The number of ordered pair(s) `(x, y)` of real numbers satisfying the equation `1+x^(2)+2x sin(cos^(-1)y)=0`, is :

Options

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

Answer

To solve the equation \(1 + x^2 + 2x \sin(\cos^{-1} y) = 0\) for the number of ordered pairs \((x, y)\) of real numbers, we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ 1 + x^2 + 2x \sin(\cos^{-1} y) = 0 \]
Rearranging gives:
\[ 1 + x^2 = -2x \sin(\cos^{-1} y) \]

### Step 2: Understand the range of \(\sin(\cos^{-1} y)\)
Recall that \(\sin(\cos^{-1} y) = \sqrt{1 - y^2}\). This is valid for \(y\) in the range \([-1, 1]\). Therefore, we can rewrite the equation as:
\[ 1 + x^2 = -2x \sqrt{1 - y^2} \]

### Step 3: Divide by \(x\) (assuming \(x \neq 0\))
If \(x \neq 0\), we can divide the entire equation by \(x\):
\[ \frac{1}{x} + x = -2\sqrt{1 - y^2} \]

### Step 4: Analyze the left-hand side
The left-hand side, \(\frac{1}{x} + x\), is a function of \(x\). To find its behavior, we can analyze it:
- As \(x \to 0^+\), \(\frac{1}{x} \to +\infty\).
- As \(x \to 0^-\), \(\frac{1}{x} \to -\infty\).
- As \(x \to +\infty\), \(\frac{1}{x} \to 0\) and \(x \to +\infty\) implies the left side approaches \(+\infty\).
- As \(x \to -\infty\), \(\frac{1}{x} \to 0\) and \(x \to -\infty\) implies the left side approaches \(-\infty\).

### Step 5: Determine the range of the left-hand side
The function \(\frac{1}{x} + x\) has a minimum value. To find the critical points, we differentiate:
\[ f(x) = \frac{1}{x} + x \]
\[ f'(x) = -\frac{1}{x^2} + 1 \]
Setting \(f'(x) = 0\) gives:
\[ 1 = \frac{1}{x^2} \implies x^2 = 1 \implies x = 1 \text{ or } x = -1 \]
Evaluating \(f(x)\) at these points:
- \(f(1) = 1 + 1 = 2\)
- \(f(-1) = -1 + 1 = 0\)

### Step 6: Analyze the right-hand side
The right-hand side is \(-2\sqrt{1 - y^2}\). The maximum value of \(-2\sqrt{1 - y^2}\) occurs when \(y = 0\):
\[ -2\sqrt{1 - 0^2} = -2 \]
The minimum value occurs when \(y = \pm 1\):
\[ -2\sqrt{1 - 1^2} = 0 \]
Thus, the right-hand side ranges from \(-2\) to \(0\).

### Step 7: Find intersections
We need to find where \(\frac{1}{x} + x\) intersects \(-2\sqrt{1 - y^2}\):
- The minimum of \(\frac{1}{x} + x\) is \(0\) at \(x = -1\).
- The maximum of \(-2\sqrt{1 - y^2}\) is \(-2\).

### Conclusion
Since the minimum value of \(\frac{1}{x} + x\) (which is \(0\)) is greater than the maximum value of \(-2\sqrt{1 - y^2}\) (which is \(-2\)), there are no real solutions for \(x\) and \(y\) that satisfy the equation.

Thus, the number of ordered pairs \((x, y)\) of real numbers satisfying the equation is:
\[ \boxed{1} \]