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 will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 1 + x^2 + 2x \sin(\cos^{-1} y) = 0 \]
Rearranging gives us:
\[ 1 + x^2 = -2x \sin(\cos^{-1} y) \]

### Step 2: Understanding the Range of \(\sin(\cos^{-1} y)\)
The function \(\cos^{-1} y\) is defined for \(y \in [-1, 1]\), and its range is \([0, \pi]\). Therefore, the sine function will take values from:
\[ \sin(0) = 0 \quad \text{to} \quad \sin(\pi) = 0 \quad \text{and} \quad \sin\left(\frac{\pi}{2}\right) = 1 \]
Thus, the range of \(\sin(\cos^{-1} y)\) is \([0, 1]\).

### Step 3: Analyzing the Right Side of the Equation
From the rearranged equation, we have:
\[ 1 + x^2 = -2x \sin(\cos^{-1} y) \]
Since \(\sin(\cos^{-1} y)\) ranges from \(0\) to \(1\), the right side \(-2x \sin(\cos^{-1} y)\) will vary depending on the value of \(x\):
- If \(x > 0\), then \(-2x \sin(\cos^{-1} y) \leq 0\).
- If \(x < 0\), then \(-2x \sin(\cos^{-1} y) \geq 0\).

### Step 4: Finding Values of \(x\)
For the left side \(1 + x^2\) to equal the right side, we need:
\[ 1 + x^2 \geq 0 \]
This is always true since \(x^2 \geq 0\) for all real \(x\).

### Step 5: Setting Up the Condition
To satisfy the equation, we need:
\[ 1 + x^2 = -2x \sin(\cos^{-1} y) \]
This implies:
\[ 1 + x^2 \geq 0 \quad \text{and} \quad -2x \sin(\cos^{-1} y) \leq 0 \quad \text{(for } x > 0\text{)} \]
Thus, \(x\) must be negative for the equality to hold.

### Step 6: Testing Specific Values of \(x\)
Let’s test \(x = -1\):
\[ 1 + (-1)^2 + 2(-1) \sin(\cos^{-1} y) = 0 \]
This simplifies to:
\[ 1 + 1 - 2 \sin(\cos^{-1} y) = 0 \implies 2 - 2 \sin(\cos^{-1} y) = 0 \implies \sin(\cos^{-1} y) = 1 \]
This occurs when \(\cos^{-1} y = \frac{\pi}{2}\), which means \(y = 0\).

### Conclusion
The only ordered pair that satisfies the equation is \((-1, 0)\). Therefore, the number of ordered pairs \((x, y)\) of real numbers satisfying the equation is:
\[ \boxed{1} \]