Number of solutions (s) of the equations
`cos^(-1) ( 1-x) - 2 cos^(-1) x = pi/2` is

Options

A. 3
B. 2
C. 1
D. 0

Answer

To solve the equation \( \cos^{-1}(1-x) - 2\cos^{-1}(x) = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Determine the Domain
The functions \( \cos^{-1}(1-x) \) and \( \cos^{-1}(x) \) are defined when their arguments lie within the range \([-1, 1]\).

1. For \( \cos^{-1}(1-x) \):
\[ -1 \leq 1 - x \leq 1 \]
This simplifies to:
\[ 0 \leq x \leq 2 \]

2. For \( \cos^{-1}(x) \):
\[ -1 \leq x \leq 1 \]

Combining these conditions, the valid range for \( x \) is:
\[ 0 \leq x \leq 1 \]

### Step 2: Evaluate at the Endpoints
Next, we will evaluate the left-hand side (LHS) of the equation at the endpoints of the interval.

1. **At \( x = 0 \)**:
\[ \text{LHS} = \cos^{-1}(1-0) - 2\cos^{-1}(0) = \cos^{-1}(1) - 2\cos^{-1}(0) \]
\[ = 0 - 2\left(\frac{\pi}{2}\right) = 0 - \pi = -\pi \]
Since \(-\pi \neq \frac{\pi}{2}\), \( x = 0 \) is not a solution.

2. **At \( x = 1 \)**:
\[ \text{LHS} = \cos^{-1}(1-1) - 2\cos^{-1}(1) = \cos^{-1}(0) - 2\cos^{-1}(1) \]
\[ = \frac{\pi}{2} - 2(0) = \frac{\pi}{2} \]
Since \(\frac{\pi}{2} = \frac{\pi}{2}\), \( x = 1 \) is a solution.

### Step 3: Analyze the Interval \( (0, 1) \)
Now, we need to check if there are any solutions in the open interval \( (0, 1) \).

- The function \( \cos^{-1}(1-x) \) is decreasing in the interval \( [0, 1] \) because as \( x \) increases, \( 1-x \) decreases, leading to an increase in the angle.
- The function \( 2\cos^{-1}(x) \) is also increasing in the interval \( [0, 1] \).

Since one function is decreasing and the other is increasing, they can intersect at most once in the interval \( (0, 1) \).

### Step 4: Conclusion
We have found:
- No solutions at \( x = 0 \)
- One solution at \( x = 1 \)
- At most one solution in the interval \( (0, 1) \)

Thus, the total number of solutions to the equation is:
\[ \text{Number of solutions} = 1 \]

### Final Answer
The number of solutions (s) of the equation is \( \boxed{1} \).

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