The number of real solution of `cot^(-1)sqrt(x(x+3))+sin^(-1)sqrt(x^(2)+3x+1)=(pi)/(2)` is /are

Options

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

Answer

To solve the equation \( \cot^{-1}(\sqrt{x(x+3)}) + \sin^{-1}(\sqrt{x^2 + 3x + 1}) = \frac{\pi}{2} \), we will proceed step by step.

### Step 1: Define the function
Let \( f(x) = \cot^{-1}(\sqrt{x(x+3)}) + \sin^{-1}(\sqrt{x^2 + 3x + 1}) \).

### Step 2: Determine the domain of the function
For \( f(x) \) to be defined, both \( \sqrt{x(x+3)} \) and \( \sqrt{x^2 + 3x + 1} \) must be defined and within their respective ranges.

1. **Condition for \( \sqrt{x(x+3)} \)**:
\[ x(x+3) \geq 0 \]
This inequality holds when \( x \leq -3 \) or \( x \geq 0 \).

2. **Condition for \( \sqrt{x^2 + 3x + 1} \)**:
\[ x^2 + 3x + 1 \geq 0 \]
To find the roots of the quadratic equation \( x^2 + 3x + 1 = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-3 \pm \sqrt{5}}{2} \]
The roots are \( x_1 = \frac{-3 - \sqrt{5}}{2} \) and \( x_2 = \frac{-3 + \sqrt{5}}{2} \). The quadratic opens upwards, so it is non-negative outside the interval \( \left( \frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2} \right) \).

### Step 3: Combine the conditions
From the conditions derived:
- \( x \leq -3 \) or \( x \geq 0 \) from \( \sqrt{x(x+3)} \).
- \( x \leq \frac{-3 - \sqrt{5}}{2} \) or \( x \geq \frac{-3 + \sqrt{5}}{2} \) from \( \sqrt{x^2 + 3x + 1} \).

### Step 4: Analyze the intervals
1. **For \( x \leq -3 \)**:
- Check if \( x \) satisfies \( x^2 + 3x + 1 \geq 0 \). Since \( -3 \) is less than \( \frac{-3 - \sqrt{5}}{2} \), this part is valid.

2. **For \( x \geq 0 \)**:
- Check if \( x \) satisfies \( x^2 + 3x + 1 \geq 0 \). This is valid for all \( x \geq 0 \).

### Step 5: Check the values at the boundaries
- At \( x = 0 \):
\[ f(0) = \cot^{-1}(0) + \sin^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \quad (\text{not a solution}) \]

- At \( x = -3 \):
\[ f(-3) = \cot^{-1}(0) + \sin^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \quad (\text{not a solution}) \]

### Step 6: Conclusion
Since there are no values of \( x \) that satisfy the equation \( f(x) = \frac{\pi}{2} \), we conclude that there are no real solutions.

**Final Answer**: The number of real solutions is **0**.