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25 Nov 2019, 03:04
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
46% (02:16) correct
54%
(02:10)
wrong
based on 662
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If \(|\frac{x}{y}| < 1\), then which of the following must be true?
I. \(\frac{|x+1|}{|y+1|}< 1\)
II. \(\frac{|x^2+1|}{|y^2+1|} < 1\)
III. \(\frac{|x-1|}{|y-1|} < 1\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
Are You Up For the Challenge: 700 Level Questions
I. \(\frac{|x+1|}{|y+1|}< 1\)
II. \(\frac{|x^2+1|}{|y^2+1|} < 1\)
III. \(\frac{|x-1|}{|y-1|} < 1\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
Are You Up For the Challenge: 700 Level Questions
05 Feb 2020, 05:09
Kudos
Bookmarks
If \(|\frac{x}{y}| < 1\), then which of the following must be true?
I. \(\frac{|x+1|}{|y+1|}< 1\)
II. \(\frac{|x^2+1|}{|y^2+1|} < 1\)
III. \(\frac{|x-1|}{|y-1|} < 1\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
Re-write \(|\frac{x}{y}| < 1\) as \(\frac{|x|}{|y|} < 1\)
Multiply by |y|: \(|x| < |y|\) (we can safely do that because |y| > 0).
\(|x| < |y|\) means that y is further from 0 than x. We can have four cases:
----y----x----0--------------
----y---------0----x---------
---------x----0---------y----
--------------0----x----y----
I. \(\frac{|x+1|}{|y+1|}< 1\) --> \(|x+1|<|y+1|\). Not necessarily true, for example consider x = 1 and y = -2.
II. \(\frac{|x^2+1|}{|y^2+1|} < 1\) --> \(|x^2+1|<|y^2+1|\). Since both x^2 + 1 and y^2 + 1 are positive, then we have \(x^2+1<y^2+1\) --> \(x^2<y^2\). Taking the square root give \(|x| < |y|\). TRUE
III. \(\frac{|x-1|}{|y-1|} < 1\) --> \(|x-1|<|y-1|\). Not necessarily true, for example consider x = -1 and y = 2.
Answer: B.
I. \(\frac{|x+1|}{|y+1|}< 1\)
II. \(\frac{|x^2+1|}{|y^2+1|} < 1\)
III. \(\frac{|x-1|}{|y-1|} < 1\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
Re-write \(|\frac{x}{y}| < 1\) as \(\frac{|x|}{|y|} < 1\)
Multiply by |y|: \(|x| < |y|\) (we can safely do that because |y| > 0).
\(|x| < |y|\) means that y is further from 0 than x. We can have four cases:
----y----x----0--------------
----y---------0----x---------
---------x----0---------y----
--------------0----x----y----
I. \(\frac{|x+1|}{|y+1|}< 1\) --> \(|x+1|<|y+1|\). Not necessarily true, for example consider x = 1 and y = -2.
II. \(\frac{|x^2+1|}{|y^2+1|} < 1\) --> \(|x^2+1|<|y^2+1|\). Since both x^2 + 1 and y^2 + 1 are positive, then we have \(x^2+1<y^2+1\) --> \(x^2<y^2\). Taking the square root give \(|x| < |y|\). TRUE
III. \(\frac{|x-1|}{|y-1|} < 1\) --> \(|x-1|<|y-1|\). Not necessarily true, for example consider x = -1 and y = 2.
Answer: B.
General Discussion
25 Nov 2019, 04:13
Kudos
Bookmarks
check with 2/3 , -2/3 and 2/-3
only \(\frac{|x^2+1|}{|y^2+1|} < 1\)
will be valid
IMO B
only \(\frac{|x^2+1|}{|y^2+1|} < 1\)
will be valid
IMO B
Bunuel
