A) \[6\le n\le 8\]
B) \[4<n\le 8\]
C) \[4\le n\le 8\]
D) \[4<n<8\]
Correct Answer: D
Solution :
| \[\sin \frac{\pi }{2n}+\cos \frac{\pi }{2n}=\frac{\sqrt{n}}{2}\] |
| \[\Rightarrow {{\sin }^{2}}\frac{\pi }{2n}+{{\cos }^{2}}\frac{\pi }{2n}+2\sin \frac{\pi }{2n}cos\frac{\pi }{2n}=\frac{n}{4}\] |
| \[\Rightarrow 1+\sin \frac{\pi }{n}=\frac{n}{4}\Rightarrow \sin \frac{\pi }{n}=\frac{n-4}{4}\] |
| For n=2 the given equation is not satisfied. Considering n>and n \[\ne \]2\[0<\sin \frac{\pi }{n}<1\Rightarrow 0<\frac{n-4}{4}<1\Rightarrow 4<n<8.\] |
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