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Matrices & Determinants: Let Vmatrix Vmatrix Constant

JEE Maths question with a full step-by-step solution.

Question
Let f(x)=∣x3sin⁡xcos⁡x6−10pp2p3∣f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}, where pp is a constant. Then d3dx3[f(x)]∣x=0\dfrac{d^3}{dx^3}[f(x)]\big|_{x=0} is:
App
Bp+p2p + p^2
Cp+p3p + p^3
DIndependent of ppcorrect
Solution
Step 1: Differentiate three times Rows 2 and 3 are constant, so each derivative only acts on row 1:
f′(x)=∣3x2cos⁡x−sin⁡x6−10pp2p3∣,f′′(x)=∣6x−sin⁡x−cos⁡x6−10pp2p3∣f'(x) = \begin{vmatrix} 3x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}, \quad f''(x) = \begin{vmatrix} 6x & -\sin x & -\cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}
f′′′(x)=∣6−cos⁡xsin⁡x6−10pp2p3∣f'''(x) = \begin{vmatrix} 6 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}
 Step 2: Evaluate at x=0x = 0
f′′′(0)=∣6−106−10pp2p3∣=0f'''(0) = \begin{vmatrix} 6 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} = 0
 Rows 1 and 2 are identical, so the determinant vanishes. The result 00 is independent of pp. Answer: (4)
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