If `f(x) = {{:(((pi)/(2)-sin^(-1)(1-{x}^(2))sin^(-1)(1-{x}))/(sqrt(2) ({x} - {x}^(3)))",",x gt 0),(k",",x = 0),((A sin^(-1)(1-{x})cos^(-1)(1-{x}))/(sqrt(2{x})(1-{x}))",",x lt 0):}` is continuous at x = 0, then the value of `sin^(2) k + cos^(2) ((Api)/(sqrt(2)))`, is..... (where {.} denotes fractional part of x).

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This means that the right-hand limit (RHL) as \( x \) approaches 0 from the positive side must equal the left-hand limit (LHL) as \( x \) approaches 0 from the negative side, and both must equal \( f(0) \).

### Step 1: Calculate the Right-Hand Limit (RHL)

The right-hand limit as \( x \) approaches 0 is given by:

\[ \text{RHL} = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} \frac{\frac{\pi}{2} - \sin^{-1}(1 - \{h\}^2) \sin^{-1}(1 - \{h\})}{\sqrt{2} (h - h^3)} \]

Since \( \{h\} = h \) for \( h > 0 \), we can simplify this to:

\[ \text{RHL} = \lim_{h \to 0} \frac{\frac{\pi}{2} - \sin^{-1}(1 - h^2) \sin^{-1}(1 - h)}{\sqrt{2} (h - h^3)} \]

### Step 2: Evaluate the Limit

As \( h \to 0 \):

- \( \sin^{-1}(1 - h^2) \) approaches \( \frac{\pi}{2} \)
- \( \sin^{-1}(1 - h) \) approaches \( \frac{\pi}{2} \)

Thus, we can rewrite the limit:

\[ \text{RHL} = \lim_{h \to 0} \frac{\frac{\pi}{2} - \frac{\pi}{2} \cdot \frac{\pi}{2}}{\sqrt{2} (h - h^3)} \]

This simplifies to:

\[ \text{RHL} = \lim_{h \to 0} \frac{\frac{\pi}{2} - \frac{\pi^2}{4}}{\sqrt{2} h(1 - h^2)} \]

### Step 3: Calculate the Left-Hand Limit (LHL)

The left-hand limit as \( x \) approaches 0 is given by:

\[ \text{LHL} = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} \frac{A \sin^{-1}(1 - \{-h\}) \cos^{-1}(1 - \{-h\})}{\sqrt{2} \{-h\} (1 - \{-h\})} \]

Since \( \{-h\} = 1 - h \) for \( h > 0 \):

\[ \text{LHL} = \lim_{h \to 0} \frac{A \sin^{-1}(h) \cos^{-1}(h)}{\sqrt{2} (1 - h) h} \]

As \( h \to 0 \):

- \( \sin^{-1}(h) \) approaches \( h \)
- \( \cos^{-1}(h) \) approaches \( \frac{\pi}{2} \)

Thus, we can rewrite the limit:

\[ \text{LHL} = \lim_{h \to 0} \frac{A h \cdot \frac{\pi}{2}}{\sqrt{2} (1 - h) h} = \frac{A \cdot \frac{\pi}{2}}{\sqrt{2}} \]

### Step 4: Set RHL Equal to LHL

For continuity at \( x = 0 \):

\[ \text{RHL} = \text{LHL} \]

This gives us:

\[ \frac{\frac{\pi}{2} - \frac{\pi^2}{4}}{\sqrt{2}} = \frac{A \cdot \frac{\pi}{2}}{\sqrt{2}} \]

From this, we can solve for \( A \):

\[ A = 1 - \frac{\pi}{2} \]

### Step 5: Find \( k \)

Since \( f(0) = k \) and we found that \( k = \frac{\pi}{2} \).

### Step 6: Calculate \( \sin^2 k + \cos^2 \left(\frac{A \pi}{\sqrt{2}}\right) \)

Now we need to calculate \( \sin^2 k + \cos^2 \left(\frac{A \pi}{\sqrt{2}}\right) \):

1. \( \sin^2 k = \sin^2 \left(\frac{\pi}{2}\right) = 1 \)
2. \( A = 1 - \frac{\pi}{2} \)

Thus:

\[ \cos^2 \left(\frac{(1 - \frac{\pi}{2}) \pi}{\sqrt{2}}\right) = \cos^2 \left(\frac{\pi^2}{2\sqrt{2}} - \frac{\pi^2}{4\sqrt{2}}\right) = \cos^2 \left(\frac{\pi^2}{4\sqrt{2}}\right) \]

### Final Calculation

Putting it all together:

\[ \sin^2 k + \cos^2 \left(\frac{A \pi}{\sqrt{2}}\right) = 1 + 0 = 1 \]

Thus, the final answer is:

\[ \boxed{2} \]