Questions — OCR MEI (4333 questions)

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OCR MEI C4 2009 January Q3
5 marks Moderate -0.8
3 Vectors \(\mathbf { a }\) and \(\mathbf { b }\) are given by \(\mathbf { a } = 2 \mathbf { i } + \mathbf { j } - \mathbf { k }\) and \(\mathbf { b } = 4 \mathbf { i } - 2 \mathbf { j } + \mathbf { k }\).
Find constants \(\lambda\) and \(\mu\) such that \(\lambda \mathbf { a } + \mu \mathbf { b } = 4 \mathbf { j } - 3 \mathbf { k }\).
OCR MEI C4 2009 January Q4
3 marks Standard +0.3
4 Prove that \(\cot \beta - \cot \alpha = \frac { \sin ( \alpha - \beta ) } { \sin \alpha \sin \beta }\).
OCR MEI C4 2009 January Q5
8 marks Standard +0.3
5
  1. Write down normal vectors to the planes \(2 x - y + z = 2\) and \(x - z = 1\).
    Hence find the acute angle between the planes.
  2. Write down a vector equation of the line through \(( 2,0,1 )\) perpendicular to the plane \(2 x - y + z = 2\). Find the point of intersection of this line with the plane.
OCR MEI C4 2009 January Q6
8 marks Standard +0.8
6
  1. Express \(\cos \theta + \sqrt { 3 } \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(\alpha\) is acute, expressing \(\alpha\) in terms of \(\pi\).
  2. Write down the derivative of \(\tan \theta\). Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { ( \cos \theta + \sqrt { 3 } \sin \theta ) ^ { 2 } } \mathrm {~d} \theta = \frac { \sqrt { 3 } } { 4 }\).
OCR MEI C4 2009 January Q7
17 marks Standard +0.3
7 Scientists can estimate the time elapsed since an animal died by measuring its body temperature.
  1. Assuming the temperature goes down at a constant rate of 1.5 degrees Fahrenheit per hour, estimate how long it will take for the temperature to drop
    (A) from \(98 ^ { \circ } \mathrm { F }\) to \(89 ^ { \circ } \mathrm { F }\),
    (B) from \(98 ^ { \circ } \mathrm { F }\) to \(80 ^ { \circ } \mathrm { F }\). In practice, rate of temperature loss is not likely to be constant. A better model is provided by Newton's law of cooling, which states that the temperature \(\theta\) in degrees Fahrenheit \(t\) hours after death is given by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k \left( \theta - \theta _ { 0 } \right)$$ where \(\theta _ { 0 } { } ^ { \circ } \mathrm { F }\) is the air temperature and \(k\) is a constant.
  2. Show by integration that the solution of this equation is \(\theta = \theta _ { 0 } + A \mathrm { e } ^ { - k t }\), where \(A\) is a constant. The value of \(\theta _ { 0 }\) is 50 , and the initial value of \(\theta\) is 98 . The initial rate of temperature loss is \(1.5 ^ { \circ } \mathrm { F }\) per hour.
  3. Find \(A\), and show that \(k = 0.03125\).
  4. Use this model to calculate how long it will take for the temperature to drop
    (A) from \(98 ^ { \circ } \mathrm { F }\) to \(89 ^ { \circ } \mathrm { F }\),
    (B) from \(98 ^ { \circ } \mathrm { F }\) to \(80 ^ { \circ } \mathrm { F }\).
  5. Comment on the results obtained in parts (i) and (iv).
OCR MEI C4 2009 January Q8
19 marks Standard +0.8
8 Fig. 8 illustrates a hot air balloon on its side. The balloon is modelled by the volume of revolution about the \(x\)-axis of the curve with parametric equations $$x = 2 + 2 \sin \theta , \quad y = 2 \cos \theta + \sin 2 \theta , \quad ( 0 \leqslant \theta \leqslant 2 \pi ) .$$ The curve crosses the \(x\)-axis at the point \(\mathrm { A } ( 4,0 )\). B and C are maximum and minimum points on the curve. Units on the axes are metres. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f61b7d80-8e21-4720-8e8c-259531c1b305-4_821_809_575_667} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
  2. Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 6 } \pi\), and find the exact coordinates of B . Hence find the maximum width BC of the balloon.
  3. (A) Show that \(y = x \cos \theta\).
    (B) Find \(\sin \theta\) in terms of \(x\) and show that \(\cos ^ { 2 } \theta = x - \frac { 1 } { 4 } x ^ { 2 }\).
    (C) Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }\).
  4. Find the volume of the balloon.
OCR MEI C4 2010 January Q1
1 marks Moderate -0.3
1 Find the first three terms in the binomial expansion of \(\frac { 1 + 2 x } { ( 1 - 2 x ) ^ { 2 } }\) in ascending powers of \(x\). State the set of values of \(x\) for which the expansion is valid.
OCR MEI C4 2010 January Q2
1 marks Standard +0.3
2 Show that \(\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }\).
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ }$$
OCR MEI C4 2010 January Q3
2 marks Moderate -0.3
3 A curve has parametric equations $$x = \mathrm { e } ^ { 2 t } , \quad y = \frac { 2 t } { 1 + t }$$
  1. Find the gradient of the curve at the point where \(t = 0\).
  2. Find \(y\) in terms of \(x\).
OCR MEI C4 2010 January Q4
2 marks Standard +0.3
4 The points A , B and C have coordinates \(( 1,3 , - 2 ) , ( - 1,2 , - 3 )\) and \(( 0 , - 8,1 )\) respectively.
  1. Find the vectors \(\overrightarrow { \mathrm { AB } }\) and \(\overrightarrow { \mathrm { AC } }\).
  2. Show that the vector \(2 \mathbf { i } - \mathbf { j } - 3 \mathbf { k }\) is perpendicular to the plane ABC . Hence find the equation of the plane ABC .
OCR MEI C4 2010 January Q5
2 marks Moderate -0.3
5
  1. Verify that the lines \(\mathbf { r } = \left( \begin{array} { r } - 5 \\ 3 \\ 4 \end{array} \right) + \lambda \left( \begin{array} { r } 3 \\ 0 \\ - 1 \end{array} \right)\) and \(\mathbf { r } = \left( \begin{array} { r } - 1 \\ 4 \\ 2 \end{array} \right) + \mu \left( \begin{array} { r } 2 \\ - 1 \\ 0 \end{array} \right)\) meet at the point (1,3,2).
  2. Find the acute angle between the lines.
OCR MEI C4 2010 January Q6
2 marks Standard +0.3
6 In Fig. 6, OAB is a thin bent rod, with \(\mathrm { OA } = a\) metres, \(\mathrm { AB } = b\) metres and angle \(\mathrm { OAB } = 120 ^ { \circ }\). The bent rod lies in a vertical plane. OA makes an angle \(\theta\) above the horizontal. The vertical height BD of B above O is \(h\) metres. The horizontal through A meets BD at C and the vertical through A meets OD at E . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{26b3b9fb-7d20-4c8d-ba15-89920534c53a-3_433_899_568_625} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a \sin \theta + b \sin \left( \theta - 60 ^ { \circ } \right) .$$
  2. Hence show that \(h = \left( a + \frac { 1 } { 2 } b \right) \sin \theta - \frac { \sqrt { 3 } } { 2 } b \cos \theta\). The rod now rotates about O , so that \(\theta\) varies. You may assume that the formulae for \(h\) in parts (i) and (ii) remain valid.
  3. Show that OB is horizontal when \(\tan \theta = \frac { \sqrt { 3 } b } { 2 a + b }\). In the case when \(a = 1\) and \(b = 2 , h = 2 \sin \theta - \sqrt { 3 } \cos \theta\).
  4. Express \(2 \sin \theta - \sqrt { 3 } \cos \theta\) in the form \(R \sin ( \theta - \alpha )\). Hence, for this case, write down the maximum value of \(h\) and the corresponding value of \(\theta\).
OCR MEI C4 2010 January Q7
2 marks Standard +0.8
7 Fig. 7 illustrates the growth of a population with time. The proportion of the ultimate (long term) population is denoted by \(x\), and the time in years by \(t\). When \(t = 0 , x = 0.5\), and as \(t\) increases, \(x\) approaches 1 . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{26b3b9fb-7d20-4c8d-ba15-89920534c53a-4_599_937_429_605} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure} One model for this situation is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ( 1 - x )$$
  1. Verify that \(x = \frac { 1 } { 1 + \mathrm { e } ^ { - t } }\) satisfies this differential equation, including the initial condition.
  2. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. An alternative model for this situation is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ^ { 2 } ( 1 - x ) ,$$ with \(x = 0.5\) when \(t = 0\) as before.
  3. Find constants \(A , B\) and \(C\) such that \(\frac { 1 } { x ^ { 2 } ( 1 - x ) } = \frac { A } { x ^ { 2 } } + \frac { B } { x } + \frac { C } { 1 - x }\).
  4. Hence show that \(t = 2 + \ln \left( \frac { x } { 1 - x } \right) - \frac { 1 } { x }\).
  5. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value.
OCR MEI C4 2010 January Q8
6 marks Easy -1.2
8 A passage of plaintext is encoded by using the Caesar cipher corresponding to a shift of 2 places followed by the Vigenere cipher with keyword ODE.
  1. The first letter in the plaintext passage is \(F\). Show that the first letter in the transmitted text is \(V\).
  2. The first four letters in the transmitted text are VFIU. What are the first four letters in the plaintext passage?
  3. The 800th letter in the transmitted text is \(W\). What is the 800th letter in the plaintext passage?
OCR MEI C4 2011 January Q1
6 marks Moderate -0.8
1
  1. Use the trapezium rule with four strips to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\), showing your working. Fig. 1 shows a sketch of \(y = \sqrt { 1 + \mathrm { e } ^ { x } }\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{f657e167-e6f8-4df2-901b-067c32835877-02_535_1074_571_532} \captionsetup{labelformat=empty} \caption{Fig. 1}
    \end{figure}
  2. Suppose that the trapezium rule is used with more strips than in part (i) to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\). State, with a reason but no further calculation, whether this would give a larger or smaller estimate.
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 2012 January Q1
5 marks Moderate -0.3
1 Express \(\frac { x + 1 } { x ^ { 2 } ( 2 x - 1 ) }\) in partial fractions.
OCR MEI C4 2012 January Q3
7 marks Moderate -0.3
3 Express \(3 \sin x + 2 \cos x\) in the form \(R \sin ( x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\).
Hence find, correct to 2 decimal places, the coordinates of the maximum point on the curve \(y = \mathrm { f } ( x )\), where $$f ( x ) = 3 \sin x + 2 \cos x , 0 \leqslant x \leqslant \pi .$$
OCR MEI C4 2012 January Q4
4 marks Moderate -0.3
4
  1. Complete the table of values for the curve \(y = \sqrt { \cos x }\).
    \(x\)0\(\frac { \pi } { 8 }\)\(\frac { \pi } { 4 }\)\(\frac { 3 \pi } { 8 }\)\(\frac { \pi } { 2 }\)
    \(y\)0.96120.8409
    Hence use the trapezium rule with strip width \(h = \frac { \pi } { 8 }\) to estimate the value of the integral \(\int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { \cos x } \mathrm {~d} x\), giving your answer to 3 decimal places. Fig. 4 shows the curve \(y = \sqrt { \cos x }\) for \(0 \leqslant x \leqslant \frac { \pi } { 2 }\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{81433914-a56f-4765-af34-990a0127f98b-02_457_750_1446_653} \captionsetup{labelformat=empty} \caption{Fig. 4}
    \end{figure}
  2. State, with a reason, whether the trapezium rule with a strip width of \(\frac { \pi } { 16 }\) would give a larger or smaller estimate of the integral.