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JEE Main 202431 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2024 · 31 January · Morning Shift · Mathematics · Determinants

If f(x) = | cases x ^ 3 & 2 x ^ 2 + 1 & 1 + 3x 3 x ^ 2 + 2 & 2x & x...

If f\left(x\right) = \left|\begin{cases}{x}^{3} & 2{x}^{2} + 1 & 1 + 3x \\ 3{x}^{2} + 2 & 2x & {x}^{3} + 6 \\ {x}^{3} - x & 4 & {x}^{2} - 2\end{cases}\right| for all x∈Rx \in \mathbb{R}, then 2f(0)+f′(0)2f\left(0\right) + {f}^{'}\left(0\right) is equal to

Options

  1. A4848
  2. B2424
  3. C4242
  4. D1818

Correct answer

C. 4242

Step-by-step solution

Given: f\left(x\right) = \left|\begin{cases}{x}^{3} & 2{x}^{2} + 1 & 1 + 3x \\ 3{x}^{2} + 2 & 2x & {x}^{3} + 6 \\ {x}^{3} - x & 4 & {x}^{2} - 2\end{cases}\right|

⇒ f\left(0\right) = \left|\begin{cases}0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & - 2\end{cases}\right|

⇒f(0)=0+4+8⇒ f\left(0\right) = 0 + 4 + 8

⇒f(0)=12⇒ f\left(0\right) = 12

⇒ {f}^{'}\left(x\right) = \left|\begin{cases}{x}^{3} & 2{x}^{2} + 1 & 1 + 3x \\ 3{x}^{2} + 2 & 2x & {x}^{3} + 6 \\ 3{x}^{2} - 1 & 0 & 2x\end{cases}\right| + \left|\begin{cases}{x}^{3} & 2{x}^{2} + 1 & 1 + 3x \\ 6x & 2 & 3{x}^{2} \\ {x}^{3} - x & 4 & {x}^{2} - 2\end{cases}\right| + \left|\begin{cases}3{x}^{2} & 4x & 3 \\ 3{x}^{2} + 2 & 2x & {x}^{3} + 6 \\ {x}^{3} - x & 4 & {x}^{2} - 2\end{cases}\right|

⇒ {f}^{'}\left(0\right) = \left|\begin{cases}0 & 1 & 1 \\ 2 & 0 & 6 \\ - 1 & 0 & 0\end{cases}\right| + \left|\begin{cases}0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & - 2\end{cases}\right| + \left|\begin{cases}0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & - 2\end{cases}\right|

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

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

⇒2f(0)+f′(0)=42⇒ 2f\left(0\right) + {f}^{'}\left(0\right) = 42

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