Let A = {2, 3, 6, 8, 9, 11} and B = {1, 4, 5, 10, 15}. Let R be a relation on defined by (a, b)R(c, d) if and only if 3ad7bc is an even integer. Then the relation R is

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let A = {2, 3, 6, 8, 9, 11} and B = {1, 4, 5, 10, 15}. Let R be a relation on $A \times B$ defined by (a, b) R(c, d) if and only if 3ad−7bc is an even integer. Then the relation R is [2024]

1 an equivalence relation.

2 reflexive and symmetric but not transitive.

3 reflexive but not symmetric.

4 transitive but not symmetric.

Ans.
(2)

We have, $(a, b) R (c, d) \Rightarrow 3 a d - 7 b c$ is an even integer.

For reflexive : $(a, b) R (a, b) \Rightarrow 3 a b - 7 a b = - 4 a b$ , which is an even integer.

For symmetric, $(a, b) R (c, d) \Rightarrow 3 a d - 7 b c$ is an even integer.

$\Rightarrow 3 a d - 7 b c + 4 b c + 4 a d$ is also an even integer

( $∵$ even + even = even number)

$\Rightarrow 7 a d - 3 b c$ is an even integer

   $\Rightarrow 3 c b - 7 d a$ is also an even integer

   $\Rightarrow (c, d) R (a, b)$

For transitive, $(a, b) R (c, d) a n d (c, d) R (e, f)$

   $\Rightarrow 3 a d - 7 b c$ and $3 c f - 7 d e$ is an even integer.

For $a = 2, b = 5, c = 6, d = 8, e = 9, f = 1$

   $3 a f - 7 b c = 3 \times 2 \times 1 - 7 \times 5 \times 9 = 6 - 315 = - 309,$ which is not an even integer.

$∴$   Given relation is not transitive.

Want Us to Email you About Special Offers & Updates?