Let n be a positive integer such that sin π2n + cos π2n = &#x

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

Options:

A . 6

\leq

n

\leq

8

B . 4 < n

\leq

8

C . 4

\leq

n < 8

D . 4 < n < 8

Answer: Option B
:
B

(b) sin $\frac{π}{2^{n}}$ + cos $\frac{π}{2^{n}}$ = $\frac{\sqrt{n}}{2}$
   $\Rightarrow$ $\sqrt{2}$ $(sin \frac{π}{2^{n}} . cos \frac{π}{4} + cos \frac{π}{2^{n}} . sin \frac{π}{4})$ = $\frac{\sqrt{n}}{2}$
   $\Rightarrow$ $\sqrt{2}$ sin $(\frac{π}{4} + \frac{π}{2^{n}})$ = $\frac{\sqrt{n}}{2}$
   Since sin $(\frac{π}{4} + \frac{π}{2^{n}}) \leq 1$
   $∴$ $\frac{\sqrt{n}}{2}$ $\leq$ $\sqrt{2}$ $\Rightarrow$ $\sqrt{2}$ $\leq$ 2 $\sqrt{2}$ $\Rightarrow$ n $\leq$ 8 .
  Again   $\frac{\sqrt{n}}{2}$ = $\sqrt{2}$ sin $(\frac{π}{4} + \frac{π}{2^{n}})$ > $\sqrt{2}$ . $\frac{1}{\sqrt{2}}$ = 1
$(∵ sin (\frac{π}{4} + \frac{π}{2^{n}}) sin \frac{π}{4})$
   $∴$ n > 4, Hence, 4 < n $\leq$ 8.

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