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26 May 2017, 04:41
Kudos
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If \(b < 0\), is \(5b(a + b) > -a^{2} - b^{2}\)?
(1) \(a + 2b > 0\)
(2) \(a + 3b > 0\)
(1) \(a + 2b > 0\)
(2) \(a + 3b > 0\)
26 May 2017, 04:41
Kudos
Bookmarks
Official Solution:
You need preliminary knowledge to solve this question.
If \(b < 0\), you get \(-3b > -2b > -b\). If you take the 1st step of the variable approach and modify the original condition and the question,
Que: is \(5b(a + b) > -a^{2} -b^{2} \rightarrow 5ab + 5b^{2} +a^{2} + b^{2} > 0, a^{2} + 5ab + 6b^{2} > 0\), or \((a + 3b) (a + 2b) > 0\)
In the original condition, there are 2 variables \((a, b)\) and 1 equation \((b < 0)\), and in order to match the number of variables to the number of equations, there must be 1 more equation. Therefore, D is most likely to be the answer.
In the case of con 1), \(a > -2b\), so \((a, b) = (3, -1)\) no, and \((a, b) = (5, -1)\) yes, hence it is not sufficient.
In the case of con 2), \(a > -3b > -2b\). This is because \(b < 0\), and it is shown in the preliminary knowledge above. If so, \(a > -3b\) and \(a > -2b\), then \(a + 3b > 0\) and \(a + 2b > 0\). Since \((a + 3b) (a + 2b) > 0\) is always yes, it is sufficient. The answer is B.
Answer: B
You need preliminary knowledge to solve this question.
If \(b < 0\), you get \(-3b > -2b > -b\). If you take the 1st step of the variable approach and modify the original condition and the question,
Que: is \(5b(a + b) > -a^{2} -b^{2} \rightarrow 5ab + 5b^{2} +a^{2} + b^{2} > 0, a^{2} + 5ab + 6b^{2} > 0\), or \((a + 3b) (a + 2b) > 0\)
In the original condition, there are 2 variables \((a, b)\) and 1 equation \((b < 0)\), and in order to match the number of variables to the number of equations, there must be 1 more equation. Therefore, D is most likely to be the answer.
In the case of con 1), \(a > -2b\), so \((a, b) = (3, -1)\) no, and \((a, b) = (5, -1)\) yes, hence it is not sufficient.
In the case of con 2), \(a > -3b > -2b\). This is because \(b < 0\), and it is shown in the preliminary knowledge above. If so, \(a > -3b\) and \(a > -2b\), then \(a + 3b > 0\) and \(a + 2b > 0\). Since \((a + 3b) (a + 2b) > 0\) is always yes, it is sufficient. The answer is B.
Answer: B
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