Bunuel wrote:
Is |n| < 1 ?
(1) n^x - n < 0
(2) x^(-1) = -2
\(\left| n \right|\,\,\mathop < \limits^? \,\,1\)
\(\left( 1 \right)\,\,{n^x} < n\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {n;x} \right) = \left( {2; -1} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {n;x} \right) = \left( {{1 \over 2};2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)
\(\left( 2 \right)\,\,{1 \over x} = - 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x = - {1 \over 2}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{INSUFF}}.\,\,\,\,\left( {{\rm{trivially}}} \right)\)
\(\left( {1 + 2} \right)\,\,\,\,{1 \over {\sqrt n }} < n\,\,\,\left\{ \matrix{
\,\,\,\mathop \Rightarrow \limits^{{\rm{implicitly}}} \,\,\,\,n > 0 \hfill \cr
\,\,\,\,\mathop \Rightarrow \limits^{\sqrt n \,\, > \,\,0} \,\,\,n\sqrt n > 1\, \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.\)
\(\left( * \right)\,\,\,\left\{ \matrix{
\,n > 0 \hfill \cr
\,\left| n \right| < 1 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,0 < n < 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{
\,0 < \sqrt n < 1 \hfill \cr
\,0 < n < 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,0 < n\sqrt n < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{impossible}}\)
The correct answer is therefore (C), indeed.
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
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