Question

If \[ f(x) = \begin{vmatrix} x^3 & 2x^2 + 1 & 1 + 3x \\ 3x^2 + 2 & 2x & x^3 + 6 \\ x^3 - x & 4 & x^2 - 2 \end{vmatrix} \] for all \( x \in \mathbb{R} \), then \( 2f(0) + f'(0) \) is equal to

• A. 48
• B. 24
• C. 42
• D. 18

Answer 1

The Correct Option is C

The problem requires evaluating the function \(f(x)\), defined as the determinant of a 3x3 matrix, and subsequently computing the expression \(2f(0) + f'(0)\).

The given matrix is:

\(x^3\)	\(2x^2 + 1\)	\(1 + 3x\)
\(3x^2 + 2\)	\(2x\)	\(x^3 + 6\)
\(x^3 - x\)	\(4\)	\(x^2 - 2\)

Step 1: Compute \(f(0)\)

Substitute \(x = 0\) into the matrix to find \(f(0)\):

• Row 1: \((0, 1, 1)\)
• Row 2: \((2, 0, 6)\)
• Row 3: \((0, 4, -2)\)

Calculate the determinant of this matrix:

\(\begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix} = 0(0(-2) - 6(4)) - 1(2(-2) - 6 \cdot 0) + 1(2 \cdot 4 - 0 \cdot 0)\)

Simplification yields:

\(0 - (-4) + 8 = 4 + 8 = 12\)

Therefore, \(f(0) = 12\).

Step 2: Compute \(f'(x)\) and \(f'(0)\)

Differentiate \(f(x)\) with respect to \(x\). Using cofactor expansion along the first row:

\(f(x) = x^3 \cdot \begin{vmatrix} 2x & x^3 + 6 \\ 4 & x^2 - 2 \end{vmatrix} - (2x^2 + 1) \cdot \begin{vmatrix} 3x^2 + 2 & x^3 + 6 \\ x^3 - x & x^2 - 2 \end{vmatrix} + (1 + 3x) \cdot \begin{vmatrix} 3x^2 + 2 & 2x \\ x^3 - x & 4 \end{vmatrix}\)

To find \(f'(0)\), we differentiate each term and evaluate at \(x=0\). This involves detailed calculation of derivatives of determinant minors.

The result of these calculations is:

\(f'(0) = 18\)

Step 3: Compute \(2f(0) + f'(0)\)

Substitute the calculated values \(f(0) = 12\) and \(f'(0) = 18\):

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

The final answer is \(42\).