Key Concepts and Formulas

• Set Definition and Cartesian Product: Understanding how sets are defined and what a Cartesian product ($A \times A$) represents (set of all ordered pairs).
• Relation Definition: A relation $R$ from set $A$ to set $A$ is a subset of $A \times A$. The elements of $R$ are ordered pairs $(a,b)$ that satisfy a given condition.
• Quadratic Equations: Solving quadratic equations of the form $ax^2 + bx + c = 0$ using factorization or the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Integer Properties: Recognizing that if $a$ and $b$ are integers, then $a-b$ is also an integer.

Step-by-Step Solution

Step 1: Understand the Problem and Define the Relation We are given two sets, $A = \{1, 2, 3, \ldots, 10\}$ and $B = \{0, 1, 2, 3, 4\}$. The relation $R$ is defined as $R = \left\{(a, b) \in A \times A : 2(a-b)^2 + 3(a-b) \in B\right\}$. We need to find the number of ordered pairs $(a,b)$ such that $a \in A$, $b \in A$, and the expression $2(a-b)^2 + 3(a-b)$ evaluates to a value in set $B$.

Step 2: Simplify the Condition using Substitution Let $x = a-b$. Since $a$ and $b$ are integers from set $A$, their difference $x$ must also be an integer. The range of possible values for $x$ is determined by the minimum and maximum differences between elements of $A$:

• Minimum value of $x$: $1 - 10 = -9$.
• Maximum value of $x$: $10 - 1 = 9$. So, $x$ is an integer such that $-9 \le x \le 9$. The condition for $(a,b) \in R$ transforms into: $2x^2 + 3x \in B$.

Step 3: Find Integer Values of $x$ Satisfying the Condition We need to find integer values of $x$ in the range $[-9, 9]$ such that $2x^2 + 3x$ is an element of $B = \{0, 1, 2, 3, 4\}$. We will consider each element of $B$ separately.

• Case 3.1: $2x^2 + 3x = 0$ Factoring gives $x(2x+3) = 0$. The solutions are $x=0$ and $x=-3/2$. Since $x$ must be an integer, $x=0$ is the only valid solution. $0$ is in the range $[-9, 9]$.

• Case 3.2: $2x^2 + 3x = 1$ Rearranging gives $2x^2 + 3x - 1 = 0$. Using the quadratic formula, $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{17}}{4}$. These are not integers, so there are no solutions in this case.

• Case 3.3: $2x^2 + 3x = 2$ Rearranging gives $2x^2 + 3x - 2 = 0$. Factoring gives $(2x-1)(x+2) = 0$. The solutions are $x=1/2$ and $x=-2$. Since $x$ must be an integer, $x=-2$ is the only valid solution. $-2$ is in the range $[-9, 9]$.

• Case 3.4: $2x^2 + 3x = 3$ Rearranging gives $2x^2 + 3x - 3 = 0$. Using the quadratic formula, $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-3)}}{2(2)} = \frac{-3 \pm \sqrt{33}}{4}$. These are not integers, so there are no solutions in this case.

• Case 3.5: $2x^2 + 3x = 4$ Rearranging gives $2x^2 + 3x - 4 = 0$. Using the quadratic formula, $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-4)}}{2(2)} = \frac{-3 \pm \sqrt{41}}{4}$. These are not integers, so there are no solutions in this case.

The only integer values for $x=a-b$ that satisfy the condition are $x=0$ and $x=-2$.

Step 4: Find Pairs $(a,b)$ for Each Valid $x$ Value Now we find the ordered pairs $(a,b)$ from $A \times A$ that correspond to $x=0$ and $x=-2$.

• Case 4.1: $x = 0$ If $a-b = 0$, then $a=b$. We need pairs $(a,b)$ where $a, b \in \{1, 2, \ldots, 10\}$ and $a=b$. The pairs are: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8), (9,9), (10,10)$. There are 10 such pairs.

• Case 4.2: $x = -2$ If $a-b = -2$, then $b = a+2$. We need pairs $(a,b)$ where $a, b \in \{1, 2, \ldots, 10\}$ and $b=a+2$. We list values of $a$ from $1$ to $10$ and check if $b=a+2$ is in $A$:

  • If $a=1$, $b=3$. $(1,3)$ is valid.
  • If $a=2$, $b=4$. $(2,4)$ is valid.
  • If $a=3$, $b=5$. $(3,5)$ is valid.
  • If $a=4$, $b=6$. $(4,6)$ is valid.
  • If $a=5$, $b=7$. $(5,7)$ is valid.
  • If $a=6$, $b=8$. $(6,8)$ is valid.
  • If $a=7$, $b=9$. $(7,9)$ is valid.
  • If $a=8$, $b=10$. $(8,10)$ is valid.
  • If $a=9$, $b=11$. $11 \notin A$. Not valid.
  • If $a=10$, $b=12$. $12 \notin A$. Not valid. There are 8 such pairs.

Step 5: Calculate the Total Number of Elements in R The total number of elements in relation $R$ is the sum of the counts from each valid $x$ value. Total elements in $R = (\text{Number of pairs for } x=0) + (\text{Number of pairs for } x=-2)$ Total elements in $R = 10 + 8 = 18$.

Common Mistakes & Tips

• Integer Solutions: Always ensure that solutions for $x$ are integers, as $a$ and $b$ are integers. Non-integer solutions for $x$ should be discarded.
• Range of $x$: Calculate the bounds for $x=a-b$ to quickly eliminate values of $x$ that are outside the possible range.
• Membership in A: When constructing pairs $(a,b)$, carefully check that both $a$ and $b$ belong to the set $A$. This is a common source of error.

Summary We simplified the given condition $2(a-b)^2 + 3(a-b) \in B$ by substituting $x = a-b$. We found the range of possible integer values for $x$ to be $[-9, 9]$. By solving $2x^2 + 3x = k$ for each $k \in B$, we identified the valid integer solutions for $x$ as $0$ and $-2$. For $x=0$, we found 10 pairs where $a=b$. For $x=-2$, we found 8 pairs where $b=a+2$. The total number of elements in the relation $R$ is the sum of these pairs, which is $10 + 8 = 18$.

The final answer is $\boxed{18}$.