Determinants Question 10
Question 10 - 2024 (31 Jan Shift 1)
If $f(x)=\left|\begin{array}{ccc}x^{3} & 2 x^{2}+1 & 1+3 x \\ 3 x^{2}+2 & 2 x & x^{3}+6 \\ x^{3}-x & 4 & x^{2}-2\end{array}\right|$ for all $x \in \mathbb{R}$, then $2 f(0)+f^{\prime}(0)$ is equal to
(1) 48
(2) 24
(3) 42
(4) 18
Show Answer
Answer (3)
Solution
$f(0)=\left|\begin{array}{ccc}0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2\end{array}\right|=12$
$f^{\prime}(x)=\left|\begin{array}{ccc}3 x^{2} & 4 x & 3 \\ 3 x^{2}+2 & 2 x & x^{3}+6 \\ x^{3}-x & 4 & x^{2}-2\end{array}\right|+$
$\left|\begin{array}{ccc}x^{3} \text { ath } & 2 x^{2}+1 & 1+3 x \\ 6 x & 2 & 3 x^{2} \\ x^{3}-x \text { thon } 4 & x^{2}-2\end{array}\right|+$ $\left|\begin{array}{ccc}x^{3} & 2 x^{2}+1 & 1+3 x \\ 3 x^{2}+2 & 2 x & x^{3}+6 \\ 3 x^{2}-1 & 0 & 2 x\end{array}\right|$
$\therefore f^{\prime}(0)=\left|\begin{array}{ccc}0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2\end{array}\right|+\left|\begin{array}{ccc}0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2\end{array}\right|+\left|\begin{array}{ccc}0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0\end{array}\right|$
$=24-6=18$
$\therefore 2 f(0)+f^{\prime}(0)=42$
