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OCR MEI C4 2012 January Q5
5 marks Standard +0.3
5 Verify that the vector \(2 \mathbf { i } - \mathbf { j } + 4 \mathbf { k }\) is perpendicular to the plane through the points \(\mathrm { A } ( 2,0,1 ) , \mathrm { B } ( 1,2,2 )\) and \(\mathrm { C } ( 0 , - 4,1 )\). Hence find the cartesian equation of the plane.
OCR MEI C4 2012 January Q6
6 marks Standard +0.3
6 Given the binomial expansion \(( 1 + q x ) ^ { p } = 1 - x + 2 x ^ { 2 } + \ldots\), find the values of \(p\) and \(q\). Hence state the set of values of \(x\) for which the expansion is valid.
OCR MEI C4 2012 January Q7
5 marks Moderate -0.3
7 Show that the straight lines with equations \(\mathbf { r } = \left( \begin{array} { l } 4 \\ 2 \\ 4 \end{array} \right) + \lambda \left( \begin{array} { l } 3 \\ 0 \\ 1 \end{array} \right)\) and \(\mathbf { r } = \left( \begin{array} { r } - 1 \\ 4 \\ 9 \end{array} \right) + \mu \left( \begin{array} { r } - 1 \\ 1 \\ 3 \end{array} \right)\) meet.
Find their point of intersection.
OCR MEI C4 2012 January Q8
18 marks Standard +0.3
8 Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2 t ^ { 2 } , y = 4 t , \quad - \sqrt { 2 } \leqslant t \leqslant \sqrt { 2 } .$$ \(\mathrm { P } \left( 2 t ^ { 2 } , 4 t \right)\) is a point on the curve with parameter \(t\). TS is the tangent to the curve at P , and PR is the line through P parallel to the \(x\)-axis. Q is the point \(( 2,0 )\). The angles that PS and QP make with the positive \(x\)-direction are \(\theta\) and \(\phi\) respectively. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{81433914-a56f-4765-af34-990a0127f98b-03_969_1262_733_388} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac { 1 } { t }\).
  2. Find the gradient of the line QP in terms of \(t\). Hence show that \(\phi = 2 \theta\), and that angle TPQ is equal to \(\theta\).
    [0pt] [The above result shows that if a lamp bulb is placed at Q , then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the \(x\)-axis.
  3. Show that the curve has cartesian equation \(y ^ { 2 } = 8 x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\).
OCR MEI C4 2012 January Q9
18 marks Standard +0.3
9 \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{81433914-a56f-4765-af34-990a0127f98b-04_269_453_255_806} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure} Fig. 9 shows a hemispherical bowl, of radius 10 cm , filled with water to a depth of \(x \mathrm {~cm}\). It can be shown that the volume of water, \(V \mathrm {~cm} ^ { 3 }\), is given by $$V = \pi \left( 10 x ^ { 2 } - \frac { 1 } { 3 } x ^ { 3 } \right) .$$ Water is poured into a leaking hemispherical bowl of radius 10 cm . Initially, the bowl is empty. After \(t\) seconds, the volume of water is changing at a rate, in \(\mathrm { cm } ^ { 3 } \mathrm {~s} ^ { - 1 }\), given by the equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = k ( 20 - x ) ,$$ where \(k\) is a constant.
  1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} x }\), and hence show that \(\pi x \frac { \mathrm {~d} x } { \mathrm {~d} t } = k\).
  2. Solve this differential equation, and hence show that the bowl fills completely after \(T\) seconds, where $$T = \frac { 50 \pi } { k } .$$ Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of \(k x \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
  3. Show that, \(t\) seconds later, \(\pi ( 20 - x ) \frac { \mathrm { d } x } { \mathrm {~d} t } = - k\).
  4. Solve this differential equation. Hence show that the bowl empties in \(3 T\) seconds.
OCR MEI C4 2013 January Q2
6 marks Moderate -0.5
2 Find the first four terms of the binomial expansion of \(\sqrt [ 3 ] { 1 - 2 x }\). State the set of values of \(x\) for which the expansion is valid.
OCR MEI C4 2013 January Q3
7 marks Moderate -0.3
3 The parametric equations of a curve are $$x = \sin \theta , \quad y = \sin 2 \theta , \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi .$$
  1. Find the exact value of the gradient of the curve at the point where \(\theta = \frac { 1 } { 6 } \pi\).
  2. Show that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4 x ^ { 4 }\).
OCR MEI C4 2013 January Q4
8 marks Standard +0.3
4 Fig. 4 shows the curve \(y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }\), and the region between the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9bceee25-35bd-448b-a4a2-1a5667be5f11-02_650_727_1176_653} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
  1. Find the exact volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
    1. Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.
      \(x\)00.511.52
      \(y\)1.92832.89644.5919
    2. The trapezium rule for \(\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x\) with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.
OCR MEI C4 2013 January Q5
6 marks Moderate -0.3
5 Solve the equation \(2 \sec ^ { 2 } \theta = 5 \tan \theta\), for \(0 \leqslant \theta \leqslant \pi\).
OCR MEI C4 2013 January Q6
5 marks Moderate -0.3
6 In Fig. 6, \(\mathrm { ABC } , \mathrm { ACD }\) and AED are right-angled triangles and \(\mathrm { BC } = 1\) unit. Angles CAB and CAD are \(\theta\) and \(\phi\) respectively. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9bceee25-35bd-448b-a4a2-1a5667be5f11-03_440_524_504_753} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Find AC and AD in terms of \(\theta\) and \(\phi\).
  2. Hence show that \(\mathrm { DE } = 1 + \frac { \tan \phi } { \tan \theta }\). Section B (36 marks)
OCR MEI C4 2013 January Q7
17 marks Standard +0.3
7 A tent has vertices ABCDEF with coordinates as shown in Fig. 7. Lengths are in metres. The \(\mathrm { O } x y\) plane is horizontal. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9bceee25-35bd-448b-a4a2-1a5667be5f11-03_547_987_1580_539} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Find the length of the ridge of the tent DE , and the angle this makes with the horizontal.
  2. Show that the vector \(\mathbf { i } - 4 \mathbf { j } + 5 \mathbf { k }\) is normal to the plane through \(\mathrm { A } , \mathrm { D }\) and E . Hence find the equation of this plane. Given that B lies in this plane, find \(a\).
  3. Verify that the equation of the plane BCD is \(x + z = 8\). Hence find the acute angle between the planes ABDE and BCD .
OCR MEI C4 2013 January Q8
19 marks Standard +0.3
8 The growth of a tree is modelled by the differential equation $$10 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 20 - h ,$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).
  1. Write down the value of \(h\) for which \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0\), and interpret this in terms of the growth of the tree.
  2. Verify that \(h = 20 \left( 1 - \mathrm { e } ^ { - 0.1 t } \right)\) satisfies this differential equation and its initial condition. The alternative differential equation $$200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 400 - h ^ { 2 }$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
  3. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } } .$$
  4. What does this solution indicate about the long-term height of the tree?
  5. After a year, the tree has grown to a height of 2 m . Which model fits this information better?
OCR MEI C4 2009 June Q1
7 marks Moderate -0.3
1 Express \(4 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
Hence solve the equation \(4 \cos \theta - \sin \theta = 3\), for \(0 \leqslant \theta \leqslant 2 \pi\).
OCR MEI C4 2009 June Q2
7 marks Moderate -0.3
2 Using partial fractions, find \(\int \frac { x } { ( x + 1 ) ( 2 x + 1 ) } \mathrm { d } x\).
[0pt] [7]
OCR MEI C4 2009 June Q4
5 marks Moderate -0.3
4 The part of the curve \(y = 4 - x ^ { 2 }\) that is above the \(x\)-axis is rotated about the \(y\)-axis. This is shown in Fig. 4. Find the volume of revolution produced, giving your answer in terms of \(\pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b4861178-720d-4803-a608-abef350efb0e-2_531_587_1204_778} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
OCR MEI C4 2009 June Q5
7 marks Moderate -0.3
5 A curve has parametric equations $$x = a t ^ { 3 } , \quad y = \frac { a } { 1 + t ^ { 2 } }$$ where \(a\) is a constant.
Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { - 2 } { 3 t \left( 1 + t ^ { 2 } \right) ^ { 2 } }\).
Hence find the gradient of the curve at the point \(\left( a , \frac { 1 } { 2 } a \right)\).
OCR MEI C4 2009 June Q6
6 marks Moderate -0.3
6 Given that \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\), show that \(\cot ^ { 2 } \theta - \cot \theta - 2 = 0\).
Hence solve the equation \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\). Section B (36 marks)
OCR MEI C4 2009 June Q7
17 marks Moderate -0.3
7 When a light ray passes from air to glass, it is deflected through an angle. The light ray ABC starts at point \(\mathrm { A } ( 1,2,2 )\), and enters a glass object at point \(\mathrm { B } ( 0,0,2 )\). The surface of the glass object is a plane with normal vector \(\mathbf { n }\). Fig. 7 shows a cross-section of the glass object in the plane of the light ray and \(\mathbf { n }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b4861178-720d-4803-a608-abef350efb0e-3_684_812_516_664} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Find the vector \(\overrightarrow { \mathrm { AB } }\) and a vector equation of the line AB . The surface of the glass object is a plane with equation \(x + z = 2\). AB makes an acute angle \(\theta\) with the normal to this plane.
  2. Write down the normal vector \(\mathbf { n }\), and hence calculate \(\theta\), giving your answer in degrees. The line BC has vector equation \(\mathbf { r } = \left( \begin{array} { l } 0 \\ 0 \\ 2 \end{array} \right) + \mu \left( \begin{array} { l } - 2 \\ - 2 \\ - 1 \end{array} \right)\). This line makes an acute angle \(\phi\) with the normal to the plane.
  3. Show that \(\phi = 45 ^ { \circ }\).
  4. Snell's Law states that \(\sin \theta = k \sin \phi\), where \(k\) is a constant called the refractive index. Find \(k\). The light ray leaves the glass object through a plane with equation \(x + z = - 1\). Units are centimetres.
  5. Find the point of intersection of the line BC with the plane \(x + z = - 1\). Hence find the distance the light ray travels through the glass object. \section*{[Question 8 is printed overleaf.]} OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations, is given to all schools that receive assessment material and is freely available to download from our public website (\href{http://www.ocr.org.uk}{www.ocr.org.uk}) after the live examination series.
    If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1PB. OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.
OCR MEI C4 2009 June Q8
19 marks Standard +0.8
8 Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for \(\pi\).
  1. Fig. 8.1 shows a regular 12 -sided polygon inscribed in a circle of radius 1 unit, centre \(\mathrm { O } . \mathrm { AB }\) is one of the sides of the polygon. \(C\) is the midpoint of \(A B\). Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{b4861178-720d-4803-a608-abef350efb0e-4_455_428_523_900} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
    \end{figure} (A) Show that \(\mathrm { AB } = 2 \sin 15 ^ { \circ }\).
    (B) Use a double angle formula to express \(\cos 30 ^ { \circ }\) in terms of \(\sin 15 ^ { \circ }\). Using the exact value of \(\cos 30 ^ { \circ }\), show that \(\sin 15 ^ { \circ } = \frac { 1 } { 2 } \sqrt { 2 - \sqrt { 3 } }\).
    (C) Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6 \sqrt { 2 - \sqrt { 3 } }\).
  2. In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{b4861178-720d-4803-a608-abef350efb0e-4_456_428_1621_900} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
    \end{figure} (A) Show that \(\mathrm { DE } = 2 \tan 15 ^ { \circ }\).
    (B) Let \(t = \tan 15 ^ { \circ }\). Use a double angle formula to express \(\tan 30 ^ { \circ }\) in terms of \(t\). Hence show that \(t ^ { 2 } + 2 \sqrt { 3 } t - 1 = 0\).
    (C) Solve this equation, and hence show that \(\pi < 12 ( 2 - \sqrt { 3 } )\).
  3. Use the results in parts (i)( \(C\) ) and (ii)( \(C\) ) to establish upper and lower bounds for the value of \(\pi\), giving your answers in decimal form. \section*{ADVANCED GCE
    MATHEMATICS (MEI)} 4754B
    Applications of Advanced Mathematics (C4) Paper B: Comprehension Candidates answer on the question paper
    Monday 1 June 2009
    OCR Supplied Materials:
    Morning
    • Insert (inserted)
    • MEI Examination Formulae and Tables (MF2)
    Duration: Up to 1 hour
    Other Materials Required:
    • Rough paper \includegraphics[max width=\textwidth, alt={}, center]{b4861178-720d-4803-a608-abef350efb0e-5_122_442_1023_1370}
    1 On lines 90 and 91, the article says "The average score for each player works out to be 0.25 points per round". Derive this figure. 2 Line 47 gives the inequality \(b > c > d > w\).
    Interpret each of the following inequalities in the context of the example from the 1st World War.
  4. \(b > w\)
  5. \(c > d\)
  6. \(\_\_\_\_\)
  7. \(\_\_\_\_\) 3 Table 3 illustrates a possible game where you always co-operate. In lines 98 and 99 the article says "Clearly the longer the game goes on the closer your average score approaches - 2 points per round and that of your opponent approaches 3 ." How many rounds have you played when your average score is - 1.999 ?
    4 A Prisoner's Dilemma game is proposed in which $$b = 6 , c = 1 , d = - 1 \text { and } w = - 3 .$$ Using the information in the article, state whether these values would allow long-term co-operation to evolve. Justify your answer.
    5 In a Prisoner's Dilemma game both players keep strictly to a Tit-for-tat strategy. You start with C and your opponent starts with D . The scoring system of \(b = 3 , c = 1 , d = - 1\) and \(w = - 2\) is used.
  8. This table shows the first 8 out of many rounds. Complete the table.
    RoundYouOpponentYour scoreOpponent's score
    1CD
    2
    3
    4
    5
    6
    7
    8
    ⋯⋯⋯⋯⋯
  9. Find your average score per round in the long run.
    6 In the article, the scoring system is \(b = 3 , c = 1 , d = - 1\) and \(w = - 2\). In Axelrod's experiment, negative numbers were avoided by taking \(b = 5 , c = 3 , d = 1\) and \(w = 0\). State the effect this change would have on
  10. the players' scores,
  11. who wins.
  12. \(\_\_\_\_\)
  13. \(\_\_\_\_\) 7 Two companies, X and Y , are the only sellers of ice cream on an island. They both have a market share of about \(50 \%\). Although their ice cream is much the same, both companies spend a lot of money on advertising.
  14. What agreement might the companies reach if they decide to co-operate?
  15. What advantage would a company hope to gain by 'defecting' from this agreement?
    RECOGNISING ACHIEVEMENT
OCR MEI C4 2011 June Q1
5 marks Moderate -0.3
1 Express \(\frac { 1 } { ( 2 x + 1 ) \left( x ^ { 2 } + 1 \right) }\) in partial fractions.
OCR MEI C4 2011 June Q2
5 marks Moderate -0.8
2 Find the first three terms in the binomial expansion of \(\sqrt [ 3 ] { 1 + 3 x }\) in ascending powers of \(x\). State the set of values of \(x\) for which the expansion is valid.
OCR MEI C4 2011 June Q3
6 marks Moderate -0.3
3 Express \(2 \sin \theta - 3 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and \(0 < \alpha < \frac { 1 } { 2 } \pi\). Hence write down the greatest and least possible values of \(1 + 2 \sin \theta - 3 \cos \theta\).
OCR MEI C4 2011 June Q4
7 marks Moderate -0.3
4 A curve has parametric equations $$x = 2 \sin \theta , \quad y = \cos 2 \theta$$
  1. Find the exact coordinates and the gradient of the curve at the point with parameter \(\theta = \frac { 1 } { 3 } \pi\).
  2. Find \(y\) in terms of \(x\).
OCR MEI C4 2011 June Q6
7 marks Challenging +1.2
6 Fig. 6 shows the region enclosed by part of the curve \(y = 2 x ^ { 2 }\), the straight line \(x + y = 3\), and the \(y\)-axis. The curve and the straight line meet at \(\mathrm { P } ( 1,2 )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0a6247c9-ba64-4a8f-9e10-83986136cf56-2_643_933_1667_607} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} The shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find, in terms of \(\pi\), the volume of the solid of revolution formed.
[0pt] [You may use the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) for the volume of a cone.] Section B (36 marks)
OCR MEI C4 2011 June Q7
18 marks Standard +0.3
7 A piece of cloth ABDC is attached to the tops of vertical poles \(\mathrm { AE } , \mathrm { BF } , \mathrm { DG }\) and CH , where \(\mathrm { E } , \mathrm { F } , \mathrm { G }\) and H are at ground level (see Fig. 7). Coordinates are as shown, with lengths in metres. The length of pole DG is \(k\) metres. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0a6247c9-ba64-4a8f-9e10-83986136cf56-3_933_1436_518_351} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Write down the vectors \(\overrightarrow { \mathrm { AB } }\) and \(\overrightarrow { \mathrm { AC } }\). Hence calculate the angle BAC .
  2. Verify that the equation of the plane ABC is \(x + y - 2 z + d = 0\), where \(d\) is a constant to be determined. Calculate the acute angle the plane makes with the horizontal plane.
  3. Given that \(\mathrm { A } , \mathrm { B } , \mathrm { D }\) and C are coplanar, show that \(k = 3\). Hence show that ABDC is a trapezium, and find the ratio of CD to AB .