A) \[2\pi \]
B) \[10\pi \]
C) \[6\pi \]
D) \[12\pi \]
Correct Answer: D
Solution :
Period of \[\sin \left( \frac{2x}{3} \right)\,=\frac{2\pi }{2/3}=3\pi \] Period of \[\sin \,\left( \frac{3x}{2} \right)=\frac{2\pi }{3/2}=\frac{4\pi }{3}\] L.C.M. of \[3\pi \]and \[\frac{4\pi }{3}\]=\[12\pi \]. Hence period is\[12\pi \].
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