Questions — Edexcel C4 (386 questions)

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Edexcel C4 2013 June Q3
7 marks Moderate -0.3
A curve \(C\) has parametric equations $$x = 2t + 5, \quad y = 3 + \frac{4}{t}, \quad t \neq 0$$
  1. Find the value of \(\frac{dy}{dx}\) at the point on \(C\) with coordinates \((9, 5)\). [4]
  2. Find a cartesian equation of the curve in the form $$y = \frac{ax + b}{cx + d}$$ where \(a\), \(b\), \(c\) and \(d\) are integers. [3]
Edexcel C4 2013 June Q4
10 marks Moderate -0.3
With respect to a fixed origin \(O\), the line \(l_1\) has vector equation $$\mathbf{r} = \begin{pmatrix} -9 \\ 8 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 5 \\ -4 \\ -3 \end{pmatrix}$$ where \(\mu\) is a scalar parameter. The point \(A\) is on \(l_1\) where \(\mu = 2\).
  1. Write down the coordinates of \(A\). [1] The acute angle between \(OA\) and \(l_1\) is \(\theta\), where \(O\) is the origin.
  2. Find the value of \(\cos \theta\). [3] The point \(B\) is such that \(\overrightarrow{OB} = 3\overrightarrow{OA}\). The line \(l_2\) passes through the point \(B\) and is parallel to the line \(l_1\).
  3. Find a vector equation of \(l_2\). [2]
  4. Find the length of \(OB\), giving your answer as a simplified surd. [1] The point \(X\) lies on \(l_2\). Given that the vector \(\overrightarrow{OX}\) is perpendicular to \(l_2\),
  5. find the length of \(OX\), giving your answer to 3 significant figures. [3]
Edexcel C4 2013 June Q5
9 marks Standard +0.3
The curve \(C\) has the equation $$\sin(\pi y) - y - x^2 y = -5, \quad x > 0$$
  1. Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). [5] The point \(P\) with coordinates \((2, 1)\) lies on \(C\). The tangent to \(C\) at \(P\) meets the \(x\)-axis at the point \(A\).
  2. Find the exact value of the \(x\)-coordinate of \(A\). [4]
Edexcel C4 2013 June Q6
11 marks Moderate -0.3
    1. Express \(\frac{7x}{(x + 3)(2x - 1)}\) in partial fractions. [3]
    2. Given that \(x > \frac{1}{2}\), find $$\int \frac{7x}{(x + 3)(2x - 1)} \, dx$$ [3]
  2. Using the substitution \(u^3 = x\), or otherwise, find $$\int \frac{1}{x + x^3} \, dx, \quad x > 0$$ [5]
Edexcel C4 2013 June Q7
10 marks Challenging +1.2
\includegraphics{figure_2} Figure 2 shows a sketch of part of the curve \(C\) with parametric equations $$x = \tan \theta, \quad y = 1 + 2\cos 2\theta, \quad 0 \leq \theta < \frac{\pi}{2}$$ The curve \(C\) crosses the \(x\)-axis at \((\sqrt{3}, 0)\). The finite shaded region \(S\) shown in Figure 2 is bounded by \(C\), the line \(x = 1\) and the \(x\)-axis. This shaded region is rotated through \(2\pi\) radians about the \(x\)-axis to form a solid of revolution.
  1. Show that the volume of the solid of revolution formed is given by the integral $$k \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (16 \cos^2 \theta - 8 + \sec^2 \theta) \, d\theta$$ where \(k\) is a constant. [5]
  2. Hence, use integration to find the exact value for this volume. [5]
Edexcel C4 2013 June Q8
9 marks Standard +0.3
\includegraphics{figure_3} Figure 3 shows a large vertical cylindrical tank containing a liquid. The radius of the circular cross-section of the tank is 40 cm. At time \(t\) minutes, the depth of liquid in the tank is \(h\) centimetres. The liquid leaks from a hole \(P\) at the bottom of the tank. The liquid leaks from the tank at a rate of \(32\pi \sqrt{h}\) cm\(^3\) min\(^{-1}\).
  1. Show that at time \(t\) minutes, the height \(h\) cm of liquid in the tank satisfies the differential equation $$\frac{dh}{dt} = -0.02\sqrt{h}$$ [4]
  2. Find the time taken, to the nearest minute, for the depth of liquid in the tank to decrease from 100 cm to 50 cm. [5]
Edexcel C4 2015 June Q1
8 marks Moderate -0.3
  1. Find the binomial expansion of $$(4 + 5x)^{\frac{1}{2}}, \quad |x| < \frac{4}{5}$$ in ascending powers of \(x\), up to and including the term in \(x^2\). Give each coefficient in its simplest form. [5]
  2. Find the exact value of \((4 + 5x)^{\frac{1}{2}}\) when \(x = \frac{1}{10}\) Give your answer in the form \(k\sqrt{2}\), where \(k\) is a constant to be determined. [1]
  3. Substitute \(x = \frac{1}{10}\) into your binomial expansion from part (a) and hence find an approximate value for \(\sqrt{2}\) Give your answer in the form \(\frac{p}{q}\) where \(p\) and \(q\) are integers. [2]
Edexcel C4 2015 June Q2
11 marks Standard +0.3
The curve \(C\) has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$
  1. Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). [5]
  2. Find the coordinates of the points on \(C\) where \(\frac{dy}{dx} = 0\) (Solutions based entirely on graphical or numerical methods are not acceptable.) [6]
Edexcel C4 2015 June Q3
8 marks Challenging +1.2
\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve with equation \(y = 4x - xe^{\frac{1}{x}}, x \geqslant 0\) The curve meets the \(x\)-axis at the origin \(O\) and cuts the \(x\)-axis at the point \(A\).
  1. Find, in terms of \(\ln 2\), the \(x\) coordinate of the point \(A\). [2]
  2. Find $$\int xe^{\frac{1}{x}} dx$$ [3]
  3. Find, by integration, the exact value for the area of \(R\). Give your answer in terms of \(\ln 2\) [3]
The finite region \(R\), shown shaded in Figure 1, is bounded by the \(x\)-axis and the curve with equation $$y = 4x - xe^{\frac{1}{x}}, x \geqslant 0$$
Edexcel C4 2015 June Q4
11 marks Standard +0.3
With respect to a fixed origin \(O\), the lines \(l_1\) and \(l_2\) are given by the equations $$l_1: \mathbf{r} = \begin{pmatrix} 5 \\ -3 \\ p \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix}, \quad l_2: \mathbf{r} = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix}$$ where \(\lambda\) and \(\mu\) are scalar parameters and \(p\) is a constant. The lines \(l_1\) and \(l_2\) intersect at the point \(A\).
  1. Find the coordinates of \(A\). [2]
  2. Find the value of the constant \(p\). [3]
  3. Find the acute angle between \(l_1\) and \(l_2\), giving your answer in degrees to 2 decimal places. [3]
The point \(B\) lies on \(l_2\) where \(\mu = 1\)
  1. Find the shortest distance from the point \(B\) to the line \(l_1\), giving your answer to 3 significant figures. [3]
Edexcel C4 2015 June Q5
6 marks Moderate -0.3
A curve \(C\) has parametric equations $$x = 4t + 3, \quad y = 4t + 8 + \frac{5}{2t}, \quad t \neq 0$$
  1. Find the value of \(\frac{dy}{dx}\) at the point on \(C\) where \(t = 2\), giving your answer as a fraction in its simplest form. [3]
  2. Show that the cartesian equation of the curve \(C\) can be written in the form $$y = \frac{x^2 + ax + b}{x - 3}, \quad x \neq 3$$ where \(a\) and \(b\) are integers to be determined. [3]
Edexcel C4 2015 June Q6
8 marks Standard +0.8
\includegraphics{figure_2} Figure 2 shows a sketch of the curve with equation \(y = \sqrt{(3-x)(x+1)}\), \(0 \leqslant x \leqslant 3\) The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis, and the \(y\)-axis.
  1. Use the substitution \(x = 1 + 2\sin\theta\) to show that $$\int_0^3 \sqrt{(3-x)(x+1)} dx = k \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2\theta d\theta$$ where \(k\) is a constant to be determined. [5]
  2. Hence find, by integration, the exact area of \(R\). [3]
Edexcel C4 2015 June Q7
13 marks Standard +0.8
  1. Express \(\frac{2}{P(P-2)}\) in partial fractions. [3]
A team of biologists is studying a population of a particular species of animal. The population is modelled by the differential equation $$\frac{dP}{dt} = \frac{1}{2}P(P-2)\cos 2t, \quad t \geqslant 0$$ where \(P\) is the population in thousands, and \(t\) is the time measured in years since the start of the study. Given that \(P = 3\) when \(t = 0\),
  1. solve this differential equation to show that $$P = \frac{6}{3 - e^{\frac{1}{2}\sin 2t}}$$ [7]
  2. find the time taken for the population to reach 4000 for the first time. Give your answer in years to 3 significant figures. [3]
Edexcel C4 2015 June Q8
10 marks Standard +0.8
\includegraphics{figure_3} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = 3^x$$ The point \(P\) lies on \(C\) and has coordinates \((2, 9)\). The line \(l\) is a tangent to \(C\) at \(P\). The line \(l\) cuts the \(x\)-axis at the point \(Q\).
  1. Find the exact value of the \(x\) coordinate of \(Q\). [4]
The finite region \(R\), shown shaded in Figure 3, is bounded by the curve \(C\), the \(x\)-axis, the \(y\)-axis and the line \(l\). This region \(R\) is rotated through \(360°\) about the \(x\)-axis.
  1. Use integration to find the exact value of the volume of the solid generated. Give your answer in the form \(\frac{p}{q}\) where \(p\) and \(q\) are exact constants. [You may assume the formula \(V = \frac{1}{3}\pi r^2 h\) for the volume of a cone.] [6]
Edexcel C4 Q1
6 marks Standard +0.3
Use the substitution \(u = 4 + 3x^2\) to find the exact value of $$\int_0^2 \frac{2x}{(4 + 3x^2)^2} \, dx .$$ [6]
Edexcel C4 Q2
8 marks Standard +0.3
A curve has equation $$x^3 - 2xy - 4x + y^3 - 51 = 0.$$ Find an equation of the normal to the curve at the point \((4, 3)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [8]
Edexcel C4 Q3
13 marks Standard +0.3
$$f(x) = \frac{1 + 14x}{(1 - x)(1 + 2x)}, \quad |x| < \frac{1}{2}.$$
  1. Express \(f(x)\) in partial fractions. [3]
  2. Hence find the exact value of \(\int_{-\frac{1}{6}}^{\frac{1}{4}} f(x) \, dx\), giving your answer in the form \(\ln p\), where \(p\) is rational. [5]
  3. Use the binomial theorem to expand \(f(x)\) in ascending powers of \(x\), up to and including the term in \(x^5\), simplifying each term. [5]
Edexcel C4 Q4
10 marks Standard +0.3
The line \(l_1\) has vector equation \(\mathbf{r} = \begin{pmatrix} 11 \\ 5 \\ 6 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix}\), where \(\lambda\) is a parameter. The line \(l_2\) has vector equation \(\mathbf{r} = \begin{pmatrix} 24 \\ 4 \\ 13 \end{pmatrix} + \mu \begin{pmatrix} 7 \\ 1 \\ 5 \end{pmatrix}\), where \(\mu\) is a parameter.
  1. Show that the lines \(l_1\) and \(l_2\) intersect. [4]
  2. Find the coordinates of their point of intersection. [2]
Given that \(\theta\) is the acute angle between \(l_1\) and \(l_2\),
  1. find the value of \(\cos \theta\). Give your answer in the form \(k\sqrt{3}\), where \(k\) is a simplified fraction. [4]
Edexcel C4 Q5
12 marks Standard +0.3
\includegraphics{figure_1} The curve shown in Fig. 1 has parametric equations $$x = \cos t, \quad y = \sin 2t, \quad 0 \leq t < 2\pi.$$
  1. Find an expression for \(\frac{dy}{dx}\) in terms of the parameter \(t\). [3]
  2. Find the values of the parameter \(t\) at the points where \(\frac{dy}{dx} = 0\). [3]
  3. Hence give the exact values of the coordinates of the points on the curve where the tangents are parallel to the \(x\)-axis. [2]
  4. Show that a cartesian equation for the part of the curve where \(0 \leq t < \pi\) is $$y = 2x\sqrt{(1 - x^2)}.$$ [3]
  5. Write down a cartesian equation for the part of the curve where \(\pi \leq t < 2\pi\). [1]
Edexcel C4 Q6
13 marks Standard +0.3
\includegraphics{figure_2} Figure 2 shows the curve with equation $$y = x^2 \sin\left(\frac{1}{2}x\right), \quad 0 < x \leq 2\pi.$$ The finite region \(R\) bounded by the line \(x = \pi\), the \(x\)-axis, and the curve is shown shaded in Fig 2.
  1. Find the exact value of the area of \(R\), by integration. Give your answer in terms of \(\pi\). [7]
The table shows corresponding values of \(x\) and \(y\).
\(x\)\(\pi\)\(\frac{5\pi}{4}\)\(\frac{3\pi}{2}\)\(\frac{7\pi}{4}\)\(2\pi\)
\(y\)\(9.8696\)\(14.247\)\(15.702\)\(G\)\(0\)
  1. Find the value of \(G\). [1]
  2. Use the trapezium rule with values of \(x^2 \sin\left(\frac{1}{2}x\right)\)
    1. at \(x = \pi\), \(x = \frac{3\pi}{2}\) and \(x = 2\pi\) to find an approximate value for the area \(R\), giving your answer to 4 significant figures,
    2. at \(x = \pi\), \(x = \frac{5\pi}{4}\), \(x = \frac{3\pi}{2}\), \(x = \frac{7\pi}{4}\) and \(x = 2\pi\) to find an improved approximation for the area \(R\), giving your answer to 4 significant figures.
    [5]
Edexcel C4 Q7
13 marks Standard +0.3
In an experiment a scientist considered the loss of mass of a collection of picked leaves. The mass \(M\) grams of a single leaf was measured at times \(t\) days after the leaf was picked. The scientist attempted to find a relationship between \(M\) and \(t\). In a preliminary model she assumed that the rate of loss of mass was proportional to the mass \(M\) grams of the leaf.
  1. Write down a differential equation for the rate of change of mass of the leaf, using this model. [2]
  2. Show, by differentiation, that \(M = 10(0.98)^t\) satisfies this differential equation. [2]
Further studies implied that the mass \(M\) grams of a certain leaf satisfied a modified differential equation $$10 \frac{dM}{dt} = -k(10M - 1), \quad (1)$$ where \(k\) is a positive constant and \(t \geq 0\). Given that the mass of this leaf at time \(t = 0\) is 10 grams, and that its mass at time \(t = 10\) is 8.5 grams,
  1. solve the modified differential equation (1) to find the mass of this leaf at time \(t = 15\). [9]
Edexcel C4 Q1
6 marks Moderate -0.8
A measure of the effective voltage, \(M\) volts, in an electrical circuit is given by $$M^2 = \int_0^1 V^2 \, dt$$ where \(V\) volts is the voltage at time \(t\) seconds. Pairs of values of \(V\) and \(t\) are given in the following table.
\(t\)00.250.50.751
\(V\)-4820737-161-29
\(V^2\)
Use the trapezium rule with five values of \(V^2\) to estimate the value of \(M\). [6]
Edexcel C4 Q2
7 marks Standard +0.3
\includegraphics{figure_1} Figure 1 shows part of a curve \(C\) with equation \(y = x^2 + 3\). The shaded region is bounded by \(C\), the \(x\)-axis and the lines \(x = 1\) and \(x = 3\). The shaded region is rotated \(360°\) about the \(x\)-axis. Using calculus, calculate the volume of the solid generated. Give your answer as an exact multiple of \(\pi\). [7]
Edexcel C4 Q3
6 marks Moderate -0.8
  1. Given that \(a^x = e^{kx}\), where \(a\) and \(k\) are constants, \(a > 0\) and \(x \in \mathbb{R}\), prove that \(k = \ln a\). [2]
  2. Hence, using the derivative of \(e^{kx}\), prove that when \(y = 2^x\), $$\frac{dy}{dx} = 2^x \ln 2.$$ [2]
  3. Hence deduce that the gradient of the curve with equation \(y = 2^x\) at the point \((2, 4)\) is \(\ln 16\). [2]
Edexcel C4 Q4
8 marks Moderate -0.3
$$f(x) = (1 + 3x)^{-1}, \quad |x| < \frac{1}{3}.$$
  1. Expand \(f(x)\) in ascending powers of \(x\) up to and including the term in \(x^3\). [3]
  2. Hence show that, for small \(x\), $$\frac{1 + x}{1 + 3x} \approx 1 - 2x + 6x^2 - 18x^3.$$ [2]
  3. Taking a suitable value for \(x\), which should be stated, use the series expansion in part \((b)\) to find an approximate value for \(\frac{101}{103}\), giving your answer to 5 decimal places. [3]