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12 Jul 2019, 02:23
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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wrong
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Is \(x^3 – 6x2 + 11x – 6 > 0\)?
(1) \(1 < x < 2\)
(2) \(x > 4\)
(1) \(1 < x < 2\)
(2) \(x > 4\)
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12 Jul 2019, 08:27
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Bunuel
I've never seen a real GMAT question where you might need to factor a proper cubic (i.e. one where x was not a factor), so I don't think this question is realistic anyway, but the two Statements clearly contradict each other, so this can't be a real GMAT question.
To factor here, you probably need to notice that x=1 is a solution to x^3 - 6x^2 + 11x - 6 = 0, so (x-1) is a factor of the left side. Then the factorization of the left side must look like (x-1)(x^2 + ax + 6), in order to get both x^3 and -6 when we expand, and then looking either at the -6x^2 or the +11x terms, you can work out that a= -5, and the factorization is (x-1)(x^2 - 5x + 6) = (x-1)(x-2)(x-3). So the sign of the cubic only changes as x passes over 1, 2 and 3, and it has a constant sign when 1 < x < 2, and also when x > 3 (so when x > 4). But with contradictory statements, there is no logically correct answer - I guess the intended answer is D, but using both statements, x cannot exist, so maybe the answer is E.
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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