A) \[x=\frac{\pi }{4},y=1\]
B) \[y=0\]
C) \[y=2\]
D) \[x=\frac{3\pi }{4}\]
Correct Answer: A
Solution :
\[\because \] \[\frac{y+\frac{1}{y}}{2}\ge \sqrt{y.\frac{1}{y}}\] \[\Rightarrow \] \[\sqrt{\left( y+\frac{1}{y} \right)}\ge \sqrt{2}\] but \[|\sin x+\cos x|\le \sqrt{2}\] which is possible only when \[y+\frac{1}{y}=2\] \[\therefore \] \[y=1\] and \[x=\frac{\pi }{4}\]
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