If `sin^(-1)x+sin^(-1)(1-x)+cos^(-1)x=0`, then x is

Options

A. 0
B. 1
C. 2
D. No real value

Answer

To solve the equation \( \sin^{-1}x + \sin^{-1}(1-x) + \cos^{-1}x = 0 \), we will follow these steps:

### Step 1: Use the Identity
We know that:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
Using this identity, we can rewrite the equation:
\[ \frac{\pi}{2} + \sin^{-1}(1-x) = 0 \]

### Step 2: Isolate \( \sin^{-1}(1-x) \)
Rearranging the equation gives:
\[ \sin^{-1}(1-x) = -\frac{\pi}{2} \]

### Step 3: Apply the Sine Function
Taking the sine of both sides, we have:
\[ 1 - x = \sin\left(-\frac{\pi}{2}\right) \]

### Step 4: Evaluate \( \sin\left(-\frac{\pi}{2}\right) \)
We know that:
\[ \sin\left(-\frac{\pi}{2}\right) = -1 \]
Thus, we can substitute this value back into the equation:
\[ 1 - x = -1 \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ 1 + 1 = x \implies x = 2 \]

### Step 6: Check the Validity of \( x \)
The values of \( x \) for \( \sin^{-1}x \) and \( \cos^{-1}x \) must lie within the domain of \([-1, 1]\). Since \( x = 2 \) is outside this range, it is not a valid solution.

### Conclusion
Thus, there is no real value of \( x \) that satisfies the original equation.