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1.Relation and Function
hard
Let $A =\{1,2,3,4, \ldots .10\}$ and $B =\{0,1,2,3,4\}$ The number of elements in the relation $R =\{( a , b )$ $\left.\in A \times A : 2( a - b )^2+3( a - b ) \in B \right\}$ is $.........$.
A
$12$
B
$14$
C
$16$
D
$18$
(JEE MAIN-2023)
Solution
$A=\{1,2,3, \ldots \ldots 10\}$
$B=\{0,1,2,3,4\}$
$R=\left\{(a, b) \in A \times A: 2(a-b)^2+3(a-b) \in B\right\}$
$\text { Now } 2(a-b)^2+3(a-b)=(a-b)(2(a-b)+3)$
$\Rightarrow a=b \text { or } a-b=-2$
Now $2(a-b)^2+3(a-b)=(a-b)(2(a-b)+3)$
$\Rightarrow a = b \text { or } a – b =-2$
When $a = b \Rightarrow 10$ order pairs
When $a-b=-2 \Rightarrow 8$ order pairs
Total $=18$
Standard 12
Mathematics
