If `f(x)=|{:( sin^2 theta , cos^(2) theta , x),(cos^(2) theta , x , sin^(2) theta ),( x , sin^(2) theta , cos^2 theta ):}| theta in (0,pi//2),` then roots of f(x)=0 are

Options

A. 1/2,-1
B. 1/2,-1,0
C. `-1//2,1,0`
D. `-1//2,-1,0`

Answer

To solve the problem, we need to find the roots of the function \( f(x) \) defined by the determinant:

\[ f(x) = \left| \begin{array}{ccc} \sin^2 \theta & \cos^2 \theta & x \\ \cos^2 \theta & x & \sin^2 \theta \\ x & \sin^2 \theta & \cos^2 \theta \end{array} \right| \]

where \( \theta \) is in the interval \( (0, \frac{\pi}{2}) \).

### Step 1: Calculate the Determinant

We will first compute the determinant \( f(x) \).

Using the determinant properties, we can perform row operations. We will add all rows together:

\[ R_1 = R_1 + R_2 + R_3 \]

This gives us:

\[ f(x) = \left| \begin{array}{ccc} x + x + 1 & x + 1 & x + 1 \\ \cos^2 \theta & x & \sin^2 \theta \\ x & \sin^2 \theta & \cos^2 \theta \end{array} \right| \]

This simplifies to:

\[ f(x) = \left| \begin{array}{ccc} 2x + 1 & x + 1 & x + 1 \\ \cos^2 \theta & x & \sin^2 \theta \\ x & \sin^2 \theta & \cos^2 \theta \end{array} \right| \]

### Step 2: Perform Column Operations

Next, we perform column operations to simplify the determinant further. We can subtract the second column from the first and the third column from the second:

\[ C_1 = C_1 - C_2, \quad C_3 = C_3 - C_2 \]

This gives us:

\[ f(x) = \left| \begin{array}{ccc} (2x + 1) - (x + 1) & x + 1 & (x + 1) - (x + 1) \\ \cos^2 \theta - x & x & \sin^2 \theta - (x) \\ x - \sin^2 \theta & \sin^2 \theta & \cos^2 \theta - (x) \end{array} \right| \]

This simplifies to:

\[ f(x) = \left| \begin{array}{ccc} x & x + 1 & 0 \\ \cos^2 \theta - x & x & \sin^2 \theta - x \\ x - \sin^2 \theta & \sin^2 \theta & \cos^2 \theta - x \end{array} \right| \]

### Step 3: Expand the Determinant

Now we can expand the determinant. The first column has a factor of \( x \), so we can factor it out:

\[ f(x) = x \cdot \left| \begin{array}{cc} x + 1 & 0 \\ \sin^2 \theta & \cos^2 \theta - x \end{array} \right| \]

Calculating this determinant gives:

\[ f(x) = x \cdot (x + 1)(\cos^2 \theta - x) - 0 \]

### Step 4: Set the Function to Zero

Now we set \( f(x) = 0 \):

\[ x \cdot (x + 1)(\cos^2 \theta - x) = 0 \]

This gives us the roots:

1. \( x = 0 \)
2. \( x + 1 = 0 \) → \( x = -1 \)
3. \( \cos^2 \theta - x = 0 \) → \( x = \cos^2 \theta \)

### Step 5: Analyze the Roots

Since \( \theta \) is in \( (0, \frac{\pi}{2}) \), \( \cos^2 \theta \) will vary between 0 and 1. Therefore, the roots are:

- \( x = 0 \)
- \( x = -1 \)
- \( x = \cos^2 \theta \) (which can take any value between 0 and 1)

### Conclusion

Thus, the roots of \( f(x) = 0 \) are:

\[ x = 0, \quad x = -1, \quad \text{and } x = \cos^2 \theta \text{ (for } \theta \in (0, \frac{\pi}{2})\text{)} \]