1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

428 questions

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CAIE P3 2021 March Q9
11 marks Standard +0.3
Let \(\text{f}(x) = \frac{e^{2x} + 1}{e^{2x} - 1}\), for \(x > 0\).
  1. The equation \(x = \text{f}(x)\) has one root, denoted by \(a\). Verify by calculation that \(a\) lies between 1 and 1.5. [2]
  2. Use an iterative formula based on the equation in part (a) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
  3. Find f\('(x)\). Hence find the exact value of \(x\) for which f\('(x) = -8\). [6]
CAIE P3 2024 November Q2
3 marks Standard +0.8
  1. By sketching a suitable pair of graphs, show that the equation \(\cot 2x = \sec x\) has exactly one root in the interval \(0 < x < \frac{1}{2}\pi\). [2]
  2. Show that if a sequence of real values given by the iterative formula $$x_{n+1} = \frac{1}{2}\tan^{-1}(\cos x_n)$$ converges, then it converges to the root in part (a). [1]
CAIE P3 2006 June Q6
8 marks Standard +0.3
  1. By sketching a suitable pair of graphs, show that the equation $$2\cot x = 1 + e^x,$$ where \(x\) is in radians, has only one root in the interval \(0 < x < \frac{1}{2}\pi\). [2]
  2. Verify by calculation that this root lies between 0.5 and 1.0. [2]
  3. Show that this root also satisfies the equation $$x = \tan^{-1}\left(\frac{2}{1 + e^x}\right).$$ [1]
  4. Use the iterative formula $$x_{n+1} = \tan^{-1}\left(\frac{2}{1 + e^{x_n}}\right),$$ with initial value \(x_1 = 0.7\), to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
CAIE P3 2010 June Q6
8 marks Standard +0.3
The curve \(y = \frac{\ln x}{x + 1}\) has one stationary point.
  1. Show that the \(x\)-coordinate of this point satisfies the equation $$x = \frac{x + 1}{\ln x},$$ and that this \(x\)-coordinate lies between 3 and 4. [5]
  2. Use the iterative formula $$x_{n+1} = \frac{x_n + 1}{\ln x_n}$$ to determine the \(x\)-coordinate correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
CAIE P3 2013 June Q2
5 marks Moderate -0.3
The sequence of values given by the iterative formula $$x_{n+1} = \frac{x_n(x_n^2 + 100)}{2(x_n^2 + 25)},$$ with initial value \(x_1 = 3.5\), converges to \(\alpha\).
  1. Use this formula to calculate \(\alpha\) correct to 4 decimal places, showing the result of each iteration to 6 decimal places. [3]
  2. State an equation satisfied by \(\alpha\) and hence find the exact value of \(\alpha\). [2]
CAIE P3 2014 June Q6
8 marks Standard +0.3
\includegraphics{figure_6} In the diagram, \(A\) is a point on the circumference of a circle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the circumference at \(B\) and \(C\). The angle \(\angle OAB\) is equal to \(x\) radians. The shaded region is bounded by \(AB\), \(AC\) and the circular arc with centre \(A\) joining \(B\) and \(C\). The perimeter of the shaded region is equal to half the circumference of the circle.
  1. Show that \(x = \cos^{-1}\left(\frac{\pi}{4 + 4x}\right)\). [3]
  2. Verify by calculation that \(x\) lies between 1 and 1.5. [2]
  3. Use the iterative formula $$x_{n+1} = \cos^{-1}\left(\frac{\pi}{4 + 4x_n}\right)$$ to determine the value of \(x\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
CAIE P3 2017 June Q10
11 marks Standard +0.3
\includegraphics{figure_10} The diagram shows the curve \(y = x^2 \cos 2x\) for \(0 \leq x \leq \frac{1}{4}\pi\). The curve has a maximum point at \(M\) where \(x = p\).
  1. Show that \(p\) satisfies the equation \(p = \frac{1}{2} \tan^{-1} \left(\frac{1}{p}\right)\). [3]
  2. Use the iterative formula \(p_{n+1} = \frac{1}{2} \tan^{-1} \left(\frac{1}{p_n}\right)\) to determine the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
  3. Find, showing all necessary working, the exact area of the region bounded by the curve and the \(x\)-axis. [5]
CAIE P3 2013 November Q6
8 marks Challenging +1.2
\includegraphics{figure_6} In the diagram, \(A\) is a point on the circumference of a circle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the circumference at \(B\) and \(C\). The angle \(OAB\) is \(\theta\) radians. The shaded region is bounded by the circumference of the circle and the arc with centre \(A\) joining \(B\) and \(C\). The area of the shaded region is equal to half the area of the circle.
  1. Show that \(\cos 2\theta = \frac{2\sin 2\theta - \pi}{4\theta}\). [5]
  2. Use the iterative formula $$\theta_{n+1} = \frac{1}{2}\cos^{-1}\left(\frac{2\sin 2\theta_n - \pi}{4\theta_n}\right),$$ with initial value \(\theta_1 = 1\), to determine \(\theta\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places. [3]
CAIE P3 2017 November Q3
6 marks Standard +0.3
The equation \(x^3 = 3x + 7\) has one real root, denoted by \(\alpha\).
  1. Show by calculation that \(\alpha\) lies between 2 and 3. [2]
Two iterative formulae, \(A\) and \(B\), derived from this equation are as follows: $$x_{n+1} = (3x_n + 7)^{\frac{1}{3}}, \quad (A)$$ $$x_{n+1} = \frac{x_n^3 - 7}{3}. \quad (B)$$ Each formula is used with initial value \(x_1 = 2.5\).
  1. Show that one of these formulae produces a sequence which fails to converge, and use the other formula to calculate \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [4]
CAIE P3 2018 November Q4
5 marks Standard +0.3
Showing all necessary working, solve the equation $$\frac{e^x + e^{-x}}{e^x + 1} = 4,$$ giving your answer correct to 3 decimal places. [5]
CAIE P3 2018 November Q5
8 marks Standard +0.3
The equation of a curve is \(y = x \ln(8 - x)\). The gradient of the curve is equal to 1 at only one point, when \(x = a\).
  1. Show that \(a\) satisfies the equation \(x = 8 - \frac{8}{\ln(8 - x)}\). [3]
  2. Verify by calculation that \(a\) lies between 2.9 and 3.1. [2]
  3. Use an iterative formula based on the equation in part (i) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
CAIE P3 2018 November Q3
7 marks Standard +0.3
  1. By sketching a suitable pair of graphs, show that the equation \(x^3 = 3 - x\) has exactly one real root. [2]
  2. Show that if a sequence of real values given by the iterative formula $$x_{n+1} = \frac{2x_n^3 + 3}{3x_n^2 + 1}$$ converges, then it converges to the root of the equation in part (i). [2]
  3. Use this iterative formula to determine the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]
Edexcel C3 Q3
6 marks Standard +0.3
The root of the equation \(f(x) = 0\), where $$f(x) = x + \ln 2x - 4$$ is to be estimated using the iterative formula \(x_{n+1} = 4 - \ln 2x_n\), with \(x_0 = 2.4\).
  1. Showing your values of \(x_1, x_2, x_3, \ldots\), obtain the value, to 3 decimal places, of the root. [4]
  2. By considering the change of sign of \(f(x)\) in a suitable interval, justify the accuracy of your answer to part (a). [2]
Edexcel C3 Q9
9 marks Standard +0.3
\includegraphics{figure_2} Figure 2 shows part of the curve \(C\) with equation \(y = f(x)\), where $$f(x) = 0.5e^x - x^2.$$ The curve \(C\) cuts the \(y\)-axis at \(A\) and there is a minimum at the point \(B\).
  1. Find an equation of the tangent to \(C\) at \(A\). [4]
The \(x\)-coordinate of \(B\) is approximately \(2.15\). A more exact estimate is to be made of this coordinate using iterations \(x_{n+1} = \ln g(x_n)\).
  1. Show that a possible form for \(g(x)\) is \(g(x) = 4x\). [3]
  2. Using \(x_{n+1} = \ln 4x_n\), with \(x_0 = 2.15\), calculate \(x_1\), \(x_2\) and \(x_3\). Give the value of \(x_3\) to 4 decimal places. [2]
Edexcel C3 Q18
11 marks Standard +0.3
\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = e^{-x} - 1\).
  1. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac{1}{2}|x - 1|\). Show the coordinates of the points where the graph meets the axes. [2]
The \(x\)-coordinate of the point of intersection of the graphs is \(\alpha\).
  1. Show that \(x = \alpha\) is a root of the equation \(x + 2e^{-x} - 3 = 0\). [3]
  2. Show that \(-1 < \alpha < 0\). [2]
The iterative formula \(x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]\) is used to solve the equation \(x + 2e^{-x} - 3 = 0\).
  1. Starting with \(x_0 = -1\), find the values of \(x_1\) and \(x_2\). [2]
  2. Show that, to 2 decimal places, \(\alpha = -0.58\). [2]
Edexcel C3 Q22
11 marks Standard +0.3
The curve with equation \(y = \ln 3x\) crosses the \(x\)-axis at the point \(P (p, 0)\).
  1. Sketch the graph of \(y = \ln 3x\), showing the exact value of \(p\). [2]
The normal to the curve at the point \(Q\), with \(x\)-coordinate \(q\), passes through the origin.
  1. Show that \(x = q\) is a solution of the equation \(x^2 + \ln 3x = 0\). [4]
  2. Show that the equation in part (b) can be rearranged in the form \(x = \frac{1}{3}e^{-x^2}\). [2]
  3. Use the iteration formula \(x_{n+1} = \frac{1}{3}e^{-x_n^2}\), with \(x_0 = \frac{1}{4}\), to find \(x_1, x_2, x_3\) and \(x_4\). Hence write down, to 3 decimal places, an approximation for \(q\). [3]
Edexcel C3 Q28
23 marks Moderate -0.3
  1. Sketch, on the same set of axes, the graphs of $$y = 2 - e^{-x} \text{ and } y = \sqrt{x}.$$ [3]
[It is not necessary to find the coordinates of any points of intersection with the axes.] Given that \(f(x) = e^{-x} + \sqrt{x} - 2, x \geq 0\),
  1. explain how your graphs show that the equation \(f(x) = 0\) has only one solution, [1]
  2. show that the solution of \(f(x) = 0\) lies between \(x = 3\) and \(x = 4\). [2]
The iterative formula \(x_{n+1} = (2 - e^{-x_n})^2\) is used to solve the equation \(f(x) = 0\).
  1. Taking \(x_0 = 4\), write down the values of \(x_1, x_2, x_3\) and \(x_4\), and hence find an approximation to the solution of \(f(x) = 0\), giving your answer to 3 decimal places. [4]
28a.
  1. Given that \(\cos(x + 30)° = 3 \cos(x - 30)°\), prove that \(\tan x° = -\frac{\sqrt{3}}{2}\). [5]
    1. Prove that \(\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan \theta\). [3]
    2. Verify that \(\theta = 180°\) is a solution of the equation \(\sin 2\theta = 2 - 2 \cos 2\theta\). [1]
    3. Using the result in part (a), or otherwise, find the other two solutions, \(0 < \theta < 360°\), of the equation using \(\sin 2\theta = 2 - 2 \cos 2\theta\). [4]
Edexcel F1 2022 January Q8
10 marks Standard +0.3
$$f(x) = 2x^{-\frac{2}{3}} + \frac{1}{2}x - \frac{1}{3x - 5} - \frac{5}{2} \quad x \neq \frac{5}{3}$$ The table below shows values of \(f(x)\) for some values of \(x\), with values of \(f(x)\) given to 4 decimal places where appropriate.
\(x\)12345
\(f(x)\)0.5\(-0.2885\)0.5834
  1. Complete the table giving the values to 4 decimal places. [2]
The equation \(f(x) = 0\) has exactly one positive root, \(\alpha\). Using the values in the completed table and explaining your reasoning,
  1. determine an interval of width one that contains \(\alpha\). [2]
  2. Hence use interval bisection twice to obtain an interval of width 0.25 that contains \(\alpha\). [3]
Given also that the equation \(f(x) = 0\) has a negative root, \(\beta\), in the interval \([-1, -0.5]\)
  1. use linear interpolation once on this interval to find an approximation for \(\beta\). Give your answer to 3 significant figures. [3]
AQA C3 2011 June Q3
5 marks Standard +0.3
The curve \(y = \cos^{-1}(2x - 1)\) intersects the curve \(y = e^x\) at a single point where \(x = \alpha\).
  1. Show that \(\alpha\) lies between 0.4 and 0.5. [2]
  2. Show that the equation \(\cos^{-1}(2x - 1) = e^x\) can be written as \(x = \frac{1}{2} + \frac{1}{2}\cos(e^x)\). [1]
  3. Use the iteration \(x_{n+1} = \frac{1}{2} + \frac{1}{2}\cos(e^{x_n})\) with \(x_1 = 0.4\) to find the values of \(x_2\) and \(x_3\), giving your answers to three decimal places. [2]
Edexcel C3 Q2
6 marks Moderate -0.3
The root of the equation f(x) = 0, where $$f(x) = x + \ln 2x - 4$$ is to be estimated using the iterative formula \(x_{n+1} = 4 - \ln 2x_n\), with \(x_0 = 2.4\).
  1. Showing your values of \(x_1, x_2, x_3, \ldots\), obtain the value, to 3 decimal places, of the root. [4]
  2. By considering the change of sign of f(x) in a suitable interval, justify the accuracy of your answer to part (a). [2]
Edexcel C3 Q4
10 marks Standard +0.2
\(\text{f}(x) = x^3 + x^2 - 4x - 1\). The equation f(x) = 0 has only one positive root, \(\alpha\).
  1. Show that f(x) = 0 can be rearranged as $$x = \sqrt{\frac{4x+1}{x+1}}, \quad x \neq -1.$$ [2]
The iterative formula \(x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}\) is used to find an approximation to \(\alpha\).
  1. Taking \(x_1 = 1\), find, to 2 decimal places, the values of \(x_2\), \(x_3\) and \(x_4\). [3]
  2. By choosing values of \(x\) in a suitable interval, prove that \(\alpha = 1.70\), correct to 2 decimal places. [3]
  3. Write down a value of \(x_1\) for which the iteration formula \(x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}\) does not produce a valid value for \(x_2\). Justify your answer. [2]
Edexcel C3 Q2
9 marks Standard +0.2
\includegraphics{figure_2} Figure 2 shows part of the curve \(C\) with equation \(y = \text{f}(x)\), where $$\text{f}(x) = 0.5e^x - x^2.$$ The curve \(C\) cuts the \(y\)-axis at \(A\) and there is a minimum at the point \(B\).
  1. Find an equation of the tangent to \(C\) at \(A\). [4]
The \(x\)-coordinate of \(B\) is approximately 2.15. A more exact estimate is to be made of this coordinate using iterations \(x_{n+1} = \ln g(x_n)\).
  1. Show that a possible form for \(g(x)\) is \(g(x) = 4x\). [3]
  2. Using \(x_{n+1} = \ln 4x_n\), with \(x_0 = 2.15\), calculate \(x_1\), \(x_2\) and \(x_3\). Give the value of \(x_3\) to 4 decimal places. [2]
Edexcel C3 Q4
10 marks Standard +0.2
  1. Sketch, on the same set of axes, the graphs of $$y = 2 - e^{-x} \text{ and } y = \sqrt{x}.$$ [3] [It is not necessary to find the coordinates of any points of intersection with the axes.] Given that f(x) = \(e^{-x} + \sqrt{x} - 2\), \(x \geq 0\),
  2. explain how your graphs show that the equation f(x) = 0 has only one solution, [1]
  3. show that the solution of f(x) = 0 lies between \(x = 3\) and \(x = 4\). [2]
The iterative formula \(x_{n+1} = (2 - e^{-x_n})^2\) is used to solve the equation f(x) = 0.
  1. Taking \(x_0 = 4\), write down the values of \(x_1\), \(x_2\), \(x_3\) and \(x_4\), and hence find an approximation to the solution of f(x) = 0, giving your answer to 3 decimal places. [4]
Edexcel C3 Q5
11 marks Standard +0.2
\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = e^{-x} - 1\).
  1. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac{1}{2}|x - 1|\). Show the coordinates of the points where the graph meets the axes. [2]
The \(x\)-coordinate of the point of intersection of the graph is \(\alpha\).
  1. Show that \(x = \alpha\) is a root of the equation \(x + 2e^{-x} - 3 = 0\). [3]
  2. Show that \(-1 < \alpha < 0\). [2]
The iterative formula \(x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]\) is used to solve the equation \(x + 2e^{-x} - 3 = 0\).
  1. Starting with \(x_0 = -1\), find the values of \(x_1\) and \(x_2\). [2]
  2. Show that, to 2 decimal places, \(\alpha = -0.58\). [2]
Edexcel C3 Q8
11 marks Standard +0.3
The curve with equation \(y = \ln 3x\) crosses the \(x\)-axis at the point \(P(p, 0)\).
  1. Sketch the graph of \(y = \ln 3x\), showing the exact value of \(p\). [2]
The normal to the curve at the point \(Q\), with \(x\)-coordinate \(q\), passes through the origin.
  1. Show that \(x = q\) is a solution of the equation \(x^2 + \ln 3x = 0\). [4]
  2. Show that the equation in part (b) can be rearranged in the form \(x = \frac{1}{3}e^{-x^2}\). [2]
  3. Use the iteration formula \(x_{n + 1} = \frac{1}{3}e^{-x_n^2}\), with \(x_0 = \frac{1}{3}\), to find \(x_1, x_2, x_3\) and \(x_4\). Hence write down, to 3 decimal places, an approximation for \(q\). [3]