AP DSC School Assistant Math Model Test 2

Concept:This problem involves differentiating a determinant function and evaluating the third derivative at a specific point. The key technique is to first expand the determinant, then differentiate term-by-term, and finally substitute the given value.Explanation:Step 1: Expand the determinant $f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}$ along the first row:$f(x) = x^3 \bigl((-1)(p^3) - (0)(p^2)\bigr) - \sin x \bigl(6(p^3) - 0(p)\bigr) + \cos x \bigl(6(p^2) - (-1)p\bigr)$Simplify: $f(x) = -p^3 x^3 - 6p^3 \sin x + (6p^2 + p) \cos x$.Step 2: Differentiate three times successively:$f'(x) = -3p^3 x^2 - 6p^3 \cos x - (6p^2 + p) \sin x$$f''(x) = -6p^3 x + 6p^3 \sin x - (6p^2 + p) \cos x$$f'''(x) = -6p^3 + 6p^3 \cos x + (6p^2 + p) \sin x$.Step 3: Evaluate $f'''(x)$ at $x = 0$:$f'''(0) = -6p^3 + 6p^3 \cos 0 + (6p^2 + p) \sin 0$Since $\cos 0 = 1$ and $\sin 0 = 0$, we get $f'''(0) = -6p^3 + 6p^3 + 0 = 0$.Thus the value is $0$, which does not contain $p$. Therefore the result is independent of $p$.Answer:D. independent of p