Let `f(x) =|(x^3, sinx,cosx),(6,-1,0),(p,p^2,p^3)|` , where p is a constant .Then `d^3/(dx^3) ` [f(x)] at x = 0 is

Options

A. p
B. `p+p^2`
C. `p+p^3`
D. independent of p

Answer

To solve the problem, we need to evaluate the third derivative of the function \( f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} \) at \( x = 0 \).

### Step 1: Calculate the Determinant \( f(x) \)

We will calculate the determinant using the method of cofactor expansion along the first row.

\[ f(x) = x^3 \begin{vmatrix} -1 & 0 \\ p^2 & p^3 \end{vmatrix} - \sin x \begin{vmatrix} 6 & 0 \\ p & p^3 \end{vmatrix} + \cos x \begin{vmatrix} 6 & -1 \\ p & p^2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -1 & 0 \\ p^2 & p^3 \end{vmatrix} = (-1)(p^3) - (0)(p^2) = -p^3 \)

2. \( \begin{vmatrix} 6 & 0 \\ p & p^3 \end{vmatrix} = (6)(p^3) - (0)(p) = 6p^3 \)

3. \( \begin{vmatrix} 6 & -1 \\ p & p^2 \end{vmatrix} = (6)(p^2) - (-1)(p) = 6p^2 + p \)

Now substituting these back into the determinant:

\[ f(x) = x^3 (-p^3) - \sin x (6p^3) + \cos x (6p^2 + p) \]

This simplifies to:

\[ f(x) = -p^3 x^3 - 6p^3 \sin x + (6p^2 + p) \cos x \]

### Step 2: Differentiate \( f(x) \) Three Times

Now we need to differentiate \( f(x) \) three times with respect to \( x \).

1. **First Derivative:**
\[ f'(x) = -3p^3 x^2 - 6p^3 \cos x + (6p^2 + p)(-\sin x) \]

2. **Second Derivative:**
\[ f''(x) = -6p^3 x - 6p^3 (-\sin x) - (6p^2 + p)(\cos x) \]
Simplifying gives:
\[ f''(x) = -6p^3 x + 6p^3 \sin x - (6p^2 + p) \cos x \]

3. **Third Derivative:**
\[ f'''(x) = -6p^3 + 6p^3 \cos x - (6p^2 + p)(-\sin x) \]
This simplifies to:
\[ f'''(x) = -6p^3 + 6p^3 \cos x + (6p^2 + p) \sin x \]

### Step 3: Evaluate \( f'''(0) \)

Now we substitute \( x = 0 \):

\[ f'''(0) = -6p^3 + 6p^3 \cos(0) + (6p^2 + p) \sin(0) \]

Since \( \cos(0) = 1 \) and \( \sin(0) = 0 \):

\[ f'''(0) = -6p^3 + 6p^3 \cdot 1 + (6p^2 + p) \cdot 0 \]

This simplifies to:

\[ f'''(0) = -6p^3 + 6p^3 = 0 \]

### Final Answer

Thus, the value of \( \frac{d^3}{dx^3} f(x) \) at \( x = 0 \) is:

\[ \boxed{0} \]