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Matrices & Determinants: Number Distinct Real Roots Equation Vmatrix Vmatrix

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

Question
The number of distinct real roots of the equation
∣x2(x−1)2(x−2)2(x−1)2(x−2)2(x−3)2(x−2)2(x−3)2(x−4)2∣=0\begin{vmatrix} x^2 & (x-1)^2 & (x-2)^2 \\ (x-1)^2 & (x-2)^2 & (x-3)^2 \\ (x-2)^2 & (x-3)^2 & (x-4)^2 \end{vmatrix} = 0
 is:
A66
B33
C22
D00correct
Solution
Step 1: Apply C1→C1−C2C_1 \to C_1 - C_2 and C2→C2−C3C_2 \to C_2 - C_3, then R1→R1−R2R_1 \to R_1 - R_2, R2→R2−R3R_2 \to R_2 - R_3 After these operations:
Δ=∣00222(2x−7)(2x−5)(2x−7)(x−4)2∣\Delta = \begin{vmatrix} 0 & 0 & 2 \\ 2 & 2 & (2x-7) \\ (2x-5) & (2x-7) & (x-4)^2 \end{vmatrix}
 Step 2: Expand along R1R_1
Δ=2∣22(2x−5)(2x−7)∣=2[2(2x−7)−2(2x−5)]=2(−4)=−8\Delta = 2 \begin{vmatrix} 2 & 2 \\ (2x-5) & (2x-7) \end{vmatrix} = 2\bigl[2(2x-7)-2(2x-5)\bigr] = 2(-4) = -8
 Step 3: Conclude The determinant equals −8-8 for all real xx. Since −8≠0-8 \neq 0, the equation has no real solutions. Answer: (4)
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