A) \[\frac{1}{2}<\alpha <\frac{1}{\sqrt{2}}\]
B) \[-\frac{1}{\sqrt{2}}<\alpha <\frac{1}{\sqrt{2}}\]
C) \[\alpha >\frac{1}{\sqrt{2}}\]
D) \[0<\alpha <\frac{1}{2}\]
Correct Answer: A
Solution :
\[\Rightarrow 2{{\alpha }^{2}}-1<0\]\[\Rightarrow -\frac{1}{\sqrt{2}}<\alpha <\frac{1}{2}\] ??..(1) |
and \[\alpha +\alpha >1\] \[\Rightarrow \alpha >\frac{1}{2}\] ???(2) |
\[\therefore \]Common solution of (1) and (2) is: |
\[\frac{1}{2}<\alpha <\frac{1}{\sqrt{2}}\] |
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