The equation of the tangent to the curve $y = \frac{(x-3)}{(x-1)(x-2)}$ at the point, where it cuts x-axis is:

$2y-x+3=0$

The correct answer is Option (3) → $2y-x+3=0$

Given curve: $y = \frac{x-3}{(x-1)(x-2)}$

Point where curve cuts x-axis: $y = 0 \Rightarrow x-3 = 0 \Rightarrow x = 3$ → Point: $(3,0)$

Differentiate using quotient rule: $y = \frac{u}{v}, \frac{dy}{dx} = \frac{u'v - uv'}{v^2}$

$u = x-3, u' = 1$

$v = (x-1)(x-2) = x^2 -3x +2, v' = 2x -3$

$\frac{dy}{dx} = \frac{(1)(x^2 -3x +2) - (x-3)(2x-3)}{(x^2 -3x +2)^2}$

Simplify numerator: $x^2 -3x +2 - (2x^2 -9x +9) = -x^2 +6x -7$

At $x = 3$: $dy/dx = \frac{-(9) + 18 -7}{(9 -9 +2)^2} = \frac{2}{4} = \frac{1}{2}$

Equation of tangent: $y - 0 = \frac{1}{2} (x - 3) \Rightarrow y = \frac{1}{2} x - \frac{3}{2}$