| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Perimeter or area constraint inequality |
| Difficulty | Moderate -0.3 This is a straightforward C1 inequality question requiring basic perimeter and area calculations from a rectilinear shape, followed by solving a simple quadratic inequality. The algebra is routine and the problem-solving demand is minimal—students just need to set up expressions and apply standard techniques. Slightly easier than average due to the guided structure and elementary geometric reasoning. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = 20x + 6\) oe | B1 | Accept unsimplified e.g. \(2x+1+2x+4x+2+2x+6x+3+4x\) or \(2(10x+3)\) |
| \(20x + 6 > 40 \Rightarrow x >\) ... | M1 | Set \(P>40\) with their linear expression and manipulate to get \(x>\)... |
| \(x > 1.7\) | A1* | cao \(x>1.7\); given answer so no errors allowed; accept \(1.7 < x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = 2x(2x+1) + 2x(6x+3) = 16x^2 + 8x\) | B1 | Correct statement in \(x\) for area (need not be simplified). Among possibilities: \(2x(2x+1)+2x(6x+3)\), \(16x^2+8x\), \(4x(6x+3)-2x(4x+2)\), \(4x(2x+1)+2x(4x+2)\) |
| \(16x^2 + 8x - 120 < 0\) | M1 | Sets quadratic \(< 120\) and collects on one side |
| Try to solve \(2x^2 + x - 15 = 0\), e.g. \((2x-5)(x+3)=0\) | M1 | Attempt to solve 3-term quadratic by factorising, formula or completing the square |
| Choose inside region | M1 | For choosing 'inside' region; must be stated, not just table or graph; implied by \(0 < x <\) upper value |
| \(-3 < x < \frac{5}{2}\) or \(0 < x < \frac{5}{2}\) (as \(x\) is a length) | A1 | Accept \(x > -3\) and \(x < 2.5\) or \((-3, 2.5)\). As \(x\) is a width accept \(0 < x < \frac{5}{2}\). Also accept \(\frac{10}{4}\) or \(2.5\) instead of \(\frac{5}{2}\). \(\leq\) would be M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1.7 < x < \frac{5}{2}\) | B1cao | Must be correct. Does not imply final M1 in (b) |
## Question 6:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = 20x + 6$ oe | B1 | Accept unsimplified e.g. $2x+1+2x+4x+2+2x+6x+3+4x$ or $2(10x+3)$ |
| $20x + 6 > 40 \Rightarrow x >$ ... | M1 | Set $P>40$ with their linear expression and manipulate to get $x>$... |
| $x > 1.7$ | A1* | cao $x>1.7$; given answer so no errors allowed; accept $1.7 < x$ |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = 2x(2x+1) + 2x(6x+3) = 16x^2 + 8x$ | B1 | Correct statement in $x$ for area (need not be simplified). Among possibilities: $2x(2x+1)+2x(6x+3)$, $16x^2+8x$, $4x(6x+3)-2x(4x+2)$, $4x(2x+1)+2x(4x+2)$ |
| $16x^2 + 8x - 120 < 0$ | M1 | Sets quadratic $< 120$ and collects on one side |
| Try to solve $2x^2 + x - 15 = 0$, e.g. $(2x-5)(x+3)=0$ | M1 | Attempt to solve 3-term quadratic by factorising, formula or completing the square |
| Choose inside region | M1 | For choosing 'inside' region; must be stated, not just table or graph; implied by $0 < x <$ upper value |
| $-3 < x < \frac{5}{2}$ or $0 < x < \frac{5}{2}$ (as $x$ is a length) | A1 | Accept $x > -3$ **and** $x < 2.5$ or $(-3, 2.5)$. As $x$ is a width accept $0 < x < \frac{5}{2}$. Also accept $\frac{10}{4}$ or $2.5$ instead of $\frac{5}{2}$. $\leq$ would be M1A0 |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1.7 < x < \frac{5}{2}$ | B1cao | Must be correct. Does not imply final M1 in (b) |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6db8acbd-7f61-46ff-8fdc-f0f4a8363aa6-08_917_1322_239_303}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the plan of a garden. The marked angles are right angles.\\
The six edges are straight lines.\\
The lengths shown in the diagram are given in metres.\\
Given that the perimeter of the garden is greater than 40 m ,
\begin{enumerate}[label=(\alph*)]
\item show that $x > 1.7$
Given that the area of the garden is less than $120 \mathrm {~m} ^ { 2 }$,
\item form and solve a quadratic inequality in $x$.
\item Hence state the range of the possible values of $x$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q6 [9]}}