| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Tangent with given gradient or condition |
| Difficulty | Standard +0.3 This is a standard tangent-to-curve problem requiring substitution to form a quadratic, using the discriminant condition (b²-4ac=0) for tangency, then finding coordinates. All steps are routine AS-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(-2x + k = \frac{2}{x-3} \Rightarrow 2x^2 - (6+k)x + 2 + 3k = 0\) | B1 | AG [1] |
| (ii) \((6+k)^2 - (4)(2)(2+3k) = 0\); \(k^2 - 12k + 20 (= 0)\); \((k-10)(k-2) = 0\); \(k = 2\) or 10 | M1, A1, A1, A1 | Apply \(b^2 - 4ac\); cao [3] |
| (iii) \(k = 2 \Rightarrow 2(x-2)^2 = 0\); \(x = 2, y = -2\) or \((2, -2)\); \(k = 10 \Rightarrow 2(x-4)^2 = 0\); \(x = 4, y = 2\) or \((4, 2)\); AB: \(y - 2 = 2(x-4)\) or \(y + 2 = 2(x-2)\) | M1, A1, M1, A1, M1, A1 | \((y = 2x - 6)\) [6] |
(i) $-2x + k = \frac{2}{x-3} \Rightarrow 2x^2 - (6+k)x + 2 + 3k = 0$ | B1 | AG [1]
(ii) $(6+k)^2 - (4)(2)(2+3k) = 0$; $k^2 - 12k + 20 (= 0)$; $(k-10)(k-2) = 0$; $k = 2$ or 10 | M1, A1, A1, A1 | Apply $b^2 - 4ac$; cao [3]
(iii) $k = 2 \Rightarrow 2(x-2)^2 = 0$; $x = 2, y = -2$ or $(2, -2)$; $k = 10 \Rightarrow 2(x-4)^2 = 0$; $x = 4, y = 2$ or $(4, 2)$; AB: $y - 2 = 2(x-4)$ or $y + 2 = 2(x-2)$ | M1, A1, M1, A1, M1, A1 | $(y = 2x - 6)$ [6]10 A straight line has equation $y = - 2 x + k$, where $k$ is a constant, and a curve has equation $y = \frac { 2 } { x - 3 }$.\\
(i) Show that the $x$-coordinates of any points of intersection of the line and curve are given by the equation $2 x ^ { 2 } - ( 6 + k ) x + ( 2 + 3 k ) = 0$.\\
(ii) Find the two values of $k$ for which the line is a tangent to the curve.
The two tangents, given by the values of $k$ found in part (ii), touch the curve at points $A$ and $B$.\\
(iii) Find the coordinates of $A$ and $B$ and the equation of the line $A B$.
\hfill \mbox{\textit{CAIE P1 2012 Q10 [10]}}