CUET · MATHS · PYQ PAPER 2025
The equation of the tangent to the curve \(y=\frac{x-3}{(x-1)(x-2)}\) at the point where it cuts the \(x\)-axis is :
- A \(x+2 y-3=0\)
- B \(2 y-x-3=0\)
- C \(2 y-x+3=0\)
- D \(3 x+2 y-3=0\)
Answer & Solution
Correct Answer
(C) \(2 y-x+3=0\)
Step-by-step Solution
Detailed explanation
\(y=0 \implies x-3=0 \implies x=3\) Point of tangency: \((3,0)\) \(\frac{dy}{dx} = \frac{d}{dx}\left(\frac{x-3}{x^2-3x+2}\right) = \frac{1 \cdot (x^2-3x+2) - (x-3)(2x-3)}{(x^2-3x+2)^2}\)…
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