Question

Let $f (x) = \begin{vmatrix} x^3 & \sin \ x & \cos \ x \\ 6 & - 1 & 0 \\ p & p^2 & p^3 \\ \end{vmatrix}$, where $p$ is constant.
Then, $ \frac{ d^3}{ dx^3} f (x) $ at $x = 0$ is

• A. $p$
• B. $p + p^2$
• C. $p + p^3$
• D. independent of $p$

Answer 1

Answer: (D)

Given, $f (x) = \begin{vmatrix}x^3 & \sin \ x & \cos \ x \\6 & - 1 & 0 \\p & p^2 & p^3 \\\end{vmatrix}$
On differentiating w.r.t. $x$, we get
$f ' (x) = \begin{vmatrix}3x^2 & \cos \ x & -\sin \ x \\6 & - 1 & 0 \\p & p^2 & p^3 \\\end{vmatrix} + \begin{vmatrix}x^3 & \sin \ x & \cos \ x \\0 & 0 & 0 \\p & p^2 & p^3 \\\end{vmatrix} + \begin{vmatrix}x^3 & \sin \ x & \cos \ x \\6 & - 1 & 0 \\0 & 0 & 0 \\\end{vmatrix} $
$\Rightarrow \ f \ ' (x) = \begin{vmatrix}x^3 & \cos \ x & -\sin \ x \\6 & - 1 & 0 \\p & p^2 & p^3 \\\end{vmatrix} $
$\Rightarrow f'' (x) = \begin{vmatrix}6x & -\sin \ x & -\cos \ x \\6 & - 1 & 0 \\p & p^2 & p^3 \\\end{vmatrix} + 0 + 0$
and $ f ''' (x) = \begin{vmatrix}6 & -\cos \ x & \sin \ x \\6 & - 1 & 0 \\p & p^2 & p^3 \\\end{vmatrix} + 0 + 0$
$\therefore f "' (0) = \begin{vmatrix}6 & - 1 & 0 \\6 & - 1 & 0 \\p & p^2 & p^3 \\\end{vmatrix} = 0 = $ independent of $p$