Questions — Edexcel (10514 questions)

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Edexcel FP2 Q5
10 marks Standard +0.8
  1. Find, in the form \(y = f(x)\), the general solution of the equation $$\frac{dy}{dx} = 2y \tan x + \sin 2x, \quad 0 < x < \frac{\pi}{2}$$ [6]
Given that \(y = 2\) at \(x = \frac{\pi}{6}\),
  1. find the value of \(y\) at \(x = \frac{\pi}{4}\), giving your answer in the form \(a + k \ln b\), where \(a\) and \(b\) are integers and \(k\) is rational. [4]
Edexcel FP2 Q6
11 marks Standard +0.8
The complex number \(z = e^{i\theta}\), where \(\theta\) is real.
  1. Use de Moivre's theorem to show that $$z^n + \frac{1}{z^n} = 2\cos n\theta$$ where \(n\) is a positive integer. [2]
  2. Show that $$\cos^n \theta = \frac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos \theta)$$ [5]
  3. Hence find all the solutions of $$\cos 5\theta + 5\cos 3\theta + 12\cos \theta = 0$$ in the interval \(0 \leq \theta < 2\pi\). [4]
Edexcel FP2 Q7
13 marks Standard +0.8
  1. Find the value of \(z\) for which \(z^{2e^x}\) is a particular integral of the differential equation $$\frac{d^2 y}{dt^2} - 6 \frac{dy}{dt} + 9y = 6e^{3t}, \quad t \geq 0$$ [5]
  2. Hence find the general solution of this differential equation. [3]
Given that when \(t = 0\), \(y = 5\) and \(\frac{dy}{dt} = 4\)
  1. find the particular solution of this differential equation, giving your solution in the form \(y = f(t)\). [5]
Edexcel FP2 Q8
13 marks Challenging +1.8
\includegraphics{figure_1} Figure 1 shows a closed curve \(C\) with equation $$r = 3(\cos 2\theta)^{\frac{1}{2}}, \quad \text{where } -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}, \frac{3\pi}{4} \leq \theta \leq \frac{5\pi}{4}.$$ The lines \(PQ\), \(SR\), \(PS\) and \(QR\) are tangents to \(C\), where \(PQ\) and \(SR\) are parallel to the initial line and \(PS\) and \(QR\) are perpendicular to the initial line. The point \(O\) is the pole.
  1. Find the total area enclosed by the curve \(C\), shown unshaded inside the rectangle in Figure 1. [4]
  2. Find the total area of the region bounded by the curve \(C\) and the four tangents, shown shaded in Figure 1. [9]
Edexcel FP2 Q1
5 marks Moderate -0.3
  1. Express \(\frac{2}{(2r + 1)(2r + 3)}\) in partial fractions. [2]
  2. Using your answer to (a), find, in terms of \(n\), $$\sum_{r=1}^n \frac{2}{(2r + 1)(2r + 3)}$$ [3]
Give your answer as a single fraction in its simplest form.
Edexcel FP2 Q2
6 marks Moderate -0.8
\(z = 5\sqrt{3} - 5i\) Find
  1. \(|z|\), [1]
  2. \(\arg(z)\), in terms of \(\pi\). [2]
$$w = 2\left[\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right]$$ Find
  1. \(\left|\frac{w}{z}\right|\), [1]
  2. \(\arg\left(\frac{w}{z}\right)\), in terms of \(\pi\). [2]
Edexcel FP2 Q3
5 marks Standard +0.8
$$\frac{d^2 y}{dx^2} + 4y - \sin x = 0$$ Given that \(y = \frac{1}{2}\) and \(\frac{dy}{dx} = \frac{1}{8}\) at \(x = 0\), find a series expansion for \(y\) in terms of \(x\), up to and including the term in \(x^5\). [5]
Edexcel FP2 Q4
7 marks Standard +0.8
  1. Given that $$z = r(\cos n\theta + i \sin n\theta), \quad r \in \mathbf{R}$$ prove, by induction, that \(z^n = r^n(\cos n\theta + i \sin n\theta)\), \(n \in \mathbf{Z}^+\). [5]
  2. Find the exact value of \(w^2\), giving your answer in the form \(a + ib\), where \(a, b \in \mathbf{R}\). [2]
Edexcel FP2 Q5
12 marks Standard +0.8
  1. Find the general solution of the differential equation $$x \frac{dy}{dx} + 2y = 4x^2$$ [5]
  2. Find the particular solution for which \(y = 5\) at \(x = 1\), giving your answer in the form \(y = f(x)\). [2]
  3. Find the exact values of the coordinates of the turning points of the curve with equation \(y = f(x)\), making your method clear. [???]
  4. Sketch the curve with equation \(y = f(x)\), showing the coordinates of the turning points. [5]
Edexcel FP2 Q6
12 marks Standard +0.8
  1. Use algebra to find the exact solutions of the equation $$|2x^2 + 6x - 5| = 5 - 2x$$ [6]
  2. On the same diagram, sketch the curve with equation \(y = |2x^2 + 6x - 5|\) and the line with equation \(y = 5 - 2x\), showing the \(x\)-coordinates of the points where the line crosses the curve. [3]
  3. Find the set of values of \(x\) for which $$|2x^2 + 6x - 5| > 5 - 2x$$ [3]
Edexcel FP2 Q7
13 marks Challenging +1.3
  1. Show that the transformation \(y = xv\) transforms the equation $$4x^2 \frac{d^2 y}{dx^2} - 8x \frac{dy}{dx} + (8 + 4x^2)y = x^4$$ [I] into the equation $$x^2 \frac{d^2 v}{dx^2} + 4v = x$$ [II] [6]
  2. Solve the differential equation (II) to find \(v\) as a function of \(x\). [6]
  3. Hence state the general solution of the differential equation (I). [1]
Edexcel FP2 2008 June Q1
Moderate -0.3
Solve the differential equation \(\frac{dy}{dx} - 3y = x\) to obtain \(y\) as a function of \(x\). (Total 5 marks)
Edexcel FP2 2008 June Q2
Standard +0.3
  1. Simplify the expression \(\frac{(x + 3)(x + 9)}{x - 1} - (3x - 5)\), giving your answer in the form \(\frac{a(x + b)(x + c)}{x - 1}\), where \(a\), \(b\) and \(c\) are integers. (4)
  2. Hence, or otherwise, solve the inequality \(\frac{(x + 3)(x + 9)}{x - 1} > 3x - 5\) (4)(Total 8 marks)
Edexcel FP2 2008 June Q3
Challenging +1.2
  1. Find the general solution of the differential equation \(3\frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = x^2\) (8)
  2. Find the particular solution for which, at \(x = 0\), \(y = 2\) and \(\frac{dy}{dx} = 3\). (6)(Total 14 marks)
Edexcel FP2 2008 June Q4
Challenging +1.2
The diagram above shows the curve \(C_1\) which has polar equation \(r = a(3 + 2\cos\theta)\), \(0 \leq \theta < 2\pi\) and the circle \(C_2\) with equation \(r = 4a\), \(0 \leq \theta < 2\pi\), where \(a\) is a positive constant.
  1. Find, in terms of \(a\), the polar coordinates of the points where the curve \(C_1\) meets the circle \(C_2\).(4)
The regions enclosed by the curves \(C_1\) and \(C_2\) overlap and this common region \(R\) is shaded in the figure.
  1. Find, in terms of \(a\), an exact expression for the area of the region \(R\).(8)
  2. In a single diagram, copy the two curves in the diagram above and also sketch the curve \(C_3\) with polar equation \(r = 2a\cos\theta\), \(0 \leq \theta < 2\pi\) Show clearly the coordinates of the points of intersection of \(C_1\), \(C_2\) and \(C_3\) with the initial line, \(\theta = 0\).(3)(Total 15 marks)
\includegraphics{figure_4}
Edexcel FP2 2008 June Q5
Standard +0.8
  1. Find, in terms of \(k\), the general solution of the differential equation $$\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 3x = kt + 5, \text{ where } k \text{ is a constant and } t > 0.$$ (7) For large values of \(t\), this general solution may be approximated by a linear function.
  2. Given that \(k = 6\), find the equation of this linear function.(2)(Total 9 marks)
Edexcel FP2 2008 June Q6
Standard +0.3
  1. Find, in the simplest surd form where appropriate, the exact values of \(x\) for which $$\frac{x}{2} + 3 = \left|\frac{4}{x}\right|.$$ (5)
  2. Sketch, on the same axes, the line with equation \(y = \frac{x}{2} + 3\) and the graph of $$y = \left|\frac{4}{x}\right|, x \neq 0.$$ (3)
  3. Find the set of values of \(x\) for which \(\frac{x}{2} + 3 > \left|\frac{4}{x}\right|\). (2)(Total 10 marks)
Edexcel FP2 2008 June Q7
Challenging +1.2
  1. Show that the substitution \(y = vx\) transforms the differential equation $$\frac{dy}{dx} = \frac{x}{y} + \frac{3y}{x}, x > 0, y > 0$$ (I) into the differential equation \(x\frac{dv}{dx} = 2v + \frac{1}{v}\). (II) (3)
  2. By solving differential equation (II), find a general solution of differential equation (I) in the form \(y = f(x)\). (7)
Given that \(y = 3\) at \(x = 1\), (c)find the particular solution of differential equation (I).(2)
Edexcel FP2 2008 June Q8
Challenging +1.3
The curve \(C\) shown in the diagram above has polar equation $$r = 4(1 - \cos\theta), 0 \leq \theta \leq \frac{\pi}{2}.$$ At the point \(P\) on \(C\), the tangent to \(C\) is parallel to the line \(\theta = \frac{\pi}{2}\).
  1. Show that \(P\) has polar coordinates \(\left(2, \frac{\pi}{3}\right)\).(5)
The curve \(C\) meets the line \(\theta = \frac{\pi}{2}\) at the point \(A\). The tangent to \(C\) at the initial line at the point \(N\). The finite region \(R\), shown shaded in the diagram above, is bounded by the initial line, the line \(\theta = \frac{\pi}{2}\), the arc \(AP\) of \(C\) and the line \(PN\).
  1. Calculate the exact area of \(R\). (8)
\includegraphics{figure_8}
Edexcel FP2 2008 June Q9
Challenging +1.8
$$(x^2 + 1)\frac{d^2y}{dx^2} = 2y^2 + (1 - 2x)\frac{dy}{dx}$$ (I)
  1. By differentiating equation (I) with respect to \(x\), show that
Edexcel FP2 2008 June Q9
Challenging +1.2
$$(x^2 + 1)\frac{d^2y}{dx^2} = 2y^2 + (1 - 2x)\frac{dy}{dx}$$ (I)
  1. By differentiating equation (I) with respect to \(x\), show that $$(x^2 + 1)\frac{d^3y}{dx^3} = (1 - 4x)\frac{d^2y}{dx^2} + (4y - 2)\frac{dy}{dx}.$$ (3) Given that \(y = 1\) and \(\frac{dy}{dx} = 1\) at \(x = 0\),
  2. find the series solution for \(y\), in ascending powers of \(x\), up to and including the term in \(x_3\).(4)
  3. Use your series to estimate the value of \(y\) at \(x = -0.5\), giving your answer to two decimal places.(1)
Edexcel FP2 2008 June Q10
Standard +0.3
The point \(P\) represents a complex number \(z\) on an Argand diagram such that $$|z - 3| = 2|z|.$$
  1. Show that, as \(z\) varies, the locus of \(P\) is a circle, and give the coordinates of the centre and the radius of the circle.(5)
The point \(Q\) represents a complex number \(z\) on an Argand diagram such that $$|z + 3| = |z - i\sqrt{3}|.$$
  1. Sketch, on the same Argand diagram, the locus of \(P\) and the locus of \(Q\) as \(z\) varies.(5)
  2. On your diagram shade the region which satisfies $$|z - 3| \geq 2|z| \text{ and } |z + 3| \geq |z - i\sqrt{3}|.$$ (2)
Edexcel FP2 2008 June Q11
Challenging +1.2
De Moivre's theorem states that \((\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta\) for \(n \in \mathbb{R}\)
  1. Use induction to prove de Moivre's theorem for \(n \in \mathbb{Z}^+\). (5)
  2. Show that \(\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\) (5)
  3. Hence show that \(2\cos\frac{\pi}{10}\) is a root of the equation $$x^4 - 5x^2 + 5 = 0$$ (3)
Edexcel FP2 Q1
5 marks Moderate -0.3
Find the set of values for which $$|x - 1| > 6x - 1.$$ [5]
Edexcel FP2 Q2
10 marks Standard +0.3
  1. Find the general solution of the differential equation $$t \frac{dv}{dt} - v = t, \quad t > 0$$ and hence show that the solution can be written in the form \(v = t(\ln t + c)\), where \(c\) is an arbitrary constant. [6]
  2. This differential equation is used to model the motion of a particle which has speed \(v\) m s\(^{-1}\) at time \(t\) s. When \(t = 2\) the speed of the particle is \(3\) m s\(^{-1}\). Find, to \(3\) significant figures, the speed of the particle when \(t = 4\). [4]