A) \[(0,\sqrt{3})\]
B) \[(\sqrt{3},\infty )\]
C) \[(-\infty ,\,-\sqrt{3})\]
D) \[(-\sqrt{3},0)\]
Correct Answer: B
Solution :
[b] \[{{y}^{2}}-y+a={{\left( y-\frac{1}{2} \right)}^{2}}+a-\frac{1}{4}\] Since \[-\sqrt{2}\le \sin x+\cos x\le \sqrt{2},\] given equation will have no real value of x for any y if \[a-\frac{1}{4}>\sqrt{2}.\] i.e., \[a\in \left( \sqrt{2}+\frac{1}{4},\infty \right)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\in (\sqrt{3},\infty )\] \[(as\,\,\sqrt{2}+1/4<\sqrt{3})\]
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