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 a constant.
The value of \( \dfrac{d^3}{dx^3} f(x) \) at \(x = 0\) is

• A. \(6p^3\)
• B. \(p + p^2\)
• C. \(p + p^3\)
• D. independent of \(p\)

Hint:

Only the first row of the determinant depends on \(x\); differentiate that row alone, three times, then check what the row becomes at \(x=0\).

Answer 1

The Correct Option is D

Step 1: Expand the determinant along the first row.
Instead of differentiating row by row, let's expand $f(x)$ directly into a plain function of $x$, using cofactor expansion along the first row.
\[ f(x) = x^3\begin{vmatrix}-1 & 0 \\ p^2 & p^3\end{vmatrix} - \sin x\begin{vmatrix}6 & 0 \\ p & p^3\end{vmatrix} + \cos x\begin{vmatrix}6 & -1 \\ p & p^2\end{vmatrix} \] Each small determinant is a simple $2\times 2$ value: $(-1)(p^3)-(0)(p^2)=-p^3$; $(6)(p^3)-(0)(p)=6p^3$; $(6)(p^2)-(-1)(p)=6p^2+p$.

Step 2: Write $f(x)$ as one explicit expression.
\[ f(x) = -p^3x^3 - 6p^3\sin x + (6p^2+p)\cos x \] This is now an ordinary function built from a cubic term, a sine term and a cosine term, so we can differentiate it directly, term by term, three times.

Step 3: Differentiate term by term, three times.
For the cubic term: $\frac{d}{dx}(-p^3x^3)=-3p^3x^2$, then $-6p^3x$, then $-6p^3$.
For the sine term: $\frac{d}{dx}(-6p^3\sin x)=-6p^3\cos x$, then $6p^3\sin x$, then $6p^3\cos x$.
For the cosine term: $\frac{d}{dx}[(6p^2+p)\cos x]=-(6p^2+p)\sin x$, then $-(6p^2+p)\cos x$, then $(6p^2+p)\sin x$.
Adding these three results gives the third derivative:
\[ f'''(x) = -6p^3 + 6p^3\cos x + (6p^2+p)\sin x \]
Step 4: Put in $x=0$.
At $x=0$: $\cos 0=1$, $\sin 0=0$.
\[ f'''(0) = -6p^3 + 6p^3(1) + (6p^2+p)(0) = -6p^3+6p^3+0 = 0 \] This matches what the row-differentiation method gave, confirming the answer through an entirely separate calculation.

Step 5: Conclusion.
Since $f'''(0)=0$ regardless of what $p$ is (the $p^3$ terms cancel exactly, and the leftover sine term is multiplied by $\sin 0=0$), the value never depends on $p$.
\[ \boxed{f'''(0) = 0,\ \text{independent of } p} \]