JEE Main 2023 — Mathematics Algebra

Given,

Set $A={1,2,3,4}$A=1,2,3,4

And relation on $A\times A$A×A as $2a+3b=4c+5d$2a+3b=4c+5d

Now, Maximum value of $2a+3b=20$2a+3b=20 at $(4,4)$(4,4)

And Minimum value of $4c+5d=9$4c+5d=9 at $(1,1)$(1,1)

So, $4c+5d$4c+5d can be equal to $9,13,14,17,18,19$9,13,14,17,18,19

Now $2a+3b$2a+3b can be $9$9 if $(a,b)=(3,1)\Rightarrow (c,d)=(1,1)$(a,b)=(3,1)⇒(c,d)=(1,1)

Similarly, $2a+3b$2a+3b can be $13$13, if $(a,b)=(2,3)\Rightarrow (c,d)=(2,1)$(a,b)=(2,3)⇒(c,d)=(2,1)

And $2a+3b$2a+3b can be $14$14, if $(a,b)=(4,2)\text{or }(1,4)$(a,b)=(4,2)or (1,4)

$$\Rightarrow (c,d)=(1,2)$$⇒(c,d)=(1,2)

Similarly, $2a+3b$2a+3b can be $17$17, if $(a,b)=(4,3)\Rightarrow$(a,b)=(4,3)⇒ $(c,d)=(3,1)$(c,d)=(3,1)

And $2a+3b$2a+3b can be $18$18 if $(a,b)=(3,4)\Rightarrow$(a,b)=(3,4)⇒

$$(c,d)=(2,2)$$(c,d)=(2,2)

Hence, there are total $6$6 elements which satisfy the given relation.