`sin^-1[x^2+1/3]+cos^-1[x^2-2/3]=x^2` the number of solution in `x in (1,-1)` where [.] is the GIF.

Options

A. 0
B. 2
C. 3
D. 4

Answer

To solve the equation \( \sin^{-1}\left(\left\lfloor x^2 + \frac{1}{3} \right\rfloor\right) + \cos^{-1}\left(\left\lfloor x^2 - \frac{2}{3} \right\rfloor\right) = x^2 \) for \( x \) in the interval \( (-1, 1) \), we will go through the following steps:

### Step 1: Determine the range of \( x^2 \)
Since \( x \) is in the interval \( (-1, 1) \), \( x^2 \) will range from \( 0 \) to \( 1 \).

### Step 2: Analyze the expressions inside the greatest integer function (GIF)
1. For \( x^2 + \frac{1}{3} \):
- The minimum value occurs at \( x^2 = 0 \): \( 0 + \frac{1}{3} = \frac{1}{3} \).
- The maximum value occurs at \( x^2 = 1 \): \( 1 + \frac{1}{3} = \frac{4}{3} \).
- Thus, \( \left\lfloor x^2 + \frac{1}{3} \right\rfloor \) can take values \( 0 \) or \( 1 \).

2. For \( x^2 - \frac{2}{3} \):
- The minimum value occurs at \( x^2 = 0 \): \( 0 - \frac{2}{3} = -\frac{2}{3} \).
- The maximum value occurs at \( x^2 = 1 \): \( 1 - \frac{2}{3} = \frac{1}{3} \).
- Thus, \( \left\lfloor x^2 - \frac{2}{3} \right\rfloor \) can take values \( -1 \) or \( 0 \).

### Step 3: Consider cases based on the values of the GIFs
#### Case 1: \( \left\lfloor x^2 + \frac{1}{3} \right\rfloor = 0 \) and \( \left\lfloor x^2 - \frac{2}{3} \right\rfloor = -1 \)
- This occurs when \( 0 \leq x^2 + \frac{1}{3} < 1 \) and \( -1 \leq x^2 - \frac{2}{3} < 0 \).
- From \( 0 \leq x^2 + \frac{1}{3} < 1 \), we have \( -\frac{1}{3} < x^2 < \frac{2}{3} \).
- From \( -1 \leq x^2 - \frac{2}{3} < 0 \), we have \( \frac{2}{3} \leq x^2 < 1 \).
- No overlap exists in this case.

#### Case 2: \( \left\lfloor x^2 + \frac{1}{3} \right\rfloor = 1 \) and \( \left\lfloor x^2 - \frac{2}{3} \right\rfloor = 0 \)
- This occurs when \( 1 \leq x^2 + \frac{1}{3} < 2 \) and \( 0 \leq x^2 - \frac{2}{3} < 1 \).
- From \( 1 \leq x^2 + \frac{1}{3} < 2 \), we have \( \frac{2}{3} \leq x^2 < \frac{5}{3} \) (but \( x^2 \) cannot exceed \( 1 \)).
- From \( 0 \leq x^2 - \frac{2}{3} < 1 \), we have \( \frac{2}{3} \leq x^2 < \frac{5}{3} \) (again, limited to \( 1 \)).
- The only valid range is \( \frac{2}{3} \leq x^2 < 1 \).

### Step 4: Solve the equation in the valid range
In this case, we have:
\[ \sin^{-1}(1) + \cos^{-1}(0) = x^2 \]
This simplifies to:
\[ \frac{\pi}{2} + \frac{\pi}{2} = x^2 \implies \pi = x^2 \]
However, \( \pi \) is greater than \( 1 \), which is outside the range of \( x^2 \).

### Conclusion
Since there are no valid solutions for \( x \) in the interval \( (-1, 1) \), we conclude that the number of solutions is:

**0**