If `a le 1/32` then the number of solution of
`(sin^(-1) x)^(3) +(cos^(-1) x)^(3) = a pi^(3)` is

Options

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

Answer

To solve the equation \((\sin^{-1} x)^3 + (\cos^{-1} x)^3 = a \pi^3\) under the condition \(a \leq \frac{1}{32}\), we will follow these steps:

### Step 1: Use the identity for the sum of cubes
We know the identity for the sum of cubes:
\[ A^3 + B^3 = (A + B)(A^2 - AB + B^2) \]
Let \(A = \sin^{-1} x\) and \(B = \cos^{-1} x\). Then, we can rewrite the equation as:
\[ (\sin^{-1} x + \cos^{-1} x)((\sin^{-1} x)^2 - \sin^{-1} x \cos^{-1} x + (\cos^{-1} x)^2) = a \pi^3 \]

### Step 2: Simplify using the known identity
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Substituting this into our equation gives:
\[ \frac{\pi}{2} \left((\sin^{-1} x)^2 - \sin^{-1} x \cos^{-1} x + (\cos^{-1} x)^2\right) = a \pi^3 \]

### Step 3: Rearranging the equation
This simplifies to:
\[ \left((\sin^{-1} x)^2 - \sin^{-1} x \cos^{-1} x + (\cos^{-1} x)^2\right) = \frac{2a \pi^2}{\pi} = 2a \pi^2 \]

### Step 4: Express \((\sin^{-1} x)^2 + (\cos^{-1} x)^2\)
Using the identity:
\[ (\sin^{-1} x)^2 + (\cos^{-1} x)^2 = \left(\frac{\pi^2}{4}\right) - 2\sin^{-1} x \cos^{-1} x \]
We can substitute this into our equation:
\[ \frac{\pi^2}{4} - 2\sin^{-1} x \cos^{-1} x - \sin^{-1} x \cos^{-1} x = 2a \pi^2 \]
This simplifies to:
\[ \frac{\pi^2}{4} - 3\sin^{-1} x \cos^{-1} x = 2a \pi^2 \]

### Step 5: Rearranging for \(\sin^{-1} x \cos^{-1} x\)
Rearranging gives:
\[ 3\sin^{-1} x \cos^{-1} x = \frac{\pi^2}{4} - 2a \pi^2 \]
Thus,
\[ \sin^{-1} x \cos^{-1} x = \frac{\pi^2}{12} - \frac{2a \pi^2}{3} \]

### Step 6: Analyzing the range of \(\sin^{-1} x \cos^{-1} x\)
The maximum value of \(\sin^{-1} x \cos^{-1} x\) occurs when \(x = \frac{1}{\sqrt{2}}\), giving:
\[ \sin^{-1} \frac{1}{\sqrt{2}} \cos^{-1} \frac{1}{\sqrt{2}} = \frac{\pi}{4} \cdot \frac{\pi}{4} = \frac{\pi^2}{16} \]

### Step 7: Setting the inequality
We need:
\[ \frac{\pi^2}{12} - \frac{2a \pi^2}{3} \leq \frac{\pi^2}{16} \]
This leads to:
\[ \frac{1}{12} - \frac{2a}{3} \leq \frac{1}{16} \]
Solving this inequality will give us the possible values for \(a\).

### Step 8: Solve for \(a\)
Multiplying through by \(48\) (the least common multiple of 12 and 16):
\[ 4 - 32a \leq 3 \]
This simplifies to:
\[ 1 \leq 32a \implies a \geq \frac{1}{32} \]

### Conclusion
Since \(a \leq \frac{1}{32}\) and \(a \geq \frac{1}{32}\), we conclude that \(a = \frac{1}{32}\). This means the equation has no solutions because the maximum value of \(\sin^{-1} x \cos^{-1} x\) cannot reach the required value for any \(x\) in the domain.

### Final Answer
Thus, the number of solutions is **0**.

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