Questions — OCR MEI (4333 questions)

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OCR MEI C4 Q3
5 marks Moderate -0.3
3 Solve the equation \(\frac { 4 x } { x + 1 } - \frac { 3 } { 2 x + 1 } = 1\).
OCR MEI C4 Q4
5 marks Standard +0.3
4 Express \(\frac { 1 } { ( 2 x + 1 ) \left( x ^ { 2 } + 1 \right) }\) in partial fractions.
OCR MEI C4 Q5
3 marks Easy -1.2
5 Express \(\frac { x } { x ^ { 2 } - 1 } + \frac { 2 } { x + 1 }\) as a single fraction, simplifying your answer.
OCR MEI C4 Q6
5 marks Moderate -0.8
6 Find the first three terms in the binomial expansion of \(\overline { 4 + x }\) in ascending powers of \(x\).
State the set of values of \(x\) for which the expansion is valid.
OCR MEI C4 Q7
8 marks Standard +0.3
7
  1. Express \(\frac { 3 } { ( y - 2 ) ( y + 1 ) }\) in partial fractions.
    [0pt] [3]
  2. Hence, given that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } ( y - 2 ) ( y + 1 )$$ show that \(\frac { y - 2 } { y + 1 } = A \mathrm { e } ^ { x ^ { 3 } }\), where \(A\) is a constant.
OCR MEI C4 Q8
3 marks Easy -1.2
8 Express \(\frac { x } { x ^ { 2 } - 4 } + \frac { 2 } { x + 2 }\) as a single fraction, simplifying your answer.
OCR MEI C4 Q9
8 marks Standard +0.3
9
  1. Find the first three non-zero terms of the binomial series expansion of \(\frac { 1 } { \sqrt { 1 + 4 x ^ { 2 } } }\), and state the set of values of \(x\) for which the expansion is valid.
  2. Hence find the first three non-zero terms of the series expansion of \(\frac { 1 - x ^ { 2 } } { \sqrt { 1 + 4 x ^ { 2 } } }\).
OCR MEI C4 Q10
8 marks Standard +0.3
10 Two students are trying to evaluate the integral \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\).
Sarah uses the trapezium rule with 2 strips, and starts by constructing the following table.
\(x\)11.52
\(\sqrt { 1 + \mathrm { e } ^ { - x } }\)1.16961.10601.0655
  1. Complete the calculation, giving your answer to 3 significant figures. Anish uses a binomial approximation for \(\sqrt { 1 + \mathrm { e } ^ { - x } }\) and then integrates this.
  2. Show that, provided \(\mathrm { e } ^ { - x }\) is suitably small, \(\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } \quad \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }\).
  3. Use this result to evaluate \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\) approximately, giving your answer to 3 significant figures.
OCR MEI C4 Q1
4 marks Moderate -0.8
1 Solve the equation \(\frac { 2 x } { x + 1 } - \frac { 1 } { x - 1 } = 1\).
OCR MEI C4 Q2
5 marks Easy -2.5
2 Express \(\frac { x + 1 } { ( 2 x - 1 ) }\) in partial fractions.
OCR MEI C4 Q3
6 marks Moderate -0.5
3 Express \(\frac { 3 x + 2 } { x \left( x ^ { 2 } + 1 \right) }\) in partial fractions.
OCR MEI C4 Q4
6 marks Moderate -0.5
4 Express \(\frac { 4 } { x \left( x ^ { 2 } + 4 \right) }\) in partial fractions.
OCR MEI C4 Q5
5 marks Moderate -0.3
5 Solve the equation \(\frac { 2 x } { x - 2 } - \frac { 4 x } { x + 1 } = 3\).
OCR MEI C4 Q6
8 marks Standard +0.3
6
  1. Express \(\frac { x } { ( 1 + x ) ( 1 - 2 x ) }\) in partial fractions.
  2. Hence use binomial expansions to show that \(\frac { x } { ( 1 + x ) ( 1 - 2 x ) } = a x + b x ^ { 2 } + \ldots\), where \(a\) and \(b\) are
    constants to be determined. State the set of values of \(x\) for which the expansion is valid.
OCR MEI C4 Q7
18 marks Standard +0.3
7 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?
OCR MEI C4 Q1
20 marks Standard +0.3
1 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.
  1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
    1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
    2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
  2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
    1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
    2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
    3. Find the greatest and least values of \(P\) predicted by this model.
OCR MEI C4 Q2
18 marks Standard +0.3
2 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 Q3
19 marks Standard +0.3
3 Some years ago an island was populated by red squirrels and there were no grey squirrels. Then grey squirrels were introduced. The population \(x\), in thousands, of red squirrels is modelled by the equation $$x = \frac { a } { 1 + k t }$$ where \(t\) is the time in years, and \(a\) and \(k\) are constants. When \(t = 0 , x = 2.5\).
  1. Show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { k x ^ { 2 } } { a }\).
  2. Given that the initial population of 2.5 thousand red squirrels reduces to 1.6 thousand after one year, calculate \(a\) and \(k\).
  3. What is the long-term population of red squirrels predicted by this model? The population \(y\), in thousands, of grey squirrels is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} t } = 2 y - y ^ { 2 }$$ When \(t = 0 , y = 1\).
  4. Express \(\frac { 1 } { 2 y - y ^ { 2 } }\) in partial fractions.
  5. Hence show by integration that \(\ln \left( \frac { y } { 2 y } \right) = 2 t\). Show that \(y = \frac { 2 } { 1 + \mathrm { e } ^ { - 2 t } }\).
  6. What is the long-term population of grey squirrels predicted by this model?
OCR MEI C4 Q1
18 marks Standard +0.3
1 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.
    [0pt] [3]
  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.
OCR MEI C4 Q2
7 marks Standard +0.3
2 A curve has parametric equations \(x = \mathrm { e } ^ { 3 t } , y = t \mathrm { e } ^ { 2 t }\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). Hence find the exact gradient of the curve at the point with parameter \(t = 1\).
  2. Find the cartesian equation of the curve in the form \(y = a x ^ { b } \ln x\), where \(a\) and \(b\) are constants to be determined.
OCR MEI C4 Q3
18 marks Standard +0.8
3 Fig. 8.1 shows an upright cylindrical barrel containing water. The water is leaking out of a hole in the side of the barrel. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{11d26af4-19d0-4310-a64e-9888285c9980-2_260_447_281_824} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
\end{figure} The height of the water surface above the hole \(t\) seconds after opening the hole is \(h\) metres, where $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A \sqrt { h }$$ and where \(A\) is a positive constant. Initially the water surface is 1 metre above the hole.
  1. Verify that the solution to this differential equation is $$h = \left( 1 - \frac { 1 } { 2 } A t \right) ^ { 2 } .$$ The water stops leaking when \(h = 0\). This occurs after 20 seconds.
  2. Find the value of \(A\), and the time when the height of the water surface above the hole is 0.5 m . Fig. 8.2 shows a similar situation with a different barrel; \(h\) is in metres. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{11d26af4-19d0-4310-a64e-9888285c9980-2_235_455_1425_820} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
    \end{figure} For this barrel, $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - B \frac { \sqrt { h } } { ( 1 + h ) ^ { 2 } } ,$$ where \(B\) is a positive constant. When \(t = 0 , h = 1\).
  3. Solve this differential equation, and hence show that $$h ^ { \frac { 1 } { 2 } } \left( 30 + 20 h + 6 h ^ { 2 } \right) = 56 - 15 B t .$$
  4. Given that \(h = 0\) when \(t = 20\), find \(B\). Find also the time when the height of the water surface above the hole is 0.5 m .
OCR MEI C4 Q4
18 marks Standard +0.3
4 The motion of a particle is modelled by the differential equation $$v \frac { \mathrm {~d} v } { \mathrm {~d} x } + 4 x = 0$$ where \(x\) is its displacement from a fixed point, and \(v\) is its velocity. Initially \(x = 1\) and \(v = 4\).
  1. Solve the differential equation to show that \(v ^ { 2 } = 20 - 4 x ^ { 2 }\). Now consider motion for which \(x = \cos 2 t + 2 \sin 2 t\), where \(x\) is the displacement from a fixed point at time \(t\).
  2. Verify that, when \(t = 0 , x = 1\). Use the fact that \(v = \frac { \mathrm { d } x } { \mathrm {~d} t }\) to verify that when \(t = 0 , v = 4\).
  3. Express \(x\) in the form \(R \cos ( 2 t - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and obtain the corresponding expression for \(v\). Hence or otherwise verify that, for this motion too, \(v ^ { 2 } = 20 - 4 x ^ { 2 }\).
  4. Use your answers to part (iii) to find the maximum value of \(x\), and the earliest time at which \(x\) reaches this maximum value.
OCR MEI C4 Q5
8 marks Moderate -0.3
5 The total value of the sales made by a new company in the first \(t\) years of its existence is denoted by \(\pounds V\). A model is proposed in which the rate of increase of \(V\) is proportional to the square root of \(V\). The constant of proportionality is \(k\).
  1. Express the model as a differential equation. Verify by differentiation that \(V = \left( \frac { 1 } { 2 } k t + c \right) ^ { 2 }\), where \(c\) is an arbitrary constant, satisfies this differential equation.
  2. The value of the company’s sales in its first year is \(\pounds 10000\), and the total value of the sales in the first two years is \(\pounds 40000\). Find \(V\) in terms of \(t\).
OCR MEI C4 Q1
8 marks Standard +0.3
1 Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { x ( x + 1 ) }\), given that when \(x = 1 , y = 1\). Your answer should express \(y\) explicitly in terms of \(x\).
OCR MEI C4 Q2
18 marks Standard +0.3
2 Water is leaking from a container. After \(t\) seconds, the depth of water in the container is \(x \mathrm {~cm}\), and the volume of water is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 1 } { 3 } x ^ { 3 }\). The rate at which water is lost is proportional to \(x\), so that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = - k x\), where \(k\) is a constant.
  1. Show that \(x \frac { \mathrm {~d} x } { \mathrm {~d} t } = - k\). Initially, the depth of water in the container is 10 cm .
  2. Show by integration that \(x = \sqrt { 100 - 2 k t }\).
  3. Given that the container empties after 50 seconds, find \(k\). Once the container is empty, water is poured into it at a constant rate of \(1 \mathrm {~cm} ^ { 3 }\) per second. The container continues to lose water as before.
  4. Show that, \(t\) seconds after starting to pour the water in, \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 - x } { x ^ { 2 } }\).
  5. Show that \(\frac { 1 } { 1 - x } - x - 1 = \frac { x ^ { 2 } } { 1 - x }\). Hence solve the differential equation in part (iv) to show that $$t = \ln \left( \frac { 1 } { 1 - x } \right) - \frac { 1 } { 2 } x ^ { 2 } - x$$
  6. Show that the depth cannot reach 1 cm .