| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2009 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Verify point and find gradient |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring substitution to find points, then applying implicit differentiation to find gradients. The steps are standard: substitute x=1, solve quadratic for y-coordinates, differentiate implicitly to get dy/dx, evaluate at both points, and write tangent equation. While it involves multiple parts, each step follows routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) EITHER: Substitute \(x = 1\) and attempt to solve 3-term quadratic in \(y\) | M1 | |
| Obtain answers \((1, 1)\) and \((1, -3)\) | A1 | |
| OR: State answers \((1, 1)\) and \((1, -3)\) | B1 + B1 | [2 marks] |
| (ii) State \(2y\frac{dy}{dx}\) as derivative of \(y^2\) | B1 | |
| State \(2y + 2x\frac{dy}{dx}\) as derivative of \(2xy\) | B1 | |
| Substitute for \(x\) and \(y\), and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain \(\frac{dy}{dx} = 0\) when \(x = 1\) and \(y = 1\) | A1 | |
| Obtain \(\frac{dy}{dx} = -2\) when \(x = 1\) and \(y = -3\) | A1√ | |
| Form the equation of the tangent at \((1, -3)\) | M1 | |
| Obtain answer \(2x + y + 1 = 0\) | A1 | [7 marks] |
**(i)** **EITHER:** Substitute $x = 1$ and attempt to solve 3-term quadratic in $y$ | M1 |
Obtain answers $(1, 1)$ and $(1, -3)$ | A1 |
**OR:** State answers $(1, 1)$ and $(1, -3)$ | B1 + B1 | [2 marks]
**(ii)** State $2y\frac{dy}{dx}$ as derivative of $y^2$ | B1 |
State $2y + 2x\frac{dy}{dx}$ as derivative of $2xy$ | B1 |
Substitute for $x$ and $y$, and solve for $\frac{dy}{dx}$ | M1 |
Obtain $\frac{dy}{dx} = 0$ when $x = 1$ and $y = 1$ | A1 |
Obtain $\frac{dy}{dx} = -2$ when $x = 1$ and $y = -3$ | A1√ |
Form the equation of the tangent at $(1, -3)$ | M1 |
Obtain answer $2x + y + 1 = 0$ | A1 | [7 marks]8 The equation of a curve is $y ^ { 2 } + 2 x y - x ^ { 2 } = 2$.\\
(i) Find the coordinates of the two points on the curve where $x = 1$.\\
(ii) Show by differentiation that at one of these points the tangent to the curve is parallel to the $x$-axis. Find the equation of the tangent to the curve at the other point, giving your answer in the form $a x + b y + c = 0$.
\hfill \mbox{\textit{CAIE P2 2009 Q8 [9]}}