the number of solutions of `cos^(- 1)(1-x)+mcos^(- 1)x=(npi)/2` where `mgt0, nleq0`

Options

A. 0
B. 1
C. 2
D. none of these

Answer

To find the number of solutions of the equation

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

where \( m > 0 \) and \( n \leq 0 \), we can follow these steps:

### Step 1: Analyze the Range of the Left-Hand Side

The function \( \cos^{-1}(y) \) is defined for \( y \) in the range \([-1, 1]\) and its output is in the range \([0, \pi]\).

- For \( \cos^{-1}(1-x) \), we need \( 1-x \) to be in the range \([0, 1]\), which gives us \( 0 \leq x \leq 1 \).
- For \( \cos^{-1}(x) \), we need \( x \) to also be in the range \([0, 1]\).

Thus, both terms \( \cos^{-1}(1-x) \) and \( m \cos^{-1}(x) \) are non-negative for \( x \) in the interval \([0, 1]\).

### Step 2: Analyze the Right-Hand Side

Since \( n \leq 0 \), we have:

\[ \frac{n\pi}{2} \leq 0 \]

This means the right-hand side of the equation is non-positive.

### Step 3: Compare Left-Hand Side and Right-Hand Side

The left-hand side \( \cos^{-1}(1-x) + m \cos^{-1}(x) \) is always non-negative (as both terms are non-negative).

- Therefore, the left-hand side is always \( \geq 0 \).
- The right-hand side is \( \leq 0 \).

### Step 4: Conclusion on Number of Solutions

Since the left-hand side is always non-negative and the right-hand side is non-positive, the only way for the two sides to be equal is if both sides are equal to zero.

This leads us to the condition:

\[ \cos^{-1}(1-x) + m \cos^{-1}(x) = 0 \]

For this to hold true, both \( \cos^{-1}(1-x) \) and \( m \cos^{-1}(x) \) must be zero.

### Step 5: Find Values of \( x \)

1. **From \( \cos^{-1}(1-x) = 0 \)**:
- This implies \( 1-x = 1 \) or \( x = 0 \).

2. **From \( m \cos^{-1}(x) = 0 \)**:
- Since \( m > 0 \), this implies \( \cos^{-1}(x) = 0 \) or \( x = 1 \).

### Step 6: Check Solutions

- For \( x = 0 \):
\[ \cos^{-1}(1-0) + m \cos^{-1}(0) = \cos^{-1}(1) + m \cdot \frac{\pi}{2} = 0 + m \cdot \frac{\pi}{2} > 0 \quad \text{(not a solution)} \]

- For \( x = 1 \):
\[ \cos^{-1}(1-1) + m \cos^{-1}(1) = \cos^{-1}(0) + m \cdot 0 = \frac{\pi}{2} + 0 > 0 \quad \text{(not a solution)} \]

### Final Conclusion

Since neither \( x = 0 \) nor \( x = 1 \) satisfies the equation, we conclude that there are **no solutions** to the given equation.

Thus, the number of solutions is **0**.

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