Let n be a positive integer such that sin π2n + cos π2n = "

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Question

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 ≤ n ≤ 8

B . 4 < n ≤ 8

C . 4 ≤ 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}

Sincesin

(\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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