Let n be a positive integer such that 2^n + 2^n = n 2 . Then - Vidyadip

MCQ

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

A
$6 \le n \le 8$
✓
$4 < n \le 8$
C
$4 \le n < 8$
D
$4 < n < 8$

✓

Answer

Correct option: B.

$4 < n \le 8$

b
(b) $\sin \frac{\pi }{{{2^n}}} + \cos \frac{\pi }{{{2^n}}} = \frac{{\sqrt n }}{2}$
==> $\sqrt 2 \left( {\sin \frac{\pi }{{{2^n}}}.\cos \frac{\pi }{4} + \cos \frac{\pi }{{{2^n}}}.\sin \frac{\pi }{4}} \right) = \frac{{\sqrt n }}{2}$
==> $\sqrt 2 \sin \left( {\frac{\pi }{4} + \frac{\pi }{{{2^n}}}} \right) = \frac{{\sqrt n }}{2}$
Since $\sin \,\left( {\frac{\pi }{4} + \frac{\pi }{{{2^n}}}} \right) \le 1$
$\therefore \;\frac{{\sqrt n }}{2} \le \sqrt 2 \Rightarrow \sqrt n \le 2\sqrt 2 \Rightarrow n \le 8$.
Again

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

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