Let `sin^(-1) x + sin^(-1) y + sin^(-1) z = (3pi)/2` for `0 le x, y, z le 1` .What is the value of `x^(1000) + y^(1001) + z^(1002)` ?

Options

A. `0`
B. `1`
C. `3`
D. `6`

Answer

To solve the problem, we need to analyze the equation given:

\[ \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2} \]

### Step 1: Understanding the Range of \(\sin^{-1}\)
The function \(\sin^{-1} t\) (inverse sine) is defined for \(t\) in the range \([-1, 1]\) and its output lies in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\). Therefore, for \(x, y, z\) in the range \([0, 1]\), we have:

\[ 0 \leq \sin^{-1} x, \sin^{-1} y, \sin^{-1} z \leq \frac{\pi}{2} \]

### Step 2: Analyzing the Sum
The maximum value of \(\sin^{-1} x + \sin^{-1} y + \sin^{-1} z\) occurs when \(x = y = z = 1\):

\[ \sin^{-1}(1) + \sin^{-1}(1) + \sin^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2} \]

This shows that the only way for the sum to equal \(\frac{3\pi}{2}\) is if each term is equal to \(\frac{\pi}{2}\).

### Step 3: Setting the Values
From the above analysis, we conclude:

\[ \sin^{-1} x = \sin^{-1} y = \sin^{-1} z = \frac{\pi}{2} \]

This implies:

\[ x = y = z = 1 \]

### Step 4: Calculating the Final Expression
Now we need to evaluate:

\[ x^{1000} + y^{1001} + z^{1002} \]

Substituting \(x = 1\), \(y = 1\), and \(z = 1\):

\[ 1^{1000} + 1^{1001} + 1^{1002} = 1 + 1 + 1 = 3 \]

### Final Answer
Thus, the value of \(x^{1000} + y^{1001} + z^{1002}\) is:

\[ \boxed{3} \]