1.09d Newton-Raphson method

7

1. The equation \(2 x ^ { 3 } + 5 x ^ { 2 } + 3 x - 132000 = 0\) has exactly one real root \(\alpha\).
   1. Show that \(\alpha\) lies in the interval \(39 < \alpha < 40\).
   2. Taking \(x _ { 1 } = 40\) as a first approximation to \(\alpha\), use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to \(\alpha\). Give your answer to two decimal places.
2. Use the formulae for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that $$\sum _ { r = 1 } ^ { n } 2 r ( 3 r + 2 ) = n ( n + p ) ( 2 n + q )$$ where \(p\) and \(q\) are integers.
3. 1. Express \(\log _ { 8 } 4 ^ { r }\) in the form \(\lambda r\), where \(\lambda\) is a rational number.
   2. By first finding a suitable cubic inequality for \(k\), find the greatest value of \(k\) for which \(\sum _ { r = k + 1 } ^ { 60 } ( 3 r + 2 ) \log _ { 8 } 4 ^ { r }\) is greater than 106060.
      [0pt] [4 marks]

4 Fig. 4 shows the graph of \(y = x ^ { 5 } - 6 \sqrt { x } + 4\). \begin{figure}[h] \end{figure} There are two roots of the equation \(x ^ { 5 } - 6 \sqrt { x } + 4 = 0\). The roots are \(\alpha\) and \(\beta\), such that \(\alpha < \beta\).

1. Show that \(0 < \alpha < 1\) and \(1 < \beta < 2\).
2. Obtain the Newton-Raphson iterative formula $$x _ { n + 1 } = x _ { n } - \frac { x _ { n } ^ { \frac { 11 } { 2 } } - 6 x _ { n } + 4 \sqrt { x _ { n } } } { 5 x _ { n } ^ { \frac { 9 } { 2 } } - 3 }$$
3. Use the iterative formula found in part (b) with a starting value of \(x _ { 0 } = 1\) to obtain \(\beta\) correct to 6 decimal places.
4. Use the iterative formula found in part (b) with a starting value of \(x _ { 0 } = 0\) to find \(x _ { 1 }\).
5. Give a geometrical explanation of why the Newton-Raphson iteration fails to find \(\alpha\) in part (d).
6. Obtain the iterative formula $$x _ { n + 1 } = \left( \frac { x _ { n } ^ { 5 } + 4 } { 6 } \right) ^ { 2 }$$
7. Use the iterative formula found in part (f) with a starting value of \(x _ { 0 } = 0\) to obtain \(\alpha\) correct to 6 decimal places.

4 Fig. 4.1 shows part of the graph of \(y = e ^ { x } - x ^ { 2 } - x - 1.1\). \begin{figure}[h] \end{figure} The equation \(\mathrm { e } ^ { x } - x ^ { 2 } - x - 1.1 = 0\) has a root \(\alpha\) such that \(1 < \alpha < 2\).
Ali is considering using the Newton-Raphson method to find \(\alpha\). Ali could use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\).

1. Without doing any calculations, explain whether Ali should use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\), or whether using either starting value would work equally well. Ali is also considering using the method of fixed point iteration to find \(\alpha\). Ali could use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\). Fig. 4.2 shows parts of the graphs of \(y = x\) and \(y = \ln \left( x ^ { 2 } + x + 1.1 \right)\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_819_1011_1818_255} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}
2. Without doing any calculations, explain whether Ali should use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\) or whether either starting value would work equally well. Ali used one of the above methods to find a sequence of approximations to \(\alpha\). These are shown, together with some further analysis in the associated spreadsheet output in Fig. 4.3. \begin{table}[h]

   		M	N	O
   	\(r\)	\(\mathrm { X } _ { \mathrm { r } }\)
   4	0	2
   5	1	1.879008	- 0.121
   6	2	1.858143	- 0.021	0.172
   7	3	1.857565	\(- 6 \mathrm { E } - 04\)	0.028
   8	4	1.857564	\(- 4 \mathrm { E } - 07\)	\(8 \mathrm { E } - 04\)
   9	5	1.857564	\(- 2 \mathrm { E } - 13\)	\(6 \mathrm { E } - 07\)

   \captionsetup{labelformat=empty} \caption{Fig. 4.3}
   \end{table} The formula in cell N5 is =M5-M4
   and the formula in cell O6 is =N6/N5
   equivalent formulae are in cells N6 to N9 and O7 to O9 respectively.
3. State what is being calculated in the following columns of the spreadsheet.
   1. Column N
   2. Column O
4. Explain whether the values in column O suggest that Ali used the Newton-Raphson method or the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \ln \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } + \mathrm { x } _ { \mathrm { n } } + 1.1 \right)\) to find this sequence of approximations to \(\alpha\).

7 The value of a function, \(\mathrm { y } = \mathrm { f } ( \mathrm { x } )\), and its gradient function, \(\frac { \mathrm { dy } } { \mathrm { dx } }\), when \(x = 2\), is given in Table 7.1. \begin{table}[h] \end{table}

1. Determine the approximate value of the error when \(f ( 2 )\) is used to estimate \(f ( 2.03 )\). The Newton-Raphson method is used to find a sequence of approximations to a root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\). The spreadsheet output showing the iterates, together with some further analysis, is shown in Table 7.2. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Table 7.2}

   	A	B	C	D
   1	r	Xr	difference	ratio
   2	0	12
   3	1	-13.1165572	-25.1165572
   4	2	1.76283279	14.87939004	-0.5924136
   5	3	2.18052157	0.41768878	0.02807163
   6	4	2.182419024	0.001897454	0.00454275
   7	5	2.182419066	\(4.13985 \mathrm { E } - 08\)	\(2.1818 \mathrm { E } - 05\)

   \end{table}
2. 1. Explain what the values in column D tell you about the order of convergence of this sequence of approximations.
   2. Without doing any further calculation, state the value of \(\alpha\) as accurately as you can, justifying the precision quoted.

5 You are given that \(g ( x ) = \frac { \sqrt [ 3 ] { x ^ { x } + 25 } } { 2 }\). Fig. 5.1 shows two values of \(x\) and the associated values of \(\mathrm { g } ( x )\). \begin{table}[h] \end{table}

1. Use the central difference method to calculate an estimate of \(\mathrm { g } ^ { \prime } ( 1.5 )\), giving your answer correct to 3 decimal places. The equation \(x ^ { x } - 8 x ^ { 3 } + 25 = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha \approx 1.5\) and \(\beta \approx 4.4\).
2. Obtain the iterative formula \(x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }\).
3. Use your answer to part (a) to explain why it is possible that the iterative formula \(x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }\) may be used to find \(\alpha\).
4. Starting with \(x _ { 0 } = 1.5\), use the iterative formula to find \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 } , x _ { 5 }\), and \(x _ { 6 }\).
5. Use your answer to part (d) to state the value of \(\alpha\) correct to 8 decimal places. Starting with \(x _ { 0 } = 4.5\) the same iterative formula is used in an attempt to find \(\beta\). The results are shown in Fig. 5.2. \begin{table}[h]

   \(n\)	\(x _ { n }\)
   0	4.5
   1	4.81826433
   2	6.27473453
   3	23.2937196
   4	\(2.0654 \mathrm { E } + 10\)
   5	\#NUM!

   \captionsetup{labelformat=empty} \caption{Fig. 5.2}
   \end{table}
6. Explain why \#NUM! is displayed in the cell for \(x _ { 5 }\).
7. On the diagram in the Printed Answer Booklet, starting with \(x _ { 0 } = 4.5\), illustrate how the iterative formula works to find \(x _ { 1 }\) and \(x _ { 2 }\).
8. Determine what happens when the relaxed iteration \(x _ { n + 1 } = ( 1 - \lambda ) x _ { n } + \lambda g \left( x _ { n } \right)\) is used to try to find \(\beta\) with \(x _ { 0 } = 4.5\), in each of the following cases.

6 The equation \(0.5 \ln x - x ^ { 2 } + x + 1 = 0\) has two roots \(\alpha\) and \(\beta\), such that \(0 < \alpha < 1\) and \(1 < \beta < 2\).

1. Use the Newton-Raphson method with \(x _ { 0 } = 1\) to obtain \(\beta\) correct to \(\mathbf { 6 }\) decimal places. Fig. 6.1 shows part of the graph of \(y = 0.5 \ln x - x ^ { 2 } + x + 1\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{945883ad-c153-4c51-83d3-978e4c769ed5-06_1112_1156_529_354} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
   \end{figure}
2. On the copy of Fig. 6.1 in the Printed Answer Booklet, illustrate the Newton-Raphson method working to obtain \(x _ { 1 }\) from \(x _ { 0 } = 1\). Beth is trying to find \(\alpha\) correct to 6 decimal places.
3. Suggest a reason why she might choose the Newton-Raphson method instead of fixed point iteration. Beth tries to find \(\alpha\) using the Newton-Raphson method with a starting value of \(x _ { 0 } = 0.5\). Her spreadsheet output is shown in Fig. 6.2. \begin{table}[h]

   \(r\)	\(\mathrm { x } _ { \mathrm { r } }\)
   0	0.5
   1	- 0.40343
   2	\#NUM!

   \captionsetup{labelformat=empty} \caption{Fig. 6.2}
   \end{table}
4. Explain how the display \#NUM! has arisen in the cell for \(x _ { 2 }\). Beth decides to use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) = \sqrt { 0.5 \ln \left( x _ { n } \right) + x _ { n } + 1 }$$
5. Determine the outcome when Beth uses this formula with \(x _ { 0 } = 0.5\).
6. Use the relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\) with \(\lambda = - 0.041\) and \(x _ { 0 } = 0.5\) to obtain \(\alpha\) correct to \(\mathbf { 6 }\) decimal places.

3 The equation \(\sinh x + x ^ { 2 } - 1 = 0\) has a root, \(\alpha\), such that \(0 < \alpha < 1\).

1. Verify that the iteration \(x _ { r + 1 } = \frac { 1 - \sinh x _ { r } } { x _ { r } }\) with \(x _ { 0 } = 1\) fails to converge to this root.
2. Use the relaxed iteration \(x _ { r + 1 } = ( 1 - \lambda ) x _ { r } + \lambda \left( \frac { 1 - \sinh x _ { r } } { x _ { r } } \right)\) with \(\lambda = \frac { 1 } { 4 }\) and \(x _ { 0 } = 1\) to find \(\alpha\) correct to 6 decimal places.

3 This question concerns the family of differential equations \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - x ^ { a } y \left( { } ^ { * } \right)\) where \(a\) is \(- 1,0\) or 1 .

1. Determine and describe geometrically the isoclines of (\textit{) when
   1. \(a = - 1\),
   2. \(a = 0\),
   3. \(a = 1\).
2. In this part of the question \(a = 0\).
   1. Write down the solution to \(( * )\) which passes through the point \(( 0 , b )\) where \(b \neq 1\).
   2. Write down the equation of the asymptote to this solution.
3. In this part of the question \(a = - 1\).
   1. Write down the solution to \(( * )\) which passes through the point \(( c , d )\) where \(c \neq 0\).
   2. Describe the relationship between \(c\) and \(d\) when the solution in part (i) has a stationary point.
4. In this part of the question \(a = 1\).
   1. The standard Runge-Kutta method of order 4 for the solution of the differential equation \(\mathrm { f } ( x , y ) = \frac { \mathrm { d } y } { \mathrm {~d} x }\) is as follows. \(k _ { 1 } = h \mathrm { f } \left( x _ { n } , y _ { n } \right)\) \(k _ { 2 } = h \mathrm { f } \left( x _ { n } + \frac { h } { 2 } , y _ { n } + \frac { k _ { 1 } } { 2 } \right)\) \(k _ { 3 } = h \mathrm { f } \left( x _ { n } + \frac { h } { 2 } , y _ { n } + \frac { k _ { 2 } } { 2 } \right)\) \(k _ { 4 } = h \mathrm { f } \left( x _ { n } + h , y _ { n } + k _ { 3 } \right)\) \(y _ { n + 1 } = y _ { n } + \frac { 1 } { 6 } \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right)\).
      Construct a spreadsheet to solve (}) in the case \(x _ { 0 } = 0\) and \(y _ { 0 } = 1.5\). State the formulae you have used in your spreadsheet.
   2. Use your spreadsheet with \(h = 0.05\) to find an approximation to the value of \(y\) when \(x = 1\).
   3. The solution to \(( * )\) in which \(x _ { 0 } = 0\) and \(y _ { 0 } = 1.5\) has a maximum point ( \(r , s\) ) with \(0 < r < 1\). Use your spreadsheet with suitable values of \(h\) to estimate \(r\) to two decimal places. Justify your answer.

4 In this question you are required to consider the family of differential equations $$\frac { d P } { d t } = r P \left( 1 - \frac { P } { K } \right) , \quad t \geqslant 0 , \quad P ( t ) \geqslant 0 \left( ^ { * } \right)$$ where \(r\) and \(K\) are positive constants. This differential equation can be used as a model for the size of a population \(P\) as a function of time \(t\).

1. 1. Determine the values of \(P\) for which
      $$\frac { d P } { d t } = 2 P ^ { 1.25 } \left( 1 - \frac { P } { 1000 } \right) ^ { 1.5 } , t \geqslant 0 , P ( t ) \geqslant 0 ( * * )$$ The diagram shows the tangent field to (**), and a solution in which \(P = 1\) when \(t = 0\), produced using a much more accurate numerical method. \includegraphics[max width=\textwidth, alt={}, center]{4715d0f0-a860-4189-802f-1d2d019e1115-4_899_1552_1763_319}
      1. The Euler method for the solution of the differential equation \(f ( t , P ) = \frac { d P } { d t }\) is as follows $$P _ { n + 1 } = P _ { n } + h f \left( t _ { n } , P _ { n } \right)$$ It is given that \(t _ { 0 } = 0\) and \(P _ { 0 } = 1\).

3 This question explores the family of differential equations \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sqrt { 1 + a x + 2 y }\) for various values of the parameter \(a\). Fig. 3 shows the tangent field in the case \(a = 1\). \begin{figure}[h] \end{figure}

1. (A) Sketch the tangent field in the case \(a = - 2\).
   (B) Explain why the tangent field is not defined for the whole coordinate plane.
   (C) Give an inequality which describes the region in which the tangent field is defined.
   (D) Find a value of \(a\) such that the region for which the tangent field is defined includes the entire \(x\)-axis.
2. (A) For the case \(a = 1\), with \(y = 1\) when \(x = 0\), construct a spreadsheet for the Runge-Kutta method of order 2 with formulae as follows, where \(\mathrm { f } ( x , y ) = \frac { \mathrm { d } y } { \mathrm {~d} x }\). $$\begin{aligned} k _ { 1 } & = h \mathrm { f } \left( x _ { n } , y _ { n } \right) \\ k _ { 2 } & = h \mathrm { f } \left( x _ { n } + h , y _ { n } + k _ { 1 } \right) \\ y _ { n + 1 } & = y _ { n } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right) \end{aligned}$$ State the formulae you have used in your spreadsheet.
   (B) Use your spreadsheet to obtain the value of \(y\) correct to 4 decimal places when \(x = 1\) for
3. (A) For the case \(a = 0\) find the analytical solution that passes through the point ( 0,1 ).
   (B) Verify that the solution in part (iii) (A) is a solution to the differential equation.
   (C) Use the solution in part (iii) (A) to find the value of \(y\) correct to 4 decimal places when \(x = 1\).
4. (A) Verify that \(y = - \frac { a } { 2 } x + \frac { a ^ { 2 } } { 8 } - \frac { 1 } { 2 }\) is a solution for all cases when \(a \leq 0\).
   (B) Show that this is the only straight line solution in these cases. \section*{Copyright Information:} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
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10 In this question you must show detailed reasoning. The equation of a curve is $$y = \frac { \sin 2 x - x } { x \sin x }$$

1. Use the small angle approximation given in the list of formulae on pages 2-3 of this question paper to show that $$\int _ { 0.01 } ^ { 0.05 } \mathrm { ydx } \approx \ln 5$$
2. Use the same small angle approximation to show that $$\frac { d y } { d x } \approx - 10000 \text { at the point where } x = 0.01 \text {. }$$ The equation \(y = 0\) has a root near \(x = 1\). Joan uses the Newton-Raphson method to find this root. The output from the spreadsheet she uses is shown in Fig. 10.1. \begin{table}[h]

   \(n\)	0	1	2	3	4	5	6	7
   \(\mathrm { x } _ { \mathrm { n } }\)	1	0.958509	0.950084	0.948261	0.94786	0.947772	0.947753	0.947748

   \captionsetup{labelformat=empty} \caption{Fig. 10.1}
   \end{table} Joan carries out some analysis of this output. The results are shown in Fig. 10.2. \begin{table}[h]

   \(x\)	\(y\)
   0.9477475	\(- 7.79967 \mathrm { E } - 07\)
   0.9477485	\(- 2.90821 \mathrm { E } - 06\)
   \(x\)	\(y\)
   0.947745	\(4.54066 \mathrm { E } - 06\)
   0.947755	\(- 1.67417 \mathrm { E } - 05\)

   \captionsetup{labelformat=empty} \caption{Fig. 10.2}
   \end{table}
3. Consider the information in Fig. 10.1 and Fig. 10.2.

10 \includegraphics[max width=\textwidth, alt={}, center]{e3942549-bfc0-432a-bf49-7d01d44af01a-7_579_764_255_651} The diagram shows the graph of \(\mathrm { f } ( x ) = \ln ( 3 x + 1 ) - x\), which has a stationary point at \(x = \alpha\). A student wishes to find the non-zero root \(\beta\) of the equation \(\ln ( 3 x + 1 ) - x = 0\) using the Newton-Raphson method.

1. (a) Determine the value of \(\alpha\).
   (b) Explain why the Newton-Raphson method will fail if \(\alpha\) is used as the initial value.
2. Show that the Newton-Raphson iterative formula for finding \(\beta\) can be written as $$x _ { n + 1 } = \frac { 3 x _ { n } - \left( 3 x _ { n } + 1 \right) \ln \left( 3 x _ { n } + 1 \right) } { 2 - 3 x _ { n } } .$$
3. Apply the iterative formula in part (ii) with initial value \(x _ { 1 } = 1\) to find the value of \(\beta\) correct to 5 significant figures. You should show the result of each iteration.
4. Use a change of sign method to verify that the value of \(\beta\) found in part (iii) is correct to 5 significant figures.

6 [Figure 1, printed on the insert, is provided for use in this question.]
The curve \(y = x ^ { 3 } + 4 x - 3\) intersects the \(x\)-axis at the point \(A\) where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.0.
2. Show that the equation \(x ^ { 3 } + 4 x - 3 = 0\) can be rearranged into the form \(x = \frac { 3 - x ^ { 3 } } { 4 }\).
   (1 mark)
3. 1. Use the iteration \(x _ { n + 1 } = \frac { 3 - x _ { n } { } ^ { 3 } } { 4 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\), giving your answer to two decimal places.
      (3 marks)
   2. The sketch on Figure 1 shows parts of the graphs of \(y = \frac { 3 - x ^ { 3 } } { 4 }\) 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.
      (3 marks)

4 [Figure 1, printed on the insert, is provided for use in this question.]

1. Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to \(\int _ { 1 } ^ { 2 } 3 ^ { x } \mathrm {~d} x\), giving your answer to three significant figures.
2. The curve \(y = 3 ^ { x }\) intersects the line \(y = x + 3\) at the point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 0.5 and 1.5.
   2. Show that the equation \(3 ^ { x } = x + 3\) can be rearranged into the form $$x = \frac { \ln ( x + 3 ) } { \ln 3 }$$
   3. Use the iteration \(x _ { n + 1 } = \frac { \ln \left( x _ { n } + 3 \right) } { \ln 3 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\) to two significant figures.
   4. The sketch on Figure 1 shows part of the graphs of \(y = \frac { \ln ( x + 3 ) } { \ln 3 }\) 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.

8 [Figure 2, printed on the insert, is provided for use in this question.]
The diagram shows a part of the graph of \(y = \mathrm { f } ( x )\), where $$f ( x ) = x ^ { 3 } - 2 x - 1$$ The point \(P\) has coordinates \(( 1 , - 2 )\). \includegraphics[max width=\textwidth, alt={}, center]{a77cc9c3-5ff6-4abc-931e-e811740267f2-05_606_565_717_740}

1. Taking \(x _ { 1 } = 1\) as a first approximation to a root of the equation \(\mathrm { f } ( x ) = 0\), use the NewtonRaphson method to find a second approximation, \(x _ { 2 }\), to the root.
2. On Figure 2, draw a straight line to illustrate the Newton-Raphson method as used in part (a). Mark \(x _ { 1 }\) and \(x _ { 2 }\) on Figure 2
3. By considering \(f ( 2 )\), show that the second approximation found in part (a) is not as good as the first approximation.
4. Taking \(x _ { 1 } = 1.6\) as a first approximation to the root, use the Newton-Raphson method to find a second approximation to the root. Give your answer to three decimal places.
   (2 marks)

7 [Figure 1, printed on the insert, is provided for use in this question.]
The diagram shows the curve $$y = x ^ { 3 } - x + 1$$ The points \(A\) and \(B\) on the curve have \(x\)-coordinates - 1 and \(- 1 + h\) respectively. \includegraphics[max width=\textwidth, alt={}, center]{a0a30197-ca11-40d9-9ccd-30281c5e0fb4-05_978_1184_676_411}

1. 1. Show that the \(y\)-coordinate of the point \(B\) is $$1 + 2 h - 3 h ^ { 2 } + h ^ { 3 }$$
   2. Find the gradient of the chord \(A B\) in the form $$p + q h + r h ^ { 2 }$$ where \(p , q\) and \(r\) are integers.
   3. Explain how your answer to part (a)(ii) can be used to find the gradient of the tangent to the curve at \(A\). State the value of this gradient.
2. The equation \(x ^ { 3 } - x + 1 = 0\) has one real root, \(\alpha\).
   1. Taking \(x _ { 1 } = - 1\) as a first approximation to \(\alpha\), use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to \(\alpha\).
   2. On Figure 1, draw a straight line to illustrate the Newton-Raphson method as used in part (b)(i). Show the points \(\left( x _ { 2 } , 0 \right)\) and \(( \alpha , 0 )\) on your diagram.

8

1. The function f is defined for all real values of \(x\) by $$\mathrm { f } ( x ) = x ^ { 3 } + x ^ { 2 } - 1$$
   1. Express \(\mathrm { f } ( 1 + h ) - \mathrm { f } ( 1 )\) in the form $$p h + q h ^ { 2 } + r h ^ { 3 }$$ where \(p , q\) and \(r\) are integers.
   2. Use your answer to part (a)(i) to find the value of \(f ^ { \prime } ( 1 )\).
2. The diagram shows the graphs of $$y = \frac { 1 } { x ^ { 2 } } \quad \text { and } \quad y = x + 1 \quad \text { for } \quad x > 0$$
   \includegraphics[max width=\textwidth, alt={}]{e44987a7-2cdf-442a-aecb-abd3e889ecd4-5_643_791_1160_596}
   The graphs intersect at the point \(P\).
   1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(\mathrm { f } ( x ) = 0\), where f is the function defined in part (a).
   2. Taking \(x _ { 1 } = 1\) as a first approximation to the root of the equation \(\mathrm { f } ( x ) = 0\), use the Newton-Raphson method to find a second approximation \(x _ { 2 }\) to the root.
      (3 marks)
3. The region enclosed by the curve \(y = \frac { 1 } { x ^ { 2 } }\), the line \(x = 1\) and the \(x\)-axis is shaded on the diagram. By evaluating an improper integral, find the area of this region.
   (3 marks)

2 Use the Newton-Raphson method to find the root of the equation \(\mathrm { e } ^ { - x } = x\) which is close to \(x = 0.5\). Give the root correct to 3 decimal places.

9 The equation \(x ^ { 3 } - x ^ { 2 } - 5 x + 10 = 0\) has exactly one real root \(\alpha\).

1. Show that the Newton-Raphson iterative formula for finding this root can be written as $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } - x _ { n } ^ { 2 } - 10 } { 3 x _ { n } ^ { 2 } - 2 x _ { n } - 5 }$$
2. Apply the iterative formula in part (a) with initial value \(x _ { 1 } = - 3\) to find \(x _ { 2 } , x _ { 3 } , x _ { 4 }\) correct to 4 significant figures.
3. Use a change of sign method to show that \(\alpha = - 2.533\) is correct to 4 significant figures.
4. Explain why the Newton-Raphson method with initial value \(x _ { 1 } = - 1\) would not converge to \(\alpha\).

8 The diagram shows a sector of a circle \(O A B\). \(C\) is the midpoint of \(O B\).
Angle \(A O B\) is \(\theta\) radians. \includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-10_700_963_536_534} 8

1. Given that the area of the triangle \(O A C\) is equal to one quarter of the area of the sector \(O A B\), show that \(\theta = 2 \sin \theta\) 8
2. Use the Newton-Raphson method with \(\theta _ { 1 } = \pi\), to find \(\theta _ { 3 }\) as an approximation for \(\theta\). Give your answer correct to five decimal places.
   8
3. Given that \(\theta = 1.89549\) to five decimal places, find an estimate for the percentage error in the approximation found in part (b).
   Turn over for the next question

14 The function f is defined by $$f ( x ) = 3 ^ { x } \sqrt { x } - 1 \quad \text { where } x \geq 0$$ 14

1. \(\quad \mathrm { f } ( x ) = 0\) has a single solution at the point \(x = \alpha\) By considering a suitable change of sign, show that \(\alpha\) lies between 0 and 1
   14
2. (i) Show that $$\mathrm { f } ^ { \prime } ( x ) = \frac { 3 ^ { x } ( 1 + x \ln 9 ) } { 2 \sqrt { x } }$$

   14 (b) (ii) Use the Newton-Raphson method with \(x _ { 1 } = 1\) to find \(x _ { 3 }\), an approximation for \(\alpha\). Give your answer to five decimal places. [2 marks] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) 14 (b) (iii) Explain why the Newton-Raphson method fails to find \(\alpha\) with \(x _ { 1 } = 0\) [2 marks] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)

10 The diagram shows a sector of a circle \(O A B\). \includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-16_758_796_360_623} The point \(C\) lies on \(O B\) such that \(A C\) is perpendicular to \(O B\).
Angle \(A O B\) is \(\theta\) radians.
10

1. Given the area of the triangle \(O A C\) is half the area of the sector \(O A B\), show that $$\theta = \sin 2 \theta$$ 10
2. Use a suitable change of sign to show that a solution to the equation $$\theta = \sin 2 \theta$$ lies in the interval given by \(\theta \in \left[ \frac { \pi } { 5 } , \frac { 2 \pi } { 5 } \right]\)

   10 The Newton-Raphson method is used to find an approximate solution to the equation

   \(\theta = \sin 2 \theta\)

   10 (c) (i) Using \(\theta _ { 1 } = \frac { \pi } { 5 }\) as a first approximation for \(\theta\) apply the Newton-Raphson method twice

   to find the value of \(\theta _ { 3 }\) Give your answer to three decimal places.
   10 (c) (ii) Explain how a more accurate approximation for \(\theta\) can be found using the Newton-Raphson method.
   10 (c) (iii) Explain why using \(\theta _ { 1 } = \frac { \pi } { 6 }\) as a first approximation in the Newton-Raphson method
   [0pt] [2 marks] does not lead to a solution for \(\theta\).

13 The function f is defined by $$\mathrm { f } ( x ) = \arccos x \text { for } 0 \leq x \leq a$$ The curve with equation \(y = \mathrm { f } ( x )\) is shown below. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-18_842_837_550_603} 13

1. State the value of \(a\) 13
2. (i) On the diagram above, sketch the curve with equation $$y = \cos x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$ and
   sketch the line with equation $$y = x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$ 13 (b) (ii) Explain why the solution to the equation $$x - \cos x = 0$$ must also be a solution to the equation $$\cos x = \arccos x$$ Question 13 continues on the next page 13
3. Use the Newton-Raphson method with \(x _ { 0 } = 0\) to find an approximate solution, \(x _ { 3 }\), to the equation $$x - \cos x = 0$$ Give your answer to four decimal places. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-21_2491_1716_219_153}

9

1. A curve passes through the point (5, 12.3) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \left( x ^ { 2 } - 9 \right) ^ { \frac { 1 } { 2 } } + \frac { 2 x y } { x ^ { 2 } - 9 } \quad x > 3$$ Use Euler's step by step method once, and then the midpoint formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right) , \quad x _ { r + 1 } = x _ { r } + h$$ once, each with a step length of 0.1 , to estimate the value of \(y\) when \(x = 5.2\) Give your answer to six significant figures.
   9
2. (i) Find the general solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \left( x ^ { 2 } - 9 \right) ^ { \frac { 1 } { 2 } } + \frac { 2 x y } { x ^ { 2 } - 9 } \quad ( x > 3 )$$ 9 (b) (ii) Given that \(y\) satisfies the differential equation in part (b)(i) and that \(y = 12.3\) when \(x = 5\), find the value of \(y\) when \(x = 5.2\) Give your answer to six significant figures.
   [0pt] [3 marks]
   9
3. Comment on the accuracy of your answer to part (a).

8. \begin{figure}[h] \end{figure} The heart rate of a horse is being monitored.
The heart rate \(H\), measured in beats per minute (bpm), is modelled by the equation $$H = 32 + 40 \mathrm { e } ^ { - 0.2 t } - 20 \mathrm { e } ^ { - 0.9 t }$$ where \(t\) minutes is the time after monitoring began.
Figure 4 is a sketch of \(H\) against \(t\). \section*{Use the equation of the model to answer parts (a) to (e).}

1. State the initial heart rate of the horse. In the long term, the heart rate of the horse approaches \(L \mathrm { bpm }\).
2. State the value of \(L\). The heart rate of the horse reaches its maximum value after \(T\) minutes.
3. Find the value of \(T\), giving your answer to 3 decimal places.
   (Solutions based entirely on calculator technology are not acceptable.) The heart rate of the horse is 37 bpm after \(M\) minutes.
4. Show that \(M\) is a solution of the equation $$t = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t } } \right)$$ Using the iteration formula $$t _ { n + 1 } = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t _ { n } } } \right) \quad \text { with } \quad t _ { 1 } = 10$$
5. 1. find, to 4 decimal places, the value of \(t _ { 2 }\)
   2. find, to 4 decimal places, the value of \(M\)