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If x > y and x < 0, then which of the following must be true
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28 Mar 2019, 13:27
28 Mar 2019, 13:27
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Question Stats:
55% (02:10) correct
44% (01:39) wrong based on 29 sessions
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If \(x > y\) and \(x < 0\), then which of the following must be true?
(I) \(\frac{1}{x} < \frac{1}{y}\)
(II) \(\frac{1}{x−1} < \frac{1}{y−1}\)
(III) \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
(I) \(\frac{1}{x} < \frac{1}{y}\)
(II) \(\frac{1}{x−1} < \frac{1}{y−1}\)
(III) \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Re: If x > y and x < 0, then which of the following must be true
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16 Jan 2020, 19:46
16 Jan 2020, 19:46
2
Seems we can safely exclude (iii) from the answer, because,
- Given x < 0, x could be -1.
- In (iii), we 1/(x+1), which if x =-1, equals, 1/(-1+1) = 1/0 = undefined and breaks the inequality.
Meanwhile, we do not have the above problem with (ii) because since x is a negative, (x-1) will become more negative and respects the initial condition of the inequality whereby x < 0.
- Given x < 0, x could be -1.
- In (iii), we 1/(x+1), which if x =-1, equals, 1/(-1+1) = 1/0 = undefined and breaks the inequality.
Meanwhile, we do not have the above problem with (ii) because since x is a negative, (x-1) will become more negative and respects the initial condition of the inequality whereby x < 0.
Re: If x > y and x < 0, then which of the following must be true
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07 Oct 2024, 12:40
07 Oct 2024, 12:40
Carcass wrote:
If \(x > y\) and \(x < 0\), then which of the following must be true?
(I) \(\frac{1}{x} < \frac{1}{y}\)
(II) \(\frac{1}{x−1} < \frac{1}{y−1}\)
(III) \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
(I) \(\frac{1}{x} < \frac{1}{y}\)
(II) \(\frac{1}{x−1} < \frac{1}{y−1}\)
(III) \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
The question mentions which one of the following *must be* true but if in case we take x=-2 and y=-3 it invalidates the *must be* condition even though in case x =-1 could cause the third inequality to be true but again, it's conditional here, could you please explain how D is correct?
