| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Curve motion: find dy/dt |
| Difficulty | Standard +0.3 This is a standard optimization problem requiring volume constraint to express h in terms of x, then differentiation to minimize surface area. The algebra is straightforward, and the second derivative test is routine. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(h = \dfrac{8}{x^2}\) | M1 A1 | Uses \(lbh = 4\); co |
| \(A = \frac{1}{2}x^2 + 2 \times \frac{1}{2}xh + 2xh + \frac{5}{4}x \times \frac{4}{5}x\) | M1 | Allow 1 error but needs the lid |
| \(A = (3/2)x^2 + 3xh\) | ||
| \(A = \dfrac{3}{2}x^2 + 3x \times \dfrac{8}{x^2}\) | M1 | For substitution of \(h\) as \(f(x)\) |
| \(A = \dfrac{3}{2}x^2 + \dfrac{24}{x}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{dA}{dx} = 3x - \dfrac{24}{x^2} = 0\) | B1 M1 | Correct derivative; sets to 0 and attempts to solve |
| \(x = 2\) | A1 | co |
| \(\dfrac{d^2A}{dx^2} = 3 + \dfrac{48}{x^3}\) | M1 | Reasonable attempt – allow 1 error |
| \(> 0\) when \(x = 2\) hence minimum | A1 | co; AG (result consistent with their \(f''\)) |
## Question 10:
**Part (i)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $h = \dfrac{8}{x^2}$ | M1 A1 | Uses $lbh = 4$; co |
| $A = \frac{1}{2}x^2 + 2 \times \frac{1}{2}xh + 2xh + \frac{5}{4}x \times \frac{4}{5}x$ | M1 | Allow 1 error but needs the lid |
| $A = (3/2)x^2 + 3xh$ | | |
| $A = \dfrac{3}{2}x^2 + 3x \times \dfrac{8}{x^2}$ | M1 | For substitution of $h$ as $f(x)$ |
| $A = \dfrac{3}{2}x^2 + \dfrac{24}{x}$ | A1 | **AG** |
**Part (ii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{dA}{dx} = 3x - \dfrac{24}{x^2} = 0$ | B1 M1 | Correct derivative; sets to 0 and attempts to solve |
| $x = 2$ | A1 | co |
| $\dfrac{d^2A}{dx^2} = 3 + \dfrac{48}{x^3}$ | M1 | Reasonable attempt – allow 1 error |
| $> 0$ when $x = 2$ hence minimum | A1 | co; **AG** (result consistent with their $f''$) |
---10\\
\includegraphics[max width=\textwidth, alt={}, center]{ae57d8f1-5a0d-426c-952d-e8b99c6aeaba-4_433_969_1475_587}
The diagram shows an open rectangular tank of height $h$ metres covered with a lid. The base of the tank has sides of length $x$ metres and $\frac { 1 } { 2 } x$ metres and the lid is a rectangle with sides of length $\frac { 5 } { 4 } x$ metres and $\frac { 4 } { 5 } x$ metres. When full the tank holds $4 \mathrm {~m} ^ { 3 }$ of water. The material from which the tank is made is of negligible thickness. The external surface area of the tank together with the area of the top of the lid is $A \mathrm {~m} ^ { 2 }$.\\
(i) Express $h$ in terms of $x$ and hence show that $A = \frac { 3 } { 2 } x ^ { 2 } + \frac { 24 } { x }$.\\
(ii) Given that $x$ can vary, find the value of $x$ for which $A$ is a minimum, showing clearly that $A$ is a minimum and not a maximum.\\[0pt]
[5]
\hfill \mbox{\textit{CAIE P1 2010 Q10 [10]}}