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16 Sep 2022, 02:56
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (01:54) correct
33%
(02:00)
wrong
based on 479
sessions
History
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Time
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Not Attempted Yet
If \((x + y)^2 < x^2\), which of the following must be true?
I. \(y(y + 2x) < 0\)
II. \(y < x\)
III. \(xy < 0\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
I. \(y(y + 2x) < 0\)
II. \(y < x\)
III. \(xy < 0\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
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25 Apr 2023, 08:01
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Official Solution:
If \((x + y)^2 < x^2\), which of the following must be true?
I. \(y(y + 2x) < 0\)
II. \(y < x\)
III. \(xy < 0\)
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
First, let's simplify the given inequality:
\(x^2 + 2xy + y^2 < x^2\)
\(2xy + y^2 < 0\)
\(y(2x + y) < 0\)
Now, let's evaluate each option:
I. \(y(y + 2x) < 0\). This option is directly true since it matches our simplified inequality.
II. \(y < x\). From \(y(2x + y) < 0\), we can infer that \(y\) and \(2x + y\) must have different signs. Let's try finding a counterexample: if \(y\) is positive, say 1, then \(2x + y\) must be negative, which can be obtained if \(x\) is, for example, -10, making \(y > x\). Hence, this option is not necessarily true.
III. \(xy < 0\). We derived that \(2xy + y^2 < 0\) earlier. Since \(y^2\) is always non-negative, \(2xy\) must be negative for the inequality to hold. This implies that \(xy\) is negative. Thus, this option is always true.
Therefore, only options I and III must be true.
Answer: E
If \((x + y)^2 < x^2\), which of the following must be true?
I. \(y(y + 2x) < 0\)
II. \(y < x\)
III. \(xy < 0\)
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
First, let's simplify the given inequality:
\(x^2 + 2xy + y^2 < x^2\)
\(2xy + y^2 < 0\)
\(y(2x + y) < 0\)
Now, let's evaluate each option:
I. \(y(y + 2x) < 0\). This option is directly true since it matches our simplified inequality.
II. \(y < x\). From \(y(2x + y) < 0\), we can infer that \(y\) and \(2x + y\) must have different signs. Let's try finding a counterexample: if \(y\) is positive, say 1, then \(2x + y\) must be negative, which can be obtained if \(x\) is, for example, -10, making \(y > x\). Hence, this option is not necessarily true.
III. \(xy < 0\). We derived that \(2xy + y^2 < 0\) earlier. Since \(y^2\) is always non-negative, \(2xy\) must be negative for the inequality to hold. This implies that \(xy\) is negative. Thus, this option is always true.
Therefore, only options I and III must be true.
Answer: E
General Discussion
16 Sep 2022, 05:56
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If \((x + y)^2 < x^2\), which of the following must be true?
\((x + y)^2 < x^2\)
\(x^2+2xy+y^2<x^2……….y(y+2x)<0\)
I. \(y(y + 2x) < 0\)
True, as above.
II. \(y < x\)
y=2 and x=-3……\((2-3)^2<(-3)^2……1<9\)
So, not necessarily true.
III. \(xy < 0\)
\(x^2+2xy+y^2<x^2……….y^2+2xy<0\)
Now, y^2 is NOT negative, so 2xy<0 or xy<0
Always true.
(E) I and III only
\((x + y)^2 < x^2\)
\(x^2+2xy+y^2<x^2……….y(y+2x)<0\)
I. \(y(y + 2x) < 0\)
True, as above.
II. \(y < x\)
y=2 and x=-3……\((2-3)^2<(-3)^2……1<9\)
So, not necessarily true.
III. \(xy < 0\)
\(x^2+2xy+y^2<x^2……….y^2+2xy<0\)
Now, y^2 is NOT negative, so 2xy<0 or xy<0
Always true.
(E) I and III only
