Consider the function
`f(x)=|{:(x^(3),sinx,cosx),(6,-1,0),(p,p^2,p^(3)):}|` where p is a constant.
What is the value of p for which `f''(0)=0` ?

Options

A. A. `-1/6" or "0`
B. B. `-1` or 0
C. C. `-1/6` or 0
D. D. `-1` or 0

Answer

To solve the problem, we need to find the value of \( p \) such that \( f''(0) = 0 \) for the function defined as:

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

### Step 1: Differentiate \( f(x) \) to find \( f'(x) \)

Using the property of determinants, we differentiate the first row while keeping the other rows constant. The derivative of \( x^3 \) is \( 3x^2 \), the derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \( -\sin x \). Thus, we have:

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

### Step 2: Differentiate \( f'(x) \) to find \( f''(x) \)

Now we differentiate \( f'(x) \) again. The first row will be differentiated again:

- The derivative of \( 3x^2 \) is \( 6x \)
- The derivative of \( \cos x \) is \( -\sin x \)
- The derivative of \( -\sin x \) is \( -\cos x \)

Thus, we have:

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

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

Substituting \( x = 0 \):

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

This determinant can be simplified by expanding along the first row:

\[ f''(0) = 0 \cdot \begin{vmatrix} -1 & 0 \\ p^2 & p^3 \end{vmatrix} - 0 \cdot \begin{vmatrix} 6 & 0 \\ p & p^3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 6 & -1 \\ p & p^2 \end{vmatrix} \]

Calculating the determinant:

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

Thus, we have:

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

### Step 4: Set \( f''(0) = 0 \)

Setting the expression equal to zero gives:

\[ - (6p^2 + p) = 0 \]

This simplifies to:

\[ 6p^2 + p = 0 \]

### Step 5: Factor the equation

Factoring out \( p \):

\[ p(6p + 1) = 0 \]

### Step 6: Solve for \( p \)

Setting each factor to zero gives:

1. \( p = 0 \)
2. \( 6p + 1 = 0 \) which leads to \( p = -\frac{1}{6} \)

### Conclusion

The values of \( p \) for which \( f''(0) = 0 \) are:

\[ p = 0 \quad \text{or} \quad p = -\frac{1}{6} \]