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 evaluate the \( n \)-th derivative of the function \( f(x) \) at \( x = 0 \) where \( n = 2m + 1 \). The function \( f(x) \) is defined as a determinant involving \( x^n \), \( \sin x \), \( \cos x \), and some constants.

### Step-by-step Solution:

1. **Understanding the Function**:
The function is given as:
\[ f(x) = \begin{vmatrix} x^n & \sin x & \cos x \\ n! & \sin\left(\frac{n\pi}{2}\right) & \cos\left(\frac{n\pi}{2}\right) \\ a & a^2 & a^3 \end{vmatrix} \]

2. **Differentiating the Determinant**:
To find \( \frac{d^n}{dx^n} f(x) \) at \( x = 0 \), we can use the property of determinants. We differentiate the first row with respect to \( x \):
- The derivative of \( x^n \) is \( n x^{n-1} \).
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( \cos x \) is \( -\sin x \).

Thus, the first derivative of the determinant becomes:
\[ f'(x) = \begin{vmatrix} n x^{n-1} & \cos x & -\sin x \\ n! & \sin\left(\frac{n\pi}{2}\right) & \cos\left(\frac{n\pi}{2}\right) \\ a & a^2 & a^3 \end{vmatrix} \]

3. **Evaluating at \( x = 0 \)**:
Now, we substitute \( x = 0 \):
- \( \sin(0) = 0 \)
- \( \cos(0) = 1 \)
- The determinant simplifies to:
\[ f(0) = \begin{vmatrix} 0 & 0 & 1 \\ n! & \sin\left(\frac{n\pi}{2}\right) & \cos\left(\frac{n\pi}{2}\right) \\ a & a^2 & a^3 \end{vmatrix} \]

Since the first row has a zero, the determinant evaluates to zero.

4. **Higher Derivatives**:
Continuing this process, we find that each time we differentiate, if we have a row of zeros, the determinant remains zero. Thus, for any odd \( n \) (where \( n = 2m + 1 \)), the \( n \)-th derivative evaluated at \( x = 0 \) will also yield zero.

5. **Conclusion**:
Therefore, the value of \( \frac{d^n}{dx^n} f(x) \) at \( x = 0 \) for \( n = 2m + 1 \) is:
\[ \boxed{0} \]