A) 2
B) 3
C) 4
D) 5
Correct Answer: D
Solution :
\[2\left( \sin \left( 2x+\frac{\pi }{6} \right)+\sin \frac{\pi }{2} \right)\] \[={{a}^{2}}+\sqrt{3}\sin 2x-\cos 2x\] \[=2\left( \sin 2x\cdot \frac{\sqrt{3}}{2}+\cos 2x\cdot \frac{1}{2}+1 \right)\] \[={{a}^{2}}+\sqrt{3}\sin 2x-\cos 2x\] \[\Rightarrow \cos 2x+2={{a}^{2}}-\cos 2x\] \[\Rightarrow 2\cos 2x={{a}^{2}}-2\] \[\Rightarrow -\,2\le {{a}^{2}}-2\le 2\] \[\Rightarrow \,0\le {{a}^{2}}\le 4\Rightarrow a\in [-2,2]\]
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