CAIE P2 2020 June — Question 3 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2020
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind dy/dx at a point
DifficultyModerate -0.3 This is a straightforward implicit differentiation question requiring application of the chain rule to find dy/dx, then substitution of given coordinates. The trigonometric functions are standard, the point is given (no need to verify it lies on the curve), and it's a single-step calculation after differentiation. Slightly easier than average due to its routine nature and clear structure.
Spec1.07s Parametric and implicit differentiation
3 The equation of a curve is \(\cos 3 x + 5 \sin y = 3\).
Find the gradient of the curve at the point \(\left( \frac { 1 } { 9 } \pi , \frac { 1 } { 6 } \pi \right)\).
Question 3:
AnswerMarks Guidance
AnswerMark Guidance
Differentiate \(\cos 3x\) to obtain \(-3\sin 3x\)B1
Differentiate \(5\sin y\) to obtain \(5\cos y \frac{dy}{dx}\)B1
Obtain \(-3\sin 3x + 5\cos y \frac{dy}{dx} = 0\)B1 OE
Substitute \(x\) and \(y\) values to find value of first derivativeM1
Obtain \(\frac{3}{5}\)A1 
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate $\cos 3x$ to obtain $-3\sin 3x$ | B1 | |
| Differentiate $5\sin y$ to obtain $5\cos y \frac{dy}{dx}$ | B1 | |
| Obtain $-3\sin 3x + 5\cos y \frac{dy}{dx} = 0$ | B1 | OE |
| Substitute $x$ and $y$ values to find value of first derivative | M1 | |
| Obtain $\frac{3}{5}$ | A1 | |

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3 The equation of a curve is $\cos 3 x + 5 \sin y = 3$.\\
Find the gradient of the curve at the point $\left( \frac { 1 } { 9 } \pi , \frac { 1 } { 6 } \pi \right)$.\\

\hfill \mbox{\textit{CAIE P2 2020 Q3 [5]}}
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