A) \[n=4\]
B) \[n=5\]
C) \[n=6\]
D) None of these
Correct Answer: C
Solution :
\[\sin \frac{\pi }{2n}+\cos \frac{\pi }{2n}=\sqrt{2}\sin \left( \frac{\pi }{4}+\frac{\pi }{2n} \right)\] \[\Rightarrow \,\,\frac{\sqrt{n}}{2}=\sqrt{2}\sin \left( \frac{\pi }{4}+\frac{\pi }{2n} \right)\] So, for \[n>1,\] \[\frac{\sqrt{n}}{2\sqrt{2}}=\sin \left( \frac{\pi }{4}+\frac{\pi }{2n} \right)>\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\] Thus, \[n>4\] Since, \[\sin \left( \frac{\pi }{4}+\frac{\pi }{2n} \right)<1\]for all \[n>2,\]we get \[\frac{\sqrt{n}}{2\sqrt{2}}<1\] or \[n<8,\]So that\[4<n<8\]. By actual verification we find that only \[n=6\]satisfies the given relation.
You need to login to perform this action.
You will be redirected in
3 sec
