The equation of the tangent to the curve \(y = \displaystyle\frac{(x-3)}{(x-1)(x-2)}\) at the point, where it cuts x-axis is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2y-x+3=0\)
The curve cuts the x-axis where y = 0, that is, where \(x - 3 = 0\). The point is (3, 0).
Writing \(y = (x - 3) \cdot \displaystyle\frac{1}{(x-1)(x-2)}\) and using the product rule:
\( \displaystyle \frac{dy}{dx} = \frac{1}{(x-1)(x-2)} + (x-3)\,\frac{d}{dx}\left[\frac{1}{(x-1)(x-2)}\right] \)
At x = 3 the second term is 0, so the slope of the tangent is
\( \displaystyle m = \frac{1}{(3-1)(3-2)} = \frac{1}{2} \)
Tangent at (3, 0):
\( \displaystyle y - 0 = \frac{1}{2}(x - 3) \)
\(2y = x - 3\)
\(2y - x + 3 = 0\)
Note: This topic was not in the CUET 2026 syllabus.