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.
• B.
• C.
• D.

Answer 1

The correct option is B

(b) 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$(\frac{\pi }{4}+\frac{\pi }{2^n})$ = $\frac{\sqrt{n}}{2}$
Since sin$(\frac{\pi }{4}+\frac{\pi }{2^n})\le 1$
$\therefore$ $\frac{\sqrt{n}}{2}$ $\le$ $\sqrt{2}$ $\Rightarrow$ $\sqrt{2}$ $\le$ 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.