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E
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Difficulty:
85%
(hard)
Question Stats:
41% (02:16) correct
59%
(01:53)
wrong
based on 175
sessions
History
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Not Attempted Yet
04 Nov 2018, 13:15
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piyush26
\(x\,\,\mathop > \limits^? \,\,y\)
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.
04 Mar 2021, 08:40
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piyush26
St. 1: x^2<y; if 0<x,y are <1; then x=0.9 and y=8.9 will give x^2<y and x>y; if x<0 , y>0 e.g x=-3 ; y=1; x^2>y x<y
St. 2: root x >y; if 0<x,y <1; x=0.89 and y=0.9 will root x> y and x<y; and x=0.9 and y=0.89 also root x>y
Combining st 1 & st 2 ; root x > y > x^2
Now we know that, root x > x^2 can happen if 0<x<1 and root x > x > x^2; (remember higher powers of x reduces value when 0<x<1;
So, now we get root x > x > x^2 and root x > y > x^2 ; but we still cant determine if x>y or x<y as there are 2 possibilities
root x > x > y> x^2 and root x > y> x> x^2 ; hence E
Important properties; x^2<y and x>y implies 0<x<1 ; y cant be -ve becuase x^2<y and hence x cant be negative
Also, x^2 > x > root x for x>1
and x^2<x<root x for x<1
