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Matrices & Determinants: Vmatrix Vmatrix

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

Question
If f(x)=∣x2−4x+62x2+4x+103x2−2x+16x−22x+23x−1123∣f(x) = \begin{vmatrix} x^2-4x+6 & 2x^2+4x+10 & 3x^2-2x+16 \\ x-2 & 2x+2 & 3x-1 \\ 1 & 2 & 3 \end{vmatrix}, then f(x)f(x) is:
AIncreasing when x∈(3,5)x \in (3, 5)
BIncreasing when x∈(−3,−5)x \in (-3, -5)
CDecreasing when x∈(3,5)x \in (3, 5)
DConstant for all real xxcorrect
Solution
Step 1: Differentiate using the row-differentiation rule
f′(x)=∣R1′R2R3∣+∣R1R2′R3∣+∣R1R2R3′∣f'(x) = \begin{vmatrix} R_1' \\ R_2 \\ R_3 \end{vmatrix} + \begin{vmatrix} R_1 \\ R_2' \\ R_3 \end{vmatrix} + \begin{vmatrix} R_1 \\ R_2 \\ R_3' \end{vmatrix}
 Step 2: Observe proportionality of derivative rows Row 1 differentiated: (2x−4, 4x+4, 6x−2)=2(x−2, 2x+2, 3x−1)=2R2(2x-4,\ 4x+4,\ 6x-2) = 2(x-2,\ 2x+2,\ 3x-1) = 2R_2. A determinant with two proportional rows is zero. Row 2 differentiated: (1,2,3)=R3(1, 2, 3) = R_3. A determinant with two identical rows is zero. Row 3 differentiated: (0,0,0)(0, 0, 0). This determinant is also zero. Step 3: Conclude f′(x)=0f'(x) = 0 for all x∈Rx \in \mathbb{R}, so ff is constant. Evaluating at x=0x = 0 gives f(0)=2f(0) = 2. Answer: (4)
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