If `x_(1),x_(2) "and" y_(1),y_(2)` are the roots of the equations
`3x^(2) -18x+9=0 "and" y^(2)-4y+2=0` the value of the determinant `|{:(x_(1)x_(2),y_(1)y_(2),1),(x_(1)+x_(2),y_(1)+y_(2),2),(sin(pix_(1)x_(2)),cos (pi//2y_(1)y_(2)),1):}|` is

Options

A. (a)0
B. (b)1
C. (c)2
D. (d)None of these

Answer

To solve the given problem, we will follow these steps:

### Step 1: Find the roots of the equations
1. **For the equation \(3x^2 - 18x + 9 = 0\)**:
- The sum of the roots \(x_1 + x_2 = -\frac{b}{a} = -\frac{-18}{3} = 6\).
- The product of the roots \(x_1 x_2 = \frac{c}{a} = \frac{9}{3} = 3\).

2. **For the equation \(y^2 - 4y + 2 = 0\)**:
- The sum of the roots \(y_1 + y_2 = -\frac{b}{a} = -\frac{-4}{1} = 4\).
- The product of the roots \(y_1 y_2 = \frac{c}{a} = \frac{2}{1} = 2\).

### Step 2: Substitute the values into the determinant
We need to evaluate the determinant:
\[ D = \begin{vmatrix} x_1 x_2 & y_1 y_2 & 1 \\ x_1 + x_2 & y_1 + y_2 & 2 \\ \sin(\pi x_1 x_2) & \cos\left(\frac{\pi}{2} y_1 y_2\right) & 1 \end{vmatrix} \]

Substituting the values we found:
- \(x_1 x_2 = 3\)
- \(y_1 y_2 = 2\)
- \(x_1 + x_2 = 6\)
- \(y_1 + y_2 = 4\)
- \(\sin(\pi x_1 x_2) = \sin(3\pi) = 0\)
- \(\cos\left(\frac{\pi}{2} y_1 y_2\right) = \cos(\pi) = -1\)

So the determinant becomes:
\[ D = \begin{vmatrix} 3 & 2 & 1 \\ 6 & 4 & 2 \\ 0 & -1 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the formula for the determinant of a 3x3 matrix:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where:
- \(a = 3\), \(b = 2\), \(c = 1\)
- \(d = 6\), \(e = 4\), \(f = 2\)
- \(g = 0\), \(h = -1\), \(i = 1\)

Calculating each part:
1. \(ei - fh = 4 \cdot 1 - 2 \cdot (-1) = 4 + 2 = 6\)
2. \(di - fg = 6 \cdot 1 - 2 \cdot 0 = 6 - 0 = 6\)
3. \(dh - eg = 6 \cdot (-1) - 4 \cdot 0 = -6 - 0 = -6\)

Now substituting back:
\[ D = 3 \cdot 6 - 2 \cdot 6 + 1 \cdot (-6) \]
\[ D = 18 - 12 - 6 = 0 \]

### Final Answer
The value of the determinant is \(0\).