Verified Solution • 2025–26 Session moderate

QJEEM24Ma8238: JEE Main 2024 Mathematics - Class 12 JEEM24Ma8238 Solution

 If f(x)=∣x3amp;2x2+1amp;1+3x3x2+2amp;2xamp;x3+6x3−xamp;4amp;x2−2∣ for all x∈R, then 2f(0)+f′(0) is equal to \text { 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| \text { for all } x \in \mathbb{R} \text {, then } 2 f(0)+f^{\prime}(0) \text { is equal to }

Multiple Choice Options

  • A 24
  • B 18
  • C 42
  • D 48

Step-by-Step Verified Solution

amp;f(0)=∣0amp;1amp;12amp;0amp;60amp;4amp;−2∣=12amp;f′(x)=∣3x2amp;4xamp;33x2+2amp;2xamp;x3+6x3−xamp;4amp;x2−2∣+\begin{aligned}& 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|+\end{aligned}

amp;∣x3amp;2x2+1amp;1+3x6xamp;2amp;3x2x3−xamp;4amp;x2−2∣+amp;∣x3amp;2x2+1amp;1+3x3x2+2amp;2xamp;x3+63x2−1amp;0amp;2x∣amp;∴f′(0)=∣0amp;0amp;32amp;0amp;60amp;4amp;−2∣+∣0amp;1amp;10amp;2amp;00amp;4amp;−2∣+∣0amp;1amp;12amp;0amp;6−1amp;0amp;0∣amp;=24−6=18amp;∴2f(0)+f′(0)=42\begin{aligned}& \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\end{aligned}

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