Let Z be the set of all integers and
`A={(a,b):a^(2)+3b^(2)=28 ,a,b in Z}`
`and B ={(a,b):a gt b in Z )`
then the number of elements is `A cap B ,` is

Options

A. 2
B. 4
C. 6
D. 5

Answer

To solve the problem, we need to find the number of elements in the intersection of sets A and B, where:

- Set A is defined by the equation \( a^2 + 3b^2 = 28 \) with \( a, b \in \mathbb{Z} \) (the set of all integers).
- Set B is defined as \( B = \{(a,b) : a > b, a, b \in \mathbb{Z}\} \).

### Step-by-Step Solution:

**Step 1: Determine the values of \( a \) and \( b \) that satisfy the equation for set A.**

We start with the equation:
\[ a^2 + 3b^2 = 28 \]

We will test integer values for \( b \) and solve for \( a \).

- **For \( b = 4 \):**
\[ a^2 + 3(4^2) = a^2 + 48 = 28 \implies a^2 = 28 - 48 = -20 \quad (\text{not possible}) \]

- **For \( b = 3 \):**
\[ a^2 + 3(3^2) = a^2 + 27 = 28 \implies a^2 = 28 - 27 = 1 \implies a = \pm 1 \]

- **For \( b = 2 \):**
\[ a^2 + 3(2^2) = a^2 + 12 = 28 \implies a^2 = 28 - 12 = 16 \implies a = \pm 4 \]

- **For \( b = 1 \):**
\[ a^2 + 3(1^2) = a^2 + 3 = 28 \implies a^2 = 28 - 3 = 25 \implies a = \pm 5 \]

- **For \( b = 0 \):**
\[ a^2 + 3(0^2) = a^2 = 28 \implies a = \pm \sqrt{28} \quad (\text{not an integer}) \]

- **For \( b = -1 \):**
\[ a^2 + 3(-1^2) = a^2 + 3 = 28 \implies a^2 = 25 \implies a = \pm 5 \]

- **For \( b = -2 \):**
\[ a^2 + 3(-2^2) = a^2 + 12 = 28 \implies a^2 = 16 \implies a = \pm 4 \]

- **For \( b = -3 \):**
\[ a^2 + 3(-3^2) = a^2 + 27 = 28 \implies a^2 = 1 \implies a = \pm 1 \]

- **For \( b = -4 \):**
\[ a^2 + 3(-4^2) = a^2 + 48 = 28 \implies a^2 = -20 \quad (\text{not possible}) \]

**Step 2: Compile the valid pairs \((a, b)\) from the calculations.**

From the above calculations, we have the following valid pairs:

- For \( b = 3 \): \( (1, 3), (-1, 3) \)
- For \( b = 2 \): \( (4, 2), (-4, 2) \)
- For \( b = 1 \): \( (5, 1), (-5, 1) \)
- For \( b = -1 \): \( (5, -1), (-5, -1) \)
- For \( b = -2 \): \( (4, -2), (-4, -2) \)
- For \( b = -3 \): \( (1, -3), (-1, -3) \)

Thus, the complete set A is:
\[ A = \{(1, 3), (-1, 3), (4, 2), (-4, 2), (5, 1), (-5, 1), (5, -1), (-5, -1), (4, -2), (-4, -2), (1, -3), (-1, -3)\} \]

**Step 3: Identify the elements in set B.**

Set B consists of pairs where \( a > b \):
- Valid pairs from set A where \( a > b \):
- \( (1, 3) \) - No
- \( (-1, 3) \) - No
- \( (4, 2) \) - Yes
- \( (-4, 2) \) - No
- \( (5, 1) \) - Yes
- \( (-5, 1) \) - No
- \( (5, -1) \) - Yes
- \( (-5, -1) \) - No
- \( (4, -2) \) - Yes
- \( (-4, -2) \) - No
- \( (1, -3) \) - Yes
- \( (-1, -3) \) - No

Thus, the valid pairs in set B are:
\[ B = \{(4, 2), (5, 1), (5, -1), (4, -2), (1, -3)\} \]

**Step 4: Find the intersection \( A \cap B \).**

Since set B is derived from set A, the intersection \( A \cap B \) will simply be the elements of set B:
\[ A \cap B = B \]

**Step 5: Count the elements in \( A \cap B \).**

The number of elements in \( A \cap B \) is:
\[ \text{Number of elements} = 5 \]

### Final Answer:
The number of elements in \( A \cap B \) is **5**.