| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation problem requiring students to find dy/dx, set it to zero for horizontal tangents, solve the resulting linear relationship, and substitute back into the original equation. While it involves multiple steps, each step follows standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State \(2y\frac{dy}{dx}\) as the derivative of \(y^2\) | B1 | |
| State \(2y + 2x\frac{dy}{dx}\), or equivalent, as derivative of \(2xy\) | B1 | |
| Equate attempted derivative of LHS to zero and set \(\frac{dy}{dx}\) equal to zero | M1 | |
| Obtain given relation \(y = -3x\) correctly | A1 | Total: 4 |
| [The M1 is dependent on at least one B1 being earned earlier] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Carry out complete method for finding \(x^2\) or \(y^2\) | M1 | |
| Obtain \(x^2 = 1\) or \(y^2 = 9\) | A1 | |
| Obtain point \((1, -3)\) | A1 | |
| Obtain second point \((-1, 3)\) | A1 | Total: 4 |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State $2y\frac{dy}{dx}$ as the derivative of $y^2$ | B1 | |
| State $2y + 2x\frac{dy}{dx}$, or equivalent, as derivative of $2xy$ | B1 | |
| Equate attempted derivative of LHS to zero and set $\frac{dy}{dx}$ equal to zero | M1 | |
| Obtain given relation $y = -3x$ correctly | A1 | Total: 4 |
| | | [The M1 is dependent on at least one B1 being earned earlier] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Carry out complete method for finding $x^2$ or $y^2$ | M1 | |
| Obtain $x^2 = 1$ or $y^2 = 9$ | A1 | |
| Obtain point $(1, -3)$ | A1 | |
| Obtain second point $(-1, 3)$ | A1 | Total: 4 |
---5 The equation of a curve is $3 x ^ { 2 } + 2 x y + y ^ { 2 } = 6$. It is given that there are two points on the curve where the tangent is parallel to the $x$-axis.\\
(i) Show by differentiation that, at these points, $y = - 3 x$.\\
(ii) Hence find the coordinates of the two points.
\hfill \mbox{\textit{CAIE P2 2006 Q5 [8]}}