Exams Joint Entrance Examination Mathematics Application of Derivatives Question 205 Question 205 of 212 Hard MCQ
\[\text { If } f(x)=\left|\[ x 3 2 x 2 + 1 1 + 3 x 3 x 2 + 2 2 x x 3 + 6 x 3 − x 4 x 2 − 2 \begin{array}{|ccccccc|}x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2\end{array} x 3 3 x 2 + 2 x 3 − x 2 x 2 + 1 2 x 4 1 + 3 x x 3 + 6 x 2 − 2 \]\right| \text { for all } x \in \mathbb{R} \text {, then } 2 f(0)+f\^{\prime}(0) \text { is equal to }\]
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\[ \begin{aligned}
& f(0)=\left|\[\begin{array}{|ccccccc|}0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2\end{array}\end{aligned} \]\right|=12 \\
\& f\^{\prime}(x)=\left|\[ 3 x 2 4 x 3 3 x 2 + 2 2 x x 3 + 6 x 3 − x 4 x 2 − 2 \begin{array}{|ccccccc|}3 x^2 & 4 x & 3 \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2\end{array} 3 x 2 3 x 2 + 2 x 3 − x 4 x 2 x 4 3 x 3 + 6 x 2 − 2 \]\right|+
\]
\[ \begin{aligned}
& \left|\[\begin{array}{|ccccccc|}x^3 & 2 x^2+1 & 1+3 x \\ 6 x & 2 & 3 x^2 \\ x^3-x & 4 & x^2-2\end{array}\end{aligned} \]\right|+ \\
\& \left|\[ x 3 2 x 2 + 1 1 + 3 x 3 x 2 + 2 2 x x 3 + 6 3 x 2 − 1 0 2 x \begin{array}{|ccccccc|}x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ 3 x^2-1 & 0 & 2 x\end{array} x 3 3 x 2 + 2 3 x 2 − 1 2 x 2 + 1 2 x 0 1 + 3 x x 3 + 6 2 x \]\right| \\
\& \therefore f\^{\prime}(0)=\left|\[ 0 0 3 2 0 6 0 4 − 2 \begin{array}{|ccccccc|}0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2\end{array} 0 2 0 0 0 4 3 6 − 2 \]\right|+\left|\[ 0 1 1 0 2 0 0 4 − 2 \begin{array}{|ccccccc|}0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2\end{array} 0 0 0 1 2 4 1 0 − 2 \]\right|+\left|\[ 0 1 1 2 0 6 − 1 0 0 \begin{array}{|ccccccc|}0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0\end{array} 0 2 − 1 1 0 0 1 6 0 \]\right| \\
\& =24-6=18 \\
\& \therefore 2 f(0)+f\^{\prime}(0)=42
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