Let `f(x)=|(4x+1,-cosx,-sinx),(6,8sinalpha,0),(12sinalpha, 16sin^(2)alpha,1+4sinalpha)|` and `f(0)=0`. If the sum of all possible values of `alpha` is `kpi` for `alpha in [0, 2pi]`, then the value of k is equal to

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem step by step, we start with the function given:

\[ f(x) = \begin{vmatrix} 4x + 1 & -\cos x & -\sin x \\ 6 & 8 \sin \alpha & 0 \\ 12 \sin \alpha & 16 \sin^2 \alpha & 1 + 4 \sin \alpha \end{vmatrix} \]

We know that \( f(0) = 0 \). Let's evaluate \( f(0) \):

1. **Substituting \( x = 0 \)**:
\[ f(0) = \begin{vmatrix} 4(0) + 1 & -\cos(0) & -\sin(0) \\ 6 & 8 \sin \alpha & 0 \\ 12 \sin \alpha & 16 \sin^2 \alpha & 1 + 4 \sin \alpha \end{vmatrix} \]
This simplifies to:
\[ f(0) = \begin{vmatrix} 1 & -1 & 0 \\ 6 & 8 \sin \alpha & 0 \\ 12 \sin \alpha & 16 \sin^2 \alpha & 1 + 4 \sin \alpha \end{vmatrix} \]

2. **Calculating the determinant**:
We can expand the determinant using the first row:
\[ f(0) = 1 \cdot \begin{vmatrix} 8 \sin \alpha & 0 \\ 16 \sin^2 \alpha & 1 + 4 \sin \alpha \end{vmatrix} - (-1) \cdot \begin{vmatrix} 6 & 0 \\ 12 \sin \alpha & 1 + 4 \sin \alpha \end{vmatrix} \]
The first determinant is:
\[ \begin{vmatrix} 8 \sin \alpha & 0 \\ 16 \sin^2 \alpha & 1 + 4 \sin \alpha \end{vmatrix} = 8 \sin \alpha (1 + 4 \sin \alpha) - 0 = 8 \sin \alpha (1 + 4 \sin \alpha) \]
The second determinant is:
\[ \begin{vmatrix} 6 & 0 \\ 12 \sin \alpha & 1 + 4 \sin \alpha \end{vmatrix} = 6(1 + 4 \sin \alpha) - 0 = 6(1 + 4 \sin \alpha) \]

3. **Putting it all together**:
Thus, we have:
\[ f(0) = 8 \sin \alpha (1 + 4 \sin \alpha) + 6(1 + 4 \sin \alpha) \]
Factor out \( (1 + 4 \sin \alpha) \):
\[ f(0) = (1 + 4 \sin \alpha)(8 \sin \alpha + 6) \]
Setting \( f(0) = 0 \):
\[ (1 + 4 \sin \alpha)(8 \sin \alpha + 6) = 0 \]

4. **Finding solutions**:
This gives us two equations:
- \( 1 + 4 \sin \alpha = 0 \) which leads to \( \sin \alpha = -\frac{1}{4} \)
- \( 8 \sin \alpha + 6 = 0 \) which leads to \( \sin \alpha = -\frac{3}{4} \)

5. **Finding values of \( \alpha \)**:
For \( \sin \alpha = -\frac{1}{4} \):
- Possible values in the interval \( [0, 2\pi] \) are:
\[ \alpha = \arcsin\left(-\frac{1}{4}\right) + 2\pi \quad \text{and} \quad \alpha = \pi - \arcsin\left(-\frac{1}{4}\right) + \pi \]

For \( \sin \alpha = -\frac{3}{4} \):
- Possible values in the interval \( [0, 2\pi] \) are:
\[ \alpha = \arcsin\left(-\frac{3}{4}\right) + 2\pi \quad \text{and} \quad \alpha = \pi - \arcsin\left(-\frac{3}{4}\right) + \pi \]

6. **Calculating the sum of all possible values of \( \alpha \)**:
The sum of all possible values of \( \alpha \) can be calculated as:
\[ \text{Sum} = \left(\pi + \arcsin\left(-\frac{1}{4}\right) + \pi + 2\pi - \arcsin\left(-\frac{1}{4}\right)\right) + \left(\pi + \arcsin\left(-\frac{3}{4}\right) + \pi + 2\pi - \arcsin\left(-\frac{3}{4}\right)\right) \]
Simplifying gives:
\[ = 6\pi \]

7. **Final result**:
Since the sum of all possible values of \( \alpha \) is \( k\pi \), we have \( k = 6 \).