Show root in interval

2 The equation \(x ^ { 3 } - 8 x - 13 = 0\) has one real root.

1. Find the two consecutive integers between which this root lies.
2. Use the iterative formula $$x _ { n + 1 } = \left( 8 x _ { n } + 13 \right) ^ { \frac { 1 } { 3 } }$$ to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of curve \(C _ { 1 }\) with equation \(y = 5 \mathrm { e } ^ { x - 1 } + 3\) and curve \(C _ { 2 }\) with equation \(y = 10 - x ^ { 2 }\) The point \(P\) lies on \(C _ { 1 }\) and has \(y\) coordinate 18

1. Find the \(x\) coordinate of \(P\), writing your answer in the form \(\ln k\), where \(k\) is a constant to be found. The curve \(C _ { 1 }\) meets the curve \(C _ { 2 }\) at \(x = \alpha\) and at \(x = \beta\), as shown in Figure 3.
2. Using a suitable interval and a suitable function that should be stated, show that to 3 decimal places \(\alpha = 1.134\) The iterative equation $$x _ { n + 1 } = - \sqrt { 7 - 5 \mathrm { e } ^ { x _ { n } - 1 } }$$ is used to find an approximation to \(\beta\). Using this iterative formula with \(x _ { 1 } = - 3\)
3. find the value of \(x _ { 2 }\) and the value of \(\beta\), giving each answer to 6 decimal places.

1. (a) On the same diagram, sketch and clearly label the graphs with equations

$$y = \mathrm { e } ^ { x } \quad \text { and } \quad y = 10 - x$$ Show on your sketch the coordinates of each point at which the graphs cut the axes.
(b) Explain why the equation \(\mathrm { e } ^ { x } - 10 + x = 0\) has only one solution.
(c) Show that the solution of the equation $$\mathrm { e } ^ { x } - 10 + x = 0$$ lies between \(x = 2\) and \(x = 3\) (d) Use the iterative formula $$x _ { n + 1 } = \ln \left( 10 - x _ { n } \right) , \quad x _ { 1 } = 2$$ to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\).
Give your answers to 4 decimal places.

7 The equation \(x ^ { 2 } = x ^ { 3 } + x - 3\) has a single solution, \(x = \alpha\) 7

1. By considering a suitable change of sign, show that \(\alpha\) lies between 1.5 and 1.6
   [0pt] [2 marks]
   7
2. Show that the equation \(x ^ { 2 } = x ^ { 3 } + x - 3\) can be rearranged into the form $$x ^ { 2 } = x - 1 + \frac { 3 } { x }$$ 7
3. Use the iterative formula $$x _ { n + 1 } = \sqrt { x _ { n } - 1 + \frac { 3 } { x _ { n } } }$$ with \(x _ { 1 } = 1.5\), to find \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to four decimal places.
   7
4. Hence, deduce an interval of width 0.001 in which \(\alpha\) lies.

4

1. Show that the equation \(x ^ { 3 } - 6 x + 2 = 0\) has a root between \(x = 0\) and \(x = 1\).
2. Use the iterative formula \(x _ { n + 1 } = \frac { 2 + x _ { n } ^ { 3 } } { 6 }\) with \(x _ { 0 } = 0.5\) to find this root correct to 4 decimal places, showing the result of each iteration.

3

1. Show that the equation \(x ^ { 2 } - \ln x - 2 = 0\) has a solution between \(x = 1\) and \(x = 2\).
2. Find an approximation to that solution using the iteration \(x _ { n + 1 } = \sqrt { 2 + \ln x _ { n } }\), giving your answer correct to 2 decimal places.

5

1. Show that the equation \(\sin x - x + 1 = 0\) has a root between 1.5 and 2 .
2. Use the iteration \(x _ { n + 1 } = 1 + \sin x _ { n }\), with a suitable starting value, to find that root correct to 2 decimal places.
3. Sketch the graphs of \(y = \sin x\) and \(y = x - 1\), on the same set of axes, for \(0 \leqslant x \leqslant \pi\).
4. Explain why the equation \(\sin x - x + 1 = 0\) has no root other than the one found in part (ii). [1]

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]

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]
2. 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]

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]

\includegraphics{figure_8} The diagram shows part of each of the curves \(y = e^{\frac{1}{3}x}\) and \(y = \sqrt[3]{(3x + 8)}\). The curves meet, as shown in the diagram, at the point \(P\). The region \(R\), shaded in the diagram, is bounded by the two curves and by the \(y\)-axis.

1. Show by calculation that the \(x\)-coordinate of \(P\) lies between 5.2 and 5.3. [3]
2. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac{2}{3} \ln(3x + 8)\). [2]
3. Use an iterative formula, based on the equation in part (ii), to find the \(x\)-coordinate of \(P\) correct to 2 decimal places. [3]
4. Use integration, and your answer to part (iii), to find an approximate value of the area of the region \(R\). [5]

\includegraphics{figure_7} The diagram shows the curve with equation \(y = \cos^{-1} x\).

1. Sketch the curve with equation \(y = 3 \cos^{-1}(x - 1)\), showing the coordinates of the points where the curve meets the axes. [3]
2. By drawing an appropriate straight line on your sketch in part (i), show that the equation \(3 \cos^{-1}(x - 1) = x\) has exactly one root. [1]
3. Show by calculation that the root of the equation \(3 \cos^{-1}(x - 1) = x\) lies between 1.8 and 1.9. [2]
4. The sequence defined by $$x_1 = 2, \quad x_{n+1} = 1 + \cos(\frac{1}{3}x_n)$$ converges to a number \(\alpha\). Find the value of \(\alpha\) correct to 2 decimal places and explain why \(\alpha\) is the root of the equation \(3 \cos^{-1}(x - 1) = x\). [5]

The equation \(2x^3 + 4x - 35 = 0\) has one real root.

1. Show by calculation that this real root lies between 2 and 3. [3]
2. Use the iterative formula $$x_{n+1} = \sqrt[3]{17.5 - 2x_n},$$ with a suitable starting value, to find the real root of the equation \(2x^3 + 4x - 35 = 0\) correct to 2 decimal places. You should show the result of each iteration. [3]

$$\text{f}(x) = x^2 + 5x - 2 \sec x, \quad x \in \mathbb{R}, \quad -\frac{\pi}{2} < x < \frac{\pi}{2}.$$

1. Show that the equation \(\text{f}(x) = 0\) has a root, \(\alpha\), such that \(1 < \alpha < 1.5\) [2]
2. Show that a suitable rearrangement of the equation \(\text{f}(x) = 0\) leads to the iterative formula $$x_{n+1} = \cos^{-1} \left( \frac{2}{x_n^2 + 5x_n} \right).$$ [3]
3. Use the iterative formula in part (ii) with a starting value of 1.25 to find \(\alpha\) correct to 3 decimal places. You should show the result of each iteration. [3]

The equation \(6\arcsin(2x - 1) - x^2 = 0\) has exactly one real root.

1. Show by calculation that the root lies between 0.5 and 0.6. [2]

In order to find the root, the iterative formula \(x_{n+1} = p + q\sin(rx_n^2)\), with initial value \(x_0 = 0.5\), is to be used.

1. Determine the values of the constants \(p\), \(q\) and \(r\). [2]
2. Hence find the root correct to 4 significant figures. Show the result of each step of the iteration process. [2]

The equation $$1 + 5x - x^4 = 0$$ has a positive root \(\alpha\).

1. Show that \(\alpha\) lies between 1 and 2. [2]
2. Use the iterative sequence based on the arrangement $$x = \sqrt[4]{1+5x}$$ with starting value 1.5 to find \(\alpha\) correct to two decimal places. [3]
3. Use the Newton-Raphson method to find \(\alpha\) correct to six decimal places. [6]

A curve \(C\) has equation \(y = f(x)\) where $$f(x) = x + 2\ln(e - x)$$

1. 1. Show that the equation of the normal to \(C\) at the point where \(C\) crosses the \(y\)-axis is given by $$y = \left(\frac{e}{2-e}\right)x + 2$$ [6 marks]
   2. Find the exact area enclosed by the normal and the coordinate axes. Fully justify your answer. [3 marks]

1. The equation \(f(x) = 0\) has one positive root, \(\alpha\).
   1. Show that \(\alpha\) lies between 2 and 3 Fully justify your answer. [3 marks]
   2. Show that the roots of \(f(x) = 0\) satisfy the equation $$x = e - e^{\frac{x}{2}}$$ [2 marks]
   3. Use the recurrence relation $$x_{n+1} = e - e^{\frac{x_n}{2}}$$ with \(x_1 = 2\) to find the values of \(x_2\) and \(x_3\) giving your answers to three decimal places. [2 marks]
   4. Figure 1 below shows a sketch of the graphs of \(y = e - e^{\frac{x}{2}}\) and \(y = x\), and the position of \(x_1\) On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x_2\) and \(x_3\) on the \(x\)-axis. [2 marks] \includegraphics{figure_9}

1. Use a change of sign to verify that the equation \(\cos x - x = 0\) has a root \(\alpha\) between \(x = 0.7\) and \(x = 0.8\). [2]
2. Sketch, on a single diagram, the curve \(y = \cos x\) and the line \(y = x\) for \(0 \leqslant x \leqslant \frac{1}{2}\pi\), giving the coordinates of all points of intersection with the coordinate axes. [2]

An iteration of the form \(x_{n+1} = \cos(x_n)\) is to be used to find \(\alpha\).

1. By considering the gradient of \(y = \cos x\), show that this iteration will converge. [3]
2. On a copy of your sketch from part (ii), illustrate how this iteration converges to \(\alpha\). [2]
3. Use a change of sign to verify that \(\alpha = 0.7391\) to 4 decimal places. [2]