If the inequality `cot^(2)x + (k +1) cot x - (k-3) < 0` is true for at least one `x in (0, pi//2)`, then `k in `.

Options

A. `(-oo,3-2sqrt(5))`
B. `(3,oo)`
C. `(-1, oo)`
D. `(-oo,3)`

Explanation

To solve the inequality \( \cot^2 x + (k + 1) \cot x - (k - 3) < 0 \) for at least one \( x \) in the interval \( (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Substitute \( t = \cot x \)
Since \( \cot x \) ranges from \( \infty \) to \( 0 \) as \( x \) goes from \( 0 \) to \( \frac{\pi}{2} \), we can rewrite the inequality in terms of \( t \):
\[ t^2 + (k + 1)t - (k - 3) < 0 \]
where \( t \in (0, \infty) \).

### Step 2: Analyze the quadratic inequality
The quadratic inequality \( t^2 + (k + 1)t - (k - 3) < 0 \) will be satisfied for at least one \( t \) if the quadratic has real roots and opens upwards (which it does since the coefficient of \( t^2 \) is positive).

### Step 3: Find the discriminant
For the quadratic to have real roots, its discriminant must be non-negative:
\[ D = (k + 1)^2 - 4(1)(-(k - 3)) \geq 0 \]
This simplifies to:
\[ (k + 1)^2 + 4(k - 3) \geq 0 \]
\[ (k + 1)^2 + 4k - 12 \geq 0 \]
\[ k^2 + 2k + 1 + 4k - 12 \geq 0 \]
\[ k^2 + 6k - 11 \geq 0 \]

### Step 4: Solve the quadratic inequality
To solve \( k^2 + 6k - 11 \geq 0 \), we first find the roots using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-11)}}{2 \cdot 1} \]
\[ = \frac{-6 \pm \sqrt{36 + 44}}{2} \]
\[ = \frac{-6 \pm \sqrt{80}}{2} \]
\[ = \frac{-6 \pm 4\sqrt{5}}{2} \]
\[ = -3 \pm 2\sqrt{5} \]
Thus, the roots are \( k_1 = -3 - 2\sqrt{5} \) and \( k_2 = -3 + 2\sqrt{5} \).

### Step 5: Determine intervals
The quadratic \( k^2 + 6k - 11 \) opens upwards, so it is non-negative outside the roots:
1. \( k \leq -3 - 2\sqrt{5} \)
2. \( k \geq -3 + 2\sqrt{5} \)

### Step 6: Consider the condition for \( k \)
Since we also need \( k > 3 \) from the earlier analysis, we combine this with our findings:
- The valid interval for \( k \) is:
\[ k \geq -3 + 2\sqrt{5} \quad \text{and} \quad k > 3 \]
Calculating \( -3 + 2\sqrt{5} \):
\[ \sqrt{5} \approx 2.236 \Rightarrow 2\sqrt{5} \approx 4.472 \Rightarrow -3 + 4.472 \approx 1.472 \]
Thus, \( -3 + 2\sqrt{5} \approx 1.472 \).

### Conclusion
The final result is that \( k \) must satisfy:
\[ k > 3 \]
Thus, the solution is:
\[ k \in (3, \infty) \]