We need to find $2f(0)+f'(0)$ where $f(x)=\begin{vmatrix} x^3 & 2x^2+1 & 1+3x \\ 3x^2+2 & 2x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{vmatrix}.$

A direct full expansion is possible, but a smarter method is to use the fact that for a determinant whose entries depend on $x$, the derivative at a point can be found by differentiating one row at a time.

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1. Compute $f(0)$

Substitute $x=0$ into the matrix: $f(0)=\begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix}.$

Expand along the first row: $f(0)=0\cdot C_{11}+1\cdot C_{12}+1\cdot C_{13}.$

Now, $C_{12}=(-1)^{1+2}\begin{vmatrix}2&6\\0&-2\end{vmatrix}=-\,(2\cdot(-2)-0\cdot6)=4,$ and $C_{13}=(-1)^{1+3}\begin{vmatrix}2&0\\0&4\end{vmatrix}=1\cdot(2\cdot4-0)=8.$

Hence, $f(0)=1\cdot 4+1\cdot 8=12.$

So, $2f(0)=24.$

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2. Compute $f'(0)$ using row-wise differentiation

Let the rows be $R_1=(x^3,\,2x^2+1,\,1+3x),$ $R_2=(3x^2+2,\,2x,\,x^3+6),$ $R_3=(x^3-x,\,4,\,x^2-2).$

Then $f'(x)=\det(R_1',R_2,R_3)+\det(R_1,R_2',R_3)+\det(R_1,R_2,R_3').$

At $x=0$: $R_1(0)=(0,1,1),\quad R_2(0)=(2,0,6),\quad R_3(0)=(0,4,-2).$

Differentiate each row: $R_1'=(3x^2,4x,3)\implies R_1'(0)=(0,0,3),$ $R_2'=(6x,2,3x^2)\implies R_2'(0)=(0,2,0),$ $R_3'=(3x^2-1,0,2x)\implies R_3'(0)=(-1,0,0).$

Thus, $f'(0)=D_1+D_2+D_3,$ where

\quad D_2=\begin{vmatrix}0&1&1\\0&2&0\\0&4&-2\end{vmatrix}, \quad D_3=\begin{vmatrix}0&1&1\\2&0&6\\-1&4&-2\end{vmatrix}.$$ ### Compute $D_1$ Expand along the first row: $$D_1=3\begin{vmatrix}2&0\\0&4\end{vmatrix}=3(8)=24.$$ ### Compute $D_2$ The first column is all zeros, so $$D_2=0.$$ ### Compute $D_3$ Expand along the first row: $$D_3=0\cdot C_{11}+1\cdot C_{12}+1\cdot C_{13}.$$ Now, $$C_{12}=(-1)^{1+2}\begin{vmatrix}2&6\\-1&-2\end{vmatrix}=-\,(2\cdot(-2)-6\cdot(-1))=-(-4+6)=-2,$$ $$C_{13}=(-1)^{1+3}\begin{vmatrix}2&0\\-1&4\end{vmatrix}=1\cdot(2\cdot4-0)=8.$$ So, $$D_3=-2+8=6.$$ Therefore, $$f'(0)=24+0+6=30.$$ --- ## 3. Compute the required value $$2f(0)+f'(0)=24+30=54.$$ --- ## 4. Check options Given options are: - A: 24 n- B: 18 - C: 42 - D: 48 Our computed value is $$54,$$ which is **not among the options**. So the stored correct answer $\text{C} = 42$ does **not** match the determinant calculation. --- ## Final Answer $$\boxed{54}$$ Hence, the stored answer appears to be incorrect.