A) \[x={{60}^{{}^\circ }}\]
B) (b)\[x={{30}^{{}^\circ }}\]
C) \[x={{45}^{{}^\circ }}\]
D) \[x={{75}^{{}^\circ }}\]
Correct Answer: B
Solution :
[b] \[\because y=sinx+\sqrt{3}.cosx\] \[=2.\left( \frac{1}{2}.\sin x+\frac{\sqrt{3}}{2}.\cos x \right)=2.\left( \cos \frac{\pi }{3}.\sin x+\sin \frac{\pi }{3}.\cos x \right)\]\[=2.in\left( x+\frac{\pi }{3} \right)\] \[\because \] y is maximum. \[\therefore sin\left( x+\frac{\pi }{3} \right)=1=\sin \frac{\pi }{2}\] \[x=\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}={{30}^{{}^\circ }}\] Hence, option [b] is correct.
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