- 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 , then is equal to
Options
- A
- B
- C
- D
Correct answer
C.
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}^{'}\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|