If `f(x)=|{:(x^(n),sinx,cosx),(n!,"sin"(npi)/(2),"cos"(npi)/(2)),(a,a^(2),a^(3)):}|`, then the value of `(d^(n))/(dx^(n))(f(x))" at "x=0" for "n=2m+1` is

Options

A. -1
B. 0
C. 1
D. independent of a

Explanation

To solve the problem, we need to find the value of the \( n \)-th derivative of the function \( f(x) \) at \( x = 0 \), where

\[ f(x) = \begin{vmatrix} x^n & \sin x & \cos x \\ n! & \sin(n\pi/2) & \cos(n\pi/2) \\ a & a^2 & a^3 \end{vmatrix} \]

for \( n = 2m + 1 \).

### Step 1: Write the determinant
The function \( f(x) \) can be expressed as a determinant of a 3x3 matrix. We will denote the rows of the determinant as \( R_1, R_2, R_3 \):

\[ R_1 = (x^n, \sin x, \cos x), \quad R_2 = (n!, \sin(n\pi/2), \cos(n\pi/2)), \quad R_3 = (a, a^2, a^3) \]

### Step 2: Differentiate the determinant
To find the \( n \)-th derivative of \( f(x) \), we will use the property of determinants. The derivative of a determinant can be computed by differentiating one row while keeping the others constant.

Using the Leibniz rule for determinants:

\[ \frac{d}{dx} \begin{vmatrix} R_1 \\ R_2 \\ R_3 \end{vmatrix} = \begin{vmatrix} \frac{dR_1}{dx} \\ R_2 \\ R_3 \end{vmatrix} + \begin{vmatrix} R_1 \\ \frac{dR_2}{dx} \\ R_3 \end{vmatrix} + \begin{vmatrix} R_1 \\ R_2 \\ \frac{dR_3}{dx} \end{vmatrix} \]

However, since \( R_2 \) and \( R_3 \) are constants with respect to \( x \), their derivatives will be zero. Thus, we only need to differentiate \( R_1 \).

### Step 3: Compute the first derivative
The first derivative of \( R_1 \) is:

\[ \frac{dR_1}{dx} = \left(n x^{n-1}, \cos x, -\sin x\right) \]

Thus, the first derivative of the determinant becomes:

\[ f'(x) = \begin{vmatrix} n x^{n-1} & \cos x & -\sin x \\ n! & \sin(n\pi/2) & \cos(n\pi/2) \\ a & a^2 & a^3 \end{vmatrix} \]

### Step 4: Compute the \( n \)-th derivative at \( x = 0 \)
We need to find the \( n \)-th derivative of \( f(x) \) at \( x = 0 \). Since \( n = 2m + 1 \) is odd, we can analyze the behavior of \( \sin(n\pi/2) \) and \( \cos(n\pi/2) \):

- For odd \( n \), \( \sin(n\pi/2) = (-1)^m \) and \( \cos(n\pi/2) = 0 \).

This means that the second row of the determinant simplifies to:

\[ (n!, (-1)^m, 0) \]

### Step 5: Evaluate the determinant at \( x = 0 \)
At \( x = 0 \):

- \( R_1 = (0, 0, 1) \)
- The determinant becomes:

\[ f(0) = \begin{vmatrix} 0 & 0 & 1 \\ n! & (-1)^m & 0 \\ a & a^2 & a^3 \end{vmatrix} \]

Calculating this determinant, we find that the first column is all zeros, which means the determinant evaluates to 0.

### Conclusion
Thus, the value of \( \frac{d^n}{dx^n} f(x) \) at \( x = 0 \) for \( n = 2m + 1 \) is:

\[ \boxed{0} \]