The number of solutions of equation cos -1(1-x) + m cos-1 x =nπ/2 is : (where m>0 ; n≤0 )

To determine the number of solutions for the equation \( \cos^{-1}(1-x) + m \cos^{-1}(x) = \frac{n\pi}{2} \) under the conditions \( m > 0 \) and \( n \leq 0 \), we need to analyze the behavior of the inverse cosine functions involved and how they interact with the parameters \( m \) and \( n \).

Understanding the Components of the Equation

The equation consists of two parts: \( \cos^{-1}(1-x) \) and \( m \cos^{-1}(x) \). Let's break these down:

• Domain of \( x \): The function \( \cos^{-1}(x) \) is defined for \( x \) in the range \([-1, 1]\). Therefore, \( 1-x \) must also fall within this range, which implies \( x \) must be in \([-1, 1]\).
• Behavior of \( \cos^{-1}(1-x) \): As \( x \) varies from -1 to 1, \( 1-x \) varies from 2 to 0. Thus, \( \cos^{-1}(1-x) \) will vary from \( \cos^{-1}(2) \) (undefined) to \( \cos^{-1}(0) = \frac{\pi}{2} \).
• Behavior of \( m \cos^{-1}(x): For \( x \) in \([-1, 1]\), \( \cos^{-1}(x) \) varies from \( \pi \) (when \( x = -1 \)) to \( 0 \) (when \( x = 1 \)). Therefore, \( m \cos^{-1}(x) \) will range from \( m\pi \) to \( 0 \).

Analyzing the Equation

Now, let's consider the left-hand side of the equation:

We can rewrite the equation as:

\( \cos^{-1}(1-x) + m \cos^{-1}(x) = \frac{n\pi}{2} \)

Given that \( n \leq 0 \), \( \frac{n\pi}{2} \) will be non-positive. This means we are looking for the sum of two non-negative terms (since both \( \cos^{-1}(1-x) \) and \( m \cos^{-1}(x) \) are non-negative) to equal a non-positive number. The only way this can happen is if both terms are zero.

Finding the Conditions for Solutions

For the left-hand side to equal zero:

• \( \cos^{-1}(1-x) = 0 \) implies \( 1-x = 1 \) or \( x = 0 \).
• \( m \cos^{-1}(x) = 0 \) implies \( \cos^{-1}(x) = 0 \) or \( x = 1 \).

However, both conditions cannot be satisfied simultaneously since \( x \) cannot be both 0 and 1 at the same time. Therefore, the only solution occurs when both terms are zero, which is not possible in the given range.

Conclusion on the Number of Solutions

Since the only way for the left-hand side to equal a non-positive number is for both terms to be zero, and this is not achievable under the given conditions, we conclude that:

The number of solutions to the equation \( \cos^{-1}(1-x) + m \cos^{-1}(x) = \frac{n\pi}{2} \) is zero.