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
$f(0)=\left|\begin{array}{ccc}0 & 1 & 1 \ 2 & 0 & 6 \ 0 & 4 & -2\end{array}\right|=12$
$f^{\prime}(x)=\left|\begin{array}{ccc}3 x^{2} & 4 x & 3 \ 3 x^{2}+2 & 2 x & x^{3}+6 \ x^{3}-x & 4 & x^{2}-2\end{array}\right|+$
$\left|\begin{array}{ccc}\mathrm{x}^{3} \text { ath } & 2 \mathrm{x}^{2}+1 & 1+3 \mathrm{x} \ 6 \mathrm{x} & 2 & 3 \mathrm{x}^{2} \ \mathrm{x}^{3}-\mathrm{x} \text { thon } 4 & \mathrm{x}^{2}-2\end{array}\right|+$ $\left|\begin{array}{ccc}\mathrm{x}^{3} & 2 \mathrm{x}^{2}+1 & 1+3 \mathrm{x} \ 3 \mathrm{x}^{2}+2 & 2 \mathrm{x} & \mathrm{x}^{3}+6 \ 3 \mathrm{x}^{2}-1 & 0 & 2 \mathrm{x}\end{array}\right|$
$\therefore \mathrm{f}^{\prime}(0)=\left|\begin{array}{ccc}0 & 0 & 3 \ 2 & 0 & 6 \ 0 & 4 & -2\end{array}\right|+\left|\begin{array}{ccc}0 & 1 & 1 \ 0 & 2 & 0 \ 0 & 4 & -2\end{array}\right|+\left|\begin{array}{ccc}0 & 1 & 1 \ 2 & 0 & 6 \ -1 & 0 & 0\end{array}\right|$
$=24-6=18$
$\therefore 2 \mathrm{f}(0)+\mathrm{f}^{\prime}(0)=42$
