`" if " f(0) = |{:(sin 0 ,,cos 0,,sin 0),(cos0,,sin0,,cos0),(cos0,, sin 0,,sin 0):}|` then

Options

A. f(0)= 0 has exactly 2 real solutions in `[0 ,pi]`
B. f(0) =0 has exactly 3 real solutions in `[0 ,pi]`
C. range of function `(f(0))/(1-sin 2 0) " is " [-sqrt(2) , sqrt(2)]`
D. range of fucntion .`(f(0))/(sin 2 0-1) " is " [-3,3]" is " [-3,3]`

Explanation

To solve the problem, we need to evaluate the determinant given by:

\[ f(0) = \begin{vmatrix} \sin 0 & \cos 0 & \sin 0 \\ \cos 0 & \sin 0 & \cos 0 \\ \cos 0 & \sin 0 & \sin 0 \end{vmatrix} \]

### Step 1: Substitute the values of \(\sin 0\) and \(\cos 0\)

We know that:
- \(\sin 0 = 0\)
- \(\cos 0 = 1\)

Substituting these values into the determinant, we get:

\[ f(0) = \begin{vmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{vmatrix} \]

### Step 2: Calculate the determinant

To calculate the determinant, we can use the formula for a \(3 \times 3\) determinant:

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Applying this to our determinant:

\[ f(0) = 0 \cdot (0 \cdot 0 - 1 \cdot 1) - 1 \cdot (1 \cdot 0 - 1 \cdot 1) + 0 \cdot (1 \cdot 0 - 0 \cdot 1) \]

This simplifies to:

\[ f(0) = 0 - 1 \cdot (0 - 1) + 0 \]

\[ f(0) = 0 + 1 + 0 = 1 \]

### Step 3: Conclusion

Thus, we find that:

\[ f(0) = 1 \]