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

Answer: (C)

42

Answer 2

$f(0) = \begin{vmatrix} 0 & 1 & 1 \\\ 2 & 0 & 6 \\\ 0 & 4 & -2 \end{vmatrix} = 12$ $f'(x) = \begin{vmatrix} 3x^2 & 4x & 3 \\\ x^3 & 2x^2 + 1 & 1 + 3x \\\ x^3 - x & 4 & x^2 - 2 \end{vmatrix} \+ \begin{vmatrix} x^3 & 2x^2 + 1 & 1 + 3x \\\ 6x & 2x & 3x^2 \\\ x^3 - x & 4 & x^2 - 2 \end{vmatrix} \+ \begin{vmatrix} 3x^2 + 2 & 2x & x^3 + 6 \\\ 3x^2 - 1 & 0 & 2x \\\ x^3 - x & 4 & x^2 - 2 \end{vmatrix}$

therefore

$\begin{vmatrix} 0 & 0 & 3 \\\ 2 & 0 & 6 \\\ 0 & 4 & -2 \end{vmatrix} \+ \begin{vmatrix} 0 & 1 & 1 \\\ 0 & 2 & 0 \\\ 0 & 4 & -2 \end{vmatrix} \+ \begin{vmatrix} 0 & 1 & 1 \\\ 2 & 0 & 6 \\\ -1 & 0 & 0 \end{vmatrix}$ $= 24 - 6 = 18$

therefore $2f(0) + f'(0) = 42$