| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show specific gradient value |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring application of the product rule and chain rule, followed by substitution to verify a given gradient. Part (i) is routine A-level technique with 5 marks. The multi-part structure adds some length but each part follows standard procedures (stationary points, vertical tangents). Slightly above average due to the implicit differentiation context, but still a textbook exercise with no novel insight required. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtain \(-4y - 4x\dfrac{dy}{dx}\) from use of the product rule | B1 | |
| Differentiate \(-2y^2\) to obtain \(-4y\dfrac{dy}{dx}\) | B1 | |
| Obtain \(2x = 0\) with no extra terms | B1 | |
| Rearrange to obtain expression for \(\dfrac{dy}{dx}\) and substitute \(x = -1,\ y = 2\) | M1 | |
| Obtain \(\dfrac{dy}{dx} = \dfrac{2x - 4y}{4x + 4y}\) OE and hence \(-\dfrac{5}{2}\) | A1 | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equate numerator of derivative to zero to produce equation in \(x\) and \(y\) | M1 | |
| Substitute into equation of curve to produce equation in \(x\) or \(y\) | M1 | |
| Obtain \(-6y^2 = 1\) or \(-\dfrac{3}{2}x^2 = 1\) OE and conclude | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use denominator of derivative equated to zero with equation of curve to produce equation in \(x\) | M1 | |
| Obtain \(3x^2 = 1\) and hence \(x = \pm\dfrac{1}{\sqrt{3}}\) | A1 | OE |
| 2 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain $-4y - 4x\dfrac{dy}{dx}$ from use of the product rule | **B1** | |
| Differentiate $-2y^2$ to obtain $-4y\dfrac{dy}{dx}$ | **B1** | |
| Obtain $2x = 0$ with no extra terms | **B1** | |
| Rearrange to obtain expression for $\dfrac{dy}{dx}$ and substitute $x = -1,\ y = 2$ | **M1** | |
| Obtain $\dfrac{dy}{dx} = \dfrac{2x - 4y}{4x + 4y}$ OE and hence $-\dfrac{5}{2}$ | **A1** | |
| | **5** | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate numerator of derivative to zero to produce equation in $x$ and $y$ | **M1** | |
| Substitute into equation of curve to produce equation in $x$ or $y$ | **M1** | |
| Obtain $-6y^2 = 1$ or $-\dfrac{3}{2}x^2 = 1$ OE and conclude | **A1** | |
| | **3** | |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use denominator of derivative equated to zero with equation of curve to produce equation in $x$ | **M1** | |
| Obtain $3x^2 = 1$ and hence $x = \pm\dfrac{1}{\sqrt{3}}$ | **A1** | OE |
| | **2** | |7 The equation of a curve is $x ^ { 2 } - 4 x y - 2 y ^ { 2 } = 1$.\\
(i) Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and show that the gradient of the curve at the point $( - 1,2 )$ is $- \frac { 5 } { 2 }$. [5]\\
(ii) Show that the curve has no stationary points.\\
(iii) Find the $x$-coordinate of each of the points on the curve at which the tangent is parallel to the $y$-axis.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE P2 2019 Q7 [10]}}