Let `"Let"(x)=|{:(,2sinx,sin^(2)x,0),(,1,2sin x,sin^(2)x),(,0,1,2sin x):}|`, then find determinent

Options

A. f(x)is independent of x
B. `f'(pi//2)=0`
C. `underset(x=2)overset(z)intf(x)dx=0`
D. tangent to the curve y=f(x)at x=0, is y=0

Answer

To solve the determinant \( f(x) = \begin{vmatrix} 2\sin x & \sin^2 x & 0 \\ 1 & 2\sin x & \sin^2 x \\ 0 & 1 & 2\sin x \end{vmatrix} \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ f(x) = \begin{vmatrix} 2\sin x & \sin^2 x & 0 \\ 1 & 2\sin x & \sin^2 x \\ 0 & 1 & 2\sin x \end{vmatrix} \]

### Step 2: Expand the Determinant
We will expand the determinant using the first row:
\[ f(x) = 2\sin x \begin{vmatrix} 2\sin x & \sin^2 x \\ 1 & 2\sin x \end{vmatrix} - \sin^2 x \begin{vmatrix} 1 & \sin^2 x \\ 0 & 2\sin x \end{vmatrix} + 0 \]

### Step 3: Calculate the 2x2 Determinants
1. For the first determinant:
\[ \begin{vmatrix} 2\sin x & \sin^2 x \\ 1 & 2\sin x \end{vmatrix} = (2\sin x)(2\sin x) - (\sin^2 x)(1) = 4\sin^2 x - \sin^2 x = 3\sin^2 x \]

2. For the second determinant:
\[ \begin{vmatrix} 1 & \sin^2 x \\ 0 & 2\sin x \end{vmatrix} = (1)(2\sin x) - (0)(\sin^2 x) = 2\sin x \]

### Step 4: Substitute Back into the Determinant
Now substitute these results back into the expression for \( f(x) \):
\[ f(x) = 2\sin x (3\sin^2 x) - \sin^2 x (2\sin x) \]
\[ = 6\sin^3 x - 2\sin^3 x = 4\sin^3 x \]

### Step 5: Final Expression
Thus, the determinant is:
\[ f(x) = 4\sin^3 x \]

### Step 6: Derivative of the Determinant
To find \( f'(x) \), we differentiate \( f(x) \):
\[ f'(x) = 4 \cdot 3\sin^2 x \cdot \cos x = 12\sin^2 x \cos x \]

### Step 7: Evaluate the Derivative at \( x = \frac{\pi}{2} \)
Now, we evaluate \( f'(\frac{\pi}{2}) \):
\[ f'(\frac{\pi}{2}) = 12\sin^2(\frac{\pi}{2}) \cos(\frac{\pi}{2}) = 12 \cdot 1^2 \cdot 0 = 0 \]

### Conclusion
Thus, we find that \( f'(\frac{\pi}{2}) = 0 \).