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eswarchethu135
GMAT 2: 640 Q49 V27
Posts: 276
03 Jan 2019, 20:34
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
22% (00:47) correct
78%
(01:03)
wrong
based on 67
sessions
History
Date
Time
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Not Attempted Yet
What is the value of x?
1) \(x^x = 1\)
2) \(x^a = 1\) where a > 0
1) \(x^x = 1\)
2) \(x^a = 1\) where a > 0
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04 Jan 2019, 09:31
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eswarchethu135
1) \(x^x = 1\)
Let x be 0, \(0^0=1\), also 1^1=0
Thus x acb be 0 or 1
Insufficient
2) \(x^a = 1\) where a > 0
\(1^a=1\) and \((-1)^a\), when a is even..
Thus x can be 1 or -1..
Insufficient
Combined
x is 1..
Sufficient
C
04 Jan 2019, 12:42
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eswarchethu135
chetan2u ´s argument for the insufficiency of statement (2) is perfect.
The SUFFICIENCY of statement (1) is proved below, although I must say the arguments are OUT of GMAT´s scope (in terms of abstraction, so to speak).
Important: 0^0 is not equal to 1, it´s an undefined mathematical expression. (If you prefer, it does not exist, from the fact that it cannot be defined properly.)
The reason people believe it is equal to 1 is related to the fact that in SOME cases it is convenient to use 0^0=1 as a CONVENTION.
(The best example I know is when dealing with multi-indexes, in the context of Partial Differential Equations, for instance.)
All that put, let´s see the reason why (1) is SUFFICIENT and, therefore, the correct answer is (A):
\({x^x} = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left| {{x^x}} \right| = 1\,\,\,\,\,\left( * \right)\)
\(x > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{\\
\,0 < x < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = 1\,\,,\,\,\,0 < x < 1\,\,\,\,\,{\rm{impossible}}\,\, \hfill \cr \\
\,x = 1\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{possible}} \hfill \cr \\
\,x > 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} > \,\,\,1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = 1\,\,,\,\,\,x > 1\,\,\,\,\,{\rm{impossible}} \hfill \cr} \right.\)
\(x < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = {1 \over {{x^{\left| x \right|}}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\mathop = \limits^{\left( * \right)} \,\,\left| {{x^x}} \right|\, = {1 \over {\,\left| {{x^{\left| x \right|}}} \right|}} = {1 \over {\left| x \right|{\,^{\left| x \right|}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| x \right|{\,^{\left| x \right|}}\,\,\, = \,\,\,1\,\,\,\,\left\{ \matrix{\\
\,\,\,\left| x \right| \ne 1\,\,\,\,\mathop \Rightarrow \limits^{{\rm{previous}}\,\,{\rm{cases}}} \,\,\,\,\,{\rm{impossible}} \hfill \cr \\
\,\,\,x = - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^x} = - 1 \ne 1 \hfill \cr} \right.\)
Regards,
Fabio.
