CAIE P3 2004 June — Question 3 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2004
SessionJune
Marks4
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 students to differentiate term-by-term, apply the product rule to -4xy, collect dy/dx terms, and substitute a point. It's slightly easier than average because it's a single-step application of a standard technique with no conceptual complications, though implicit differentiation itself is a mid-level A-level skill.
Spec1.07s Parametric and implicit differentiation

3 Find the gradient of the curve with equation $$2 x ^ { 2 } - 4 x y + 3 y ^ { 2 } = 3$$ at the point \(( 2,1 )\).

Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
EITHER: State \(6y\frac{dy}{dx}\) as the derivative of \(3y^2\)B1
State \(\pm 4x\frac{dy}{dx} \pm 4y\) as the derivative of \(-4xy\)B1
Equate attempted derivative of LHS to zero and solve for \(\frac{dy}{dx}\)M1 M1 conditional on at least one B mark. Allow any combination of signs for second B1.
Obtain answer 2A1
OR: Obtain correct expression for \(y\) in terms of \(x\)B1
Differentiate using chain ruleM1 M1 conditional on reasonable attempt at solving the quadratic in \(y\)
Obtain derivative in any correct formA1
Substitute \(x = 2\) and obtain answer 2 onlyA1 Total: 4
## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| **EITHER:** State $6y\frac{dy}{dx}$ as the derivative of $3y^2$ | B1 | |
| State $\pm 4x\frac{dy}{dx} \pm 4y$ as the derivative of $-4xy$ | B1 | |
| Equate attempted derivative of LHS to zero and solve for $\frac{dy}{dx}$ | M1 | M1 conditional on at least one B mark. Allow any combination of signs for second B1. |
| Obtain answer 2 | A1 | |
| **OR:** Obtain correct expression for $y$ in terms of $x$ | B1 | |
| Differentiate using chain rule | M1 | M1 conditional on reasonable attempt at solving the quadratic in $y$ |
| Obtain derivative in any correct form | A1 | |
| Substitute $x = 2$ and obtain answer 2 only | A1 | **Total: 4** |

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3 Find the gradient of the curve with equation

$$2 x ^ { 2 } - 4 x y + 3 y ^ { 2 } = 3$$

at the point $( 2,1 )$.

\hfill \mbox{\textit{CAIE P3 2004 Q3 [4]}}