Key Concepts and Formulas

• Determinant Differentiation: If the elements of a determinant are functions of $x$, the derivative of the determinant can be found by differentiating each row (or column) separately while keeping the others constant and summing the resulting determinants.
• Determinant Evaluation: The determinant of a 3x3 matrix $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$ is given by $a(ei - fh) - b(di - fg) + c(dh - eg)$.
• Determinant Property: If a row or column of a determinant consists entirely of zeros, the value of the determinant is zero.

Step-by-Step Solution

Step 1: Calculate $f(0)$

We need to find the value of the function $f(x)$ at $x=0$. We substitute $x=0$ into the given determinant:

$f(0)=\left|\begin{array}{ccc} (0)^3 & 2(0)^2+1 & 1+3(0) \\ 3(0)^2+2 & 2(0) & (0)^3+6 \\ (0)^3-(0) & 4 & (0)^2-2 \end{array}\right|$

Simplifying the elements, we get:

$f(0)=\left|\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{array}\right|$

Now, we evaluate this determinant. We can expand along the first column (C1), which contains two zeros:

$f(0) = 0 \cdot C_{11} + (-1)^{2+1}(2) \cdot \left|\begin{array}{cc} 1 & 1 \\ 4 & -2 \end{array}\right| + 0 \cdot C_{31}$ where $C_{ij}$ is the cofactor of the element in the $i$-th row and $j$-th column.

$f(0) = -2 \cdot ((1)(-2) - (1)(4))$ $f(0) = -2 \cdot (-2 - 4)$ $f(0) = -2 \cdot (-6)$ $f(0) = 12$

Step 2: Calculate $f'(x)$

We use the rule for differentiating a determinant. $f'(x)$ is the sum of three determinants, each obtained by differentiating one row of $f(x)$ and leaving the other two rows unchanged.

First, we find the derivatives of each row with respect to $x$:

$R_1(x) = \left[ x^3, \quad 2x^2+1, \quad 1+3x \right]$ $R'_1(x) = \left[ 3x^2, \quad 4x, \quad 3 \right]$

$R_2(x) = \left[ 3x^2+2, \quad 2x, \quad x^3+6 \right]$ $R'_2(x) = \left[ 6x, \quad 2, \quad 3x^2 \right]$

$R_3(x) = \left[ x^3-x, \quad 4, \quad x^2-2 \right]$ $R'_3(x) = \left[ 3x^2-1, \quad 0, \quad 2x \right]$

Now we can write $f'(x)$ as:

$f'(x) = \left|\begin{array}{ccc} 3 x^2 & 4 x & 3 \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right| + \left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 6 x & 2 & 3 x^2 \\ x^3-x & 4 & x^2-2 \end{array}\right| + \left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ 3 x^2-1 & 0 & 2 x \end{array}\right|$

Step 3: Calculate $f'(0)$

We substitute $x=0$ into each of the three determinants that form $f'(x)$.

First Determinant (Differentiating R1):

Substitute $x=0$:

$\left|\begin{array}{ccc} 3(0)^2 & 4(0) & 3 \\ 3(0)^2+2 & 2(0) & (0)^3+6 \\ (0)^3-(0) & 4 & (0)^2-2 \end{array}\right| = \left|\begin{array}{ccc} 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{array}\right|$

Expand along the first row (R1):

$= 0 \cdot C_{11} + 0 \cdot C_{12} + 3 \cdot \left|\begin{array}{cc} 2 & 0 \\ 0 & 4 \end{array}\right|$ $= 3 \cdot ((2)(4) - (0)(0))$ $= 3 \cdot (8 - 0) = 24$

Second Determinant (Differentiating R2):

Substitute $x=0$:

$\left|\begin{array}{ccc} (0)^3 & 2(0)^2+1 & 1+3(0) \\ 6(0) & 2 & 3(0)^2 \\ (0)^3-(0) & 4 & (0)^2-2 \end{array}\right| = \left|\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2 \end{array}\right|$

Since the first column consists entirely of zeros, the value of this determinant is 0.

Third Determinant (Differentiating R3):

Substitute $x=0$:

$\left|\begin{array}{ccc} (0)^3 & 2(0)^2+1 & 1+3(0) \\ 3(0)^2+2 & 2(0) & (0)^3+6 \\ 3(0)^2-1 & 0 & 2(0) \end{array}\right| = \left|\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0 \end{array}\right|$

Expand along the third row (R3):

$= -1 \cdot \left|\begin{array}{cc} 1 & 1 \\ 0 & 6 \end{array}\right| - 0 + 0$ $= -1 \cdot ((1)(6) - (1)(0))$ $= -1 \cdot (6) = -6$

Therefore, $f'(0) = 24 + 0 + (-6) = 18$.

Step 4: Calculate $2f(0) + f'(0)$

We have $f(0) = 12$ and $f'(0) = 18$. Therefore:

$2f(0) + f'(0) = 2(12) + 18 = 24 + 18 = 42$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when expanding determinants, especially when calculating cofactors. A single sign error can lead to an incorrect answer.
• Determinant Properties: Remember that if a row or column of a determinant is all zeros, the determinant is zero. This can save time in calculations.
• Organization: Keep your work organized, especially when differentiating determinants. Writing out the derivatives of each row separately can help avoid errors.

Summary

We first evaluated $f(0)$ by substituting $x=0$ into the original determinant. Next, we found $f'(x)$ by differentiating the determinant using the rule of differentiating each row separately. We then evaluated $f'(0)$ by substituting $x=0$ into each of the three determinants that make up $f'(x)$. Finally, we calculated $2f(0) + f'(0)$, which gave us the answer 42.

Final Answer

The final answer is \boxed{42}, which corresponds to option (C).