JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Sets Relations and Functions

Let

A = {1, 2, 3, 4, 5}

. Let

R

be a relation on

A

defined by

x R y

if and only

4 x \leq 5 y

. Let

m

be the number of elements in

R

and

n

be the minimum number of elements from

A \times A

that are required to be added to

R

to make it a symmetric relation. Then

m + n

is equal to:

Answer: (B)

Solution

R = {(x, y) : 4 x \leq 5 y, x, y \in {1, \dots, 5}}

. Counting by

x

x = 1

gives all 5 values of

y

;

x = 2

gives

y \geq 2

(4);

x = 3

gives

y \geq 3

(3);

x = 4

gives

y \geq 4

(2);

x = 5

gives

y \geq 4

(2). Total

m = 16

. All 5 diagonal pairs lie in

R

. Among the 11 off-diagonal pairs,

(4, 5)

and

(5, 4)

are already mutual, so only the remaining 9 need their reverse added for symmetry:

n = 9

. Hence

m + n = 25

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option