Please prove mathematically that when y=0, the defined set C is not a subset.  Why is the vector (0,0) not a subset of C?  PS: The set A lies in the first quadrant and the set B (like a mirror image) is in the second quadrant. a+b must include one common point (0,0) right? Where am I going wrong here?

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 94E
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The question is attached in the image. The correct answer is obviously A or B. The correct answer according to university records is B.

Please prove mathematically that when y=0, the defined set C is not a subset. 

Why is the vector (0,0) not a subset of C? 

PS: The set A lies in the first quadrant and the set B (like a mirror image) is in the second quadrant. a+b must include one common point (0,0) right? Where am I going wrong here?

21. If
{(r, y) E R² | x > 0, y > 0, xy > 1},
{(x, y) E R² | ¤ < 0, y > 0, xy < -1} and
{a +b| a € A, b € B}
A
=
В
=
=
then
A. {(x, y) E R² | x = 0, y > 0} is a subset of C
B. {(r, y) E R² | x = 0, y > 0} is a subset of C
C. {(x, y) E R² | x > 0, y = 0} is a subset of C
D. {(r, y) € R² | x > 0, y = 0} is a subset of C
Transcribed Image Text:21. If {(r, y) E R² | x > 0, y > 0, xy > 1}, {(x, y) E R² | ¤ < 0, y > 0, xy < -1} and {a +b| a € A, b € B} A = В = = then A. {(x, y) E R² | x = 0, y > 0} is a subset of C B. {(r, y) E R² | x = 0, y > 0} is a subset of C C. {(x, y) E R² | x > 0, y = 0} is a subset of C D. {(r, y) € R² | x > 0, y = 0} is a subset of C
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