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Matrices & Determinants: Vmatrix 12x Vmatrix Coefficient

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

Question
If f(x)=∣4x−4(x−2)2x38x−42(x−22)2(x+1)312x−43(x−23)2(x−1)3∣f(x) = \begin{vmatrix} 4x-4 & (x-2)^2 & x^3 \\ 8x-4\sqrt{2} & (x-2\sqrt{2})^2 & (x+1)^3 \\ 12x-4\sqrt{3} & (x-2\sqrt{3})^2 & (x-1)^3 \end{vmatrix}, then the coefficient of xx in f(x)f(x) is:
A64(5−2−3)64(5-\sqrt{2}-\sqrt{3})
B64(5+2−3)64(5+\sqrt{2}-\sqrt{3})
C64(5+2+3)64(5+\sqrt{2}+\sqrt{3})
DNone of thesecorrect
Solution
Step 1: Express the coefficient of xx as f′(0)f'(0)
f′(0)=D1+D2+D3f'(0) = D_1 + D_2 + D_3
 where D1,D2,D3D_1, D_2, D_3 arise from differentiating columns 1, 2, 3 respectively and evaluating at x=0x=0. Step 2: Show D1=D2=0D_1 = D_2 = 0 at x=0x = 0 At x=0x=0: column 1 of D1D_1 is (4,8,12)(4, 8, 12) and column 2 becomes (4,8,12)(4, 8, 12) (since (0−2)2=4(0-2)^2=4, (0−22)2=8(0-2\sqrt{2})^2=8, (0−23)2=12(0-2\sqrt{3})^2=12). Columns 1 and 2 coincide, so D1=0D_1=0. Similarly for D2D_2, columns 1 and 2 coincide at x=0x=0, so D2=0D_2=0. Step 3: Compute D3D_3
D3=∣−440−4283−43123∣D_3 = \begin{vmatrix} -4 & 4 & 0 \\ -4\sqrt{2} & 8 & 3 \\ -4\sqrt{3} & 12 & 3 \end{vmatrix}
 Expanding along R1R_1:
=−4∣83123∣−4∣−423−433∣+0= -4\begin{vmatrix}8&3\\12&3\end{vmatrix} - 4\begin{vmatrix}-4\sqrt{2}&3\\-4\sqrt{3}&3\end{vmatrix} + 0
=−4(24−36)−4(−122+123)=48+482−483=48(1+2−3)= -4(24-36) - 4(-12\sqrt{2}+12\sqrt{3}) = 48 + 48\sqrt{2} - 48\sqrt{3} = 48(1+\sqrt{2}-\sqrt{3})
 Answer: (4)
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