A) \[\left] \frac{1}{2},1 \right[\]
B) \[\left[ -1,\infty \right[\]
C) \[\left[ 1,\infty \right[\]
D) None of these
Correct Answer: A
Solution :
[a] Given, \[f(x)=\frac{1}{\sqrt{2x-1}}-\sqrt{1-{{x}^{2}}}=p(x)-q(x)\] Where \[p(x)=\frac{1}{\sqrt{2x-1}}\] and \[q(x)=\sqrt{1-{{x}^{2}}}\] Now, Domain of \[p(x)\] exist when \[2x-1\ne 0\] \[\Rightarrow x=\frac{1}{2}\] And \[2x-1>0\] \[\Rightarrow x=\frac{1}{2}\] And \[x>\frac{1}{2}\therefore x\in \left( \frac{1}{2},\infty \right)\] And domain of q(x) exists when \[1-{{x}^{2}}\ge 0\] \[\Rightarrow {{x}^{2}}\le 1\Rightarrow \left| x \right|\le 1\therefore -1\le x\le 1\] \[\therefore \] Common domain is\[\left] \frac{1}{2},1 \right[\]
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