JEE Mains · Maths · STD 11 - 3. trignometrical ratios,functions and identities
If \(2 \sin ^3 x+\sin 2 x \cos x+4 \sin x-4=0\) has exactly \(3\) solutions in the interval \(\left[0, \frac{\mathrm{n} \pi}{2}\right], \mathrm{n} \in \mathrm{N}\), then the roots of the equation \(x^2+n x+(n-3)=0\) belong to :
- A \((0, \infty)\)
- B \((-\infty, 0)\)
- C \(\left(-\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2}\right)\)
- D \(\mathrm{Z}\)
Answer & Solution
Correct Answer
(B) \((-\infty, 0)\)
Step-by-step Solution
Detailed explanation
\( 2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0\) \( 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \) \( 6 \sin x-4=0 \) \( \sin x=\frac{2}{3} \) \({n}=5 \text { (in the given interval) } \) \( x^2+5 x+2=0\) \( x=\frac{-5 \pm \sqrt{17}}{2}\)…
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