Questions — OCR MEI C4 (354 questions)

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OCR MEI C4 2014 June Q3
5 marks Moderate -0.3
Fig. 3 shows the curve \(y = x^3 + \sqrt{(\sin x)}\) for \(0 \leqslant x \leqslant \frac{\pi}{4}\). \includegraphics{figure_3}
  1. Use the trapezium rule with 4 strips to estimate the area of the region bounded by the curve, the \(x\)-axis and the line \(x = \frac{\pi}{4}\), giving your answer to 3 decimal places. [4]
  2. Suppose the number of strips in the trapezium rule is increased. Without doing further calculations, state, with a reason, whether the area estimate increases, decreases, or it is not possible to say. [1]
OCR MEI C4 2014 June Q4
8 marks Moderate -0.3
  1. Show that \(\cos(\alpha + \beta) = \frac{1 - \tan \alpha \tan \beta}{\sec \alpha \sec \beta}\). [3]
  2. Hence show that \(\cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha}\). [2]
  3. Hence or otherwise solve the equation \(\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1}{2}\) for \(0° \leqslant \theta \leqslant 180°\). [3]
OCR MEI C4 2014 June Q5
7 marks Standard +0.3
A curve has parametric equations \(x = e^{2t}, y = te^{2t}\).
  1. Find \(\frac{dy}{dx}\) in terms of \(t\). Hence find the exact gradient of the curve at the point with parameter \(t = 1\). [4]
  2. Find the cartesian equation of the curve in the form \(y = ax^b \ln x\), where \(a\) and \(b\) are constants to be determined. [3]
OCR MEI C4 2014 June Q6
6 marks Standard +0.8
Fig. 6 shows the region enclosed by the curve \(y = (1 + 2x^2)^{\frac{1}{2}}\) and the line \(y = 2\). \includegraphics{figure_6} This region is rotated about the \(y\)-axis. Find the volume of revolution formed, giving your answer as a multiple of \(\pi\). [6]
OCR MEI C4 2014 June Q7
18 marks Standard +0.3
Fig. 7 shows a tetrahedron ABCD. The coordinates of the vertices, with respect to axes Oxyz, are A(-3, 0, 0), B(2, 0, -2), C(0, 4, 0) and D(0, 4, 5). \includegraphics{figure_7}
  1. Find the lengths of the edges AB and AC, and the size of the angle CAB. Hence calculate the area of triangle ABC. [7]
    1. Verify that 4i - 3j + 10k is normal to the plane ABC. [2]
    2. Hence find the equation of this plane. [2]
  3. Write down a vector equation for the line through D perpendicular to the plane ABC. Hence find the point of intersection of this line with the plane ABC. [5]
The volume of a tetrahedron is \(\frac{1}{3} \times \text{area of base} \times \text{height}\).
  1. Find the volume of the tetrahedron ABCD. [2]
OCR MEI C4 2014 June Q8
18 marks Standard +0.8
Fig. 8.1 shows an upright cylindrical barrel containing water. The water is leaking out of a hole in the side of the barrel. \includegraphics{figure_8.1} The height of the water surface above the hole \(t\) seconds after opening the hole is \(h\) metres, where $$\frac{dh}{dt} = -A\sqrt{h}$$ 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 $$h = \left(1 - \frac{1}{2}At\right)^2.$$ [3]
The water stops leaking when \(h = 0\). This occurs after 20 seconds.
  1. Find the value of \(A\), and the time when the height of the water surface above the hole is 0.5 m. [4]
Fig. 8.2 shows a similar situation with a different barrel; \(h\) is in metres. \includegraphics{figure_8.2} For this barrel, $$\frac{dh}{dt} = -B\frac{\sqrt{h}}{(1+h)^2},$$ where \(B\) is a positive constant. When \(t = 0\), \(h = 1\).
  1. Solve this differential equation, and hence show that $$h^{\frac{1}{2}}(30 + 20h + 6h^2) = 56 - 15Bt.$$ [7]
  2. 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. [4]
OCR MEI C4 Q1
18 marks Moderate -0.3
In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac{dM}{dt} = \frac{M}{t(1+t^2)}.$$
  1. Find \(\int \frac{t}{1+t^2} dt\). [3]
  2. Find constants \(A\), \(B\) and \(C\) such that $$\frac{1}{t(1+t^2)} = \frac{A}{t} + \frac{Bt+C}{1+t^2}.$$ [5]
  3. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac{Kt}{\sqrt{1+t^2}},$$ where \(K\) is a constant. [6]
  4. When \(t = 1\), \(M = 25\). Calculate \(K\). What is the mass of the chemical in the long term? [4]
OCR MEI C4 Q2
19 marks Standard +0.3
The growth of a tree is modelled by the differential equation $$10\frac{dh}{dt} = 20 - h,$$ 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 \(\frac{dh}{dt} = 0\), and interpret this in terms of the growth of the tree. [1]
  2. Verify that \(h = 20(1 - e^{-0.1t})\) satisfies this differential equation and its initial condition. [5]
The alternative differential equation $$200\frac{dh}{dt} = 400 - h^2$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
  1. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac{20(1 - e^{-0.2t})}{1 + e^{-0.2t}}.$$ [9]
  2. What does this solution indicate about the long-term height of the tree? [1]
  3. After a year, the tree has grown to a height of 2 m. Which model fits this information better? [3]
OCR MEI C4 Q3
18 marks Standard +0.3
\includegraphics{figure_3} Fig. 9 shows a hemispherical bowl, of radius 10 cm, filled with water to a depth of \(x\) cm. It can be shown that the volume of water, \(V\) cm\(^3\), is given by $$V = \pi(10x^2 - \frac{1}{3}x^3).$$ 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 cm\(^3\) s\(^{-1}\), given by the equation $$\frac{dV}{dt} = k(20 - x),$$ where \(k\) is a constant.
  1. Find \(\frac{dV}{dx}\), and hence show that \(\pi x \frac{dx}{dt} = k\). [4]
  2. Solve this differential equation, and hence show that the bowl fills completely after \(T\) seconds, where \(T = \frac{50\pi}{k}\). [5]
Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of \(kx\) cm\(^3\) s\(^{-1}\).
  1. Show that, \(t\) seconds later, \(\pi(20 - x) \frac{dx}{dt} = -k\). [3]
  2. Solve this differential equation. Hence show that the bowl empties in 37 seconds. [6]
OCR MEI C4 Q4
18 marks Standard +0.3
A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v\) m s\(^{-1}\). Its terminal (long-term) velocity is 5 m s\(^{-1}\). A model of the particle's motion is proposed. In this model, \(v = 5(1 - e^{-2t})\).
  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. [3]
  2. Verify that \(v\) satisfies the differential equation \(\frac{dv}{dt} = 10 - 2v\). [3]
In a second model, \(v\) satisfies the differential equation $$\frac{dv}{dt} = 10 - 0.4v^2.$$ As before, when \(t = 0\), \(v = 0\).
  1. Show that this differential equation may be written as $$\frac{10}{(5-v)(5+v)} \frac{dv}{dt} = 4.$$ Using partial fractions, solve this differential equation to show that $$t = \frac{1}{4} \ln\left(\frac{5+v}{5-v}\right).$$ [8] This can be re-arranged to give \(v = \frac{5(1-e^{-4t})}{1+e^{-4t}}\). [You are not required to show this result.]
  2. Verify that this model also gives a terminal velocity of 5 m s\(^{-1}\). Calculate the velocity after 0.5 seconds as given by this model. [3]
The velocity of the particle after 0.5 seconds is measured as 3 m s\(^{-1}\).
  1. Which of the two models fits the data better? [1]
OCR MEI C4 Q1
20 marks Standard +0.3
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. \includegraphics{figure_7} One model for this situation is given by the differential equation $$\frac{dx}{dt} = x(1-x).$$
  1. Verify that \(x = \frac{1}{1+e^{-t}}\) satisfies this differential equation, including the initial condition. [6]
  2. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. [3]
An alternative model for this situation is given by the differential equation $$\frac{dx}{dt} = x^2(1-x),$$ with \(x = 0.5\) when \(t = 0\) as before.
  1. 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]
  2. Hence show that \(t = 2 + \ln\left(\frac{x}{1-x}\right) - \frac{1}{x}\). [5]
  3. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. [2]
OCR MEI C4 Q2
17 marks Standard +0.3
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
    1. from 98°F to 89°F,
    2. from 98°F to 80°F. [2]
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 $$\frac{d\theta}{dt} = -k(\theta - \theta_0),$$ where \(\theta_0\)°F is the air temperature and \(k\) is a constant.
  1. Show by integration that the solution of this equation is \(\theta = \theta_0 + Ae^{-kt}\), where \(A\) is a constant. [5]
The value of \(\theta_0\) is 50, and the initial value of \(\theta\) is 98. The initial rate of temperature loss is 1.5°F per hour.
  1. Find \(A\), and show that \(k = 0.03125\). [4]
  2. Use this model to calculate how long it will take for the temperature to drop
    1. from 98°F to 89°F,
    2. from 98°F to 80°F. [5]
  3. Comment on the results obtained in parts (i) and (iv). [1]
OCR MEI C4 Q3
19 marks Standard +0.3
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 $$x = \frac{a}{1 + kt},$$ where \(t\) is the time in years, and \(a\) and \(k\) are constants. When \(t = 0\), \(x = 2.5\).
  1. Show that \(\frac{dx}{dt} = -\frac{kx^2}{a}\). [3]
  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]
  3. What is the long-term population of red squirrels predicted by this model? [1]
The population \(y\), in thousands, of grey squirrels is modelled by the differential equation $$\frac{dy}{dt} = 2y - y^2.$$ When \(t = 0\), \(y = 1\).
  1. Express \(\frac{1}{2y - y^2}\) in partial fractions. [4]
  2. Hence show by integration that \(\ln\left(\frac{y}{2-y}\right) = 2t\). Show that \(y = \frac{2}{1 + e^{-2t}}\). [7]
  3. What is the long-term population of grey squirrels predicted by this model? [1]
OCR MEI C4 Q4
18 marks Standard +0.3
A curve has equation $$x^2 + 4y^2 = k^2,$$ where \(k\) is a positive constant.
  1. Verify that $$x = k\cos\theta, \quad y = \frac{k}{2}\sin\theta,$$ are parametric equations for the curve. [3]
  2. Hence or otherwise show that \(\frac{dy}{dx} = -\frac{x}{4y}\). [3]
  3. Fig. 8 illustrates the curve for a particular value of \(k\). Write down this value of \(k\). [1]
\includegraphics{figure_8}
  1. Copy Fig. 8 and on the same axes sketch the curves for \(k = 1\), \(k = 3\) and \(k = 4\). [3]
On a map, the curves represent the contours of a mountain. A stream flows down the mountain. Its path on the map is always at right angles to the contour it is crossing.
  1. Explain why the path of the stream is modelled by the differential equation $$\frac{dy}{dx} = \frac{4y}{x}.$$ [2]
  2. Solve this differential equation. Given that the path of the stream passes through the point (2, 1), show that its equation is \(y = \frac{x^4}{16}\). [6]
OCR MEI C4 Q1
7 marks Moderate -0.3
Using partial fractions, find \(\int \frac{x}{(x+1)(2x+1)} dx\). [7]
OCR MEI C4 Q2
8 marks Standard +0.3
  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 \(\pi\). [4]
  2. Write down the derivative of \(\tan \theta\). Hence show that \(\int_0^{\frac{\pi}{6}} \frac{1}{(\cos \theta + \sqrt{3} \sin \theta)^2} d\theta = \frac{\sqrt{3}}{4}\). [4]
OCR MEI C4 Q3
18 marks Standard +0.3
In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac{dM}{dt} = -\frac{M}{2(1 + \frac{t}{2})}$$
  1. Find \(\int \frac{1}{1 + \frac{t}{2}} dt\) [3]
  2. Find constants \(A\), \(B\) and \(C\) such that $$\frac{1}{t(1 + \frac{t}{2})} = \frac{A}{t} + \frac{Bt + C}{1 + \frac{t}{2}}$$ [5]
  3. Use integration, together with your results in parts (i) and (ii), to show that $$M \sim \frac{K}{.1 + \frac{t}{2}}$$ where \(K\) is a constant. [6]
  4. When \(t = 1\), \(M = 25\). Calculate \(K\) What is the mass of the chemical in the long term? [4]
OCR MEI C4 Q4
19 marks Standard +0.3
The growth of a tree is modelled by the differential equation $$10\frac{dh}{dt} = 20 - h$$ 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 \(\frac{dh}{dt} = 0\), and interpret this in terms of the growth of the tree. [1]
  2. Verify that \(h = 20(1 - e^{-0.1t})\) satisfies this differential equation and its initial condition. [5]
The alternative differential equation $$200\frac{dh}{dt} = 400 - h^2$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
  1. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac{20(1 - e^{-0.2t})}{1 + e^{-0.2t}}$$ [9]
  2. What does this solution indicate about the long-term height of the tree? [1]
  3. After a year, the tree has grown to a height of 2m. Which model fits this information better? [3]
OCR MEI C4 Q5
7 marks Standard +0.3
  1. Find the first three non-zero terms of the binomial expansion of \(\frac{1}{\sqrt{4-x^2}}\) for \(|x| < 2\). [4]
  2. Use this result to find an approximation for \(\int_0^1 \frac{1}{\sqrt{4-x^2}} dx\), rounding your answer to 4 significant figures. [2]
  3. Given that \(\int \frac{1}{\sqrt{4-x^2}} dx = \arcsin\left(\frac{1}{2}x\right) + c\), evaluate \(\int_0^1 \frac{1}{\sqrt{4-x^2}} dx\), rounding your answer to 4 significant figures. [1]
OCR MEI C4 Q1
6 marks Moderate -0.3
Given that \(\cosec^2 \theta - \cot \theta = 3\), show that \(\cot^2 \theta - \cot \theta - 2 = 0\). Hence solve the equation \(\cosec^2 \theta - \cot \theta = 3\) for \(0° \leqslant \theta \leqslant 180°\). [6]
OCR MEI C4 Q2
19 marks Standard +0.3
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 O. AB is one of the sides of the polygon. C is the midpoint of AB. Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon. \includegraphics{figure_1}
    1. Show that \(\text{AB} = 2 \sin 15°\). [2]
    2. Use a double angle formula to express \(\cos 30°\) in terms of \(\sin 15°\). Using the exact value of \(\cos 30°\), show that \(\sin 15° = \frac{1}{2}\sqrt{2 - \sqrt{3}}\). [4]
    3. Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6\sqrt{2 - \sqrt{3}}\). [2]
  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. \includegraphics{figure_2}
    1. Show that \(\text{DE} = 2 \tan 15°\). [2]
    2. Let \(t = \tan 15°\). Use a double angle formula to express \(\tan 30°\) in terms of \(t\). Hence show that \(t^2 + 2\sqrt{3}t - 1 = 0\). [3]
    3. Solve this equation, and hence show that \(\pi < 12(2 - \sqrt{3})\). [4]
  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. [2]
OCR MEI C4 Q3
7 marks Moderate -0.3
Express \(\sin \theta - 3 \cos \theta\) in the form \(R \sin (\theta - \alpha)\), where \(R\) and \(\alpha\) are constants to be determined, and \(0° < \alpha < 90°\). Hence solve the equation \(\sin \theta - 3 \cos \theta = 1\) for \(0° \leqslant \theta \leqslant 360°\). [7]
OCR MEI C4 Q4
16 marks Standard +0.3
\includegraphics{figure_3} In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations $$x = 10 \cos \theta + 5 \cos 2\theta, \quad y = 10 \sin \theta + 5 \sin 2\theta, \quad (0 \leqslant \theta < 2\pi),$$ where \(x\) and \(y\) are in metres.
  1. Show that \(\frac{\text{d}y}{\text{d}x} = -\frac{\cos \theta + \cos 2\theta}{\sin \theta + \sin 2\theta}\). Verify that \(\frac{\text{d}y}{\text{d}x} = 0\) when \(\theta = \frac{1}{3}\pi\). Hence find the exact coordinates of the highest point A on the path of C. [6]
  2. Express \(x^2 + y^2\) in terms of \(\theta\). Hence show that $$x^2 + y^2 = 125 + 100 \cos \theta.$$ [4]
  3. Using this result, or otherwise, find the greatest and least distances of C from O. [2]
You are given that, at the point B on the path vertically above O, $$2 \cos^2 \theta + 2 \cos \theta - 1 = 0.$$
  1. Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures. [4]
OCR MEI C4 Q5
7 marks Standard +0.3
Show that \(\cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan \theta}\). Hence solve the equation $$\cot 2\theta = 1 + \tan \theta \quad \text{for } 0° < \theta < 360°.$$ [7]
OCR MEI C4 Q1
18 marks Standard +0.3
The upper and lower surfaces of a coal seam are modelled as planes ABC and DEF, as shown in Fig. 8. All dimensions are metres. \includegraphics{figure_1} Relative to axes \(Ox\) (due east), \(Oy\) (due north) and \(Oz\) (vertically upwards), the coordinates of the points are as follows. A: \((0, 0, -15)\) \quad B: \((100, 0, -30)\) \quad C: \((0, 100, -25)\) D: \((0, 0, -40)\) \quad E: \((100, 0, -50)\) \quad F: \((0, 100, -35)\)
  1. Verify that the cartesian equation of the plane ABC is \(3x + 2y + 20z + 300 = 0\). [3]
  2. Find the vectors \(\overrightarrow{DE}\) and \(\overrightarrow{DF}\). Show that the vector \(2\mathbf{i} - \mathbf{j} + 20\mathbf{k}\) is perpendicular to each of these vectors. Hence find the cartesian equation of the plane DEF. [6]
  3. By calculating the angle between their normal vectors, find the angle between the planes ABC and DEF. [4]
It is decided to drill down to the seam from a point R \((15, 34, 0)\) in a line perpendicular to the upper surface of the seam. This line meets the plane ABC at the point S.
  1. Write down a vector equation of the line RS. Find the coordinates of S. [5]