A) \[2\pi n(n-1)\]
B) \[4n\,(n-1)\]
C) \[2n\,(n-1)\]
D) None of these
Correct Answer: C
Solution :
\[f(x)=\sin \left( \frac{\pi x}{n-1} \right)+\cos \left( \frac{\pi x}{n} \right)\] Period of \[\sin \left( \frac{\pi x}{n-1} \right)=\frac{2\pi }{\left( \frac{\pi }{n-1} \right)}=2\left( n-1 \right)\] and period of \[\cos \left( \frac{\pi x}{n} \right)=\frac{2\pi }{\left( \frac{\pi }{n} \right)}=2n\] Hence period of \[f(x)\] is L.C.M. of \[2\,n\] and \[2(n-1)\] \[\Rightarrow 2n(n-1)\].
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