JEE Main 2024 — Mathematics Algebra

Given: $f(x)=|\begin{matrix}{x}^{3} & 2{x}^{2}+1 & 1+3x \\ 3{x}^{2}+2 & 2x & {x}^{3}+6 \\ {x}^{3}-x & 4 & {x}^{2}-2\end{matrix}|$f(x)=∣x33x2+2x3−x​2x2+12x4​1+3xx3+6x2−2​∣

$$\Rightarrow f(0)=|\begin{matrix}0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2\end{matrix}|$$⇒f(0)=∣020​104​16−2​∣

$$\Rightarrow f(0)=0+4+8$$⇒f(0)=0+4+8

$$\Rightarrow f(0)=12$$⇒f(0)=12

$$\Rightarrow {f}^{'}(x)=|\begin{matrix}{x}^{3} & 2{x}^{2}+1 & 1+3x \\ 3{x}^{2}+2 & 2x & {x}^{3}+6 \\ 3{x}^{2}-1 & 0 & 2x\end{matrix}|+|\begin{matrix}{x}^{3} & 2{x}^{2}+1 & 1+3x \\ 6x & 2 & 3{x}^{2} \\ {x}^{3}-x & 4 & {x}^{2}-2\end{matrix}|+|\begin{matrix}3{x}^{2} & 4x & 3 \\ 3{x}^{2}+2 & 2x & {x}^{3}+6 \\ {x}^{3}-x & 4 & {x}^{2}-2\end{matrix}|$$⇒f′(x)=∣x33x2+23x2−1​2x2+12x0​1+3xx3+62x​∣+∣x36xx3−x​2x2+124​1+3x3x2x2−2​∣+∣3x23x2+2x3−x​4x2x4​3x3+6x2−2​∣

$$\Rightarrow {f}^{'}(0)=|\begin{matrix}0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0\end{matrix}|+|\begin{matrix}0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2\end{matrix}|+|\begin{matrix}0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2\end{matrix}|$$⇒f′(0)=∣02−1​100​160​∣+∣000​124​10−2​∣+∣020​004​36−2​∣

$$\Rightarrow {f}^{'}(0)=0-(0+6)+0+0+0+0+3(8)$$⇒f′(0)=0−(0+6)+0+0+0+0+3(8)

$$\Rightarrow {f}^{'}(0)=-6+24=18$$⇒f′(0)=−6+24=18

$$\Rightarrow 2f(0)+{f}^{'}(0)=42$$⇒2f(0)+f′(0)=42