| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Line tangent to curve, find constant |
| Difficulty | Moderate -0.3 This is a straightforward quadratic problem requiring standard techniques: solving simultaneous equations for part (i) and using the discriminant condition (b²-4ac=0) for tangency in part (ii). While it involves multiple steps, each step uses routine AS-level methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x^2 - 4x + 4 = x \Rightarrow x^2 - 5x + 4 = 0\) \((x-1)(x-4)(= 0)\) or other valid method | M1 M1 | Eliminate \(y\) to reach 3-term quadratic Attempt solution |
| \((1, 1), (4, 4)\) | A1 | ft dependent on 1st M1 |
| Mid-point \(= (2\frac{1}{2}, 2\frac{1}{2})\) | A1√ |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(x^2 - (4+m)x + 4 = 0 \Rightarrow (4+m)^2 - 4(4) = 0\) | M1 DM1 | Applying \(b^2 - 4ac = 0\) Attempt solution |
| \(4 + m = \pm 4\) or \(m(8+m) = 0\) | A1 | Ignore \(m = 0\) in addition |
| \(m = -8\) | M1 | Sub non-zero \(m\) and attempt to solve |
| \(x^2 + 4x + 4 = 0\) \(x = -2, y = 16\) | A1 A1 | Ignore (2, 0) solution from \(m = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Alt (ii) \(2x - 4 = m\) or \(x^2 - 4x + 4 = (2x - 4)x\) | M1 | OR \(2x - 4 = m\) Sub \(x = \frac{m+4}{2}, y = \frac{m(m+4)}{2}\) into quad |
| \(m = -8\) (from resulting quad \(m(m+8)=0\)) \(x = -2\) \(y = 16\) | DM1 A1 A1 | \(m = -8\) from resulting quad \(m(m+8)=0\) \(x = -2\) \(y = 16\) |
(i) $x^2 - 4x + 4 = x \Rightarrow x^2 - 5x + 4 = 0$ $(x-1)(x-4)(= 0)$ or other valid method | M1 M1 | Eliminate $y$ to reach 3-term quadratic Attempt solution
$(1, 1), (4, 4)$ | A1 | ft dependent on 1st M1
Mid-point $= (2\frac{1}{2}, 2\frac{1}{2})$ | A1√ |
[4]
(ii) $x^2 - (4+m)x + 4 = 0 \Rightarrow (4+m)^2 - 4(4) = 0$ | M1 DM1 | Applying $b^2 - 4ac = 0$ Attempt solution
$4 + m = \pm 4$ or $m(8+m) = 0$ | A1 | Ignore $m = 0$ in addition
$m = -8$ | M1 | Sub non-zero $m$ and attempt to solve
$x^2 + 4x + 4 = 0$ $x = -2, y = 16$ | A1 A1 | Ignore (2, 0) solution from $m = 0$
[5]
**Alt (ii)** $2x - 4 = m$ or $x^2 - 4x + 4 = (2x - 4)x$ | M1 | OR $2x - 4 = m$ Sub $x = \frac{m+4}{2}, y = \frac{m(m+4)}{2}$ into quad
$m = -8$ (from resulting quad $m(m+8)=0$) $x = -2$ $y = 16$ | DM1 A1 A1 | $m = -8$ from resulting quad $m(m+8)=0$ $x = -2$ $y = 16$7 A curve has equation $y = x ^ { 2 } - 4 x + 4$ and a line has equation $y = m x$, where $m$ is a constant.\\
(i) For the case where $m = 1$, the curve and the line intersect at the points $A$ and $B$. Find the coordinates of the mid-point of $A B$.\\
(ii) Find the non-zero value of $m$ for which the line is a tangent to the curve, and find the coordinates of the point where the tangent touches the curve.
\hfill \mbox{\textit{CAIE P1 2013 Q7 [9]}}