Functions Question 355
Question: A relation R is defined over the set of nonnegative integers as $ xRy\Rightarrow x^{2}+y^{2}=36 $ what is R?
Options:
A) $ {(0,6)} $
B) $ {(6,0)(\sqrt{11},5),(3,3,\sqrt{3}) $
C) $ {(6,0)(0,6)} $
D) $ (\sqrt{11},5),(2,4\sqrt{2}),(5\sqrt{11}),(4\sqrt{2}2)} $
Show Answer
Answer:
Correct Answer: C
Solution:
[b] R is defined over the set of non-negative integers, $ x^{2}+y^{2}=36 $
$ \Rightarrow y=\sqrt{36-x^{2}}=\sqrt{(6-x)(6+x)},x=0 $ or 6 for x = 0, y = 6 and for x = 6, y = 0 So, y is 6 or 0 so, $ R={(6,0),(0,6)} $
BETA
