A) \[\frac{1}{2}<\alpha <\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{2}<\alpha <\frac{1}{3}\]
C) \[\frac{1}{3}<\alpha <\frac{1}{\sqrt{3}}\]
D) \[\frac{1}{4}<\alpha <\frac{1}{2}\]
Correct Answer: C
Solution :
The point should lies on the opposite side of the origin of the line\[~x+y-1=0\] Then, \[\alpha +\alpha -1>0\] \[\Rightarrow \] \[2\alpha \ge 1\Rightarrow \alpha >\frac{1}{2}\] ... (i) Also, \[({{\alpha }^{2}}+{{\alpha }^{2}})<1\] \[\Rightarrow \] \[\left( \frac{-1}{\sqrt{2}} \right)<\alpha <\left( \frac{1}{\sqrt{2}} \right)\] ? (ii) From Eqs. (i) and (ii), we get \[\frac{1}{2}<\alpha <\frac{1}{\sqrt{2}}\]You need to login to perform this action.
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