1.09a Sign change methods: locate roots

4

1. By sketching a suitable pair of graphs, show that the equation $$\ln x = 4 - \frac { 1 } { 2 } x$$ has exactly one real root, \(\alpha\).
2. Verify by calculation that \(4.5 < \alpha < 5.0\).
3. Use the iterative formula \(x _ { n + 1 } = 8 - 2 \ln x _ { n }\) to find \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6

1. By sketching a suitable pair of graphs, show that the equation \(\cot \frac { 1 } { 2 } x = 1 + \mathrm { e } ^ { - x }\) has exactly one root in the interval \(0 < x \leqslant \pi\).
2. Verify by calculation that this root lies between 1 and 1.5.
3. Use the iterative formula \(x _ { n + 1 } = 2 \tan ^ { - 1 } \left( \frac { 1 } { 1 + \mathrm { e } ^ { - x _ { n } } } \right)\) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

10 The constant \(a\) is such that \(\int _ { 1 } ^ { a } x ^ { 2 } \ln x \mathrm {~d} x = 4\).

1. Show that \(a = \left( \frac { 35 } { 3 \ln a - 1 } \right) ^ { \frac { 1 } { 3 } }\).
2. Verify by calculation that \(a\) lies between 2.4 and 2.8.
3. 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.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

9 The constant \(a\) is such that \(\int _ { 0 } ^ { a } x \mathrm { e } ^ { - 2 x } \mathrm {~d} x = \frac { 1 } { 8 }\).

1. Show that \(a = \frac { 1 } { 2 } \ln ( 4 a + 2 )\).
2. Verify by calculation that \(a\) lies between 0.5 and 1 .
3. Use an iterative formula based on the equation in (a) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6 The equation \(\cot \frac { 1 } { 2 } x = 3 x\) has one root in the interval \(0 < x < \pi\), denoted by \(\alpha\).

1. Show by calculation that \(\alpha\) lies between 0.5 and 1 .
2. Show that, if a sequence of positive values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( x _ { n } + 4 \tan ^ { - 1 } \left( \frac { 1 } { 3 x _ { n } } \right) \right)$$ converges, then it converges to \(\alpha\).
3. Use this iterative formula to calculate \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6

1. By sketching a suitable pair of graphs, show that the equation \(\operatorname { cosec } \frac { 1 } { 2 } x = \mathrm { e } ^ { x } - 3\) has exactly one root, denoted by \(\alpha\), in the interval \(0 < x < \pi\).
2. Verify by calculation that \(\alpha\) lies between 1 and 2 .
3. Show that if a sequence of values in the interval \(0 < x < \pi\) given by the iterative formula $$x _ { n + 1 } = \ln \left( \operatorname { cosec } \frac { 1 } { 2 } x _ { n } + 3 \right)$$ converges, then it converges to \(\alpha\).
4. Use this iterative formula with an initial value of 1.4 to determine \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
5. State the minimum number of calculated iterations needed with this initial value to determine \(\alpha\) correct to 2 decimal places.

5

1. It is given that the equation \(\mathrm { e } ^ { 2 x } = 5 + \cos 3 x\) has only one root.
   Show by calculation that this root lies in the interval \(0.7 < x < 0.8\).
2. Show that if a sequence of values in the interval \(0.7 < x < 0.8\) given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( 5 + \cos 3 x _ { n } \right)$$ converges then it converges to the root of the equation in part (a).
3. Use this iterative formula to determine the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

4 Find \(\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 3 } \pi } x \sec ^ { 2 } x \mathrm {~d} x\). Give your answer in a simplified exact form.

7 \includegraphics[max width=\textwidth, alt={}, center]{8c26235b-c78c-40d8-9e8e-213dc1311186-10_627_611_255_767} The diagram shows a circle with centre \(O\) and radius \(r\). The angle of the minor sector \(A O B\) of the circle is \(x\) radians. The area of the major sector of the circle is 3 times the area of the shaded region.

1. Show that \(x = \frac { 3 } { 4 } \sin x + \frac { 1 } { 2 } \pi\).
2. Show by calculation that the root of the equation in (a) lies between 2 and 2.5.
3. Use an iterative formula based on the equation in (a) to calculate this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

7 \includegraphics[max width=\textwidth, alt={}, center]{446573d3-73b1-482a-a3f6-1abddfdd90d0-10_620_517_260_774} The diagram shows the curve \(\mathrm { y } = \mathrm { xe } ^ { 2 \mathrm { x } } - 5 \mathrm { x }\) and its minimum point \(M\), where \(x = \alpha\).

1. Show that \(\alpha\) satisfies the equation \(\alpha = \frac { 1 } { 2 } \ln \left( \frac { 5 } { 1 + 2 \alpha } \right)\).
2. Verify by calculation that \(\alpha\) lies between 0.4 and 0.5.
3. Use an iterative formula based on the equation in part (a) to determine \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

8

1. By sketching a suitable pair of graphs, show that the equation $$\sqrt { x } = \mathrm { e } ^ { x } - 3$$ has only one root.
2. Show by calculation that this root lies between 1 and 2 .
3. Show that, if a sequence of values given by the iterative formula $$x _ { n + 1 } = \ln \left( 3 + \sqrt { x _ { n } } \right)$$ converges, then it converges to the root of the equation in (a).
4. Use the iterative formula to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6

1. By sketching a suitable pair of graphs, show that the equation $$\cot x = 2 - \cos x$$ has one root in the interval \(0 < x \leqslant \frac { 1 } { 2 } \pi\).
2. Show by calculation that this root lies between 0.6 and 0.8 .
3. Use the iterative formula \(x _ { n + 1 } = \tan ^ { - 1 } \left( \frac { 1 } { 2 - \cos x _ { n } } \right)\) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

5

1. By sketching a suitable pair of graphs, show that the equation \(2 + \mathrm { e } ^ { - 0.2 x } = \ln ( 1 + x )\) has only one root.
2. Show by calculation that this root lies between 7 and 9 . \includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-08_2716_40_109_2009}
3. Use the iterative formula $$x _ { n + 1 } = \exp \left( 2 + \mathrm { e } ^ { - 0.2 x _ { n } } \right) - 1$$ to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. \(\left[ \exp ( x ) \right.\) is an alternative notation for \(\left. \mathrm { e } ^ { x } .\right]\) \includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-10_481_789_262_639} The diagram shows the curve \(y = \sin 2 x ( 1 + \sin 2 x )\), for \(0 \leqslant x \leqslant \frac { 3 } { 4 } \pi\), and its minimum point \(M\). The shaded region bounded by the curve that lies above the \(x\)-axis and the \(x\)-axis itself is denoted by \(R\).

7. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation $$y = 2 \cos 3 x - 3 x + 4 \quad x > 0$$ where \(x\) is measured in radians. The curve crosses the \(x\)-axis at the point \(P\), as shown in Figure 3.
Given that the \(x\) coordinate of \(P\) is \(\alpha\),

1. show that \(\alpha\) lies between 0.8 and 0.9 The iteration formula $$x _ { n + 1 } = \frac { 1 } { 3 } \arccos \left( 1.5 x _ { n } - 2 \right)$$ can be used to find an approximate value for \(\alpha\).
2. Using this iteration formula with \(x _ { 1 } = 0.8\) find, to 4 decimal places, the value of
   1. \(X _ { 2 }\)
   2. \(X _ { 5 }\) The point \(Q\) and the point \(R\) are local minimum points on the curve, as shown in Figure 3.
      Given that the \(x\) coordinates of \(Q\) and \(R\) are \(\beta\) and \(\lambda\) respectively, and that they are the two smallest values of \(x\) at which local minima occur,
3. find, using calculus, the exact value of \(\beta\) and the exact value of \(\lambda\).

6. $$\mathrm { f } ( x ) = x \cos \left( \frac { x } { 3 } \right) \quad x > 0$$

1. Find \(\mathrm { f } ^ { \prime } ( x )\)
2. Show that the equation \(\mathrm { f } ^ { \prime } ( x ) = 0\) can be written as $$x = k \arctan \left( \frac { k } { x } \right)$$ where \(k\) is an integer to be found.
3. Starting with \(x _ { 1 } = 2.5\) use the iteration formula $$x _ { n + 1 } = k \arctan \left( \frac { k } { x _ { n } } \right)$$ with the value of \(k\) found in part (b), to calculate the values of \(x _ { 2 }\) and \(x _ { 6 }\) giving your answers to 3 decimal places.
4. Using a suitable interval and a suitable function that should be stated, show that a root of \(\mathrm { f } ^ { \prime } ( x ) = 0\) is 2.581 correct to 3 decimal places.
   In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation $$y = 6 \ln ( 2 x + 3 ) - \frac { 1 } { 2 } x ^ { 2 } + 4 \quad x > - \frac { 3 } { 2 }$$ The curve cuts the negative \(x\)-axis at the point \(P\), as shown in Figure 1.

1. Show that the \(x\) coordinate of \(P\) lies in the interval \([ - 1.25 , - 1.2 ]\) The curve cuts the positive \(x\)-axis at the point \(Q\), also shown in Figure 1.
   Using the iterative formula $$x _ { n + 1 } = \sqrt { 12 \ln \left( 2 x _ { n } + 3 \right) + 8 } \text { with } x _ { 1 } = 6$$
2. 1. find, to 4 decimal places, the value of \(x _ { 2 }\)
   2. find, by continued iteration, the \(x\) coordinate of \(Q\). Give your answer to 4 decimal places. The curve has a maximum turning point at \(M\), as shown in Figure 1.
3. Using calculus and showing each stage of your working, find the \(x\) coordinate of \(M\).

1. A curve has equation \(y = \mathrm { f } ( x )\) where

$$\mathrm { f } ( x ) = x ^ { 4 } - 5 x ^ { 2 } + 4 x - 7 \quad x \in \mathbb { R }$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval [2,3]
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \sqrt [ 3 ] { \frac { 5 x ^ { 2 } - 4 x + 7 } { x } }$$ The iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { \frac { 5 x _ { n } ^ { 2 } - 4 x _ { n } + 7 } { x _ { n } } }$$ is used to find \(\alpha\)
3. Starting with \(x _ { 1 } = 2\) and using the iterative formula,
   1. find, to 4 decimal places, the value of \(x _ { 2 }\)
   2. find, to 4 decimal places, the value of \(\alpha\)

9. \begin{figure}[h] \end{figure} In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 5 shows the curve with equation $$y = \frac { 1 + 2 \cos x } { 1 + \sin x } \quad - \frac { \pi } { 2 } < x < \frac { 3 \pi } { 2 }$$ The point \(M\), shown in Figure 5, is the minimum point on the curve.

1. Show that the \(x\) coordinate of \(M\) is a solution of the equation $$2 \sin x + \cos x = - 2$$
2. Hence find, to 3 significant figures, the \(x\) coordinate of \(M\).

1. $$g ( x ) = x ^ { 6 } + 2 x - 1000$$

1. Show that \(\mathrm { g } ( x ) = 0\) has a root \(\alpha\) in the interval [3,4] Using the iteration formula $$x _ { n + 1 } = \sqrt [ 6 ] { 1000 - 2 x _ { n } } \quad \text { with } x _ { 1 } = 3$$
2. 1. find, to 4 decimal places, the value of \(x _ { 2 }\)
   2. find, by repeated iteration, the value of \(\alpha\). Give your answer 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.

5. \begin{figure}[h] \end{figure} The profit made by a company, \(\pounds P\) million, \(t\) years after the company started trading, is modelled by the equation $$P = \frac { 4 t - 1 } { 10 } + \frac { 3 } { 4 } \ln \left[ \frac { t + 1 } { ( 2 t + 1 ) ^ { 2 } } \right]$$ The graph of \(P\) against \(t\) is shown in Figure 2. According to the model,

1. show that exactly one year after it started trading, the company had made a loss of approximately £ 830000 A manager of the company wants to know the value of \(t\) for which \(P = 0\)
2. Show that this value of \(t\) occurs in the interval [6,7]
3. Show that the equation \(P = 0\) can be expressed in the form $$t = \frac { 1 } { 4 } + \frac { 15 } { 8 } \ln \left[ \frac { ( 2 t + 1 ) ^ { 2 } } { t + 1 } \right]$$
4. Using the iteration formula $$t _ { n + 1 } = \frac { 1 } { 4 } + \frac { 15 } { 8 } \ln \left[ \frac { \left( 2 t _ { n } + 1 \right) ^ { 2 } } { t _ { n } + 1 } \right] \text { with } t _ { 1 } = 6$$ find the value of \(t _ { 2 }\) and the value of \(t _ { 6 }\), giving your answers to 3 decimal places.
5. Hence find, according to the model, how many months it takes in total, from when the company started trading, for it to make a profit.
   (2)

1. A curve has equation \(y = \mathrm { f } ( x )\) where

$$\mathrm { f } ( x ) = x ^ { 2 } - 5 x + \mathrm { e } ^ { x } \quad x \in \mathbb { R }$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval [1,2] The iterative formula $$x _ { n + 1 } = \sqrt { 5 x _ { n } - \mathrm { e } ^ { x _ { n } } }$$ with \(x _ { 1 } = 1\) is used to find an approximate value for the root \(\alpha\).
2. 1. Find the value of \(x _ { 2 }\) to 4 decimal places.
   2. Find, by repeated iteration, the value of \(\alpha\), giving your answer to 4 decimal places.

2. $$f ( x ) = x ^ { 3 } + 3 x ^ { 2 } + 4 x - 12$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \sqrt { \left( \frac { 4 ( 3 - x ) } { ( 3 + x ) } \right) } \quad x \neq - 3$$ The equation \(x ^ { 3 } + 3 x ^ { 2 } + 4 x - 12 = 0\) has a single root which is between 1 and 2
2. Use the iteration formula $$x _ { n + 1 } = \sqrt { \left( \frac { 4 \left( 3 - x _ { n } \right) } { \left( 3 + x _ { n } \right) } \right) } \quad n \geqslant 0$$ with \(x _ { 0 } = 1\) to find, to 2 decimal places, the value of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) The root of \(\mathrm { f } ( x ) = 0\) is \(\alpha\).
3. By choosing a suitable interval, prove that \(\alpha = 1.272\) to 3 decimal places.
   

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5. $$f ( x ) = - x ^ { 3 } + 4 x ^ { 2 } - 6$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\)
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \sqrt { \left( \frac { 6 } { 4 - x } \right) }$$
3. Starting with \(x _ { 1 } = 1.5\) use the iteration \(x _ { n + 1 } = \sqrt { \left( \frac { 6 } { 4 - x _ { n } } \right) }\) to calculate the values of \(x _ { 2 }\), \(x _ { 3 }\) and \(x _ { 4 }\) giving all your answers to 4 decimal places.
4. Using a suitable interval, show that 1.572 is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.

3. $$\mathrm { f } ( x ) = \frac { x ^ { 2 } } { 4 } + \ln ( 2 x ) , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x ^ { 2 } }$$ The equation \(\mathrm { f } ( x ) = 0\) has a root near 0.5
2. Starting with \(x _ { 1 } = 0.5\) use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x _ { n } ^ { 2 } }$$ to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
3. Using a suitable interval, show that 0.473 is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.