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20 May 2017, 10:35
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
94% (01:14) correct
6%
(01:57)
wrong
based on 238
sessions
History
Date
Time
Result
Not Attempted Yet
For all real numbers \(x\) and \(y\), let x#y = \((xy)^2 − x + y^2\) . What is the value of y that makes x # y equal to \(–x\) for all values of \(x\) ?
(A) 0
(B) 2
(C) 5
(D) 7
(E) 10
(A) 0
(B) 2
(C) 5
(D) 7
(E) 10
Source: Nova GMAT
Difficulty Level: 500
Difficulty Level: 500
20 May 2017, 10:59
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We have to make x#y = -x
or (x^2)(y^2) -x + y^2 = -x
y^2 (x^2 + 1) = 0
This equation will obviously be true for all values of x if y^2 = 0 or if y=0
Hence A
or (x^2)(y^2) -x + y^2 = -x
y^2 (x^2 + 1) = 0
This equation will obviously be true for all values of x if y^2 = 0 or if y=0
Hence A
20 May 2017, 12:05
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The only possibility for the equation,
when x#y = \((xy)^2 − x + y^2\) = -x is when \((xy)^2 + y^2 = 0\).
Only one possibility can give us a 0 for the expression at all possible values of x(y=0)
If y =0, the expression will always yield a zero and hence the value of x#y will always be -x
Therefore, Option A is the correct answer.
when x#y = \((xy)^2 − x + y^2\) = -x is when \((xy)^2 + y^2 = 0\).
Only one possibility can give us a 0 for the expression at all possible values of x(y=0)
If y =0, the expression will always yield a zero and hence the value of x#y will always be -x
Therefore, Option A is the correct answer.
