उत्तर (3)

समाधान

$f(0)=\left|\begin{array}{ccc}0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2\end{array}\right|=12$

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

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

$=24-6=18$

$\therefore 2 f(0)+f^{\prime}(0)=42$