The total number of ordered pairs `(x , y)` satisfying `|x|+|y|=2,sin((pix^2)/3)=1,` is equal to 2 (b) 3 (c) 4 (d) 6

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To solve the problem of finding the total number of ordered pairs \((x, y)\) satisfying the equations \(|x| + |y| = 2\) and \(\sin\left(\frac{\pi x^2}{3}\right) = 1\), we will follow these steps:

### Step 1: Analyze the first equation
The first equation is given by:
\[ |x| + |y| = 2 \]
This equation represents a diamond shape in the coordinate plane with vertices at \((2, 0)\), \((0, 2)\), \((-2, 0)\), and \((0, -2)\).

### Step 2: Analyze the second equation
The second equation is:
\[ \sin\left(\frac{\pi x^2}{3}\right) = 1 \]
The sine function equals 1 at angles of the form:
\[ \frac{\pi x^2}{3} = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
This simplifies to:
\[ x^2 = \frac{3}{2} + 6n \]
Thus, we can express \(x\) as:
\[ x = \pm \sqrt{\frac{3}{2} + 6n} \]

### Step 3: Find valid values of \(n\)
We need to find integer values of \(n\) such that \(x\) remains real and satisfies the first equation.

1. **For \(n = 0\)**:
\[ x^2 = \frac{3}{2} \implies x = \pm \sqrt{\frac{3}{2}} \]
Substitute \(x\) into \(|x| + |y| = 2\):
\[ |y| = 2 - \sqrt{\frac{3}{2}} \quad \text{or} \quad |y| = 2 + \sqrt{\frac{3}{2}} \]
The first case gives two values for \(y\) (positive and negative), while the second case is not valid since \(|y|\) cannot exceed 2.

2. **For \(n = 1\)**:
\[ x^2 = \frac{3}{2} + 6 = \frac{15}{2} \implies x = \pm \sqrt{\frac{15}{2}} \]
Substitute \(x\) into \(|x| + |y| = 2\):
\[ |y| = 2 - \sqrt{\frac{15}{2}} \quad \text{or} \quad |y| = 2 + \sqrt{\frac{15}{2}} \]
Again, the first case gives two values for \(y\) (positive and negative), while the second case is not valid.

3. **For \(n = 2\)**:
\[ x^2 = \frac{3}{2} + 12 = \frac{27}{2} \implies x = \pm \sqrt{\frac{27}{2}} \]
Substitute \(x\) into \(|x| + |y| = 2\):
\[ |y| = 2 - \sqrt{\frac{27}{2}} \quad \text{or} \quad |y| = 2 + \sqrt{\frac{27}{2}} \]
Again, the first case gives two values for \(y\) (positive and negative), while the second case is not valid.

### Step 4: Count the valid pairs
From the calculations:
- For \(n = 0\): 2 valid pairs \((\sqrt{\frac{3}{2}}, y)\) and \((- \sqrt{\frac{3}{2}}, y)\).
- For \(n = 1\): 2 valid pairs \((\sqrt{\frac{15}{2}}, y)\) and \((- \sqrt{\frac{15}{2}}, y)\).
- For \(n = 2\): 2 valid pairs \((\sqrt{\frac{27}{2}}, y)\) and \((- \sqrt{\frac{27}{2}}, y)\).

### Conclusion
Adding these valid pairs gives us a total of \(2 + 2 + 2 = 6\) valid ordered pairs \((x, y)\).

Thus, the total number of ordered pairs \((x, y)\) satisfying the given equations is:
\[ \boxed{6} \]