OCR MEI C4 2008 June — Question 9 18 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2008
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeSeparable variables - partial fractions
DifficultyStandard +0.3 This is a standard C4 differential equations question with clear scaffolding through five parts. Parts (i)-(ii) involve straightforward integration of an exponential function. Parts (iii)-(v) follow a routine procedure: partial fractions decomposition, separation of variables, and integration—all standard techniques for this level. The context is applied but the mathematical steps are textbook exercises with no novel problem-solving required.
Spec1.02y Partial fractions: decompose rational functions4.10a General/particular solutions: of differential equations4.10c Integrating factor: first order equations
9 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).
  1. Find \(v\) in terms of \(t\).
  2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
  3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
  4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
  5. According to this model, what is the speed of the skydiver in the long term? RECOGNISING ACHIEVEMENT \section*{ADVANCED GCE} \section*{4754/01B} \section*{MATHEMATICS (MEI)} Applications of Advanced Mathematics (C4) Paper B: Comprehension
    WEDNESDAY 21 MAY 2008
    Afternoon
    Time: Up to 1 hour
    Additional materials: Rough paper
    MEI Examination Formulae and Tables (MF 2) \section*{Candidate Forename}
    \includegraphics[max width=\textwidth, alt={}]{8ad99e2a-4cef-40b3-af8d-673b97536227-05_125_547_986_516}
    This document consists of \(\mathbf { 6 }\) printed pages, \(\mathbf { 2 }\) blank pages and an insert. 1 Complete these Latin square puzzles.
  6. 213
    3
  7. \includegraphics[max width=\textwidth, alt={}, center]{8ad99e2a-4cef-40b3-af8d-673b97536227-06_391_419_836_854} 2 In line 51, the text says that the Latin square
    1234
    3142
    2413
    4321
    could not be the solution to a Sudoku puzzle.
    Explain this briefly.
    3 On lines 114 and 115 the text says "It turns out that there are 16 different ways of filling in the remaining cells while keeping to the Sudoku rules. One of these ways is shown in Fig.10." Complete the grid below with a solution different from that given in Fig. 10.
    1234
    4 Lines 154 and 155 of the article read "There are three other embedded Latin squares in Fig. 14; one of them is illustrated in Fig. 16." Indicate one of the other two embedded Latin squares on this copy of Fig. 14.
    4231
    24
    42
    2413
    5 The number of \(9 \times 9\) Sudokus is given in line 121 .
    Without doing any calculations, explain why you would expect 9! to be a factor of this number.
    6 In the table below, \(M\) represents the maximum number of givens for which a Sudoku puzzle may have no unique solution (Investigation 3 in the article). \(s\) is the side length of the Sudoku grid and \(b\) is the side length of its blocks.
    Block side
    length, \(b\)
    Sudoku,
    \(s \times s\)
    		\(M\)
    1\(1 \times 1\)-
    2\(4 \times 4\)12
    3\(9 \times 9\)
    4\(16 \times 16\)
    5
  8. Complete the table.
  9. Give a formula for \(M\) in terms of \(b\).
    7 A man is setting a Sudoku puzzle and starts with this solution.
    123456789
    456897312
    789312564
    231564897
    564978123
    897123645
    312645978
    645789231
    978231456
    He then removes some of the numbers to give the puzzles in parts (i) and (ii). In each case explain briefly, and without trying to solve the puzzle, why it does not have a unique solution.
    [0pt] [2,2]
  10. 12469
    4891
    86
    2147
    647812
    8924
    16497
    64791
    982146
  11. 123456789
    456897312
    789564
    231564897
    564978123
    897645
    312645978
    645789231
    978456
  12. \(\_\_\_\_\)
  13. \(\_\_\_\_\)
Question 9(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(v = \int 10e^{-\frac{1}{2}t}\,dt\)M1 separate variables and intend to integrate
\(= -20e^{-\frac{1}{2}t}+c\)A1 \(-20e^{-\frac{1}{2}t}\)
when \(t=0,\ v=0\): \(0=-20+c \Rightarrow c=20\)M1 finding \(c\)
so \(v = 20-20e^{-\frac{1}{2}t}\)A1 cao
Question 9(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
As \(t\to\infty\), \(e^{-\frac{1}{2}t}\to 0 \Rightarrow v\to 20\)M1
So long term speed is \(20\ \text{m s}^{-1}\)A1 ft (for their \(c>0\), found)
Question 9(iii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{1}{(w-4)(w+5)} = \frac{A}{w-4}+\frac{B}{w+5} = \frac{A(w+5)+B(w-4)}{(w-4)(w+5)}\)M1 cover up, substitution or equating coeffs
\(w=4\): \(1=9A \Rightarrow A=\frac{1}{9}\)A1 \(\frac{1}{9}\)
\(w=-5\): \(1=-9B \Rightarrow B=-\frac{1}{9}\)A1 \(-\frac{1}{9}\)
\(\frac{1}{(w-4)(w+5)} = \frac{1}{9(w-4)}-\frac{1}{9(w+5)}\)[4]
Question 9(iv):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{dw}{dt} = -\frac{1}{2}(w-4)(w+5)\)
\(\Rightarrow \int\frac{dw}{(w-4)(w+5)} = \int-\frac{1}{2}\,dt\)M1 separating variables
\(\Rightarrow \int\left[\frac{1}{9(w-4)}-\frac{1}{9(w+5)}\right]dw = \int-\frac{1}{2}\,dt\)M1 substituting their partial fractions
\(\Rightarrow \frac{1}{9}\ln(w-4)-\frac{1}{9}\ln(w+5) = -\frac{1}{2}t+c\)A1ft integrating correctly (condone absence of \(c\))
\(\Rightarrow \frac{1}{9}\ln\frac{w-4}{w+5} = -\frac{1}{2}t+c\)
When \(t=0,\ w=10\): \(c=\frac{1}{9}\ln\frac{6}{15}=\frac{1}{9}\ln\frac{2}{5}\)M1 correctly evaluating \(c\) (at any stage)
\(\Rightarrow \ln\frac{w-4}{w+5} = -\frac{9}{2}t+\ln\frac{2}{5}\)M1 combining lns (at any stage)
\(\Rightarrow \frac{w-4}{w+5} = e^{-\frac{9}{2}t+\ln\frac{2}{5}} = \frac{2}{5}e^{-\frac{9}{2}t} = 0.4e^{-4.5t}\)E1 www
Question 9(v):
AnswerMarks Guidance
Answer/WorkingMark Guidance
As \(t\to\infty\), \(e^{-4.5t}\to 0 \Rightarrow w-4\to 0\)M1
So long term speed is \(4\ \text{m s}^{-1}\)A1
Comprehension Question 1(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\begin{array}{c c
Comprehension Question 1(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\begin{array}{c c
Comprehension Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Dividing the grid into four \(2\times2\) blocks — lines drawn on diagram or reference to \(2\times2\) blocksM1
One (or more) block does not contain all 4 of the symbols 1, 2, 3 and 4E1 oe
Comprehension Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Valid completed \(4\times4\) Latin square with first row \(1,2,3,4\)B1 Row 2 correct
B1Rest correct
Comprehension Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Correct completed Sudoku (two possible solutions shown)B2
Comprehension Question 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
In the top row there are 9 ways of allocating a symbol to the left cell, then 8 for the next, 7 for the next and so on down to 1 for the right cellM1
\(9\times8\times7\times6\times5\times4\times3\times2\times1 = 9!\) ways; so there must be \(9!\times\) the number of ways of completing the rest of the puzzleE1
Comprehension Question 6(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(b=5\): \(25\times25\)B1
\(M\) values: \(77,\ 252\) and \(621\)B1
Comprehension Question 6(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(M = b^4 - 4\)B1 \(b^4\)
B1\(-4\) 
## Question 9(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = \int 10e^{-\frac{1}{2}t}\,dt$ | M1 | separate variables and intend to integrate |
| $= -20e^{-\frac{1}{2}t}+c$ | A1 | $-20e^{-\frac{1}{2}t}$ |
| when $t=0,\ v=0$: $0=-20+c \Rightarrow c=20$ | M1 | finding $c$ |
| so $v = 20-20e^{-\frac{1}{2}t}$ | A1 | cao |

---

## Question 9(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| As $t\to\infty$, $e^{-\frac{1}{2}t}\to 0 \Rightarrow v\to 20$ | M1 | |
| So long term speed is $20\ \text{m s}^{-1}$ | A1 | ft (for their $c>0$, found) |

---

## Question 9(iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{(w-4)(w+5)} = \frac{A}{w-4}+\frac{B}{w+5} = \frac{A(w+5)+B(w-4)}{(w-4)(w+5)}$ | M1 | cover up, substitution or equating coeffs |
| $w=4$: $1=9A \Rightarrow A=\frac{1}{9}$ | A1 | $\frac{1}{9}$ |
| $w=-5$: $1=-9B \Rightarrow B=-\frac{1}{9}$ | A1 | $-\frac{1}{9}$ |
| $\frac{1}{(w-4)(w+5)} = \frac{1}{9(w-4)}-\frac{1}{9(w+5)}$ | [4] | |

---

## Question 9(iv):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dw}{dt} = -\frac{1}{2}(w-4)(w+5)$ | | |
| $\Rightarrow \int\frac{dw}{(w-4)(w+5)} = \int-\frac{1}{2}\,dt$ | M1 | separating variables |
| $\Rightarrow \int\left[\frac{1}{9(w-4)}-\frac{1}{9(w+5)}\right]dw = \int-\frac{1}{2}\,dt$ | M1 | substituting their partial fractions |
| $\Rightarrow \frac{1}{9}\ln(w-4)-\frac{1}{9}\ln(w+5) = -\frac{1}{2}t+c$ | A1ft | integrating correctly (condone absence of $c$) |
| $\Rightarrow \frac{1}{9}\ln\frac{w-4}{w+5} = -\frac{1}{2}t+c$ | | |
| When $t=0,\ w=10$: $c=\frac{1}{9}\ln\frac{6}{15}=\frac{1}{9}\ln\frac{2}{5}$ | M1 | correctly evaluating $c$ (at any stage) |
| $\Rightarrow \ln\frac{w-4}{w+5} = -\frac{9}{2}t+\ln\frac{2}{5}$ | M1 | combining lns (at any stage) |
| $\Rightarrow \frac{w-4}{w+5} = e^{-\frac{9}{2}t+\ln\frac{2}{5}} = \frac{2}{5}e^{-\frac{9}{2}t} = 0.4e^{-4.5t}$ | E1 | www |

---

## Question 9(v):

| Answer/Working | Mark | Guidance |
|---|---|---|
| As $t\to\infty$, $e^{-4.5t}\to 0 \Rightarrow w-4\to 0$ | M1 | |
| So long term speed is $4\ \text{m s}^{-1}$ | A1 | |

---

## Comprehension Question 1(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{array}{|c|c|c|}2&1&3\\3&\mathbf{2}&\mathbf{1}\\1&\mathbf{3}&\mathbf{2}\end{array}$ | B1 | cao |

## Comprehension Question 1(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{array}{|c|c|c|}2&3&1\\3&1&2\\1&2&3\end{array}$ | B1 | cao |

---

## Comprehension Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Dividing the grid into four $2\times2$ blocks — lines drawn on diagram or reference to $2\times2$ blocks | M1 | |
| One (or more) block does not contain all 4 of the symbols 1, 2, 3 and 4 | E1 | oe |

---

## Comprehension Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Valid completed $4\times4$ Latin square with first row $1,2,3,4$ | B1 | Row 2 correct |
| | B1 | Rest correct |

---

## Comprehension Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct completed Sudoku (two possible solutions shown) | B2 | |

---

## Comprehension Question 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| In the top row there are 9 ways of allocating a symbol to the left cell, then 8 for the next, 7 for the next and so on down to 1 for the right cell | M1 | |
| $9\times8\times7\times6\times5\times4\times3\times2\times1 = 9!$ ways; so there must be $9!\times$ the number of ways of completing the rest of the puzzle | E1 | |

---

## Comprehension Question 6(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $b=5$: $25\times25$ | B1 | |
| $M$ values: $77,\ 252$ and $621$ | B1 | |

## Comprehension Question 6(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $M = b^4 - 4$ | B1 | $b^4$ |
| | B1 | $-4$ |
9 A skydiver drops from a helicopter. Before she opens her parachute, her speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ after time $t$ seconds is modelled by the differential equation

$$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$

When $t = 0 , v = 0$.\\
(i) Find $v$ in terms of $t$.\\
(ii) According to this model, what is the speed of the skydiver in the long term?

She opens her parachute when her speed is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Her speed $t$ seconds after this is $w \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and is modelled by the differential equation

$$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$

(iii) Express $\frac { 1 } { ( w - 4 ) ( w + 5 ) }$ in partial fractions.\\
(iv) Using this result, show that $\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }$.\\
(v) According to this model, what is the speed of the skydiver in the long term?

RECOGNISING ACHIEVEMENT

\section*{ADVANCED GCE}
\section*{4754/01B}
\section*{MATHEMATICS (MEI)}
Applications of Advanced Mathematics (C4) Paper B: Comprehension\\
WEDNESDAY 21 MAY 2008\\
Afternoon\\
Time: Up to 1 hour\\
Additional materials: Rough paper\\
MEI Examination Formulae and Tables (MF 2)

\section*{Candidate Forename}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{8ad99e2a-4cef-40b3-af8d-673b97536227-05_125_547_986_516}
\end{center}

This document consists of $\mathbf { 6 }$ printed pages, $\mathbf { 2 }$ blank pages and an insert.

1 Complete these Latin square puzzles.\\
(i)

\begin{center}
\begin{tabular}{ | l | l | l | }
\hline
2 & 1 & 3 \\
\hline
3 &  &  \\
\hline
 &  &  \\
\hline
\end{tabular}
\end{center}

(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{8ad99e2a-4cef-40b3-af8d-673b97536227-06_391_419_836_854}

2 In line 51, the text says that the Latin square

\begin{center}
\begin{tabular}{ | l | l | l | l | }
\hline
1 & 2 & 3 & 4 \\
\hline
3 & 1 & 4 & 2 \\
\hline
2 & 4 & 1 & 3 \\
\hline
4 & 3 & 2 & 1 \\
\hline
\end{tabular}
\end{center}

could not be the solution to a Sudoku puzzle.\\
Explain this briefly.\\

3 On lines 114 and 115 the text says "It turns out that there are 16 different ways of filling in the remaining cells while keeping to the Sudoku rules. One of these ways is shown in Fig.10."

Complete the grid below with a solution different from that given in Fig. 10.

\begin{center}
\begin{tabular}{ | l | l | l | l | }
\hline
1 & 2 & 3 & 4 \\
\hline
 &  &  &  \\
\hline
 &  &  &  \\
\hline
 &  &  &  \\
\hline
\end{tabular}
\end{center}

4 Lines 154 and 155 of the article read "There are three other embedded Latin squares in Fig. 14; one of them is illustrated in Fig. 16."

Indicate one of the other two embedded Latin squares on this copy of Fig. 14.

\begin{center}
\begin{tabular}{ | l | l | l | l | }
\hline
4 & 2 & 3 & 1 \\
\hline
 &  & 2 & 4 \\
\hline
 &  & 4 & 2 \\
\hline
2 & 4 & 1 & 3 \\
\hline
\end{tabular}
\end{center}

5 The number of $9 \times 9$ Sudokus is given in line 121 .\\
Without doing any calculations, explain why you would expect 9! to be a factor of this number.\\

6 In the table below, $M$ represents the maximum number of givens for which a Sudoku puzzle may have no unique solution (Investigation 3 in the article). $s$ is the side length of the Sudoku grid and $b$ is the side length of its blocks.

\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
\begin{tabular}{ c }
Block side \\
length, $b$ \\
\end{tabular} & \begin{tabular}{ c }
Sudoku, \\
$s \times s$ \\
\end{tabular} & $M$ \\
\hline
1 & $1 \times 1$ & - \\
\hline
2 & $4 \times 4$ & 12 \\
\hline
3 & $9 \times 9$ &  \\
\hline
4 & $16 \times 16$ &  \\
\hline
5 &  &  \\
\hline
\end{tabular}
\end{center}

(i) Complete the table.\\
(ii) Give a formula for $M$ in terms of $b$.\\

7 A man is setting a Sudoku puzzle and starts with this solution.

\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | l | }
\hline
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
4 & 5 & 6 & 8 & 9 & 7 & 3 & 1 & 2 \\
\hline
7 & 8 & 9 & 3 & 1 & 2 & 5 & 6 & 4 \\
\hline
2 & 3 & 1 & 5 & 6 & 4 & 8 & 9 & 7 \\
\hline
5 & 6 & 4 & 9 & 7 & 8 & 1 & 2 & 3 \\
\hline
8 & 9 & 7 & 1 & 2 & 3 & 6 & 4 & 5 \\
\hline
3 & 1 & 2 & 6 & 4 & 5 & 9 & 7 & 8 \\
\hline
6 & 4 & 5 & 7 & 8 & 9 & 2 & 3 & 1 \\
\hline
9 & 7 & 8 & 2 & 3 & 1 & 4 & 5 & 6 \\
\hline
\end{tabular}
\end{center}

He then removes some of the numbers to give the puzzles in parts (i) and (ii). In each case explain briefly, and without trying to solve the puzzle, why it does not have a unique solution.\\[0pt]
[2,2]\\
(i)

\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | l | }
\hline
1 & 2 &  & 4 &  & 6 &  &  & 9 \\
\hline
4 &  &  & 8 & 9 &  &  & 1 &  \\
\hline
 & 8 &  &  &  &  &  & 6 &  \\
\hline
2 &  & 1 &  &  & 4 &  &  & 7 \\
\hline
 & 6 & 4 &  & 7 & 8 & 1 & 2 &  \\
\hline
8 & 9 &  &  & 2 &  &  & 4 &  \\
\hline
 & 1 &  & 6 & 4 &  & 9 & 7 &  \\
\hline
6 & 4 &  & 7 &  & 9 &  &  & 1 \\
\hline
9 &  & 8 & 2 &  & 1 & 4 &  & 6 \\
\hline
\end{tabular}
\end{center}

(ii)

\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | l | }
\hline
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
4 & 5 & 6 & 8 & 9 & 7 & 3 & 1 & 2 \\
\hline
7 & 8 & 9 &  &  &  & 5 & 6 & 4 \\
\hline
2 & 3 & 1 & 5 & 6 & 4 & 8 & 9 & 7 \\
\hline
5 & 6 & 4 & 9 & 7 & 8 & 1 & 2 & 3 \\
\hline
8 & 9 & 7 &  &  &  & 6 & 4 & 5 \\
\hline
3 & 1 & 2 & 6 & 4 & 5 & 9 & 7 & 8 \\
\hline
6 & 4 & 5 & 7 & 8 & 9 & 2 & 3 & 1 \\
\hline
9 & 7 & 8 &  &  &  & 4 & 5 & 6 \\
\hline
\end{tabular}
\end{center}

(i) $\_\_\_\_$\\

(ii) $\_\_\_\_$\\

\hfill \mbox{\textit{OCR MEI C4 2008 Q9 [18]}}
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