Let $f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}$ where $p$ is a constant. The value of $\frac{d^3}{dx^3}f(x)$ at $x = 0$ is
Determinant Expansion
The given function is a determinant:
$f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}$Expand the determinant using cofactor expansion along the first row:
$f(x) = x^3 \begin{vmatrix} -1 & 0 \\ p^2 & p^3 \end{vmatrix} - \sin x \begin{vmatrix} 6 & 0 \\ p & p^3 \end{vmatrix} + \cos x \begin{vmatrix} 6 & -1 \\ p & p^2 \end{vmatrix}$Calculate the $2 \times 2$ determinants:
$f(x) = x^3 ((-1)(p^3) - (0)(p^2)) - \sin x ((6)(p^3) - (0)(p)) + \cos x ((6)(p^2) - (-1)(p))$ $f(x) = x^3 (-p^3) - \sin x (6p^3) + \cos x (6p^2 + p)$Simplify the expression for $f(x)$:
$f(x) = -p^3 x^3 - 6p^3 \sin x + (6p^2 + p) \cos x$Third Derivative Calculation
We need to find the third derivative of $f(x)$ with respect to $x$. Let's differentiate term by term:
- Derivative of $-p^3 x^3$: $\frac{d^3}{dx^3}(-p^3 x^3) = -p^3 \cdot (3 \cdot 2 \cdot 1) = -6p^3$
- Derivative of $-6p^3 \sin x$: $\frac{d^3}{dx^3}(-6p^3 \sin x) = -6p^3 \frac{d^3}{dx^3}(\sin x) = -6p^3 (-\cos x) = 6p^3 \cos x$
- Derivative of $(6p^2 + p) \cos x$: $\frac{d^3}{dx^3}((6p^2 + p) \cos x) = (6p^2 + p) \frac{d^3}{dx^3}(\cos x) = (6p^2 + p) (\sin x)$
Combine the derivatives to find $f'''(x)$:
$f'''(x) = -6p^3 + 6p^3 \cos x + (6p^2 + p) \sin x$Derivative Evaluation at x = 0
Now, evaluate the third derivative $f'''(x)$ at $x = 0$. Recall that $\cos(0) = 1$ and $\sin(0) = 0$.
$f'''(0) = -6p^3 + 6p^3 \cos(0) + (6p^2 + p) \sin(0)$ $f'''(0) = -6p^3 + 6p^3 (1) + (6p^2 + p) (0)$ $f'''(0) = -6p^3 + 6p^3 + 0$ $f'''(0) = 0$Result Analysis
The value of the third derivative $\frac{d^3}{dx^3}f(x)$ at $x=0$ is $0$. Since $0$ does not contain the variable $p$, the result is independent of $p$.
Important Questions from Determinants
- An $n \times n$ square matrix $A$ satisfies $A^T = A^{-1}$. The determinant of this matrix may take which of the following value(s)?
The determinant of the matrix $\begin{pmatrix} 1 & p & q + r \\ 1 & q & r + p \\ 1 & r & p + q \end{pmatrix}$ is ______.
How many matrices of different orders are possible with elements comprising all prime numbers less than 30?
If A is a square matrix of order 3 and det A = 5, then what is det [(2A) -1 ] equal to ?
If (a - b) (b - c) (c - a) = 2 and abc = 6, then what is the value of \(\begin{vmatrix}a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}?\)
