d
(d) \(x = 3 - \sqrt 5 \)

\(\sqrt x = \sqrt {3 - \sqrt 5 } = {1 \over {\sqrt 2 }}\,.\sqrt {6 - 2\sqrt 5 } = {1 \over {\sqrt 2 }}(\sqrt 5 - 1)\)         \(3x - 2 = 9 - 3\sqrt 5 - 2 = 7 - 3\sqrt 5 = {{14 - 6\sqrt 5 } \over 2}\)

= \({{{{(3 - \sqrt 5 )}^2}} \over 2}\);

\( \Rightarrow \) \(\sqrt 2 + \sqrt {3x - 2} = \sqrt 5 \,.\,\sqrt x \);

\(\therefore {{\sqrt x } \over {\sqrt 2 + \sqrt {3x - 2} }} = {1 \over {\sqrt 5 }}\).