A-LevelMathematicsDifferentiation9709/32 · Oct/Nov 2021 · Q98 Marks

The equation of a curve is ye²ˣ – y²eˣ = 2.

(a)Show that dy/dx = (2yeˣ - y²) / (2y - eˣ).[4]
(b)Find the exact coordinates of the point on the curve where the tangent is parallel to the y-axis.[4]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

[B1] State correct derivative of ye²ˣ with respect to x
[B1] State correct derivative of y²eˣ with respect to x
[M1] Equate attempted derivative of the LHS to zero and solve for dy/dx
[A1] Obtain dy/dx = (2yeˣ - y²) / (2y - eˣ)
Alternative method for Question 9(a)
[B1] Other rearrangements are possible e.g. y = 2e⁻²ˣ + y²e⁻ˣ (d/dx (y²e⁻ˣ) = 2ye⁻ˣ dy/dx – y²e⁻ˣ)
[B1] ⇒ dy/dx = -4e⁻ˣ + 2ye⁻ˣ dy/dx - y²e⁻ˣ
[M1] Solve for dy/dx
[A1] Obtain dy/dx = (2yeˣ - y²) / (2y - eˣ)
[*M1] Equate denominator to zero and substitute for y or for eˣ in the equation of the curve
[DM1] Obtain equation of the form ae³ˣ = b or cy³ = d
[A1] Obtain x = ln 2
[A1] Obtain y = 1

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Topic: Differentiation, A-Level Mathematics (syllabus 9709). Structured question, 8 marks.
Source: A-Level Mathematics, Oct/Nov 2021 Paper 3 Variant 2, question 9.

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