| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Intersection points of line-curve |
| Difficulty | Moderate -0.3 This is a standard intersection problem requiring equating expressions, solving quadratics, and using the discriminant condition for tangency. Part (i) is algebraic manipulation, part (ii) uses sum of roots, and part (iii) applies b²-4ac=0. All techniques are routine for P1 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f 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 |
| \(x^2 - x + 3 = 3x + a \rightarrow x^2 - 4x + (3-a) = 0\) | B1 [1] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5 + (3-a) = 0 \rightarrow a = 8\) | B1 | Sub \(x = -1\) into (i) |
| \(x^2 - 4x - 5 = 0 \rightarrow x = 5\) | B1 [2] | OR B2 for \(x=5\) www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(16 - 4(3-a) = 0\) (applying \(b^2 - 4ac = 0\)) | M1 | OR \(dy/dx = 2x-1 \rightarrow 2x-1=3\), \(x=2\) |
| \(a = -1\) | A1 | \(y = 2^2 - 2 + 3 \rightarrow y = 5\) |
| \((x-2)^2 = 0 \rightarrow x = 2\) | A1 | \(5 = 6 + a \rightarrow a = -1\) |
| \(y = 5\) | A1 [4] |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - x + 3 = 3x + a \rightarrow x^2 - 4x + (3-a) = 0$ | B1 [1] | AG |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5 + (3-a) = 0 \rightarrow a = 8$ | B1 | Sub $x = -1$ into (i) |
| $x^2 - 4x - 5 = 0 \rightarrow x = 5$ | B1 [2] | OR B2 for $x=5$ www |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $16 - 4(3-a) = 0$ (applying $b^2 - 4ac = 0$) | M1 | OR $dy/dx = 2x-1 \rightarrow 2x-1=3$, $x=2$ |
| $a = -1$ | A1 | $y = 2^2 - 2 + 3 \rightarrow y = 5$ |
| $(x-2)^2 = 0 \rightarrow x = 2$ | A1 | $5 = 6 + a \rightarrow a = -1$ |
| $y = 5$ | A1 [4] | |
---6 A curve has equation $y = x ^ { 2 } - x + 3$ and a line has equation $y = 3 x + a$, where $a$ is a constant.\\
(i) Show that the $x$-coordinates of the points of intersection of the line and the curve are given by the equation $x ^ { 2 } - 4 x + ( 3 - a ) = 0$.\\
(ii) For the case where the line intersects the curve at two points, it is given that the $x$-coordinate of one of the points of intersection is - 1 . Find the $x$-coordinate of the other point of intersection.\\
(iii) For the case where the line is a tangent to the curve at a point $P$, find the value of $a$ and the coordinates of $P$.
\hfill \mbox{\textit{CAIE P1 2015 Q6 [7]}}