A) \[x=\frac{\pi }{12}\]
B) \[x=\frac{\pi }{6}\]
C) \[x=\frac{\pi }{3}\]
D) \[x=\frac{\pi }{2}\]
Correct Answer: A
Solution :
Now,\[\cos \left( x+\frac{\pi }{6} \right)+\sin \left( x+\frac{\pi }{6} \right)\] \[=\sqrt{2}\left[ \frac{1}{\sqrt{2}}\cos \left( x+\frac{\pi }{6} \right)+\frac{1}{\sqrt{2}}\sin \left( x+\frac{\pi }{6} \right) \right]\] \[=\sqrt{2}\cos \left( x+\frac{\pi }{6}-\frac{\pi }{4} \right)\] \[=\sqrt{2}\cos \left( x-\frac{\pi }{12} \right)\] For maximum value \[x=\frac{\pi }{12}\]
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