The number of solution of the equation `2sin^(-1)((2x)/(1+x^(2)))-pi x^(3)=0` is equal to

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the equation \(2 \sin^{-1}\left(\frac{2x}{1+x^2}\right) - \pi x^3 = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate the inverse sine function:
\[ 2 \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi x^3 \]
Dividing both sides by 2 gives:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{\pi}{2} x^3 \]

### Step 2: Finding the Range of the Inverse Sine Function
The range of the function \(\sin^{-1}(y)\) is \([- \frac{\pi}{2}, \frac{\pi}{2}]\). Therefore, we need to ensure that:
\[ -\frac{\pi}{2} \leq \frac{\pi}{2} x^3 \leq \frac{\pi}{2} \]
This simplifies to:
\[ -1 \leq x^3 \leq 1 \]
Thus, \(x\) must be in the range \([-1, 1]\).

### Step 3: Analyzing the Function \(y = \frac{2x}{1+x^2}\)
Next, we analyze the function \(y = \frac{2x}{1+x^2}\). This function is defined for all real \(x\) and we need to find its range:
- When \(x = 0\), \(y = 0\).
- When \(x = 1\), \(y = 1\).
- When \(x = -1\), \(y = -1\).

To find its maximum and minimum, we can differentiate:
\[ y' = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]
Setting \(y' = 0\) gives \(x = 1\) and \(x = -1\) as critical points. Evaluating \(y\) at these points:
- \(y(1) = 1\)
- \(y(-1) = -1\)

Thus, the range of \(y = \frac{2x}{1+x^2}\) is \([-1, 1]\).

### Step 4: Finding Solutions
Now we need to find the number of solutions to the equation:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{\pi}{2} x^3 \]
We check the values of \(x\):
1. **For \(x = 0\)**:
\[ \sin^{-1}(0) = 0 \quad \text{and} \quad \frac{\pi}{2} \cdot 0^3 = 0 \]
So, \(x = 0\) is a solution.

2. **For \(x = 1\)**:
\[ \sin^{-1}(1) = \frac{\pi}{2} \quad \text{and} \quad \frac{\pi}{2} \cdot 1^3 = \frac{\pi}{2} \]
So, \(x = 1\) is a solution.

3. **For \(x = -1\)**:
\[ \sin^{-1}(-1) = -\frac{\pi}{2} \quad \text{and} \quad \frac{\pi}{2} \cdot (-1)^3 = -\frac{\pi}{2} \]
So, \(x = -1\) is a solution.

### Conclusion
The solutions we found are \(x = -1\), \(x = 0\), and \(x = 1\). Therefore, the total number of solutions to the equation is **3**.

### Final Answer
The number of solutions of the equation is **3**.