A) 2
B) 3
C) 4
D) 5
Correct Answer: A
Solution :
For\[n=2,\,\,f(x)=\frac{\sin \,2x}{\sin \,\left( \frac{x}{2} \right)}=\frac{4\,\sin \,\left( \frac{x}{2} \right)\,\cos \,\left( \frac{x}{2} \right)\,\cos x}{\sin \,\left( \frac{x}{2} \right)}\] \[=4\,\cos \,\left( \frac{x}{2} \right)\,\cos x\] The period of \[\cos x=2\pi \] and that of \[\cos \frac{x}{2}\] is \[4\pi ,\] so period of \[\frac{\sin \,\,2x}{\sin \,\left( \frac{x}{2} \right)}\] is \[4\pi .\] For \[n=3,\,\,\frac{\sin \,\left\{ 3\,(x+4\pi ) \right\}}{\sin \,\left\{ \frac{(x+4\pi )}{3} \right\}}=\frac{\sin \,3x}{\sin \,\left( \frac{x}{3}+\frac{4\pi }{3} \right)}\ne \frac{\sin \,3x}{\sin \,\left( \frac{x}{3} \right)}\] So, \[4\pi \] is not the period for \[n=3.\] Similarly, we can see that \[4\pi \] is not the period of \[\frac{\sin \,nx}{\sin \left( \frac{x}{n} \right)}\] for \[n=4\] and 5 also.
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