The number of real solution of `cot^(-1)sqrt(x(x+4))+cos^(-1)sqrt(x^(2)+4x+1)=(pi)/(2)` is equal to

Options

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

Answer

To solve the equation

\[ \cot^{-1}(\sqrt{x(x+4)}) + \cos^{-1}(\sqrt{x^2 + 4x + 1}) = \frac{\pi}{2}, \]

we will analyze the conditions under which the expressions are defined and find the number of real solutions.

### Step 1: Analyze the first term \( \cot^{-1}(\sqrt{x(x+4)}) \)

For \( \sqrt{x(x+4)} \) to be defined, we need:

\[ x(x+4) \geq 0. \]

This inequality holds when:

1. \( x \leq 0 \) (both factors are non-positive)
2. \( x \geq 0 \) (both factors are non-negative)

The critical points are \( x = 0 \) and \( x = -4 \). We can analyze the intervals:

- For \( x < -4 \): \( x(x+4) > 0 \)
- For \( -4 \leq x < 0 \): \( x(x+4) \leq 0 \)
- For \( x \geq 0 \): \( x(x+4) \geq 0 \)

Thus, \( \sqrt{x(x+4)} \) is defined for \( x \in (-\infty, -4] \cup [0, \infty) \).

### Step 2: Analyze the second term \( \cos^{-1}(\sqrt{x^2 + 4x + 1}) \)

The expression \( \sqrt{x^2 + 4x + 1} \) must also be defined, which requires:

\[ x^2 + 4x + 1 \geq 0. \]

The roots of the quadratic equation \( x^2 + 4x + 1 = 0 \) can be found using the quadratic formula:

\[ x = \frac{-4 \pm \sqrt{16 - 4}}{2} = -2 \pm \sqrt{3}. \]

The quadratic opens upwards, so it is non-negative outside the roots:

- \( x \leq -2 - \sqrt{3} \) or \( x \geq -2 + \sqrt{3} \).

### Step 3: Find the range for \( \sqrt{x^2 + 4x + 1} \)

Next, we need \( \sqrt{x^2 + 4x + 1} \) to be in the range [0, 1] for the \( \cos^{-1} \) function:

\[ 0 \leq x^2 + 4x + 1 \leq 1. \]

This gives us two inequalities:

1. \( x^2 + 4x + 1 \geq 0 \) (already established).
2. \( x^2 + 4x + 1 \leq 1 \) simplifies to \( x^2 + 4x \leq 0 \), or \( x(x + 4) \leq 0 \).

### Step 4: Combine the conditions

From the analysis, we have:

1. \( x(x + 4) \geq 0 \) (from the first term).
2. \( x(x + 4) \leq 0 \) (from the second term).

The only way both conditions can hold simultaneously is when \( x(x + 4) = 0 \).

### Step 5: Solve for \( x \)

Setting \( x(x + 4) = 0 \) gives us:

\[ x = 0 \quad \text{or} \quad x = -4. \]

### Step 6: Verify the solutions

1. For \( x = 0 \):
\[ \cot^{-1}(0) + \cos^{-1}(1) = \frac{\pi}{2} + 0 = \frac{\pi}{2}. \]

2. For \( x = -4 \):
\[ \cot^{-1}(\sqrt{0}) + \cos^{-1}(0) = \frac{\pi}{2} + \frac{\pi}{2} = \frac{\pi}{2}. \]

Both values satisfy the original equation.

### Conclusion

The number of real solutions to the equation is:

\[ \boxed{2}. \]