A) \[n=6\]
B) \[n=2\]
C) \[n=1\]
D) \[n=3,4,5\]
Correct Answer: A
Solution :
\[\sin \left( \frac{\pi }{2n} \right)+\cos \left( \frac{\pi }{2n} \right)=\frac{\sqrt{n}}{2}\] \[\Rightarrow \] \[\sqrt{2}\left\{ \frac{1}{\sqrt{2}}.\cos \left( \frac{\pi }{2n} \right)+\frac{1}{\sqrt{2}}.\sin \left( \frac{\pi }{2n} \right) \right\}\] \[=\frac{\sqrt{n}}{2}\] \[\Rightarrow \] \[\sqrt{2}\left\{ \cos \left( \frac{\pi }{4}-\frac{\pi }{2n} \right) \right\}=\frac{\sqrt{n}}{2}\] \[\Rightarrow \] \[\cos \left( \frac{\pi }{4}-\frac{\pi }{2n} \right)=\frac{\pi }{2\sqrt{2}}\] ?(i) When \[n=6,\,LHS=cos\left( \frac{\pi }{6} \right)=\frac{\sqrt{3}}{2}\] and \[RHS=\frac{\sqrt{6}}{2\sqrt{2}}=\frac{\sqrt{3}}{2}\] Eq. (i) is not satisfied for \[n=1,2,3,4,5,\]
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