1.09d Newton-Raphson method

166 questions

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Pre-U Pre-U 9794/1 2012 Specimen Q7
4 marks Standard +0.3
7 Given that the equation \(x = 2 - \frac { 1 } { ( x + 1 ) ^ { 2 } }\) has a root between \(x = 1\) and \(x = 2\), use the Newton-Raphson formula with \(x _ { 0 } = 2\) to find this root correct to 3 decimal places.
Pre-U Pre-U 9794/1 2013 June Q8
6 marks Moderate -0.3
8
  1. Let \(\mathrm { f } ( x ) = x ^ { 3 } - x - 1\). Use a sign change method to show that the equation \(x ^ { 3 } - x - 1 = 0\) has a root between \(x = 1\) and \(x = 2\).
  2. By taking \(x = 1\) as a first approximation to this root, use the Newton-Raphson formula to find this root correct to 3 decimal places.
Pre-U Pre-U 9794/1 2014 June Q7
4 marks Standard +0.3
7 Taking \(x = 2\) as a first approximation, use the Newton-Raphson process to find a root of the equation \(\frac { 1 } { x ^ { 2 } } - 0.119 - 0.018 x = 0\). Give your answer correct to 3 significant figures.
Pre-U Pre-U 9794/1 2016 Specimen Q8
4 marks Moderate -0.3
8 Given that the equation \(x ^ { 3 } + 2 x - 7 = 0\) has a root between \(x = 1\) and \(x = 2\), use the Newton-Raphson formula with \(x _ { \mathrm { o } } = 1\) to find this root correct to 3 decimal places.
Pre-U Pre-U 9795/2 2016 Specimen Q6
12 marks Standard +0.8
6 The lengths of time, in years, that sales representatives for a certain company keep their company cars may be modelled by the distribution with probability density function \(\mathrm { f } ( x )\), where $$f ( x ) = \left\{ \begin{array} { c c } \frac { 4 } { 27 } x ^ { 2 } ( 3 - x ) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{array} \right.$$
  1. Draw a sketch of this probability density function.
  2. Calculate the mean and the mode of \(X\).
  3. Comment briefly on the values obtained in part (ii) in relation to the sketch in part (i).
  4. Show that the lower quartile \(\mathrm { Q } _ { 1 }\) of \(X\) satisfies the equation \(\mathrm { Q } _ { 1 } { } ^ { 4 } - 4 \mathrm { Q } _ { 1 } { } ^ { 3 } + 6.75 = 0\), and use an appropriate numerical method to find the value of \(\mathrm { Q } _ { 1 }\) correct to 2 decimal places, showing full details of your method.
Pre-U Pre-U 9794/1 2017 June Q7
5 marks Standard +0.3
7
  1. Describe the transformation which maps the graph of \(y = \ln x\) onto the graph of \(y = \ln ( 1 + x )\).
  2. By sketching the curves \(y = \ln ( 1 + x )\) and \(y = 4 - x\) on a single diagram, show that the equation $$\ln ( 1 + x ) = 4 - x$$ has exactly one root.
  3. Use the Newton-Raphson method with \(x _ { 0 } = 2\) to find the root of the equation \(\ln ( 1 + x ) = 4 - x\) correct to 3 decimal places. Show the result of each iteration.
Pre-U Pre-U 9794/1 2018 June Q9
12 marks Standard +0.8
9
  1. Find the coordinates of the stationary point of the curve with equation $$y = \ln x - k x , \text { where } k > 0 \text { and } x > 0$$ and determine its nature.
  2. Hence show that the equation \(\ln x - k x = 0\) has real roots if \(0 < k \leqslant \frac { 1 } { \mathrm { e } }\).
  3. In the particular case that \(k = \frac { 1 } { 3 }\), the equation \(\ln x - k x = 0\) has two roots, one of which is near \(x = 5\). Use the Newton-Raphson process to find, correct to 3 significant figures, the root of the equation \(\ln x - \frac { 1 } { 3 } x = 0\) which is near \(x = 5\).
  4. Show that the equation \(\ln x - k x = 0\) has one real root if \(k \leqslant 0\).
  5. Explain why the equation \(\ln x - k x = 0\) has two distinct real roots if \(0 < k < \frac { 1 } { \mathrm { e } }\).
Pre-U Pre-U 9795/2 2019 Specimen Q6
2 marks Standard +0.8
6 The lengths of time, in years, that sales representatives for a certain company keep their company cars may be modelled by the distribution with probability density function \(\mathrm { f } ( x )\), where $$f ( x ) = \left\{ \begin{array} { c c } \frac { 4 } { 27 } x ^ { 2 } ( 3 - x ) & 0 \leqslant x \leqslant 3 , \\ 0 & \text { otherwise } . \end{array} \right.$$
  1. Draw a sketch of this probability density function.
  2. Calculate the mean and the mode of \(X\).
  3. Comment briefly on the values obtained in part (b) in relation to the sketch in part (a).
  4. Show that the lower quartile \(\mathrm { Q } _ { 1 }\) of \(X\) satisfies the equation \(\mathrm { Q } _ { 1 } { } ^ { 4 } - 4 \mathrm { Q } _ { 1 } { } ^ { 3 } + 6.75 = 0\), and use an appropriate numerical method to find the value of \(\mathrm { Q } _ { 1 }\) correct to 2 decimal places, showing full details of your method.
Pre-U Pre-U 9795/2 2020 Specimen Q6
2 marks Standard +0.3
6 The lengths of time, in years, that sales representatives for a certain company keep their company cars may be modelled by the distribution with probability density function \(\mathrm { f } ( x )\), where $$f ( x ) = \left\{ \begin{array} { c c } \frac { 4 } { 27 } x ^ { 2 } ( 3 - x ) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise. } \end{array} \right.$$
  1. Draw a sketch of this probability density function.
  2. Calculate the mean and the mode of \(X\).
  3. Comment briefly on the values obtained in part (b) in relation to the sketch in part (a).
  4. Show that the lower quartile \(\mathrm { Q } _ { 1 }\) of \(X\) satisfies the equation \(\mathrm { Q } _ { 1 } { } ^ { 4 } - 4 \mathrm { Q } _ { 1 } { } ^ { 3 } + 6.75 = 0\), and use an appropriate numerical method to find the value of \(\mathrm { Q } _ { 1 }\) correct to 2 decimal places, showing full details of your method.
Pre-U Pre-U 9794/1 Specimen Q6
6 marks Moderate -0.3
6 The equation \(x ^ { 3 } - x - 1 = 0\) has exactly one real root in the interval \(0 \leq x \leq 3\).
  1. Denoting this root by \(\alpha\), find the integer \(n\) such that \(n < \alpha < n + 1\).
  2. Taking \(n\) as a first approximation, use the Newton-Raphson method to find \(\alpha\), correct to 2 decimal places. You must show the result of each iteration correct to an appropriate degree of accuracy.
Pre-U Pre-U 9794/1 Specimen Q9
16 marks Challenging +1.8
9 The cubic polynomial \(x ^ { 3 } + a x ^ { 2 } + b x + c\), where \(a , b\) and \(c\) are real, is denoted by \(\mathrm { p } ( x )\).
  1. Give a reason why the equation \(\mathrm { p } ( x ) = 0\) has at least one real root.
  2. Suppose that the curve with equation \(y = \mathrm { p } ( x )\) has a local minimum point and a local maximum point with \(y\)-coordinates \(y _ { \text {min } }\) and \(y _ { \text {max } }\) respectively.
    1. Prove that if \(y _ { \text {min } } y _ { \text {max } } < 0\), then the equation \(\mathrm { p } ( x ) = 0\) has three real roots.
    2. Comment on the number of distinct real roots of the equation \(\mathrm { p } ( x ) = 0\) in the case \(y _ { \text {min } } y _ { \text {max } } = 0\).
    3. Suppose instead that the equation \(\mathrm { p } ( x ) = 0\) has only one real root for all values of \(c\). Prove that \(a ^ { 2 } \leqslant 3 b\).
    4. The iterative scheme $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 1 } { 3 x _ { n } ^ { 2 } + 1 } , \quad x _ { 0 } = 0$$ converges to a root of the cubic equation \(\mathrm { p } ( x ) = 0\).
      (a) Find \(\mathrm { p } ( x )\).
      (b) Find the limit of the iteration, correct to 4 decimal places.
    5. Determine the rate of convergence of the iterative scheme.
CAIE P2 2024 March Q5
12 marks Standard +0.3
\includegraphics{figure_5} The diagram shows part of the curve with equation \(y = \frac{x^3}{x + 2}\). At the point \(P\), the gradient of the curve is 6.
  1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \sqrt[3]{12x + 12}\). [4]
  2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 3.8 and 4.0. [2]
  3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Show the result of each iteration to 5 significant figures. [3]
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 FP1 Q5
9 marks Moderate -0.3
$$\text{f}(x) = 3\sqrt{x} + \frac{18}{\sqrt{x}} - 20.$$
  1. Show that the equation f\((x) = 0\) has a root \(a\) in the interval \([1.1, 1.2]\). [2]
  2. Find \(f'(x)\). [3]
  3. Using \(x_0 = 1.1\) as a first approximation to \(a\), apply the Newton-Raphson procedure once to f\((x)\) to find a second approximation to \(a\), giving your answer to 3 significant figures. [4]
Edexcel FP1 2013 June Q3
10 marks Moderate -0.3
$$f(x) = \frac{1}{2}x^4 - x^3 + x - 3$$
  1. Show that the equation \(f(x) = 0\) has a root \(\alpha\) between \(x = 2\) and \(x = 2.5\) [2]
  2. Starting with the interval \([2, 2.5]\) use interval bisection twice to find an interval of width \(0.125\) which contains \(\alpha\). [3]
The equation \(f(x) = 0\) has a root \(\beta\) in the interval \([-2, -1]\).
  1. Taking \(-1.5\) as a first approximation to \(\beta\), apply the Newton-Raphson process once to \(f(x)\) to obtain a second approximation to \(\beta\). Give your answer to 2 decimal places. [5]
Edexcel FP1 Q3
7 marks Standard +0.3
$$f(x) = x^3 + x - 3.$$ The equation \(f(x) = 0\) has a root, \(\alpha\), between 1 and 2.
  1. By considering \(f'(x)\), show that \(\alpha\) is the only real root of the equation \(f(x) = 0\). [3]
  2. Taking 1.2 as your first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(f(x)\) to obtain a second approximation to \(\alpha\). Give your answer to 3 significant figures. [2]
  3. Prove that your answer to part (b) gives the value of \(\alpha\) correct to 3 significant figures. [2]
Edexcel FP1 Q37
11 marks Standard +0.3
$$f (x) = x^3 + 8x - 19.$$
  1. Show that the equation \(f(x) = 0\) has only one real root. [3]
  2. Show that the real root of \(f(x) = 0\) lies between 1 and 2. [2]
  3. Obtain an approximation to the real root of \(f(x) = 0\) by performing two applications of the Newton-Raphson procedure to \(f(x)\) , using \(x = 2\) as the first approximation. Give your answer to 3 decimal places. [4]
  4. By considering the change of sign of \(f(x)\) over an appropriate interval, show that your answer to part (c) is accurate to 3 decimal places. [2]
OCR FP2 2009 January Q5
8 marks Challenging +1.2
\includegraphics{figure_5} The diagram shows the curve with equation \(y = f(x)\), where $$f(x) = 2x^3 - 9x^2 + 12x - 4.36.$$ The curve has turning points at \(x = 1\) and \(x = 2\) and crosses the \(x\)-axis at \(x = \alpha\), \(x = \beta\) and \(x = \gamma\), where \(0 < \alpha < \beta < \gamma\).
  1. The Newton-Raphson method is to be used to find the roots of the equation \(f(x) = 0\), with \(x_1 = k\).
    1. To which root, if any, would successive approximations converge in each of the cases \(k < 0\) and \(k = 1\)? [2]
    2. What happens if \(1 < k < 2\)? [2]
  2. Sketch the curve with equation \(y^2 = f(x)\). State the coordinates of the points where the curve crosses the \(x\)-axis and the coordinates of any turning points. [4]
OCR FP2 2010 January Q1
5 marks Standard +0.3
It is given that \(f(x) = x^2 - \sin x\).
  1. The iteration \(x_{n+1} = \sqrt{\sin x_n}\), with \(x_1 = 0.875\), is to be used to find a real root, \(\alpha\), of the equation \(f(x) = 0\). Find \(x_2, x_3\) and \(x_4\), giving the answers correct to 6 decimal places. [2]
  2. The error \(e_n\) is defined by \(e_n = \alpha - x_n\). Given that \(\alpha = 0.876726\), correct to 6 decimal places, find \(e_3\) and \(e_4\). Given that \(g(x) = \sqrt{\sin x}\), use \(e_3\) and \(e_4\) to estimate \(g'(\alpha)\). [3]
OCR FP2 2010 January Q3
7 marks Moderate -0.3
\includegraphics{figure_3} A curve with no stationary points has equation \(y = f(x)\). The equation \(f(x) = 0\) has one real root \(\alpha\), and the Newton-Raphson method is to be used to find \(\alpha\). The tangent to the curve at the point \((x_1, f(x_1))\) meets the \(x\)-axis where \(x = x_2\) (see diagram).
  1. Show that \(x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}\). [3]
  2. Describe briefly, with the help of a sketch, how the Newton-Raphson method, using an initial approximation \(x = x_1\), gives a sequence of approximations approaching \(\alpha\). [2]
  3. Use the Newton-Raphson method, with a first approximation of 1, to find a second approximation to the root of \(x^2 - 2\sinh x + 2 = 0\). [2]
OCR FP2 2012 January Q5
11 marks Standard +0.8
It is given that \(f(x) = x^3 - k\), where \(k > 0\), and that \(\alpha\) is the real root of the equation \(f(x) = 0\). Successive approximations to \(\alpha\), using the Newton-Raphson method, are denoted by \(x_1, x_2, \ldots, x_n, \ldots\).
  1. Show that \(x_{n+1} = \frac{2x_n^3 + k}{3x_n^2}\). [2]
  2. Sketch the graph of \(y = f(x)\), giving the coordinates of the intercepts with the axes. Show on your sketch how it is possible for \(|x_2 - x_1|\) to be greater than \(|x_1|\). [3]
It is now given that \(k = 100\) and \(x_1 = 5\).
  1. Write down the exact value of \(\alpha\) and find \(x_2\) and \(x_3\) correct to 5 decimal places. [3]
  2. The error \(e_n\) is defined by \(e_n = \alpha - x_n\). By finding \(e_1\), \(e_2\) and \(e_3\), verify that \(e_3 \approx \frac{e_2^2}{e_1}\). [3]
OCR H240/03 2019 June Q3
7 marks Moderate -0.3
  1. In this question you must show detailed reasoning. Solve the inequality \(|x - 2| \leqslant |2x - 6|\). [4]
  2. Give full details of a sequence of two transformations needed to transform the graph of \(y = |x - 2|\) to the graph of \(y = |2x - 6|\). [3]
OCR H240/03 2020 November Q4
11 marks Standard +0.3
A curve has equation \(y = 2\ln(k - 3x) + x^2 - 3x\), where \(k\) is a positive constant.
  1. Given that the curve has a point of inflection where \(x = 1\), show that \(k = 6\). [5] It is also given that the curve intersects the \(x\)-axis at exactly one point.
  2. Show by calculation that the \(x\)-coordinate of this point lies between 0.5 and 1.5. [2]
  3. Use the Newton-Raphson method, with initial value \(x_0 = 1\), to find the \(x\)-coordinate of the point where the curve intersects the \(x\)-axis, giving your answer correct to 5 decimal places. Show the result of each iteration to 6 decimal places. [3]
  4. By choosing suitable bounds, verify that your answer to part (c) is correct to 5 decimal places. [1]
OCR H240/03 2022 June Q5
14 marks Standard +0.3
In this question you must show detailed reasoning. \includegraphics{figure_5} The diagram shows the curve with equation \(y = \frac{2x - 3}{4x^2 + 1}\). The tangent to the curve at the point \(P\) has gradient 2.
  1. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$4x^3 + 3x - 3 = 0.$$ [5]
  2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.5 and 1. [2]
  3. Show that the iteration $$x_{n+1} = \frac{3 - 4x_n^3}{3}$$ cannot converge to the \(x\)-coordinate of \(P\) whatever starting value is used. [2]
  4. Use the Newton-Raphson method, with initial value 0.5, to determine the coordinates of \(P\) correct to 5 decimal places. [5]
AQA Paper 1 2024 June Q14
10 marks Standard +0.3
  1. The equation $$x^3 = e^{6-2x}$$ has a single solution, \(x = \alpha\) By considering a suitable change of sign, show that \(\alpha\) lies between 0 and 4 [2 marks]
  2. Show that the equation \(x^3 = e^{6-2x}\) can be rearranged to give $$x = 3 - \frac{3}{2}\ln x$$ [3 marks]
    1. Use the iterative formula $$x_{n+1} = 3 - \frac{3}{2}\ln x_n$$ with \(x_1 = 4\), to find \(x_2\), \(x_3\) and \(x_4\) Give your answers to three decimal places. [2 marks]
    2. Figure 1 below shows a sketch of parts of the graphs of $$y = 3 - \frac{3}{2}\ln x \quad \text{and} \quad y = x$$ On Figure 1, draw a staircase or cobweb diagram to show how convergence takes place. Label, on the \(x\)-axis, the positions of \(x_2\), \(x_3\) and \(x_4\) [2 marks]
      [diagram]
    3. Explain why the iterative formula $$x_{n+1} = 3 - \frac{3}{2}\ln x_n$$ fails to converge to \(\alpha\) when the starting value is \(x_1 = 0\) [1 mark]