Question

Let n be a positive integer such that $sin\frac{\pi }{2^n}+cos\frac{\pi }{2^n}=\frac{\sqrt{n}}{2}$. Then

• A. $6\le n\le 8$
• B. 4 < $n\le 8$
• C. $4\le$ n < 8
• D. 4 < n < 8

Answer 1

The correct option is B

4 < $n\le 8$

$sin\frac{\pi }{2^n}+cos\frac{\pi }{2^n}=\frac{\sqrt{n}}{2}\Rightarrow \sqrt{2}(sin\frac{\pi }{2^n}.cos\frac{\pi }{4}+cos\frac{\pi }{2^n}.sin\frac{\pi }{4})=\frac{\sqrt{n}}{2}\Rightarrow \sqrt{2}sin(\frac{\pi }{4}+\frac{\pi }{2^n})=\frac{\sqrt{n}}{2}$

$Sincesin(\frac{\pi }{4}+\frac{\pi }{2^n})\le 1\therefore \frac{\sqrt{n}}{2}\le \sqrt{2}\Rightarrow \sqrt{n}\le 2\sqrt{2}\Rightarrow n\le 8.Again\frac{\sqrt{n}}{2}=\sqrt{2}sin(\frac{\pi }{4}+\frac{\pi }{2^n})>\sqrt{2}.\frac{1}{\sqrt{2}}=1(\because sin(\frac{\pi }{4}+\frac{\pi }{2^n})>sin\frac{\pi }{4})$

$\therefore$ n > 4, Hence, 4 < n $\le$ 8