The line ax + by = 7 is a tangent to the curve \(y = \displaystyle\frac{x-7}{(x-2)(x-3)}\) at the point where it cuts the x-axis
A. The y-intercept of the line is −0.7
B. b = −7
C. a = 1
D. the line passes through the point (−13, −1)
E. b = −20
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → C, D, E Only
The curve cuts the x-axis where y = 0, that is, x − 7 = 0. So the point is (7, 0).
By the quotient rule,
\( \displaystyle \frac{dy}{dx} = \frac{(x-2)(x-3) - (x-7)\,\frac{d}{dx}\big[(x-2)(x-3)\big]}{(x-2)^2(x-3)^2} \)
At x = 7 the factor x − 7 is 0, so the second term drops out:
\( \displaystyle \left.\frac{dy}{dx}\right|_{x=7} = \frac{1}{(7-2)(7-3)} = \frac{1}{20} \)
The tangent at (7, 0):
\( \displaystyle y - 0 = \frac{1}{20}(x - 7) \)
\(20y = x - 7\)
\(x - 20y = 7\)
So the tangent is x − 20y = 7, with a = 1, b = −20 and y-intercept −7/20 = −0.35.
Checking the statements
(A) The y-intercept of the line is −0.7 — Incorrect. Putting x = 0 gives y = −7/20 = −0.35.
(B) b = −7 — Incorrect. In x − 20y = 7, b = −20.
(C) a = 1 — Correct. The coefficient of x in x − 20y = 7 is 1.
(D) the line passes through the point (−13, −1) — Correct. −13 − 20(−1) = 7.
(E) b = −20 — Correct. The coefficient of y in x − 20y = 7 is −20.
Note: This topic was not in the CUET 2026 syllabus.