Is x > y? (1) x^2 < y (2) x^(1/2) >y

\(x\,\,\mathop > \limits^? \,\,y\)

We could go straight for the (1+2) BIFURCATION, to guarantee the correct answer is (E).

I will bifurcate each one separately first, for didactic reasons.

\(\left( 1 \right)\,\,{x^2} < y\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {0;1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{1 \over 2};{1 \over 3}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\sqrt x > y\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {1;0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{1 \over 4};{1 \over 3}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,{x^2} < y < \sqrt x \,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\,\left( {x;y} \right) = \left( {{1 \over 4};{1 \over 4}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{1 \over 4};{1 \over 8}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.