A) 1
B) 4
C) 3
D) 2
Correct Answer: A
Solution :
[a] We have, \[2{{(si{{n}^{-1}}y)}^{2}}=\frac{4n+1}{16}{{\pi }^{2}}\] \[\Rightarrow 0\le \frac{4n+1}{32}{{\pi }^{2}}\le \frac{{{\pi }^{2}}}{4}\] Also, \[2(co{{s}^{-1}}x)=\frac{4n-1}{16}{{\pi }^{2}}\Rightarrow -\frac{1}{4}\le n\le \frac{7}{4}\] Also, \[2(co{{s}^{-1}}x)=\frac{4n-1}{16}{{\pi }^{2}}\] \[\Rightarrow 0\le \frac{4n-1}{32}{{\pi }^{2}}\le \pi \Rightarrow \frac{1}{4}\le n\,\le \frac{8}{\pi }+\frac{1}{4}\Rightarrow n=1\]
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