Questions C4 (1219 questions)

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OCR C4 2005 June Q4
7 marks Standard +0.3
  1. Show that the substitution \(x = \tan \theta\) transforms \(\int \frac{1}{(1 + x^2)^2} dx\) to \(\int \cos^2 \theta d\theta\). [3]
  2. Hence find the exact value of \(\int_0^1 \frac{1}{(1 + x^2)^2} dx\). [4]
OCR C4 2005 June Q5
7 marks Moderate -0.3
\(ABCD\) is a parallelogram. The position vectors of \(A\), \(B\) and \(C\) are given respectively by $$\mathbf{a} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}, \quad \mathbf{b} = 3\mathbf{i} - 2\mathbf{j}, \quad \mathbf{c} = \mathbf{i} - \mathbf{j} - 2\mathbf{k}.$$
  1. Find the position vector of \(D\). [3]
  2. Determine, to the nearest degree, the angle \(ABC\). [4]
OCR C4 2005 June Q6
8 marks Standard +0.3
The equation of a curve is \(xy^2 = 2x + 3y\).
  1. Show that \(\frac{dy}{dx} = \frac{2 - y^2}{2xy - 3}\). [5]
  2. Show that the curve has no tangents which are parallel to the \(y\)-axis. [3]
OCR C4 2005 June Q7
10 marks Standard +0.3
A curve is given parametrically by the equations $$x = t^2, \quad y = \frac{1}{t}.$$
  1. Find \(\frac{dy}{dx}\) in terms of \(t\), giving your answer in its simplest form. [3]
  2. Show that the equation of the tangent at the point \(P\left(4, -\frac{1}{4}\right)\) is \(x - 16y = 12\). [3]
  3. Find the value of the parameter at the point where the tangent at \(P\) meets the curve again. [4]
OCR C4 2005 June Q8
11 marks Standard +0.3
  1. Given that \(\frac{3x + 4}{(1 + x)(2 + x)^2} \equiv \frac{A}{1 + x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2}\), find \(A\), \(B\) and \(C\). [5]
  2. Hence or otherwise expand \(\frac{3x + 4}{(1 + x)(2 + x)^2}\) in ascending powers of \(x\), up to and including the term in \(x^2\). [5]
  3. State the set of values of \(x\) for which the expansion in part (ii) is valid. [1]
OCR C4 2005 June Q9
13 marks Standard +0.3
Newton's law of cooling states that the rate at which the temperature of an object is falling at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant. A container of hot liquid is placed in a room which has a constant temperature of \(20°C\). At time \(t\) minutes later, the temperature of the liquid is \(\theta°C\).
  1. Explain how the information above leads to the differential equation $$\frac{d\theta}{dt} = -k(\theta - 20),$$ where \(k\) is a positive constant. [2]
  2. The liquid is initially at a temperature of \(100°C\). It takes 5 minutes for the liquid to cool from \(100°C\) to \(68°C\). Show that $$\theta = 20 + 80e^{-(\frac{k}{5} \ln \frac{5}{3})t}.$$ [8]
  3. Calculate how much longer it takes for the liquid to cool by a further \(32°C\). [3]
OCR C4 2006 June Q1
4 marks Moderate -0.3
Find the gradient of the curve \(4x^2 + 2xy + y^2 = 12\) at the point \((1, 2)\). [4]
OCR C4 2006 June Q2
7 marks Moderate -0.8
  1. Expand \((1 - 3x)^{-2}\) in ascending powers of \(x\), up to and including the term in \(x^2\). [3]
  2. Find the coefficient of \(x^2\) in the expansion of \(\frac{(1 + 2x)^2}{(1 - 3x)^2}\) in ascending powers of \(x\). [4]
OCR C4 2006 June Q3
8 marks Moderate -0.3
  1. Express \(\frac{3 - 2x}{x(3 - x)}\) in partial fractions. [3]
  2. Show that \(\int_1^2 \frac{3 - 2x}{x(3 - x)} dx = 0\). [4]
  3. What does the result of part (ii) indicate about the graph of \(y = \frac{3 - 2x}{x(3 - x)}\) between \(x = 1\) and \(x = 2\)? [1]
OCR C4 2006 June Q4
8 marks Standard +0.3
The position vectors of three points \(A\), \(B\) and \(C\) relative to an origin \(O\) are given respectively by $$\overrightarrow{OA} = 7\mathbf{i} + 3\mathbf{j} - 3\mathbf{k},$$ $$\overrightarrow{OB} = 4\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}$$ and $$\overrightarrow{OC} = 5\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}.$$
  1. Find the angle between \(AB\) and \(AC\). [6]
  2. Find the area of triangle \(ABC\). [2]
OCR C4 2006 June Q5
8 marks Standard +0.3
A forest is burning so that, \(t\) hours after the start of the fire, the area burnt is \(A\) hectares. It is given that, at any instant, the rate at which this area is increasing is proportional to \(A^2\).
  1. Write down a differential equation which models this situation. [2]
  2. After 1 hour, 1000 hectares have been burnt; after 2 hours, 2000 hectares have been burnt. Find after how many hours 3000 hectares have been burnt. [6]
OCR C4 2006 June Q6
8 marks Standard +0.3
  1. Show that the substitution \(u = e^x + 1\) transforms \(\int \frac{e^{2x}}{e^x + 1} dx\) to \(\int \frac{u - 1}{u} du\). [3]
  2. Hence show that \(\int_0^1 \frac{e^{2x}}{e^x + 1} dx = e - 1 - \ln\left(\frac{e + 1}{2}\right)\). [5]
OCR C4 2006 June Q7
8 marks Standard +0.3
Two lines have vector equations $$\mathbf{r} = \mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + \lambda(3\mathbf{i} + \mathbf{j} + a\mathbf{k})$$ and $$\mathbf{r} = -8\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + \mu(\mathbf{i} - 2\mathbf{j} - \mathbf{k}),$$ where \(a\) is a constant.
  1. Given that the lines are skew, find the value that \(a\) cannot take. [6]
  2. Given instead that the lines intersect, find the point of intersection. [2]
OCR C4 2006 June Q8
9 marks Standard +0.8
  1. Show that \(\int \cos^2 6x dx = \frac{1}{2}x + \frac{1}{24}\sin 12x + c\). [3]
  2. Hence find the exact value of \(\int_0^{\frac{\pi}{12}} x\cos^2 6x dx\). [6]
OCR C4 2006 June Q9
12 marks Standard +0.3
A curve is given parametrically by the equations $$x = 4\cos t, \quad y = 3\sin t,$$ where \(0 \leq t \leq \frac{1}{2}\pi\).
  1. Find \(\frac{dy}{dx}\) in terms of \(t\). [3]
  2. Show that the equation of the tangent at the point \(P\), where \(t = p\), is $$3x\cos p + 4y\sin p = 12.$$ [3]
  3. The tangent at \(P\) meets the \(x\)-axis at \(R\) and the \(y\)-axis at \(S\). \(O\) is the origin. Show that the area of triangle \(ORS\) is \(\frac{6}{\sin 2p}\). [3]
  4. Write down the least possible value of the area of triangle \(ORS\), and give the corresponding value of \(p\). [3]
OCR MEI C4 2012 January Q1
5 marks Moderate -0.5
Express \(\frac{x+1}{x^2(2x-1)}\) in partial fractions. [5]
OCR MEI C4 2012 January Q2
4 marks Moderate -0.3
Solve, correct to 2 decimal places, the equation \(\cot 2\theta = 3\) for \(0° < \theta < 180°\). [4]
OCR MEI C4 2012 January Q3
7 marks Moderate -0.3
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 \(y = f(x)\), where $$f(x) = 3\sin x + 2\cos x, \quad 0 < x < \pi.$$ [7]
OCR MEI C4 2012 January Q4
4 marks Moderate -0.3
  1. Complete the table of values for the curve \(y = \sqrt{\cos x}\).
    \(x\)0\(\frac{\pi}{6}\)\(\frac{\pi}{4}\)\(\frac{3\pi}{8}\)\(\frac{\pi}{2}\)
    \(y\)0.96120.8409
    Hence use the trapezium rule with strip width \(h = \frac{\pi}{8}\) to estimate the value of the integral \(\int_0^{\frac{\pi}{2}} \sqrt{\cos x} \, dx\), giving your answer to 3 decimal places. [3] Fig. 4 shows the curve \(y = \sqrt{\cos x}\) for \(0 \leq x \leq \frac{\pi}{2}\). \includegraphics{figure_4}
  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. [1]
OCR MEI C4 2012 January Q5
5 marks Moderate -0.8
Verify that the vector \(2\mathbf{i} - \mathbf{j} + 4\mathbf{k}\) is perpendicular to the plane through the points A(2, 0, 1), B(1, 2, 2) and C(0, -4, 1). Hence find the cartesian equation of the plane. [5]
OCR MEI C4 2012 January Q6
6 marks Standard +0.3
Given the binomial expansion \((1 + qx)^p = 1 - x + 2x^2 + \ldots\), find the values of \(p\) and \(q\). Hence state the set of values of \(x\) for which the expansion is valid. [6]
OCR MEI C4 2012 January Q7
5 marks Moderate -0.3
Show that the straight lines with equations \(\mathbf{r} = \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix} -1 \\ 4 \\ 9 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix}\) meet. Find their point of intersection. [5]
OCR MEI C4 2012 January Q8
18 marks Standard +0.3
Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2t^2, y = 4t, \quad -\sqrt{2} < t < \sqrt{2}.$$ P\((2t^2, 4t)\) is a point on the curve with parameter \(t\). TS is the tangent to the curve at P, and PR is the line through P parallel to the \(x\)-axis. Q is the point (2, 0). The angles that PS and QP make with the positive \(x\)-direction are \(\theta\) and \(\phi\) respectively. \includegraphics{figure_8}
  1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac{1}{t}\). [3]
  2. Find the gradient of the line QP in terms of \(t\). Hence show that \(\phi = 2\theta\), and that angle TPQ is equal to \(\theta\). [8]
[The above result shows that if a lamp bulb is placed at Q, then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the \(x\)-axis.
  1. Show that the curve has cartesian equation \(y^2 = 8x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\). [7]
OCR MEI C4 2012 January Q9
18 marks Standard +0.3
\includegraphics{figure_9} 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 \(k\) 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 \(3T\) seconds. [6]
OCR MEI C4 2009 June Q1
7 marks Moderate -0.3
Express \(4\cos\theta - \sin\theta\) in the form \(R\cos(\theta + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{1}{2}\pi\). Hence solve the equation \(4\cos\theta - \sin\theta = 3\), for \(0 \leq \theta \leq 2\pi\). [7]