, \( \tan^{-1}(x) \) has a range of \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), capturing angles from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• **Applications**: They are particularly useful in solving equations or evaluating situations in trigonometry where the angle is unknown but the ratio of sides is known.

For instance, in the given problem of the equation, \( \tan^{-1} \) and \( \cos^{-1} \) are applied to determine angles that satisfy a sum equal to \( \pi \), prompting the need to analyze the inputs to these functions to ensure they are within specified domains.

Domain and Range

Understanding domain and range is essential for working with functions, including inverse trigonometric functions.

• **Domain**: This is the set of all possible input values (\( x \)) that a function can accept. For \( \tan^{-1}(\sqrt{x^2-3x+2}) \), the domain is determined by where the expression under the square root is non-negative, i.e., \( x^2 - 3x + 2 \geq 0 \).
• **Range**: This is the set of all possible output values. As mentioned earlier, for \( \tan^{-1} \,\), the range is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). For \( \cos^{-1} \,\), it's \([0, \pi]\).

In the exercise, calculating the intersection of domains from two trigonometric expressions is critical. Through intersecting the domains of \( \tan^{-1} \) and \( \cos^{-1} \), the solution space for \( x \) is narrowed down to \([2, 3]\). This selection guarantees the values of \( x \) satisfy both trigonometric conditions given by the original equation.

Solving Equations

Solving equations involving trigonometric functions requires assessing both the algebraic and trigonometric conditions. Here's a step-by-step way to think about it:

• **Understanding the Problem**: Recognize the form of the equation and the trigonometric properties involved. Here, the equation requires us to find \( x \) such that \( \tan^{-1}(\cdot) + \cos^{-1}(\cdot) = \pi \).
• **Determining Valid Values for \( x \)**: We first analyze the separate expressions. Each trigonometric function has conditions for its arguments, necessitating that they lie within certain intervals. The valid \( x \) values that satisfy both expressions were narrowed down to \([2, 3]\).
• **Validation**: Verify solutions in the final interval of interest. For \( x = 2 \), calculation shows that \( \tan^{-1}(\sqrt{0}) + \cos^{-1}(\sqrt{0}) = \pi \), validating this as a solution.

By ensuring step-by-step analysis and validation, one can solve complex trigonometric equations and confirm the solutions conclude in the given context. The ability to reconcile the algebraic manipulation with trigonometric rules often determines the success in solving such problems.