| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Optimization with constraint |
| Difficulty | Standard +0.3 This is a standard optimization problem requiring constraint substitution, differentiation, and second derivative test. The constraint equation is straightforward (perimeter = 2x + 2πr = 400), and part (i) guides students through the area formula derivation. Part (ii) requires routine differentiation of A with respect to r, setting dA/dr = 0, and applying the second derivative test. The conclusion that x = 0 (no straight sections) emerges naturally from the algebra. While it involves multiple steps, each is a standard A-level technique with no novel insight required, making it slightly easier than average. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(A = 2vr + \pi r^2\) | B1 | |
| \(2x + 2\pi r = 400\) (\(\Rightarrow x = 200 - \pi r\)) | B1 | |
| \(A = 400r - \pi r^2\) | M1A1 | Subst & simplify to AG (www) |
| (ii) \(\frac{dA}{dr} = 400 - 2\pi r\) | B1 | Differentiate |
| \(= 0\) | M1 | Set to zero and attempt to find \(r\) |
| \(r = \frac{200}{\pi}\) oe | A1 | |
| \(x = 0 \Rightarrow\) no straight sections | A1 | AG |
| \(\frac{d^2A}{dr^2} = -2\pi\) (\(< 0\)) Max | B1 | Dep on \(-2\pi\), or use of other valid reason |
(i) $A = 2vr + \pi r^2$ | B1 |
$2x + 2\pi r = 400$ ($\Rightarrow x = 200 - \pi r$) | B1 |
$A = 400r - \pi r^2$ | M1A1 | Subst & simplify to AG (www) | [4]
(ii) $\frac{dA}{dr} = 400 - 2\pi r$ | B1 | Differentiate
$= 0$ | M1 | Set to zero and attempt to find $r$
$r = \frac{200}{\pi}$ oe | A1 |
$x = 0 \Rightarrow$ no straight sections | A1 | AG
$\frac{d^2A}{dr^2} = -2\pi$ ($< 0$) Max | B1 | Dep on $-2\pi$, or use of other valid reason | [5]8\\
\includegraphics[max width=\textwidth, alt={}, center]{02da6b6a-6db1-4bc3-ad4e-537e4f61dcac-3_365_663_1813_740}
The inside lane of a school running track consists of two straight sections each of length $x$ metres, and two semicircular sections each of radius $r$ metres, as shown in the diagram. The straight sections are perpendicular to the diameters of the semicircular sections. The perimeter of the inside lane is 400 metres.\\
(i) Show that the area, $A \mathrm {~m} ^ { 2 }$, of the region enclosed by the inside lane is given by $A = 400 r - \pi r ^ { 2 }$.\\
(ii) Given that $x$ and $r$ can vary, show that, when $A$ has a stationary value, there are no straight sections in the track. Determine whether the stationary value is a maximum or a minimum. [5]
\hfill \mbox{\textit{CAIE P1 2013 Q8 [9]}}