Questions C4 (1219 questions)

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OCR MEI C4 2011 January Q2
5 marks Moderate -0.3
2 A curve is defined parametrically by the equations $$x = \frac { 1 } { 1 + t } , \quad y = \frac { 1 - t } { 1 + 2 t }$$ Find \(t\) in terms of \(x\). Hence find the cartesian equation of the curve, giving your answer as simply as possible.
OCR MEI C4 2011 January Q3
7 marks Moderate -0.3
3 Find the first three terms in the binomial expansion of \(\frac { 1 } { ( 3 - 2 x ) ^ { 3 } }\) in ascending powers of \(x\). State the set of values of \(x\) for which the expansion is valid.
OCR MEI C4 2011 January Q4
7 marks Moderate -0.3
4 The points A , B and C have coordinates \(( 2,0 , - 1 ) , ( 4,3 , - 6 )\) and \(( 9,3 , - 4 )\) respectively.
  1. Show that AB is perpendicular to BC .
  2. Find the area of triangle ABC .
OCR MEI C4 2011 January Q5
3 marks Moderate -0.8
5 Show that \(\frac { \sin 2 \theta } { 1 + \cos 2 \theta } = \tan \theta\).
OCR MEI C4 2011 January Q6
8 marks Standard +0.3
6
  1. Find the point of intersection of the line \(\mathbf { r } = \left( \begin{array} { r } - 8 \\ - 2 \\ 6 \end{array} \right) + \lambda \left( \begin{array} { r } - 3 \\ 0 \\ 1 \end{array} \right)\) and the plane \(2 x - 3 y + z = 11\).
  2. Find the acute angle between the line and the normal to the plane. Section B (36 marks)
OCR MEI C4 2011 January Q7
18 marks Standard +0.3
7 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).
  1. Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
  2. Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\). In a second model, \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when \(t = 0 , v = 0\).
  3. Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
  4. Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  5. Which of the two models fits the data better?
OCR MEI C4 2011 January Q8
18 marks Standard +0.3
8 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at \(\alpha\) to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all \(\beta\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f657e167-e6f8-4df2-901b-067c32835877-04_684_872_461_278} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure} In the following, all lengths are in metres.
  1. Find AC in terms of \(\alpha\), and hence show that \(\mathrm { GF } = 10 \sec \alpha \tan \beta\).
  2. Show that \(\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )\). $$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$ Similarly, it can be shown that \(\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }\). [You are not required to derive this result.]
    You are now given that \(\alpha = 45 ^ { \circ }\) and that \(\tan \beta = t\).
  3. Find CE and CD in terms of \(t\). Hence show that \(\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }\).
  4. Show that \(\mathrm { GF } = 10 \sqrt { 2 } t\). For a certain value of \(\beta , \mathrm { DE } = 2 \mathrm { GF }\).
  5. Show that \(t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }\). Hence find this value of \(\beta\).
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 2015 June Q2
7 marks Moderate -0.3
2 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
OCR MEI C4 2015 June Q3
8 marks Standard +0.3
3
  1. Find the first three terms of the binomial expansion of \(\frac { 1 } { \sqrt [ 3 ] { 1 - 2 x } }\). State the set of values of \(x\) for which
    the expansion is valid. the expansion is valid.
  2. Hence find \(a\) and \(b\) such that \(\frac { 1 - 3 x } { \sqrt [ 3 ] { 1 - 2 x } } = 1 + a x + b x ^ { 2 } + \ldots\).
OCR MEI C4 2015 June Q4
8 marks Moderate -0.3
4 You are given that \(\mathrm { f } ( x ) = \cos x + \lambda \sin x\) where \(\lambda\) is a positive constant.
  1. Express \(\mathrm { f } ( x )\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving \(R\) and \(\alpha\) in terms of \(\lambda\).
  2. Given that the maximum value (as \(x\) varies) of \(\mathrm { f } ( x )\) is 2 , find \(R , \lambda\) and \(\alpha\), giving your answers in exact form.
OCR MEI C4 2015 June Q5
8 marks Standard +0.3
5 A curve has parametric equations \(x = \sec \theta , y = 2 \tan \theta\).
  1. Given that the derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta\).
  2. Verify that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4\). Fig. 5 shows the region enclosed by the curve and the line \(x = 2\). This region is rotated through \(180 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{132ae754-bd4c-4819-80ef-4823ac2ead4f-02_545_853_1738_607} \captionsetup{labelformat=empty} \caption{Fig. 5}
    \end{figure}
  3. Find the volume of revolution produced, giving your answer in exact form.
OCR MEI C4 2015 June Q6
18 marks Standard +0.3
6 Fig. 6 shows a lean-to greenhouse ABCDHEFG . With respect to coordinate axes Oxyz , the coordinates of the vertices are as shown. All distances are in metres. Ground level is the plane \(z = 0\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{132ae754-bd4c-4819-80ef-4823ac2ead4f-03_785_1283_424_392} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Verify that the equation of the plane through \(\mathrm { A } , \mathrm { B }\) and E is \(x + 6 y + 12 = 0\). Hence, given that F lies in this plane, show that \(a = - 2 \frac { 1 } { 3 }\).
  2. (A) Show that the vector \(\left( \begin{array} { r } 1 \\ - 6 \\ 0 \end{array} \right)\) is normal to the plane DHC.
    (B) Hence find the cartesian equation of this plane.
    (C) Given that G lies in the plane DHC , find \(b\) and the length FG .
  3. Find the angle EFB . A straight wire joins point H to a point P which is half way between E and F . Q is a point two-thirds of the way along this wire, so that \(\mathrm { HQ } = 2 \mathrm { QP }\).
  4. Find the height of Q above the ground. \section*{Question 7 begins on page 4.}
OCR MEI C4 2015 June Q7
18 marks Standard +0.3
7 A drug is administered by an intravenous drip. The concentration, \(x\), of the drug in the blood is measured as a fraction of its maximum level. The drug concentration after \(t\) hours is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k \left( 1 + x - 2 x ^ { 2 } \right) ,$$ where \(0 \leqslant x < 1\), and \(k\) is a positive constant. Initially, \(x = 0\).
  1. Express \(\frac { 1 } { ( 1 + 2 x ) ( 1 - x ) }\) in partial fractions.
  2. Hence solve the differential equation to show that \(\frac { 1 + 2 x } { 1 - x } = \mathrm { e } ^ { 3 k t }\).
  3. After 1 hour the drug concentration reaches \(75 \%\) of its maximum value and so \(x = 0.75\). Find the value of \(k\), and the time taken for the drug concentration to reach \(90 \%\) of its maximum value.
  4. Rearrange the equation in part (ii) to show that \(x = \frac { 1 - \mathrm { e } ^ { - 3 k t } } { 1 + 2 \mathrm { e } ^ { - 3 k t } }\). Verify that in the long term the drug concentration approaches its maximum value. \section*{END OF QUESTION PAPER} \section*{Tuesday 16 J une 2015 - Afternoon} \section*{A2 GCE MATHEMATICS (MEI)} 4754/01B Applications of Advanced Mathematics (C4) Paper B: Comprehension \section*{QUESTION PAPER} \section*{Candidates answer on the Question Paper.} \section*{OCR supplied materials:}
    • Insert (inserted)
    • MEI Examination Formulae and Tables (MF2)
    \section*{Other materials required:}
    • Scientific or graphical calculator
    • Rough paper
    Duration: Up to 1 hour \includegraphics[max width=\textwidth, alt={}, center]{132ae754-bd4c-4819-80ef-4823ac2ead4f-05_117_495_1014_1308} PLEASE DO NOT WRITE IN THIS SPACE 2 In line 79 it says "For most journeys, more than half the journey time is composed of load time and transfer time". For what percentage of the journey time for the round trip made by car A in Table 4 is the car stationary?
    \includegraphics[max width=\textwidth, alt={}]{132ae754-bd4c-4819-80ef-4823ac2ead4f-07_645_1746_388_164}
    3 Using the expression on line 51, work out the answer to the question on lines 39 and 40 for the case where there are 10 upper floors and 7 people. Give your answer to 2 decimal places.
    \includegraphics[max width=\textwidth, alt={}]{132ae754-bd4c-4819-80ef-4823ac2ead4f-07_488_1746_1233_164}
    4 In lines 89 and 90 it says "... on average there will be approximately 8 stops per trip. A round trip with 8 stops would take between 188 and 200 seconds". Explain how the figure of 188 seconds has been derived. 5
  5. Referring to Strategy 3 and lines 99 to 101, complete the table below for car C .
  6. Calculate the time car C will take to transport all the people who work on floors 7 and 8 , and return to the ground floor.
    5
  		7. \includegraphics[max width=\textwidth, alt={}]{132ae754-bd4c-4819-80ef-4823ac2ead4f-08_1095_816_484_700}
    68 people make independent visits to any one of the upper floors of a building with 10 upper floors. What is the probability that at least one of the visitors goes to the top floor?
    6
    7 On lines 94 and 95 it says "Table 4 gives the timings for round trips in which the cars are required to stop at every floor they serve; Table 2 suggests this is a common occurrence in this case". Explain how Table 2 is used to make this claim. \includegraphics[max width=\textwidth, alt={}, center]{132ae754-bd4c-4819-80ef-4823ac2ead4f-09_1093_1740_1238_166} END OF QUESTION PAPER
OCR MEI C4 Q1
Easy -2.5
1 Explain why the number 1836.108 for the ratio Rest mass of electron would be suitable for communication with other civilisations whereas neither the rest mass of the proton nor that of the electron would be.
OCR MEI C4 Q2
Standard +0.8
2 A civilisation which works in base 5 sends out the first 6 digits of \(\pi\) as 3.032 32. Convert this to base 10.
OCR MEI C4 Q3
Easy -1.8
3 Complete this table to show the next 3 values of the iteration $$x _ { n + 1 } = k x _ { n } \left( 1 - x _ { n } \right)$$ in the case when \(k = 3.2\) and \(x _ { 0 } = 0.5\). Give your answers to calculator accuracy.
\(n\)\(x _ { n }\)
00.5
10.8
20.512
3
4
5
OCR MEI C4 Q4
Moderate -0.5
4 Justify the statement that the equation in line 83, $$\frac { \phi } { 1 } = \frac { 1 } { \phi - 1 }$$ has the solution \(\phi = \frac { 1 \pm \sqrt { 5 } } { 2 }\).
OCR MEI C4 Q5
Standard +0.8
5 Justify the statement in line 87 that $$\frac { 1 } { \phi } = \frac { \sqrt { 5 } - 1 } { 2 }$$