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Step-by-Step Solution
Step 1: Identify the Sets and the Required Condition
We have two sets:
• A = \{1,2,3,4,\ldots,10\}
• B = \{0,1,2,3,4\} .
The relation R is defined on A \times A such that for an ordered pair (a, b) to be in R , the expression
2(a - b)^2 + 3(a - b) must lie in the set B .
Step 2: Introduce d = a - b
Let d = a - b . Then the condition to be satisfied becomes
2d^2 + 3d \in \{0,1,2,3,4\} .
Step 3: Check Possible Integer Values of d
We test integer values of d to see when 2d^2 + 3d belongs to \{0,1,2,3,4\} .
Case 1: d = 0
2(0)^2 + 3(0) = 0 . Since 0 \in B , all pairs with a - b = 0 qualify, i.e., a = b .
Those pairs are (1,1), (2,2), \dots, (10,10) .
Number of such pairs is 10 .
Case 2: d = 1
2(1)^2 + 3(1) = 2 + 3 = 5 . Since 5 \notin B , no pairs qualify.
Case 3: d = -1
2(-1)^2 + 3(-1) = 2 - 3 = -1 . Since -1 \notin B , no pairs qualify.
Case 4: d = 2
2(2)^2 + 3(2) = 8 + 6 = 14 . Since 14 \notin B , no pairs qualify.
Case 5: d = -2
2(-2)^2 + 3(-2) = 8 - 6 = 2 . Since 2 \in B , all pairs with a - b = -2 qualify.
This implies a = b - 2 . For a to be in A , b - 2 must also be in \{1,2,\ldots,10\} . Hence
b \in \{3,4,5,6,7,8,9,10\} .
The valid pairs are (1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9), (8,10) .
Number of such pairs is 8 .
For |d| > 2 , 2d^2 + 3d exceeds the values in B , so further values do not contribute any valid pairs.
Step 4: Count All Valid Pairs
Summing the valid pairs:
Number from d = 0 is 10 .
Number from d = -2 is 8 .
Therefore, the total number of elements in R is 10 + 8 = 18 .
Final Answer
The number of elements in the relation R is \boxed{18} .
