[JEE Advanced 1994] let n be a positive integer with sin pi 2n cos pi 2n sqrt
Let $n$ be a positive integer with $\sin\frac\pi{2n}+\cos\frac\pi{2n}=\frac{\sqrt n}2$. Then
(a) $6\le n\le8$
(b) $4<n\le8$
(c) $4\le n\le8$
(d) $4<n<8$
1 Answer
VAVidaara Admin
✓ Vidaara Team
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· 2d ago
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Correct answer: (d) $4<n<8$
Squaring: $1+\sin\frac\pi n=\frac n4\Rightarrow\sin\frac\pi n=\frac n4-1\in[0,1)$, so $4\le n<8$; testing integer values that actually satisfy gives $4<n<8$.
JEE Advanced 1994 · Trigonometry — verified solution by the Vidaara Team.
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