Which of the following are NOT correct regarding the equation of tangent and normal to the curve \(y =\displaystyle\frac{x-11}{(x-2)(x-3)}\) at the point, where it cuts the x-axis?
(A) The point of contact is (11, 0).
(B) The equation of tangent is \(x - 72y-11 = 0\).
(C) The equation of normal is \(72x+y-11=0\)
(D) The slope of the tangent at the given point of contact is \(\displaystyle\frac{1}{88}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) and (D) only
The curve cuts the x-axis where y = 0, that is, where \(x - 11 = 0\):
\(x = 11\), so the point is (11, 0).
Let \(u = x - 11\) and \(v = (x-2)(x-3)\). By the quotient rule,
\( \displaystyle \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \)
At x = 11, \(u = 0\) and \(u' = 1\), so the term \(uv'\) vanishes:
\( \displaystyle \left.\frac{dy}{dx}\right|_{x=11} = \frac{1\cdot v(11)}{v(11)^2} = \frac{1}{v(11)} \)
\( \displaystyle = \frac{1}{(11-2)(11-3)} = \frac{1}{72} \)
Tangent at (11, 0) with slope \(\frac{1}{72}\):
\( \displaystyle 72y = x - 11 \)
\( \displaystyle x - 72y - 11 = 0 \)
Normal at (11, 0) with slope \(-72\):
\( \displaystyle y = -72(x - 11) \)
\( \displaystyle 72x + y - 792 = 0 \)
So the point of contact is (11, 0), the slope of the tangent is \(\frac{1}{72}\), the tangent is \(x - 72y - 11 = 0\) and the normal is \(72x + y - 792 = 0\).
Checking the statements
(A) The point of contact is (11, 0). — Correct. y = 0 gives x = 11.
(B) The equation of tangent is \(x - 72y-11 = 0\). — Correct. It comes from \(72y = x - 11\).
(C) The equation of normal is \(72x+y-11=0\) — Incorrect. The normal is \(72x + y - 792 = 0\).
(D) The slope of the tangent at the given point of contact is \(\frac{1}{88}\). — Incorrect. The slope is \(\frac{1}{72}\).
The statements that are NOT correct are (C) and (D).
Note: This topic was not in the CUET 2026 syllabus.