Questions C4 (1219 questions)

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OCR C4 Q6
10 marks Standard +0.3
Relative to a fixed origin, \(O\), the line \(l\) has the equation $$\mathbf{r} = \begin{pmatrix} 1 \\ p \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -1 \\ q \end{pmatrix},$$ where \(p\) and \(q\) are constants and \(\lambda\) is a scalar parameter. Given that the point \(A\) with coordinates \((-5, 9, -9)\) lies on \(l\),
  1. find the values of \(p\) and \(q\), [3]
  2. show that the point \(B\) with coordinates \((25, -1, 11)\) also lies on \(l\). [2]
The point \(C\) lies on \(l\) and is such that \(OC\) is perpendicular to \(l\).
  1. Find the coordinates of \(C\). [3]
  2. Find the ratio \(AC : CB\) [2]
OCR C4 Q7
11 marks Standard +0.3
  1. Use the substitution \(x = 2 \sin u\) to evaluate $$\int_0^{\sqrt{3}} \frac{1}{\sqrt{4-x^2}} \, dx.$$ [6]
  2. Evaluate $$\int_0^{\frac{\pi}{2}} x \cos x \, dx.$$ [5]
OCR C4 Q8
12 marks Moderate -0.3
The rate of increase in the number of bacteria in a culture, \(N\), at time \(t\) hours is proportional to \(N\).
  1. Write down a differential equation connecting \(N\) and \(t\). [1]
Given that initially there are \(N_0\) bacteria present in a culture,
  1. Show that \(N = N_0 e^{kt}\), where \(k\) is a positive constant. [6]
Given also that the number of bacteria present doubles every six hours,
  1. find the value of \(k\), [3]
  2. Find how long it takes for the number of bacteria to increase by a factor of ten, giving your answer to the nearest minute. [2]
OCR C4 Q1
4 marks Moderate -0.5
Express $$\frac{5x}{(x-4)(x+1)} + \frac{3}{(x-2)(x+1)}$$ as a single fraction in its simplest form. [4]
OCR C4 Q2
5 marks Standard +0.3
A curve has the equation $$x^2 + 2xy^2 + y = 4.$$ Find an expression for \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). [5]
OCR C4 Q3
5 marks Standard +0.3
Evaluate $$\int_0^{\frac{\pi}{4}} \sin 2x \cos x \, dx.$$ [5]
OCR C4 Q4
7 marks Standard +0.3
A curve has parametric equations $$x = \cos 2t, \quad y = \cosec t, \quad 0 < t < \frac{\pi}{2}.$$ The point \(P\) on the curve has \(x\)-coordinate \(\frac{1}{2}\).
  1. Find the value of the parameter \(t\) at \(P\). [2]
  2. Show that the tangent to the curve at \(P\) has the equation $$y = 2x + 1.$$ [5]
OCR C4 Q5
8 marks Standard +0.8
  1. Express \(\frac{2 + 20x}{1 + 2x - 8x^2}\) as a sum of partial fractions. [3]
  2. Hence find the series expansion of \(\frac{2 + 20x}{1 + 2x - 8x^2}\), \(|x| < \frac{1}{4}\), in ascending powers of \(x\) up to and including the term in \(x^3\), simplifying each coefficient. [5]
OCR C4 Q6
8 marks Standard +0.3
Use the substitution \(x = 2 \tan u\) to show that $$\int_0^2 \frac{x^2}{x^2 + 4} \, dx = \frac{1}{2}(4 - \pi).$$ [8]
OCR C4 Q7
9 marks Standard +0.3
A straight road passes through villages at the points \(A\) and \(B\) with position vectors \((9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k})\) and \((4\mathbf{j} + \mathbf{k})\) respectively, relative to a fixed origin. The road ends at a junction at the point \(C\) with another straight road which lies along the line with equation $$\mathbf{r} = (2\mathbf{i} + 16\mathbf{j} - \mathbf{k}) + t(-5\mathbf{i} + 3\mathbf{j}),$$ where \(t\) is a scalar parameter.
  1. Find the position vector of \(C\). [5]
Given that 1 unit on each coordinate axis represents 200 metres,
  1. find the distance, in kilometres, from the village at \(A\) to the junction at \(C\). [4]
OCR C4 Q8
12 marks Challenging +1.2
  1. Find \(\int \tan^2 x \, dx\). [3]
  2. Show that $$\int \tan x \, dx = \ln |\sec x| + c,$$ where \(c\) is an arbitrary constant. [4]
\includegraphics{figure_8} The diagram shows part of the curve with equation \(y = x^{\frac{1}{2}} \tan x\). The shaded region bounded by the curve, the \(x\)-axis and the line \(x = \frac{\pi}{3}\) is rotated through \(360°\) about the \(x\)-axis.
  1. Show that the volume of the solid formed is \(\frac{1}{18}\pi^2(6\sqrt{3} - \pi) - \pi \ln 2\). [5]
OCR C4 Q9
14 marks Standard +0.3
An entomologist is studying the population of insects in a colony. Initially there are 300 insects in the colony and in a model, the entomologist assumes that the population, \(P\), at time \(t\) weeks satisfies the differential equation $$\frac{dP}{dt} = kP,$$ where \(k\) is a constant.
  1. Find an expression for \(P\) in terms of \(k\) and \(t\). [5]
Given that after one week there are 360 insects in the colony,
  1. find the value of \(k\) to 3 significant figures. [2]
Given also that after two and three weeks there are 440 and 600 insects respectively,
  1. comment on suitability of the modelling assumption. [2]
An alternative model assumes that $$\frac{dP}{dt} = P(0.4 - 0.25 \cos 0.5t).$$
  1. Using the initial data, \(P = 300\) when \(t = 0\), solve this differential equation. [3]
  2. Compare the suitability of the two models. [2]
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]