| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Line and curve intersection |
| Difficulty | Standard +0.3 This is a standard coordinate geometry question requiring intersection of line and curve (solving a quadratic), distance formula, midpoint formula, and discriminant condition for tangency. All techniques are routine A-level methods with no novel insight required, but involves multiple steps and the tangency condition pushes it slightly above average difficulty. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02f Solve quadratic equations: including in a function of unknown1.02q Use intersection points: of graphs to solve equations1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 + 3x + 4 = 2x + 6 \Rightarrow x^2 + x - 2 (= 0)\) | M1 | 3-term simplification |
| \((x-1)(x+2) = 0 \to (1, 8), (-2, 2)\) | DM1A1 | DM1 for attempted solution for \(x\) |
| \(AB = \sqrt{3^2 + 6^2} = 6.71\) or \(\sqrt{45}\) or \(3\sqrt{5}\) | B1 | cao (\(\sqrt{45}\) from wrong points scores B0) |
| \(\left(-\frac{1}{2}, 5\right)\) | B1\(\sqrt{}\) | [5] Ft *their* coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 + (3-k)x + 2k - 6 (= 0)\) | M1 | Simplified to 3-term quadratic |
| \((3-k)^2 - 4(2k - 6) = 0\) | DM1 | Apply \(b^2 - 4ac = 0\) as function of \(k\) only |
| \((3-k)(11-k) = 0\) | DM1 | Attempt factorisation or use formula. Both correct |
| \(k = 3\) or \(11\) | A1 | [4] NB Alternative methods for (ii) possible |
## Question 9:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 + 3x + 4 = 2x + 6 \Rightarrow x^2 + x - 2 (= 0)$ | M1 | 3-term simplification |
| $(x-1)(x+2) = 0 \to (1, 8), (-2, 2)$ | DM1A1 | DM1 for attempted solution for $x$ |
| $AB = \sqrt{3^2 + 6^2} = 6.71$ or $\sqrt{45}$ or $3\sqrt{5}$ | B1 | cao ($\sqrt{45}$ from wrong points scores B0) |
| $\left(-\frac{1}{2}, 5\right)$ | B1$\sqrt{}$ | [5] Ft *their* coordinates |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 + (3-k)x + 2k - 6 (= 0)$ | M1 | Simplified to 3-term quadratic |
| $(3-k)^2 - 4(2k - 6) = 0$ | DM1 | Apply $b^2 - 4ac = 0$ as function of $k$ only |
| $(3-k)(11-k) = 0$ | DM1 | Attempt factorisation or use formula. Both correct |
| $k = 3$ or $11$ | A1 | [4] NB Alternative methods for **(ii)** possible |
---9 A line has equation $y = k x + 6$ and a curve has equation $y = x ^ { 2 } + 3 x + 2 k$, where $k$ is a constant.\\
(i) For the case where $k = 2$, the line and the curve intersect at points $A$ and $B$. Find the distance $A B$ and the coordinates of the mid-point of $A B$.\\
(ii) Find the two values of $k$ for which the line is a tangent to the curve.
\hfill \mbox{\textit{CAIE P1 2011 Q9 [9]}}