Question
Let $$n$$ be a positive integer such that $$\sin \frac{\pi }{{2n}} + \cos \frac{\pi }{{2n}} = \frac{{\sqrt n }}{2}.$$ Then
A.
$$6 \leqslant n \leqslant 8$$
B.
$$4 < n \leqslant 8$$
C.
$$4 \leqslant n \leqslant 8$$
D.
$$4 < n < 8$$
Answer :
$$4 < n < 8$$
Solution :
$$\eqalign{
& \sin \frac{\pi }{{2n}} + \cos \frac{\pi }{{2n}} = \frac{{\sqrt n }}{2} \cr
& \Rightarrow \,\,{\sin ^2}\frac{\pi }{{2n}} + {\cos ^2}\frac{\pi }{{2n}} + 2\sin \frac{\pi }{{2n}}\cos \frac{\pi }{{2n}} = \frac{n}{4} \cr
& \Rightarrow \,\,1 + \sin \frac{\pi }{4} = \frac{n}{4} \cr
& \Rightarrow \,\,\sin \frac{\pi }{4} = \frac{{n - 4}}{4} \cr} $$
For $$n = 2$$ the given equation is not satisfied.
Considering $$n > 1$$ and $$n \ne 2$$
$$\eqalign{
& 0 < \sin \frac{\pi }{4} < 1 \cr
& \Rightarrow \,\,0 < \frac{{n - 4}}{4} < 1 \cr
& \Rightarrow \,\,4 < n < 8. \cr} $$