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26 Nov 2021, 02:20
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If \(0 > x > y\), then which of the following must be true?
I. \(x^2 – y^2 < 0\)
II. \(\frac{1}{x^2} < 1\)
III. \(\frac{x + y}{x} > 0\)
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
I. \(x^2 – y^2 < 0\)
II. \(\frac{1}{x^2} < 1\)
III. \(\frac{x + y}{x} > 0\)
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
26 Nov 2021, 02:20
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Official Solution:
If \(0 > x > y\), then which of the following must be true?
I. \(x^2 – y^2 < 0\)
II. \(\frac{1}{x^2} < 1\)
III. \(\frac{x + y}{x} > 0\)
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
We are given that \(0 > x > y\), so both \(x\) and \(y\) are negative and \(y\) is further from 0 than \(x\).
Check the options:
I. \(x^2 – y^2 < 0\):
• \(x^2 < y^2\)
• \(|x| < |y|\)
• The above must be true because we know that \(y\) is further from 0 than \(x\). Or, since both \(x\) and \(y\) are negative, then \(|x| < |y|\) means \(-x < -y\), which gives \(x > y\) and we know from the stem that this is true.
II. \(\frac{1}{x^2} < 1\)
• The above is not necessarily true. For example, consider \(x=-\frac{1}{2}\)
III. \(\frac{x + y}{x} > 0\)
• \(\frac{x}{x}+\frac{y}{x} > 0\)
• \(\frac{y}{x} > -1\). Since both \(x\) and \(y\) are negative, then \(\frac{y}{x} > 0 >-1\), so this option must be true.
Answer: D
If \(0 > x > y\), then which of the following must be true?
I. \(x^2 – y^2 < 0\)
II. \(\frac{1}{x^2} < 1\)
III. \(\frac{x + y}{x} > 0\)
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
We are given that \(0 > x > y\), so both \(x\) and \(y\) are negative and \(y\) is further from 0 than \(x\).
Check the options:
I. \(x^2 – y^2 < 0\):
• \(x^2 < y^2\)
• \(|x| < |y|\)
• The above must be true because we know that \(y\) is further from 0 than \(x\). Or, since both \(x\) and \(y\) are negative, then \(|x| < |y|\) means \(-x < -y\), which gives \(x > y\) and we know from the stem that this is true.
II. \(\frac{1}{x^2} < 1\)
• The above is not necessarily true. For example, consider \(x=-\frac{1}{2}\)
III. \(\frac{x + y}{x} > 0\)
• \(\frac{x}{x}+\frac{y}{x} > 0\)
• \(\frac{y}{x} > -1\). Since both \(x\) and \(y\) are negative, then \(\frac{y}{x} > 0 >-1\), so this option must be true.
Answer: D
18 Nov 2025, 23:21
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