| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Optimisation via quadratic model |
| Difficulty | Standard +0.3 This is a straightforward applied quadratic optimization problem requiring coordinate geometry to establish the area formula (which is given to show), then completing the square or differentiation to find the minimum. The setup is clear, the algebra is routine, and finding a minimum of a quadratic is a standard technique, making this slightly easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = 2400 - 20(60-2x) - x(40-x) - 30x\) | M1 | Needs attempts at all areas |
| \(\rightarrow A = x^2 - 30x + 1200\) | A1 | co, answer given (could be trapezium \(-\) triangle) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dA}{dx} = 2x - 30\) or \((x-15)^2 + 975\) | B1 | co, either method okay |
| \(= 0\) when \(x = 15\) or Min at \(x = 15\) | M1 | Sets differential to \(0\) + solution |
| \(\rightarrow A = 975\) | A1 | co |
| [3] |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = 2400 - 20(60-2x) - x(40-x) - 30x$ | M1 | Needs attempts at all areas |
| $\rightarrow A = x^2 - 30x + 1200$ | A1 | co, answer given (could be trapezium $-$ triangle) |
| **[2]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dA}{dx} = 2x - 30$ or $(x-15)^2 + 975$ | B1 | co, either method okay |
| $= 0$ when $x = 15$ or Min at $x = 15$ | M1 | Sets differential to $0$ + solution |
| $\rightarrow A = 975$ | A1 | co |
| **[3]** | | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{11bfe5bd-604c-43e5-81e7-4c1f5676bcbb-2_485_755_751_696}
The diagram shows a plan for a rectangular park $A B C D$, in which $A B = 40 \mathrm {~m}$ and $A D = 60 \mathrm {~m}$. Points $X$ and $Y$ lie on $B C$ and $C D$ respectively and $A X , X Y$ and $Y A$ are paths that surround a triangular playground. The length of $D Y$ is $x \mathrm {~m}$ and the length of $X C$ is $2 x \mathrm {~m}$.\\
(i) Show that the area, $A \mathrm {~m} ^ { 2 }$, of the playground is given by
$$A = x ^ { 2 } - 30 x + 1200$$
(ii) Given that $x$ can vary, find the minimum area of the playground.
\hfill \mbox{\textit{CAIE P1 2012 Q3 [5]}}