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26 May 2017, 04:41
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If \(x \ne 0\), is \(|x| < 1\)?
(1) \(\frac{x^{2}}{|x|} > x\)
(2) \(\frac{x}{|x|} < x\)
(1) \(\frac{x^{2}}{|x|} > x\)
(2) \(\frac{x}{|x|} < x\)
26 May 2017, 04:41
Kudos
Bookmarks
Official Solution:
For inequality questions, you need preliminary knowledge.
(1) First, if the range of the question includes the range of condition, you should use the fact that the condition is sufficient.
(2) Secondly, you should remember \(|A| = A\) (when \(A > 0\)), \(|0| = 0\), \(|A| = -A\) (when \(A < 0\)).
If you use the 1st step of the variable approach and change the question and the original condition, you get \(|x| < 1\), then is \(-1 < x < 1\). There is 1 variable, so in order to match the number of variables to the number of equations, there must be 1 equation as well. Therefore, D is most likely to be the answer.
In the case of con 1), \(\frac{x^{2}}{|x|} > x\), \(\frac{|x|^{2}}{|x|} > x\), \(|x| > x\), or \(x < 0\), and the range of the questions does not include the range of condition, hence it is not sufficient.
In the case of con 2),
(1) If \(x > 0\), from \(\frac{x}{|x|} < x \rightarrow \frac{x}{x} < x \rightarrow 1 < x\), you get \(1 < x\).
(2) If \(x < 0\), from \(\frac{x}{|x|} < x \rightarrow \frac{x}{-x} < x \rightarrow -1 < x\), you get \(-1 < x < 0\).
In this case, the range of the question does not include the range of the condition, hence it is not sufficient.
By solving con 1) & con 2), it becomes \(-1 < x < 0\), so the range of the questions includes the range of condition, hence it is sufficient.
Answer: C
For inequality questions, you need preliminary knowledge.
(1) First, if the range of the question includes the range of condition, you should use the fact that the condition is sufficient.
(2) Secondly, you should remember \(|A| = A\) (when \(A > 0\)), \(|0| = 0\), \(|A| = -A\) (when \(A < 0\)).
If you use the 1st step of the variable approach and change the question and the original condition, you get \(|x| < 1\), then is \(-1 < x < 1\). There is 1 variable, so in order to match the number of variables to the number of equations, there must be 1 equation as well. Therefore, D is most likely to be the answer.
In the case of con 1), \(\frac{x^{2}}{|x|} > x\), \(\frac{|x|^{2}}{|x|} > x\), \(|x| > x\), or \(x < 0\), and the range of the questions does not include the range of condition, hence it is not sufficient.
In the case of con 2),
(1) If \(x > 0\), from \(\frac{x}{|x|} < x \rightarrow \frac{x}{x} < x \rightarrow 1 < x\), you get \(1 < x\).
(2) If \(x < 0\), from \(\frac{x}{|x|} < x \rightarrow \frac{x}{-x} < x \rightarrow -1 < x\), you get \(-1 < x < 0\).
In this case, the range of the question does not include the range of the condition, hence it is not sufficient.
By solving con 1) & con 2), it becomes \(-1 < x < 0\), so the range of the questions includes the range of condition, hence it is sufficient.
Answer: C
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