Let A=1,2,3,4, .,10 and B=0,1,2,3,4.

Question

Let $A={1,2,3,4,\ldots .,10}$ and $B={0,1,2,3,4}$. The number of elements in the relation $R={(a,b)\in A	imes A:2(a-b{)}^{2}+3(a-b)\in B}$ is \_\_\_\_\_\_\_\_\_\_\_.

Answer: 18

Accepted Answer

Given sets :

A={1,2,3,4, ............,10}

B={0,1,2,3,4}

We are looking for pairs $(a,b)\in A	imes A$ such that :

$2(a-b{)}^{2}+3(a-b)\in B$

Let's break down the relation :

Case 1 : $a-b=0$

$2(a-b{)}^{2}+3(a-b)=0$

Pairs : $(1,1),(2,2),(3,3),\ldots ,(10,10)$ which gives 10 pairs.

Case 2 : $a-b=1$

$2(a-b{)}^{2}+3(a-b)=2(1)+3(1)=5$

But 5 is not in B, so no pairs for this case.

Case 3 : $a-b=-1$

$2(a-b{)}^{2}+3(a-b)=2(1)-3(1)=-1$

This value is not in B, so no pairs for this case.

Case 4 : $a-b=2$

$2(a-b{)}^{2}+3(a-b)=2(4)+3(2)=8+6=14$

Again, 14 is not in B, so no pairs for this case.

Case 5 : $a-b=-2$

$2(a-b{)}^{2}+3(a-b)=2(4)-3(2)=8-6=2$

Pairs : $(1,3),(2,4),(3,5),(4,6),(5,7),(6,8),(7,9),(8,10)$ which gives 8 pairs.

For any other $a-b$ value, the quadratic will grow larger than the maximum value in B, so we don't need to consider them.

In total, we have $10+8=18$ pairs in the relation $R$.

Therefore, the number of elements in the relation $R$ is 18.

Solutions · 1