If f(x) = left|begin{array}{ccc} x^3 & 2x^2+1 | vedclass

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(C) First,we calculate $f(0)$ by substituting $x=0$ into the determinant:$f(0) = \left|\begin{array}{ccc} 0 & 1 & 1 \ 2 & 0 & 6 \ 0 & 4 & -2 \end{ar

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$42$

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(C) First,we calculate $f(0)$ by substituting $x=0$ into the determinant:$f(0) = \left|\begin{array}{ccc} 0 & 1 & 1 \ 2 & 0 & 6 \ 0 & 4 & -2 \end{array}ight| = 0(0-24) - 1(-4-0) + 1(8-0) = 4 + 8 = 12$.Next,we find $f'(x)$ using the property of differentiation of a determinant:$f'(x) = \left|\begin{array}{ccc} 3x^2 & 4x & 3 \ 3x^2+2 & 2x & x^3+6 \ x^3-x & 4 & x^2-2 \end{array}ight| + \left|\begin{array}{ccc} x^3 & 2x^2+1 & 1+3x \ 6x & 2 & 3x^2 \ x^3-x & 4 & x^2-2 \end{array}ight| + \left|\begin{array}{ccc} x^3 & 2x^2+1 & 1+3x \ 3x^2+2 & 2x & x^3+6 \ 3x^2-1 & 0 & 2x \end{array}ight|$.Now,evaluate $f'(0)$ by substituting $x=0$:$f'(0) = \left|\begin{array}{ccc} 0 & 0 & 3 \ 2 & 0 & 6 \ 0 & 4 & -2 \end{array}ight| + \left|\begin{array}{ccc} 0 & 1 & 1 \ 0 & 2 & 0 \ 0 & 4 & -2 \end{array}ight| + \left|\begin{array}{ccc} 0 & 1 & 1 \ 2 & 0 & 6 \ -1 & 0 & 0 \end{array}ight|$.Calculating each determinant:First: $3(8-0) = 24$.Second: $0$ (since the first column is all zeros).Third: $-1(6-0) = -6$.So,$f'(0) = 24 + 0 - 6 = 18$.Finally,$2f(0) + f'(0) = 2(12) + 18 = 24 + 18 = 42$.

If $f(x) = \left|\begin{array}{ccc} x^3 & 2x^2+1 & 1+3x \\ 3x^2+2 & 2x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right|$ for all $x \in R$,then $2f(0) + f'(0)$ is equal to

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