CAIE P2 2018 June — Question 5 6 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind tangent equation at point
DifficultyStandard +0.3 This is a straightforward implicit differentiation problem requiring students to find where the curve crosses the y-axis (x=0, giving y=2), differentiate implicitly using product and chain rules, substitute to find the gradient, then write the tangent equation. While it involves multiple techniques, each step is routine and the question structure is standard for A-level, making it slightly easier than average.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation
5 A curve has equation $$y ^ { 3 } \sin 2 x + 4 y = 8$$ Find the equation of the tangent to the curve at the point where it crosses the \(y\)-axis.
Question 5:
AnswerMarks Guidance
AnswerMarks Guidance
Use product rule to differentiate first term obtaining form \(k_1 y^2 \frac{dy}{dx}\sin 2x + k_2 y^3 \cos 2x\)M1
Obtain correct \(3y^2\frac{dy}{dx}\sin 2x + 2y^3\cos 2x\)A1
State \(3y^2\frac{dy}{dx}\sin 2x + 2y^3\cos 2x + 4\frac{dy}{dx} = 0\)A1
Identify \(x = 0\), \(y = 2\) as relevant pointB1
Find equation of tangent through \((0, 2)\) with numerical gradientM1 Dependent on previous M1
Obtain \(y = -4x + 2\) or equivalentA1 
## Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use product rule to differentiate first term obtaining form $k_1 y^2 \frac{dy}{dx}\sin 2x + k_2 y^3 \cos 2x$ | M1 | |
| Obtain correct $3y^2\frac{dy}{dx}\sin 2x + 2y^3\cos 2x$ | A1 | |
| State $3y^2\frac{dy}{dx}\sin 2x + 2y^3\cos 2x + 4\frac{dy}{dx} = 0$ | A1 | |
| Identify $x = 0$, $y = 2$ as relevant point | B1 | |
| Find equation of tangent through $(0, 2)$ with numerical gradient | M1 | Dependent on previous M1 |
| Obtain $y = -4x + 2$ or equivalent | A1 | |
5 A curve has equation

$$y ^ { 3 } \sin 2 x + 4 y = 8$$

Find the equation of the tangent to the curve at the point where it crosses the $y$-axis.\\

\hfill \mbox{\textit{CAIE P2 2018 Q5 [6]}}
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