JEE Main On 16 April 2018 Question 26
Question: Let N denote the set of all natural numbers. Define two binary relations on N as $ R _1={(x,y)\in N\times N:2x+y=10} $ and $ R _2={(x,y)\in N\times N:x+2y=10}. $ Then. [JEE Main 16-4-2018]
Options:
A) Both $ R _1 $ and $ R _2 $ are transitive relations
B) Both $ R _1 $ and $ R _2 $ are symmetric relations
C) Range of $ R _2 $ is $ {1,2,3,4} $
D) Range of $ R _1 $ is $ {2,4,8} $
Show Answer
Answer:
Correct Answer: C
Solution:
Define two binary relations on N as $ R _1={(x,y)\in N\times N:2x+y=10} $ and $ R _2={(x,y)\in N\times N:x+2y=10} $
From, $ R _1,2x+y=10 $ and $ x,y\in N $ So, possible values for $ x $ and $ y $ are: $ x=1,y=8 $ i.e (1, 8) $ x=2,y=6 $
i.e $ (2,6) $ $ x=3,y=4 $ i.e (3, 4) $ x=4,y=2 $ i.e $ (4,2) $ $ R _1={(1,8),(2,6),(3,4),(4,2)} $
Therefore, Range of $ R _1 $ is $ {2,4,6,8} $ $ R _1 $ is not symmetric Also, $ R _1 $ is not transitive because
$ (3,4),(4,2)\in R _1 $ but $ (3,2)\cancel{\in }R _1 $ Thus, options A, B and D are incorrect. From $ R _2,x+2y=10 $ , and $ x,y\in N $
So, possible values for $ x $ and $ y $ are: $ x=8,y=1i.e,(8,1) $ $ x=6,y=2i.e,(6,2) $ $ x=4,y=3i.e,(4,3) $ $ x=2,y=4i.e,(2,4) $ $ R _2={(8,1),(6,2),(4,3),(2,4)} $
Therefore, Range of $ R _2 $ is $ {1,2,3,4} $ $ R _2 $ is not symmetric Hence, option C is correct.
