A-LevelMathematicsDifferentiation9709/22 · Oct/Nov 2020 · Q59 Marks

The equation of a curve is 2e²ˣy - y³ + 4 = 0.

(a)Show that dy/dx = 4e²ˣy 3y²-2e²ˣ[4]
(b)The curve passes through the point (0, 2). Find the equation of the tangent to the curve at this point, giving your answer in the form ax + by + c = 0.[3]
(c)Show that the curve has no stationary points.[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

[1] Use product rule to differentiate 2e²ˣy Must be in the form k₁ye²ˣ + k₂e²ˣ dy/dx
[1] Obtain 4e²ˣy + 2e²ˣ dy/dx
[1] Differentiate -y³ to obtain -3y² dy/dx
[1] Obtain dy/dx = 4e²ˣy / (3y²-2e²ˣ)
[1] Substitute 0 and 2 to find gradient of tangent
[1] Attempt to find equation of tangent through (0, 2) with numerical gradient
[1] Obtain 4x-5y+10=0 or equivalent of required form
[1] Equate numerator of derivative to zero and state that at least one of e²ˣ and y cannot be zero
[1] Complete argument

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Topic: Differentiation, A-Level Mathematics (syllabus 9709). Structured question, 9 marks.
Source: A-Level Mathematics, Oct/Nov 2020 Paper 2 Variant 2, question 5.

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