Let n be positive integer such that sin-2 n -cos-2 n -n-2- ThenA- 6- n -8B- 6 - n - 8C- 4- n - 8D- 4- n -8

The correct option is D 4 lt; n lt; 8 sinπ 2n+cosπ 2n=√(2) sin(π 2n+π 4) or, sin(π 2n+π 4)=√(n)/2√(2) Since π 4π 2n + π 43π 4 for n 1 or, 1√(2)√(n) 2√(2 ...

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Standard XII

Mathematics

Using Monotonicity to Find the Range of a Function

Let n be posi...

Question

Let n be positive integer such that $s i n \frac{π}{2 n} + c o s \frac{π}{2 n} = \frac{\sqrt{n}}{2}$ . Then

6 ≤ n ≤ 8

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4 < n ≤ 8

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6 < n < 8

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4 < n < 8

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Solution

The correct option is D
4 < n < 8

$s i n \frac{π}{2 n} + c o s \frac{π}{2 n} = \sqrt{2} s i n (\frac{π}{2 n} + \frac{π}{4})$
or, $s i n (\frac{π}{2 n} + \frac{π}{4}) = \frac{\sqrt{n}}{2 \sqrt{2}}$
Since $\frac{π}{4} < \frac{π}{2 n} + \frac{π}{4} < \frac{3 π}{4} f o r n > 1$
or, $\frac{1}{\sqrt{2}} < \frac{\sqrt{n}}{2 \sqrt{2}} \leq 1$
or, $2 < \sqrt{n} \leq 2 \sqrt{2} o r, 4 < n \leq 8$ .
If n=1, L.H.S. = 1, R.H.S. $\frac{1}{2}$
Similarly for $n = 8, s i n (\frac{π}{16} + \frac{π}{4}) \neq 1$
$∴ 4 < n < 8$

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