| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent with given gradient |
| Difficulty | Moderate -0.3 This is a straightforward tangent problem requiring implicit differentiation of a simple curve, substituting a given gradient (1/4), and solving simultaneous equations. The algebraic manipulation is routine and the question structure is standard, making it slightly easier than average but still requiring proper technique. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Eliminates \(x\) or \(y \rightarrow y^2 - 4y + c - 3 = 0\) or \(x^2 + (2c-16)x + c^2 - 48 = 0\) | M1 | Eliminates \(x\) or \(y\) completely to a quadratic |
| Uses \(b^2 = 4ac \rightarrow 4c - 28 = 0\) | M1 | Uses discriminant \(= 0\). (\(c\) the only variable). Any valid method (may be seen in part (i)) |
| \(c = 7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2\sqrt{(x+3)}} = \frac{1}{4}\) | M1 | |
| Solving | M1 | |
| \(c = 7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Uses \(c = 7\), \(y^2 - 4y + 4 = 0\) | M1 | Ignore \((1,-2)\), \(c=-9\) |
| \((1, 2)\) | A1 |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Eliminates $x$ or $y \rightarrow y^2 - 4y + c - 3 = 0$ or $x^2 + (2c-16)x + c^2 - 48 = 0$ | M1 | Eliminates $x$ or $y$ completely to a quadratic |
| Uses $b^2 = 4ac \rightarrow 4c - 28 = 0$ | M1 | Uses discriminant $= 0$. ($c$ the only variable). Any valid method (may be seen in part (i)) |
| $c = 7$ | A1 | |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{2\sqrt{(x+3)}} = \frac{1}{4}$ | M1 | |
| Solving | M1 | |
| $c = 7$ | A1 | |
## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Uses $c = 7$, $y^2 - 4y + 4 = 0$ | M1 | Ignore $(1,-2)$, $c=-9$ |
| $(1, 2)$ | A1 | |
---2 The line $4 y = x + c$, where $c$ is a constant, is a tangent to the curve $y ^ { 2 } = x + 3$ at the point $P$ on the curve.\\
(i) Find the value of $c$.\\
(ii) Find the coordinates of $P$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q2 [5]}}