<p>Let \(f(x) = \begin{vmatrix} x^2 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}\) where \(p\) is a constant. Then \(\dfrac{d^3}{dx^3}[f(x)]\) at \(x = 0\) is</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) is a determinant where only the first row depends on x, so the third derivative at x=0 depends only on the third derivative of the first row. Use the property that d³/dx³[f(x)] = det of matrix with third row derivatives replacing first row.
<p><strong>Step 1:</strong> Recognize that f(x) has only the first row depending on x. Use the property:</p><p>$$\frac{d}{dx}\begin{vmatrix} a(x) & b(x) & c(x) \\ d & e & f \\ g & h & i \end{vmatrix} = \begin{vmatrix} a'(x) & b'(x) & c'(x) \\ d & e & f \\ g & h & i \end{vmatrix}$$</p><p><strong>Step 2:</strong> Find derivatives of the first row elements:</p><p>• $\frac{d}{dx}(x^2) = 2x$, $\frac{d^2}{dx^2}(x^2) = 2$, $\frac{d^3}{dx^3}(x^2) = 0$</p><p>• $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d^2}{dx^2}(\sin x) = -\sin x$, $\frac{d^3}{dx^3}(\sin x) = -\cos x$</p><p>• $\frac{d}{dx}(\cos x) = -\sin x$, $\frac{d^2}{dx^2}(\cos x) = -\cos x$, $\frac{d^3}{dx^3}(\cos x) = \sin x$</p><p><strong>Step 3:</strong> Therefore:</p><p>$$\frac{d^3f}{dx^3} = \begin{vmatrix} 0 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}$$</p><p><strong>Step 4:</strong> Evaluate at x = 0:</p><p>$$\frac{d^3f}{dx^3}\bigg|_{x=0} = \begin{vmatrix} 0 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}$$</p><p><strong>Step 5:</strong> Expand along the first row (only one non-zero element at position [1,2]):</p><p>$$= -(-1)\begin{vmatrix} 6 & 0 \\ p & p^3 \end{vmatrix} = 1 \cdot (6p^3 - 0) = 6p^3$$</p><p>∴ Answer: <strong>D</strong> (6p³)</p>
Correct Answer: D