The domain of $$f\left( x \right) = \frac{1}{{\sqrt {2x - 1} }} - \sqrt {1 - {x^2}} $$      is :

Solution :
$$\eqalign{ & {\text{Given,}}\,\,f\left( x \right) = \frac{1}{{\sqrt {2x - 1} }} - \sqrt {1 - {x^2}} = p\left( x \right) - q\left( x \right) \cr & {\text{where }}p\left( x \right) = \frac{1}{{\sqrt {2x - 1} }}{\text{ and }}q\left( x \right){\text{ = }}\sqrt {1 - {x^2}} \cr & {\text{Now, domain of }}p\left( x \right)\,{\text{exist when}}\,2x - 1 \ne 0 \cr & \Rightarrow x = \frac{1}{2}{\text{ and }}2x - 1 > 0 \cr & \Rightarrow x = \frac{1}{2}{\text{ and }}x > \frac{1}{2} \cr & \therefore \,x\, \in \left( {\frac{1}{2},\,\infty } \right) \cr & {\text{and domain of }}q\left( x \right)\,{\text{exists when }}1 - {x^2} \geqslant 0 \cr & \Rightarrow {x^2} \leqslant 1 \cr & \Rightarrow \left| x \right| \leqslant {\text{1}} \cr & \therefore \, - 1 \leqslant x \leqslant 1 \cr & \therefore \,{\text{Common domain is }}\left] {\frac{1}{2},\,1} \right[ \cr} $$