| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Area with optimization or parameters |
| Difficulty | Standard +0.3 This is a straightforward optimization problem requiring finding the equation of line ST, expressing area in terms of x, then differentiating to find a maximum. All steps are standard AS-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Sim triangles \(\frac{y}{16 - x} = \frac{12}{16}\) (or trig) \(\to y = 12 - \frac{3}{4}x\). \(A = xy = 12x - \frac{3}{4}x^2\). | M1, A1, A1 [3] | Trig, similarity or eqn of line (could also come from eqn of line). ag – check working. |
| (ii) \(\frac{dA}{dx} = 12 - \frac{6x}{4} = 0\) when \(x = 8, \to A = 48\). This is a Maximum. From –ve quadratic or 2nd differential. | B1, M1 A1, B1 [4] | Sets to 0 + solution. Can be deduced without any working. Allow even if '48' incorrect. |
(i) Sim triangles $\frac{y}{16 - x} = \frac{12}{16}$ (or trig) $\to y = 12 - \frac{3}{4}x$. $A = xy = 12x - \frac{3}{4}x^2$. | M1, A1, A1 [3] | Trig, similarity or eqn of line (could also come from eqn of line). ag – check working.
(ii) $\frac{dA}{dx} = 12 - \frac{6x}{4} = 0$ when $x = 8, \to A = 48$. This is a Maximum. From –ve quadratic or 2nd differential. | B1, M1 A1, B1 [4] | Sets to 0 + solution. Can be deduced without any working. Allow even if '48' incorrect.6\\
\includegraphics[max width=\textwidth, alt={}, center]{d5f66324-e1fc-40e1-98e7-625187e24d3d-3_465_663_1160_740}
In the diagram, $S$ is the point ( 0,12 ) and $T$ is the point ( 16,0 ). The point $Q$ lies on $S T$, between $S$ and $T$, and has coordinates $( x , y )$. The points $P$ and $R$ lie on the $x$-axis and $y$-axis respectively and $O P Q R$ is a rectangle.\\
(i) Show that the area, $A$, of the rectangle $O P Q R$ is given by $A = 12 x - \frac { 3 } { 4 } x ^ { 2 }$.\\
(ii) Given that $x$ can vary, find the stationary value of $A$ and determine its nature.
\hfill \mbox{\textit{CAIE P1 2013 Q6 [7]}}