| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| 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 question requiring standard technique to find dy/dx, then solving dy/dx=0 for stationary points. The algebra is routine: differentiate implicitly, rearrange for dy/dx (part i is shown), set numerator to zero giving y=3x/2, substitute back into original equation to find coordinates. Slightly above average difficulty due to implicit differentiation being less routine than explicit cases, but still a standard textbook exercise with no novel insight required. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| (i) State \(4y\frac{dy}{dx}\) as derivative of \(2y^2\), or equivalent | B1 |
| State \(4y + 4x\frac{dy}{dx}\) as derivative of \(4xy\), or equivalent | B1 |
| Equate derivative of LHS to zero and solve for \(\frac{dy}{dx}\) | M1 |
| Obtain given answer correctly | A1 |
| [4] | |
| (ii) State or imply that the coordinates satisfy \(3x - 2y = 0\) | B1 |
| Obtain an equation in \(x^2\) (or \(y^2\)) | M1 |
| Solve and obtain \(x^2 = 4\) (or \(y^2 = 9\)) | A1 |
| State answer \((2, 3)\) | A1 |
| State answer \((-2, -3)\) | A1 |
| [5] |
**(i)** State $4y\frac{dy}{dx}$ as derivative of $2y^2$, or equivalent | B1 |
State $4y + 4x\frac{dy}{dx}$ as derivative of $4xy$, or equivalent | B1 |
Equate derivative of LHS to zero and solve for $\frac{dy}{dx}$ | M1 |
Obtain given answer correctly | A1 |
| | [4] |
**(ii)** State or imply that the coordinates satisfy $3x - 2y = 0$ | B1 |
Obtain an equation in $x^2$ (or $y^2$) | M1 |
Solve and obtain $x^2 = 4$ (or $y^2 = 9$) | A1 |
State answer $(2, 3)$ | A1 |
State answer $(-2, -3)$ | A1 |
| | [5] |7 The equation of a curve is
$$3 x ^ { 2 } - 4 x y + 2 y ^ { 2 } - 6 = 0$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 x - 2 y } { 2 x - 2 y }$.\\
(ii) Find the coordinates of each of the points on the curve where the tangent is parallel to the $x$-axis.
\hfill \mbox{\textit{CAIE P2 2012 Q7 [9]}}