Edexcel C1 2013 June — Question 8 8 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2013
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypePerimeter or area constraint inequality
DifficultyModerate -0.8 This is a straightforward C1 inequality question requiring basic algebraic manipulation. Part (a) involves setting up and solving a linear inequality from a perimeter constraint (routine), part (b) requires forming and solving a quadratic inequality using factorization, and part (c) combines the results. All steps are standard textbook exercises with no novel problem-solving required.
Spec1.02g Inequalities: linear and quadratic in single variable1.02h Express solutions: using 'and', 'or', set and interval notation
  1. A rectangular room has a width of \(x \mathrm {~m}\).
The length of the room is 4 m longer than its width. Given that the perimeter of the room is greater than 19.2 m ,
  1. show that \(x > 2.8\) Given also that the area of the room is less than \(21 \mathrm {~m} ^ { 2 }\),
    1. write down an inequality, in terms of \(x\), for the area of the room.
    2. Solve this inequality.
  3. Hence find the range of possible values for \(x\).
Question 8:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Length is \(x + 4\)B1 May be implied
\(x + x + x + 4 + x + 4 > 19.2 \Rightarrow x > ..\)M1 \(2x + 2(x \pm 4) > 19.2\) and proceeds to \(x >\). Attempts 2 widths + 2 lengths \(> 19.2\) leading to \(x > ...\)
\(x > 2.8\)A1(*) Achieves \(x > 2.8\) with no errors
Part (b)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x(x+4) < 21\)B1 cao
Part (b)(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x^2 + 4x - 21 < 0\), \((x+7)(x-3) < 0 \Rightarrow x = ...\)M1 Multiply out lhs, produce 3TQ \(= 0\) and attempt to solve leading to \(x = ...\)
Either \(-7 < x < 3\) or \(0 < x < 3\)M1A1 M1: Attempts the 'inside' for their critical values. A1: Accept either \(-7 < x < 3\) or \(0 < x < 3\) or (\(x > -7\) and \(x < 3\)) or (\(x > 0\) and \(x < 3\))
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(2.8 < x < 3\)B1ft Follow through their answers to (a) and (b), provided "their 3" \(> 2.8\) 
## Question 8:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Length is $x + 4$ | B1 | May be implied |
| $x + x + x + 4 + x + 4 > 19.2 \Rightarrow x > ..$ | M1 | $2x + 2(x \pm 4) > 19.2$ and proceeds to $x >$. Attempts 2 widths + 2 lengths $> 19.2$ leading to $x > ...$ |
| $x > 2.8$ | A1(*) | Achieves $x > 2.8$ with no errors |

### Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x(x+4) < 21$ | B1 | cao |

### Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 + 4x - 21 < 0$, $(x+7)(x-3) < 0 \Rightarrow x = ...$ | M1 | Multiply out lhs, produce 3TQ $= 0$ and attempt to solve leading to $x = ...$ |
| Either $-7 < x < 3$ or $0 < x < 3$ | M1A1 | M1: Attempts the 'inside' for their critical values. A1: Accept either $-7 < x < 3$ or $0 < x < 3$ or ($x > -7$ and $x < 3$) or ($x > 0$ and $x < 3$) |

### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2.8 < x < 3$ | B1ft | Follow through their answers to (a) and (b), provided "their 3" $> 2.8$ |

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\begin{enumerate}
  \item A rectangular room has a width of $x \mathrm {~m}$.
\end{enumerate}

The length of the room is 4 m longer than its width.

Given that the perimeter of the room is greater than 19.2 m ,\\
(a) show that $x > 2.8$

Given also that the area of the room is less than $21 \mathrm {~m} ^ { 2 }$,\\
(b) (i) write down an inequality, in terms of $x$, for the area of the room.\\
(ii) Solve this inequality.\\
(c) Hence find the range of possible values for $x$.\\

\hfill \mbox{\textit{Edexcel C1 2013 Q8 [8]}}
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