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 \((\sin^{-1} x)^3 + (\cos^{-1} x)^3 = a \pi^3\) given that \(a \leq \frac{1}{32}\), we will follow these steps:

### Step 1: Use the identity for the sum of cubes
We know that \(A^3 + B^3 = (A + B)(A^2 - AB + B^2)\). Here, let \(A = \sin^{-1} x\) and \(B = \cos^{-1} x\).

### Step 2: Apply the identity
Using the identity, we have:
\[ (\sin^{-1} x)^3 + (\cos^{-1} x)^3 = (\sin^{-1} x + \cos^{-1} x)((\sin^{-1} x)^2 - \sin^{-1} x \cos^{-1} x + (\cos^{-1} x)^2) \]
Since \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\), we can substitute this into our equation:
\[ = \frac{\pi}{2} \left((\sin^{-1} x)^2 + (\cos^{-1} x)^2 - \sin^{-1} x \cos^{-1} x\right) \]

### Step 3: Simplify the expression
Now, we know:
\[ (\sin^{-1} x)^2 + (\cos^{-1} x)^2 = \left(\frac{\pi}{2}\right)^2 - 2 \sin^{-1} x \cos^{-1} x \]
Thus, we can rewrite:
\[ (\sin^{-1} x)^3 + (\cos^{-1} x)^3 = \frac{\pi}{2} \left(\frac{\pi^2}{4} - 2 \sin^{-1} x \cos^{-1} x - \sin^{-1} x \cos^{-1} x\right) \]
This simplifies to:
\[ = \frac{\pi}{2} \left(\frac{\pi^2}{4} - 3 \sin^{-1} x \cos^{-1} x\right) \]

### Step 4: Set the equation equal to \(a \pi^3\)
We set this equal to \(a \pi^3\):
\[ \frac{\pi}{2} \left(\frac{\pi^2}{4} - 3 \sin^{-1} x \cos^{-1} x\right) = a \pi^3 \]

### Step 5: Simplify the equation
Dividing both sides by \(\pi\) (assuming \(\pi \neq 0\)):
\[ \frac{1}{2} \left(\frac{\pi^2}{4} - 3 \sin^{-1} x \cos^{-1} x\right) = a \pi^2 \]
Multiplying both sides by 2 gives:
\[ \frac{\pi^2}{4} - 3 \sin^{-1} x \cos^{-1} x = 2a \pi^2 \]
Rearranging gives:
\[ 3 \sin^{-1} x \cos^{-1} x = \frac{\pi^2}{4} - 2a \pi^2 \]

### Step 6: Solve for \(\sin^{-1} x \cos^{-1} x\)
This can be rewritten as:
\[ \sin^{-1} x \cos^{-1} x = \frac{\pi^2}{12} - \frac{2}{3} a \pi^2 \]

### Step 7: Analyze the range of \(\sin^{-1} x \cos^{-1} x\)
The maximum value of \(\sin^{-1} x \cos^{-1} x\) occurs at \(x = \frac{1}{\sqrt{2}}\):
\[ \sin^{-1} \left(\frac{1}{\sqrt{2}}\right) = \cos^{-1} \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \]
Thus, the maximum value is:
\[ \frac{\pi}{4} \cdot \frac{\pi}{4} = \frac{\pi^2}{16} \]

### Step 8: Set the inequality
We need:
\[ \frac{\pi^2}{12} - \frac{2}{3} a \pi^2 \leq \frac{\pi^2}{16} \]
This simplifies to:
\[ \frac{1}{12} - \frac{2}{3} a \leq \frac{1}{16} \]

### Step 9: Solve for \(a\)
Multiplying through by 48 (the least common multiple of 12 and 16):
\[ 4 - 32a \leq 3 \]
This leads to:
\[ 1 \leq 32a \quad \Rightarrow \quad a \geq \frac{1}{32} \]

### Step 10: Conclusion
Since \(a \leq \frac{1}{32}\) and \(a \geq \frac{1}{32}\), we conclude that there is exactly one solution to the equation.

### Final Answer
The number of solutions is **1**.

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