Matrices & DeterminantshardFree

Matrices & Determinants: Vmatrix 12x Vmatrix Coefficient

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

Question

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

x

f(x)

is:

64(5-\sqrt{2}-\sqrt{3})

64(5+\sqrt{2}-\sqrt{3})

64(5+\sqrt{2}+\sqrt{3})

DNone of thesecorrect

Solution

Step 1: Express the coefficient of

x

f'(0)

f'(0) = D_1 + D_2 + D_3

where

D_1, D_2, D_3

arise from differentiating columns 1, 2, 3 respectively and evaluating at

x=0

. Step 2: Show

D_1 = D_2 = 0

x = 0

x=0

: column 1 of

D_1

(4, 8, 12)

and column 2 becomes

(4, 8, 12)

(since

(0-2)^2=4

(0-2\sqrt{2})^2=8

(0-2\sqrt{3})^2=12

). Columns 1 and 2 coincide, so

D_1=0

. Similarly for

D_2

, columns 1 and 2 coincide at

x=0

, so

D_2=0

. Step 3: Compute

D_3

D_3 = \begin{vmatrix} -4 & 4 & 0 \\ -4\sqrt{2} & 8 & 3 \\ -4\sqrt{3} & 12 & 3 \end{vmatrix}

Expanding along

R_1

= -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(-12\sqrt{2}+12\sqrt{3}) = 48 + 48\sqrt{2} - 48\sqrt{3} = 48(1+\sqrt{2}-\sqrt{3})

Answer: (4)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Mathsquestions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.