Question:
If \( 2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0 \) has exactly 3 solutions in the interval \( \left[ 0, \frac{n \pi}{2} \right] \), \( n \in \mathbb{N} \), then the roots of the equation \( x^2 + nx + (n - 3) = 0 \) belong to:
If \( 2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0 \) has exactly 3 solutions in the interval \( \left[ 0, \frac{n \pi}{2} \right] \), \( n \in \mathbb{N} \), then the roots of the equation \( x^2 + nx + (n - 3) = 0 \) belong to:
Updated On: Mar 20, 2025
- \( (0, \infty) \)
- \( (-\infty, 0) \)
- \( \left( -\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2} \right) \)
- \( \mathbb{Z} \)
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The Correct Option is B
Solution and Explanation
Rewrite the given equation and analyze solution intervals to find values of \( n \) for which the quadratic equation \( x^2 + nx + (n - 3) = 0 \) has roots in the desired interval.
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