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If the inequality cot^(2)x + (k +1) cot ...

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 `.

A

`(-oo,3-2sqrt(5))`

B

`(3,oo)`

C

`(-1, oo)`

D

`(-oo,3)`

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To solve the inequality \( \cot^2 x + (k + 1) \cot x - (k - 3) < 0 \) for at least one \( x \in (0, \frac{\pi}{2}) \), we will follow these steps: ### Step 1: Substitute \( \cot x \) with \( t \) Let \( t = \cot x \). As \( x \) varies from \( 0 \) to \( \frac{\pi}{2} \), \( t \) varies from \( \infty \) to \( 0 \). Therefore, we rewrite the inequality as: \[ t^2 + (k + 1)t - (k - 3) < 0 \] ### Step 2: Analyze the quadratic inequality The quadratic inequality \( t^2 + (k + 1)t - (k - 3) < 0 \) can be analyzed using the discriminant. For the quadratic to have real roots, the discriminant must be non-negative: \[ D = (k + 1)^2 - 4 \cdot 1 \cdot (-(k - 3)) \geq 0 \] This simplifies to: \[ D = (k + 1)^2 + 4(k - 3) \geq 0 \] \[ D = k^2 + 2k + 1 + 4k - 12 \geq 0 \] \[ D = k^2 + 6k - 11 \geq 0 \] ### Step 3: Solve the quadratic inequality for \( k \) Now we need to find the roots of the quadratic equation \( k^2 + 6k - 11 = 0 \) 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 4: Determine the intervals The quadratic \( k^2 + 6k - 11 \) opens upwards (since the coefficient of \( k^2 \) is positive). Therefore, the inequality \( k^2 + 6k - 11 \geq 0 \) holds for: \[ k \leq -3 - 2\sqrt{5} \quad \text{or} \quad k \geq -3 + 2\sqrt{5} \] ### Step 5: Find the numerical values Calculating \( -3 + 2\sqrt{5} \): \[ \sqrt{5} \approx 2.236 \implies 2\sqrt{5} \approx 4.472 \] Thus, \[ -3 + 2\sqrt{5} \approx -3 + 4.472 \approx 1.472 \] ### Conclusion The values of \( k \) for which the inequality \( \cot^2 x + (k + 1) \cot x - (k - 3) < 0 \) holds for at least one \( x \in (0, \frac{\pi}{2}) \) are: \[ k \in (-\infty, -3 - 2\sqrt{5}] \cup [-3 + 2\sqrt{5}, \infty) \]

To solve the inequality \( \cot^2 x + (k + 1) \cot x - (k - 3) < 0 \) for at least one \( x \in (0, \frac{\pi}{2}) \), we will follow these steps: ### Step 1: Substitute \( \cot x \) with \( t \) Let \( t = \cot x \). As \( x \) varies from \( 0 \) to \( \frac{\pi}{2} \), \( t \) varies from \( \infty \) to \( 0 \). Therefore, we rewrite the inequality as: \[ t^2 + (k + 1)t - (k - 3) < 0 \] ...
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