| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2017 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard technique to find dy/dx, evaluate at a point for the tangent, then solve an equation to show no solution exists. The algebra is routine and the methods are textbook exercises, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(4y + 4x\frac{dy}{dx}\) as derivative of \(4xy\) | B1 | |
| Obtain \(4y\frac{dy}{dx}\) as derivative of \(2y^2\) | B1 | |
| State \(2x + 4y + 4x\frac{dy}{dx} + 4y\frac{dy}{dx} = 0\) | B1 | 3rd B1 may be implied by later work |
| Substitute \(x = -1\), \(y = 3\) to find gradient of line | \*M1 | dep at least one B1 |
| Form equation of tangent through \((-1, 3)\) with numerical gradient | DM1 | dep \*M |
| Obtain \(5x + 4y - 7 = 0\) or equivalent of required form | A1 | Allow any 3 term integer form for A1 |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(\frac{dy}{dx} = \frac{1}{2}\) to find relation between \(x\) and \(y\) | \*M1 | dep at least one B1 in part (i), must be linear |
| Obtain \(4x + 6y = 0\) or equivalent | A1 | |
| Substitute for \(x\) or \(y\) in equation of curve | DM1 | dep on \*M |
| Obtain \(-\frac{7}{4}y^2 = 7\) or \(-\frac{7}{9}x^2 = 7\) or equivalent and conclude appropriately | A1 | |
| 4 |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $4y + 4x\frac{dy}{dx}$ as derivative of $4xy$ | **B1** | |
| Obtain $4y\frac{dy}{dx}$ as derivative of $2y^2$ | **B1** | |
| State $2x + 4y + 4x\frac{dy}{dx} + 4y\frac{dy}{dx} = 0$ | **B1** | 3rd **B1** may be implied by later work |
| Substitute $x = -1$, $y = 3$ to find gradient of line | **\*M1** | dep at least one **B1** |
| Form equation of tangent through $(-1, 3)$ with numerical gradient | **DM1** | dep \*M |
| Obtain $5x + 4y - 7 = 0$ or equivalent of required form | **A1** | Allow any 3 term integer form for **A1** |
| | **6** | |
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $\frac{dy}{dx} = \frac{1}{2}$ to find relation between $x$ and $y$ | **\*M1** | dep at least one **B1** in part **(i)**, must be linear |
| Obtain $4x + 6y = 0$ or equivalent | **A1** | |
| Substitute for $x$ or $y$ in equation of curve | **DM1** | dep on \*M |
| Obtain $-\frac{7}{4}y^2 = 7$ or $-\frac{7}{9}x^2 = 7$ or equivalent and conclude appropriately | **A1** | |
| | **4** | |7 The equation of a curve is $x ^ { 2 } + 4 x y + 2 y ^ { 2 } = 7$.\\
(i) Find the equation of the tangent to the curve at the point $( - 1,3 )$. Give your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers.\\
(ii) Show that there is no point on the curve at which the gradient is $\frac { 1 } { 2 }$.\\
\hfill \mbox{\textit{CAIE P2 2017 Q7 [10]}}