Matrices & DeterminantsmediumFree

Matrices & Determinants: Let Vmatrix Vmatrix Constant

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

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. Then

\dfrac{d^3}{dx^3}[f(x)]\big|_{x=0}

is:

p

p + p^2

p + p^3

DIndependent of

p

correct

Solution

Step 1: Differentiate three times Rows 2 and 3 are constant, so each derivative only acts on row 1:

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) = \begin{vmatrix} 6 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}

Step 2: Evaluate at

x = 0

f'''(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

0

is independent of

p

. Answer: (4)

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