A) [2, 6]
B) [3, 7]
C) R
D) [1, 4]
Correct Answer: A
Solution :
\[\cos 2x+\alpha \sin x=2\alpha -7\] \[\Rightarrow 2{{\sin }^{2}}x-\alpha \sin x+2\alpha -8=0\] \[{{\sin }^{2}}x-\frac{\alpha }{2}\sin x+\alpha -4=0\] \[\Rightarrow \sin x=2\](rejected) or \[\sin x=\frac{\alpha -4}{2}\] \[\Rightarrow \left| \frac{\alpha -4}{2} \right|\le 1\]\[\Rightarrow \alpha \in [2,6]\]
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