Correct option : (a) 2
\(\cfrac{\frac{1}{\sqrt9}-\frac{1}{\sqrt{11}}} {\frac{1}{\sqrt9}+\frac{1}{\sqrt{11}}} \times \frac{10+\sqrt{99}}{x} =\frac12\)
\(\Rightarrow \cfrac{\left(\frac{1}{\sqrt9}-\frac{1}{\sqrt{11}}\right) \left(\frac{1}{\sqrt9}-\frac{1}{\sqrt{11}}\right)} {\left(\frac{1}{\sqrt9}+\frac{1}{\sqrt{11}}\right)\left(\frac{1}{\sqrt9}-\frac{1}{\sqrt{11}}\right)} \times \frac{10+\sqrt{99}}{x} =\frac12\)
\(\Rightarrow \cfrac{\frac19+\frac{1}{11}-\frac{2}{\sqrt{99}}} {\frac19 -\frac{1}{11}} \times \frac{10+\sqrt{99}}{x} =\frac12\)
(\(\because \) \((a-b)^2 =a^2+b^2-2ab\) & \((a+b)(a-b)=a^2 -b^2\))
\(\Rightarrow \cfrac{\frac{11+9-2\sqrt{99}}{99}}{\frac{11-9}{99}} \times \frac{10+\sqrt{99}}{x} =\frac12\)
\(\Rightarrow \frac{2(10-\sqrt{99})}{2} \times \frac{10+\sqrt{99}}{x} =\frac12\)
\(\Rightarrow \frac{100-99}{x} =\frac12\) (\(\because\) \((a-b)(a+b)=a^2-b^2\))
\(\therefore x =2.\)
