Functions Question 366
Question: A relation R is defined in the set Z of integers as follows $ (x,y)\in R $ iff $ x^{2}+y^{2}=9. $ Which of the following is false?
Options:
A) $ R={(0,3),(0,-3),(3,0),(-3,0)} $
B) Domain of $ R={-3,0,3} $
C) Range of $ R={-3,0,3} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ x^{2}+y^{2}=9\Rightarrow y^{2}=9-x^{2} $
$ \Rightarrow y=\pm \sqrt{9-x^{2}} $ $ x=0\Rightarrow y=\pm \sqrt{9-0}=\pm 3\in Z $ $ x=\pm 1\Rightarrow y=\pm \sqrt{9-1}=\pm \sqrt{8}\notin Z $ $ x=\pm 2\Rightarrow y=\pm \sqrt{9-4}=\pm \sqrt{5}\notin Z $ $ x=\pm 3\Rightarrow y=\pm \sqrt{9-9}=0\in Z $ $ x=\pm 4\Rightarrow y=\pm \sqrt{9-16}=\pm \sqrt{-7}\notin Z $ And so on,
$ \therefore R={(0,3),(0,-3),(3,0),(-3,0)} $ Domain of $ R={x:(x,y)\in R}={0,3,-3} $ Range of $ R={y:(x,y)\in R}={3,-3,0}. $
BETA
