1.09a Sign change methods: locate roots

4. $$f ( x ) = 2 ^ { x } - 6 x$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [4,5]. Using the end points of this interval find, by linear interpolation, an approximation to \(\alpha\).

$$\mathrm { f } ( x ) = 2 \sin \left( x ^ { 2 } \right) + x - 2 , \quad 0 \leqslant x < 2 \pi$$

1. Show that \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between \(x = 0.75\) and \(x = 0.85\) The equation \(\mathrm { f } ( x ) = 0\) can be written as \(x = [ \arcsin ( 1 - 0.5 x ) ] ^ { \frac { 1 } { 2 } }\).
2. Use the iterative formula $$x _ { n + 1 } = \left[ \arcsin \left( 1 - 0.5 x _ { n } \right) \right] ^ { \frac { 1 } { 2 } } , \quad x _ { 0 } = 0.8$$ to find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 5 decimal places.
3. Show that \(\alpha = 0.80157\) is correct to 5 decimal places.

5. \includegraphics[max width=\textwidth, alt={}, center]{5dd332a5-56d9-407a-8ff6-fa59294b358d-2_520_787_246_479} The diagram shows the graph of \(y = \mathrm { f } ( x )\). The graph has a minimum at \(\left( \frac { \pi } { 2 } , - 1 \right)\), a maximum at \(\left( \frac { 3 \pi } { 2 } , - 5 \right)\) and an asymptote with equation \(x = \pi\).

1. Showing the coordinates of any stationary points, sketch the graph of \(y = | \mathrm { f } ( x ) |\). Given that $$\mathrm { f } : x \rightarrow a + b \operatorname { cosec } x , \quad x \in \mathbb { R } , \quad 0 < x < 2 \pi , \quad x \neq \pi$$
2. find the values of the constants \(a\) and \(b\),
3. find, to 2 decimal places, the \(x\)-coordinates of the points where the graph of \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis.

7 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-3_465_748_1133_717} 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.
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.
3. Show by calculation that the root of the equation \(3 \cos ^ { - 1 } ( x - 1 ) = x\) lies between 1.8 and 1.9 .
4. The sequence defined by $$x _ { 1 } = 2 , \quad x _ { n + 1 } = 1 + \cos \left( \frac { 1 } { 3 } x _ { n } \right)$$ 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 \(3 \cos ^ { - 1 } ( x - 1 ) = x\).

8 \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-3_588_915_954_614} The diagram shows part of each of the curves \(y = e ^ { \frac { 1 } { 5 } x }\) and \(y = \sqrt [ 3 ] { } ( 3 x + 8 )\). 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.
2. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 5 } { 3 } \ln ( 3 x + 8 )\).
3. Use an iterative formula, based on the equation in part (ii), to find the \(x\)-coordinate of \(P\) correct to 2 decimal places.
4. Use integration, and your answer to part (iii), to find an approximate value of the area of the region \(R\). \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-4_625_647_264_749} The function f is defined by \(\mathrm { f } ( x ) = \sqrt { } ( m x + 7 ) - 4\), where \(x \geqslant - \frac { 7 } { m }\) and \(m\) is a positive constant. The diagram shows the curve \(y = \mathrm { f } ( x )\).

1. A sequence of transformations maps the curve \(y = \sqrt { } x\) to the curve \(y = \mathrm { f } ( x )\). Give details of these transformations.
2. Explain how you can tell that f is a one-one function and find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
3. It is given that the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) do not meet. Explain how it can be deduced that neither curve meets the line \(y = x\), and hence determine the set of possible values of \(m\). [5]

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

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

2 It is given that \(\mathrm { f } ( x ) = x ^ { 2 } - \tan ^ { - 1 } x\).

1. Show by calculation that the equation \(\mathrm { f } ( x ) = 0\) has a root in the interval \(0.8 < x < 0.9\).
2. Use the Newton-Raphson method, with a first approximation 0.8, to find the next approximation to this root. Give your answer correct to 3 decimal places.

3.(a)(i)Write down the binomial series expansion of $$\left( 1 + \frac { 2 } { n } \right) ^ { n } \quad n \in \mathbb { N } , n > 2$$ in powers of \(\left( \frac { 2 } { n } \right)\) up to and including the term in \(\left( \frac { 2 } { n } \right) ^ { 3 }\) (ii)Hence prove that,for \(n \in \mathbb { N } , n \geqslant 3\) $$\left( 1 + \frac { 2 } { n } \right) ^ { n } \geqslant \frac { 19 } { 3 } - \frac { 6 } { n }$$ (b)Use the binomial series expansion of \(\left( 1 - \frac { x } { 4 } \right) ^ { \frac { 1 } { 2 } }\) to show that \(\sqrt { 3 } < \frac { 7 } { 4 }\) $$\mathrm { f } ( x ) = \left( 1 + \frac { 2 } { x } \right) ^ { x } - 3 ^ { \frac { x } { 6 } } \quad x \in \mathbb { R } , x > 0$$ Given that the function \(\mathrm { f } ( x )\) is continuous and that \(\sqrt [ 6 ] { 3 } > \frac { 6 } { 5 }\) (c)prove that \(\mathrm { f } ( x ) = 0\) has a root in the interval[9,10]

1.(i) $$f ( x ) = x ^ { 3 } + 4 x - 6$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval[1,1.5]
2. Taking 1.5 as a first approximation,apply the Newton Raphson process twice to \(\mathrm { f } ( x )\) to obtain an approximate value of \(\alpha\) .Give your answer to 3 decimal places. Show your working clearly.
   (ii) $$g ( x ) = 4 x ^ { 2 } + x - \tan x$$ where \(x\) is measured in radians. The equation \(\mathrm { g } ( x ) = 0\) has a single root \(\beta\) in the interval[1.4,1.5]
   Use linear interpolation on the values at the end points of this interval to obtain an approximation to \(\beta\) .Give your answer to 3 decimal places.

6 \includegraphics[max width=\textwidth, alt={}, center]{33a2b09d-0df9-48d6-9ee9-e0a1ec345f41-3_524_720_246_676} The diagram shows the curve \(y = x ^ { 4 } - 8 x\).

1. By sketching a second curve on the copy of the diagram, show that the equation $$x ^ { 4 } + x ^ { 2 } - 8 x - 9 = 0$$ has two real roots. State the equation of the second curve.
2. The larger root of the equation \(x ^ { 4 } + x ^ { 2 } - 8 x - 9 = 0\) is denoted by \(\alpha\).
   1. Show by calculation that \(2.1 < \alpha < 2.2\).
   2. Use an iterative process based on the equation $$x = \sqrt [ 4 ] { 9 + 8 x - x ^ { 2 } } ,$$ with a suitable starting value, to find \(\alpha\) correct to 3 decimal places. Give the result of each step of the iterative process.

6 \includegraphics[max width=\textwidth, alt={}, center]{00a4be37-c095-4d9c-a1cd-d03b8ab1d411-3_553_579_274_726} The diagram shows the curve \(y = 8 \sin ^ { - 1 } \left( x - \frac { 3 } { 2 } \right)\). The end-points \(A\) and \(B\) of the curve have coordinates ( \(a , - 4 \pi\) ) and ( \(b , 4 \pi\) ) respectively.

1. State the values of \(a\) and \(b\).
2. It is required to find the root of the equation \(8 \sin ^ { - 1 } \left( x - \frac { 3 } { 2 } \right) = x\).
   1. Show by calculation that the root lies between 1.7 and 1.8.
   2. In order to find the root, the iterative formula $$x _ { n + 1 } = p + \sin \left( q x _ { n } \right) ,$$ with a suitable starting value, is to be used. Determine the values of the constants \(p\) and \(q\) and hence find the root correct to 4 significant figures. Show the result of each step of the iteration process.

11 The curve \(C\) has polar equation $$r = \frac { a } { 1 + \theta }$$ where \(a\) is a positive constant and \(0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(r\) decreases as \(\theta\) increases.
2. The point \(P\) of \(C\) is further from the initial line than any other point of \(C\). Show that, at \(P\), $$\tan \theta = 1 + \theta$$ and verify that this equation has a root between 1.1 and 1.2.
3. Draw a sketch of \(C\).
4. Find the area of the region bounded by the initial line, the line \(\theta = \frac { 1 } { 2 } \pi\) and \(C\), leaving your answer in terms of \(\pi\) and \(a\).

10 \includegraphics[max width=\textwidth, alt={}, center]{05bec6d6-b526-4b6f-86f3-39aa38cbf5c6-6_405_661_251_703} The diagram shows a sector \(A O B\) of a circle with centre \(O\) and radius 6 cm .
The angle \(A O B\) is \(\theta\) radians.
The area of the segment bounded by the chord \(A B\) and the \(\operatorname { arc } A B\) is \(7.2 \mathrm {~cm} ^ { 2 }\).

1. Show that \(\theta = 0.4 + \sin \theta\).
2. Let \(\mathrm { F } ( \theta ) = 0.4 + \sin \theta\). By considering the value of \(\mathrm { F } ^ { \prime } ( \theta )\) where \(\theta = 1.2\), explain why using an iterative method based on the equation in part (a) will converge to the root, assuming that 1.2 is sufficiently close to the root.
3. Use the iterative formula \(\theta _ { n + 1 } = 0.4 + \sin \theta _ { n }\) with a starting value of 1.2 to find the value of \(\theta\) correct to 4 significant figures.
   You should show the result of each iteration.
4. Use a change of sign method to show that the value of \(\theta\) found in part (c) is correct to 4 significant figures.

10
The diagram shows part of the curve \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { x } } { 4 x ^ { 2 } - 1 } + 2\). The equation \(\mathrm { f } ( x ) = 0\) has a positive root \(\alpha\) close to \(x = 0.3\).

1. Explain why using the sign change method with \(x = 0\) and \(x = 1\) will fail to locate \(\alpha\).
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as \(x = \frac { 1 } { 4 } \sqrt { \left( 4 - 2 \mathrm { e } ^ { x } \right) }\).
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 4 } \sqrt { \left( 4 - 2 \mathrm { e } ^ { x _ { n } } \right) }\) with a starting value of \(x _ { 1 } = 0.3\) to find the value of \(\alpha\) correct to \(\mathbf { 4 }\) significant figures, showing the result of each iteration.
4. An alternative iterative formula is \(x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)\), where \(\mathrm { F } \left( x _ { n } \right) = \ln \left( 2 - 8 x _ { n } ^ { 2 } \right)\). By considering \(\mathrm { F } ^ { \prime } ( 0.3 )\) explain why this iterative formula will not find \(\alpha\).

2 The diagram shows part of the graph of \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x )\) is a cubic polynomial in \(x\). \includegraphics[max width=\textwidth, alt={}, center]{7298e7b9-ad52-480c-bc2b-8289aeab9ebb-04_437_620_909_274} Explain why one of the roots of the equation \(\mathrm { f } ( x ) = 0\) cannot be found by the sign change method.

6. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation $$f ( x ) = 4 \cos 2 x - 2 x + 1 \quad x > 0$$ and where \(x\) is measured in radians.
The curve crosses the \(x\)-axis at the point \(A\), as shown in figure 1 .
Given that \(x\)-coordinate of \(A\) is \(\alpha\) a. show that \(\alpha\) lies between 0.7 and 0.8 Given that \(x\)-coordinates of \(B\) and \(C\) are \(\beta\) and \(\gamma\) respectively and they are two smallest values of \(x\) at which local maxima occur
b. find, using calculus, the value of \(\beta\) and the value of \(\gamma\), giving your answers to 3 significant figures.
c. taking \(x _ { 0 } = 0.7\) or 0.8 as a first approximation to \(\alpha\), apply the Newton-Raphson method once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Show, your method and give your answer to 2 significant figures.

The curve with equation \(y = 2 \ln ( 8 - x )\) meets the line \(y = x\) at a single point, \(x = \alpha\).

1. Show that \(3 < \alpha < 4\) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{b5f50f17-9f1b-4b4c-baf3-e50de5f2ea9c-08_666_1061_445_502} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows the graph of \(y = 2 \ln ( 8 - x )\) and the graph of \(y = x\).
   A student uses the iteration formula $$x _ { n + 1 } = 2 \ln \left( 8 - x _ { n } \right) , \quad n \in \mathbb { N }$$ in an attempt to find an approximation for \(\alpha\).
   Using the graph and starting with \(x _ { 1 } = 4\)
2. determine whether or not this iteration formula can be used to find an approximation for \(\alpha\), justifying your answer.

2. \begin{figure}[h] \end{figure} Figure 1 shows a plot of part of the curve with equation \(y = \cos x\) where \(x\) is measured in radians. Diagram 1, on the opposite page, is a copy of Figure 1.

1. Use Diagram 1 to show why the equation $$\cos x - 2 x - \frac { 1 } { 2 } = 0$$ has only one real root, giving a reason for your answer. Given that the root of the equation is \(\alpha\), and that \(\alpha\) is small,
2. use the small angle approximation for \(\cos x\) to estimate the value of \(\alpha\) to 3 decimal places.
   \includegraphics[max width=\textwidth, alt={}]{91a2f26a-add2-4b58-997d-2ae229548217-05_664_1452_246_333}
   \section*{Diagram 1}

3. $$\mathrm { f } ( x ) = x + \tan \left( \frac { 1 } { 2 } x \right) \quad \pi < x < \frac { 3 \pi } { 2 }$$ Given that the equation \(\mathrm { f } ( x ) = 0\) has a single root \(\alpha\)

1. show that \(\alpha\) lies in the interval [3.6, 3.7]
2. Find \(\mathrm { f } ^ { \prime } ( x )\)
3. Using 3.7 as a first approximation for \(\alpha\), apply the Newton-Raphson method once to obtain a second approximation for \(\alpha\). Give your answer to 3 decimal places.

1. The sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by

$$u _ { n + 1 } = k - \frac { 24 } { u _ { n } } \quad u _ { 1 } = 2$$ where \(k\) is an integer.
Given that \(u _ { 1 } + 2 u _ { 2 } + u _ { 3 } = 0\)

1. show that $$3 k ^ { 2 } - 58 k + 240 = 0$$
2. Find the value of \(k\), giving a reason for your answer.
3. Find the value of \(u _ { 3 }\)

7. Figure 2 Figure 2 shows a sketch of a triangle \(A B C\).
Given \(\overrightarrow { A B } = 2 \mathbf { i } + 3 \mathbf { j } + \mathbf { k }\) and \(\overrightarrow { B C } = \mathbf { i } - 9 \mathbf { j } + 3 \mathbf { k }\),
show that \(\angle B A C = 105.9 ^ { \circ }\) to one decimal place.

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = 8 \sin \left( \frac { 1 } { 2 } x \right) - 3 x + 9 \quad x > 0$$ and \(x\) is measured in radians.
The point \(P\), shown in Figure 2, is a local maximum point on the curve.
Using calculus and the sketch in Figure 2,

1. find the \(x\) coordinate of \(P\), giving your answer to 3 significant figures. The curve crosses the \(x\)-axis at \(x = \alpha\), as shown in Figure 2 .
   Given that, to 3 decimal places, \(f ( 4 ) = 4.274\) and \(f ( 5 ) = - 1.212\)
2. explain why \(\alpha\) must lie in the interval \([ 4,5 ]\)
3. Taking \(x _ { 0 } = 5\) as a first approximation to \(\alpha\), apply the Newton-Raphson method once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Show your method and give your answer to 3 significant figures.

2 By considering a change of sign, show that the equation \(\mathrm { e } ^ { x } - 5 x ^ { 3 } = 0\) has a root between 0 and 1 .

8 Kareem wants to solve the equation \(\sin 4 x + \mathrm { e } ^ { - x } + 0.75 = 0\). He uses his calculator to create the following table of values for \(\mathrm { f } ( x ) = \sin 4 x + \mathrm { e } ^ { - x } + 0.75\). He argues that because \(\mathrm { f } ( 6 )\) is the first negative value in the table, there is a root of the equation between 5 and 6 .

1. Comment on the validity of his argument. The diagram shows the graph of \(y = \sin 4 x + e ^ { - x } + 0.75\). \includegraphics[max width=\textwidth, alt={}, center]{4fac72cb-85cb-48d9-8817-899ef3f80a0f-07_538_1260_920_244}
2. Explain why Kareem failed to find other roots between 0 and 6 . Kareem decides to use the Newton-Raphson method to find the root close to 3 .
3. 1. Determine the iterative formula he should use for this equation.
   2. Use the Newton-Raphson method with \(x _ { 0 } = 3\) to find a root of the equation \(\mathrm { f } ( x ) = 0\). Show three iterations and give your answer to a suitable degree of accuracy. Kareem uses the Newton-Raphson method with \(x _ { 0 } = 5\) and also with \(x _ { 0 } = 6\) to try to find the root which lies between 5 and 6 . He produces the following tables.

      \(x _ { 0 }\)	5
      \(x _ { 1 }\)	3.97288
      \(x _ { 2 }\)	4.12125

      \(x _ { 0 }\)	6
      \(x _ { 1 }\)	6.09036
      \(x _ { 2 }\)	6.07110

4. 1. For the iteration beginning with \(x _ { 0 } = 5\), represent the process on the graph in the Printed Answer Booklet.
   2. Explain why the method has failed to find the root which lies between 5 and 6 .
   3. Explain how Kareem can adapt his method to find the root between 5 and 6 .

9 This question is about the equation \(\mathrm { f } ( x ) = 0\), where \(\mathrm { f } ( x ) = x ^ { 4 } - x - \frac { 1 } { 3 x - 2 }\).
Fig. 9.1 shows the curve \(y = f ( x )\).
Fig. 9.1 \includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-06_940_929_518_239}

1. Show, by calculation, that the equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\).
2. Fig. 9.2 shows part of a spreadsheet being used to find a root of the equation. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Fig. 9.2}

   	A	B
   1	\(x\)	\(f ( x )\)
   2	1.5	3.1625
   3	1.25	0.619977679
   4	1.125	- 0.250466087
   5

   \end{table} Write down a suitable number to use as the next value of \(x\) in the spreadsheet.
3. Determine a root of the equation \(\mathrm { f } ( x ) = 0\). Give your answer correct to \(\mathbf { 1 }\) decimal place.
4. Fig. 9.3 shows a similar spreadsheet being used to search for another root of \(\mathrm { f } ( x ) = 0\). \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Fig. 9.3}

   	A	B
   1	x	f(x)
   2	0	0.5
   3	1	-1
   4	0.5	1.5625
   5	0.75	-4.4336
   6	0.6	4.5296
   7	0.7	-10.4599
   8	0.65	19.5285
   9	0.675	-40.4674
   10	0.6625	79.5301
   11	0.66875	-160.4687
   10

   \end{table}
   1. Explain why it looks from rows 2 and 3 of the spreadsheet as if there is a root between 0 and 1.
   2. Explain why this process will not find a root between 0 and 1 .