| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Optimisation via quadratic model |
| Difficulty | Moderate -0.8 This is a straightforward applied quadratic question requiring interpretation of completed square form. Part (a) involves substitution, part (b) solving a simple quadratic inequality, and part (c) reading the vertex from completed square form. All techniques are routine for AS-level with no problem-solving insight required. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(P = 100 - 6.25(15-9)^2\) | M1 | Substitutes \(x=15\) into \(P = 100 - 6.25(x-9)^2\) and attempts to calculate |
| \(= -125\), not sensible as company would make a loss | A1 | Finds \(P = -125\) or states \(P < 0\) AND explains company would make a loss. Condone \(P=-125\) with explanation of loss of £125 rather than £125,000. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(P > 80 \Rightarrow (x-9)^2 < 3.2\) or \(P = 80 \Rightarrow (x-9)^2 = 3.2\) | M1 | Uses \(P\)...80 where ... is any inequality or "=" in \(P = 100 - 6.25(x-9)^2\) and proceeds to \((x-9)^2\)...\(k\) where \(k>0\) |
| \(\Rightarrow 9 - \sqrt{3.2} < x < 9 + \sqrt{3.2}\) | dM1 | Award for solving to find two positive values for \(x\). Allow decimal answers. FYI correct answers: \(x = 9 \pm \sqrt{3.2}\) |
| Minimum Price \(= £7.22\) | A1 | Deduces minimum Price \(= £7.22\). (£7.21 not acceptable) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) Maximum profit \(= £100\,000\) | B1 | Accept "100 thousand pound(s)" with units |
| (ii) Selling price \(= £9\) | B1 | SC1: Missing units in (b) and (c) penalise once only, withhold final mark. SC2: If both correct but wrong order, score SC B1 B0 |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $P = 100 - 6.25(15-9)^2$ | M1 | Substitutes $x=15$ into $P = 100 - 6.25(x-9)^2$ and attempts to calculate |
| $= -125$, not sensible as company would make a loss | A1 | Finds $P = -125$ or states $P < 0$ AND explains company would make a loss. Condone $P=-125$ with explanation of loss of £125 rather than £125,000. |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $P > 80 \Rightarrow (x-9)^2 < 3.2$ or $P = 80 \Rightarrow (x-9)^2 = 3.2$ | M1 | Uses $P$...80 where ... is any inequality or "=" in $P = 100 - 6.25(x-9)^2$ and proceeds to $(x-9)^2$...$k$ where $k>0$ |
| $\Rightarrow 9 - \sqrt{3.2} < x < 9 + \sqrt{3.2}$ | dM1 | Award for solving to find two positive values for $x$. Allow decimal answers. FYI correct answers: $x = 9 \pm \sqrt{3.2}$ |
| Minimum Price $= £7.22$ | A1 | Deduces minimum Price $= £7.22$. (£7.21 not acceptable) |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) Maximum profit $= £100\,000$ | B1 | Accept "100 thousand pound(s)" with units |
| (ii) Selling price $= £9$ | B1 | SC1: Missing units in (b) and (c) penalise once only, withhold final mark. SC2: If both correct but wrong order, score SC B1 B0 |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f7935caa-6626-4ba8-87ef-e9bb59e1ac3e-12_599_1084_292_486}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A company makes a particular type of children's toy.\\
The annual profit made by the company is modelled by the equation
$$P = 100 - 6.25 ( x - 9 ) ^ { 2 }$$
where $P$ is the profit measured in thousands of pounds and $x$ is the selling price of the toy in pounds.
A sketch of $P$ against $x$ is shown in Figure 1.\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item explain why $\pounds 15$ is not a sensible selling price for the toy.
Given that the company made an annual profit of more than $\pounds 80000$
\item find, according to the model, the least possible selling price for the toy.
The company wishes to maximise its annual profit.\\
State, according to the model,
\item \begin{enumerate}[label=(\roman*)]
\item the maximum possible annual profit,
\item the selling price of the toy that maximises the annual profit.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2018 Q6 [7]}}