| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2009 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Moderate -0.5 This is a straightforward integration problem with an additional constraint. Part (i) requires understanding that perpendicular tangents have gradients whose product is -1, leading to a simple algebraic equation. Part (ii) is routine integration of a polynomial followed by finding the constant using a given point. Both parts are standard textbook exercises requiring only basic techniques with no novel insight. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
M1: $\frac{dy}{dx} = k – 2x$
(i) M1: At $x = 2$, $m = (k − 4)$; $x = 3$ $m = (k − 6)$
M1: $(k − 4)(k − 6) = −1$ — Uses $m_1 m_2 = −1$ with gradients $f(k)$
DM1: Solution of quadratic = 0
A1: $\to k = 5$ (watch for fortuitous answers)
[4]
(ii) B1: $y = kx – x^2 (+ c)$ — For integration without $c$
M1: Substitutes $(4, 9)$ — Realises need to substitute for $x$ and $y$
A1: $\to c = 5$ (nb If $k = 5$ is fortuitous, loses last A1)
[3]6 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = k - 2 x$, where $k$ is a constant.\\
(i) Given that the tangents to the curve at the points where $x = 2$ and $x = 3$ are perpendicular, find the value of $k$.\\
(ii) Given also that the curve passes through the point $( 4,9 )$, find the equation of the curve.
\hfill \mbox{\textit{CAIE P1 2009 Q6 [7]}}