1.09d Newton-Raphson method

6 A curve has equation \(y = x ^ { 3 } \mathrm { e } ^ { 0.2 x }\) where \(x \geqslant 0\). At the point \(P\) on the curve, the gradient of the curve is 15 .

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \sqrt { \frac { 75 \mathrm { e } ^ { - 0.2 x } } { 15 + x } }\).
2. Use the equation in part (a) to show by calculation that the \(x\)-coordinate of \(P\) lies between 1.7 and 1.8.
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(P\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

5 The equation of a curve is \(y = 6 x \mathrm { e } ^ { \frac { 1 } { 3 } x }\). At the point on the curve with \(x\)-coordinate \(p\), the gradient of the curve is 40 .

1. Show that \(p = 3 \ln \left( \frac { 20 } { p + 3 } \right)\).
2. Show by calculation that \(3.3 < p < 3.5\).
3. Use an iterative formula based on the equation in part (i) to find the value of \(p\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

4 \includegraphics[max width=\textwidth, alt={}, center]{76371b0f-0145-4cc4-a147-27bcd749816a-2_339_1395_1089_374} The diagram shows a semicircle \(A C B\) with centre \(O\) and radius \(r\). The tangent at \(C\) meets \(A B\) produced at \(T\). The angle \(B O C\) is \(x\) radians. The area of the shaded region is equal to the area of the semicircle.

1. Show that \(x\) satisfies the equation $$\tan x = x + \pi$$
2. Use the iterative formula \(x _ { n + 1 } = \tan ^ { - 1 } \left( x _ { n } + \pi \right)\) to determine \(x\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

10

1. It is given that \(2 \tan 2 x + 5 \tan ^ { 2 } x = 0\). Denoting \(\tan x\) by \(t\), form an equation in \(t\) and hence show that either \(t = 0\) or \(t = \sqrt [ 3 ] { } ( t + 0.8 )\).
2. It is given that there is exactly one real value of \(t\) satisfying the equation \(t = \sqrt [ 3 ] { } ( t + 0.8 )\). Verify by calculation that this value lies between 1.2 and 1.3 .
3. Use the iterative formula \(t _ { n + 1 } = \sqrt [ 3 ] { } \left( t _ { n } + 0.8 \right)\) to find the value of \(t\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
4. Using the values of \(t\) found in previous parts of the question, solve the equation $$2 \tan 2 x + 5 \tan ^ { 2 } x = 0$$ for \(- \pi \leqslant x \leqslant \pi\).

6 \includegraphics[max width=\textwidth, alt={}, center]{436d891d-92ee-4076-8369-db756d413979-2_435_597_1516_776} The diagram shows the curves \(y = \mathrm { e } ^ { 2 x - 3 }\) and \(y = 2 \ln x\). When \(x = a\) the tangents to the curves are parallel.

1. Show that \(a\) satisfies the equation \(a = \frac { 1 } { 2 } ( 3 - \ln a )\).
2. Verify by calculation that this equation has a root between 1 and 2 .
3. Use the iterative formula \(a _ { n + 1 } = \frac { 1 } { 2 } \left( 3 - \ln a _ { n } \right)\) to calculate \(a\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.

9 The complex number \(u\) is given by $$u = \frac { 3 + \mathrm { i } } { 2 - \mathrm { i } }$$

1. Express \(u\) in the form \(x + \mathrm { i } y\), where \(x\) and \(y\) are real.
2. Find the modulus and argument of \(u\).
3. Sketch an Argand diagram showing the point representing the complex number \(u\). Show on the same diagram the locus of the point representing the complex number \(z\) such that \(| z - u | = 1\).
4. Using your diagram, calculate the least value of \(| z |\) for points on this locus.

3 The sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 3 x _ { n } } { 4 } + \frac { 15 } { x _ { n } ^ { 3 } }$$ with initial value \(x _ { 1 } = 3\), converges to \(\alpha\).

1. Use this iterative formula to find \(\alpha\) correct to 2 decimal places, giving the result of each iteration to 4 decimal places.
2. State an equation satisfied by \(\alpha\) and hence find the exact value of \(\alpha\).

8 \includegraphics[max width=\textwidth, alt={}, center]{7fe27759-d014-4bc6-8391-342d9df8280e-3_397_750_255_699} The diagram shows the curve \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x ^ { 2 } } \sqrt { } \left( 1 + 2 x ^ { 2 } \right)\) for \(x \geqslant 0\), and its maximum point \(M\).

1. Find the exact value of the \(x\)-coordinate of \(M\).
2. The sequence of values given by the iterative formula $$x _ { n + 1 } = \sqrt { } \left( \ln \left( 4 + 8 x _ { n } ^ { 2 } \right) \right) ,$$ with initial value \(x _ { 1 } = 2\), converges to a certain value \(\alpha\). State an equation satisfied by \(\alpha\) and hence show that \(\alpha\) is the \(x\)-coordinate of a point on the curve where \(y = 0.5\).
3. Use the iterative formula to determine \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

8 \includegraphics[max width=\textwidth, alt={}, center]{346e8866-ca23-4ea6-81bf-bf62502a16d1-3_397_750_255_699} The diagram shows the curve \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x ^ { 2 } } \sqrt { } \left( 1 + 2 x ^ { 2 } \right)\) for \(x \geqslant 0\), and its maximum point \(M\).

1. Find the exact value of the \(x\)-coordinate of \(M\).
2. The sequence of values given by the iterative formula $$x _ { n + 1 } = \sqrt { } \left( \ln \left( 4 + 8 x _ { n } ^ { 2 } \right) \right) ,$$ with initial value \(x _ { 1 } = 2\), converges to a certain value \(\alpha\). State an equation satisfied by \(\alpha\) and hence show that \(\alpha\) is the \(x\)-coordinate of a point on the curve where \(y = 0.5\).
3. Use the iterative formula to determine \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6 \includegraphics[max width=\textwidth, alt={}, center]{6694ccc1-c8b1-42a7-8b21-829a89af74c9-08_732_807_258_667} The diagram shows the curve with equation \(y = \frac { 8 + x ^ { 3 } } { 2 - 5 x }\). The maximum point is denoted by \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and determine the gradient of the curve at the point where the curve crosses the \(x\)-axis.
2. Show that the \(x\)-coordinate of the point \(M\) satisfies the equation \(x = \sqrt { } \left( 0.6 x + 4 x ^ { - 1 } \right)\).
3. Use an iterative formula, based on the equation in part (ii), to find the \(x\)-coordinate of \(M\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

9 \includegraphics[max width=\textwidth, alt={}, center]{3149080d-ad1a-4d2e-8e20-eb9977ced619-14_558_686_260_726} The diagram shows the curves \(y = \cos x\) and \(y = \frac { k } { 1 + x }\), where \(k\) is a constant, for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The curves touch at the point where \(x = p\).

1. Show that \(p\) satisfies the equation \(\tan p = \frac { 1 } { 1 + p }\).
2. Use the iterative formula \(p _ { n + 1 } = \tan ^ { - 1 } \left( \frac { 1 } { 1 + p _ { n } } \right)\) to determine the value of \(p\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
3. Hence find the value of \(k\) correct to 2 decimal places.

6

1. By sketching a suitable pair of graphs, show that the equation \(x ^ { 5 } = 2 + x\) has exactly one real root.
2. Show that if a sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 4 x _ { n } ^ { 5 } + 2 } { 5 x _ { n } ^ { 4 } - 1 }$$ converges, then it converges to the root of the equation in part (a).
3. Use the iterative formula with initial value \(x _ { 1 } = 1.5\) to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. \(7 \quad\) Let \(\mathrm { f } ( x ) = \frac { 2 } { ( 2 x - 1 ) ( 2 x + 1 ) }\).

10 \includegraphics[max width=\textwidth, alt={}, center]{c1fbc9ef-2dc6-43c3-bc58-179f683c9acf-18_471_686_276_717} The curve \(y = x \sqrt { \sin x }\) has one stationary point in the interval \(0 < x < \pi\), where \(x = a\) (see diagram).

1. Show that \(\tan a = - \frac { 1 } { 2 } a\).
2. Verify by calculation that \(a\) lies between 2 and 2.5.
3. Show that if a sequence of values in the interval \(0 < x < \pi\) given by the iterative formula \(x _ { n + 1 } = \pi - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x _ { n } \right)\) converges, then it converges to \(a\), the root of the equation in part (a). [2]
4. Use the iterative formula given in part (c) 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.

11 The equation of a curve is \(y = \sqrt { \tan x }\), for \(0 \leqslant x < \frac { 1 } { 2 } \pi\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\tan x\), and verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) when \(x = \frac { 1 } { 4 } \pi\).
   The value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is also 1 at another point on the curve where \(x = a\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{87be326f-f638-43e9-a654-b7b53d5141ef-18_605_492_1493_822}
2. Show that \(t ^ { 3 } + t ^ { 2 } + 3 t - 1 = 0\), where \(t = \tan a\).
3. Use the iterative formula $$a _ { n + 1 } = \tan ^ { - 1 } \left( \frac { 1 } { 3 } \left( 1 - \tan ^ { 2 } a _ { n } - \tan ^ { 3 } a _ { n } \right) \right)$$ to determine \(a\) correct to 2 decimal places, giving 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. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \sqrt { 3 + 4 \mathrm { e } ^ { x ^ { 2 } } } \quad x \geqslant 0$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer in simplest form. The point \(P\) with \(x\) coordinate \(\alpha\) lies on \(C\).
   Given that the tangent to \(C\) at \(P\) passes through the origin, as shown in Figure 3,
2. show that \(x = \alpha\) is a solution of the equation $$4 x ^ { 2 } e ^ { x ^ { 2 } } - 4 e ^ { x ^ { 2 } } - 3 = 0$$
3. Hence show that \(\alpha\) lies between 1 and 2
4. Show that the equation in part (b) can be written in the form $$x = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x ^ { 2 } } }$$ The iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x _ { n } ^ { 2 } } }$$ with \(x _ { 1 } = 1\) is used to find an approximation for \(\alpha\).
5. Use the iteration formula to find, to 4 decimal places, the value of
   1. \(X _ { 3 }\)
   2. \(\alpha\)

7. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = 2 x ( 1 + x ) \ln x , \quad x > 0$$ The curve has a minimum turning point at \(A\).

1. Find f'(x)
2. Hence show that the \(x\) coordinate of \(A\) is the solution of the equation $$x = \mathrm { e } ^ { - \frac { 1 + x } { 1 + 2 x } }$$
3. Use the iteration formula $$x _ { n + 1 } = \mathrm { e } ^ { - \frac { 1 + x _ { n } } { 1 + 2 x _ { n } } } , \quad x _ { 0 } = 0.46$$ to find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 4 decimal places.
4. Use your answer to part (c) to estimate the coordinates of \(A\) to 2 decimal places.

11. $$y = \left( 2 x ^ { 2 } - 3 \right) \tan \left( \frac { 1 } { 2 } x \right) , \quad 0 < x < \pi$$

1. Find the exact value of \(x\) when \(y = 0\) Given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = \alpha\),
2. show that $$2 \alpha ^ { 2 } - 3 + 4 \alpha \sin \alpha = 0$$ The iterative formula $$x _ { n + 1 } = \frac { 3 } { \left( 2 x _ { n } + 4 \sin x _ { n } \right) }$$ can be used to find an approximation for \(\alpha\).
3. Taking \(x _ { 1 } = 0.7\), find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving each answer to 4 decimal places.
4. By choosing a suitable interval, show that \(\alpha = 0.7283\) to 4 decimal places.
   \includegraphics[max width=\textwidth, alt={}]{29b56d51-120a-4275-a761-8b8aed7bca54-38_2253_50_314_1977}

1. (a) By writing \(\sec x\) as \(\frac { 1 } { \cos x }\), show that \(\frac { \mathrm { d } ( \sec x ) } { \mathrm { d } x } = \sec x \tan x\).

Given that \(y = \mathrm { e } ^ { 2 x } \sec 3 x\),
(b) find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). The curve with equation \(y = \mathrm { e } ^ { 2 x } \sec 3 x , - \frac { \pi } { 6 } < x < \frac { \pi } { 6 }\), has a minimum turning point at \(( a , b )\).
(c) Find the values of the constants \(a\) and \(b\), giving your answers to 3 significant figures.

7. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \left( x ^ { 2 } + 3 x + 1 \right) \mathrm { e } ^ { x ^ { 2 } }$$ The curve cuts the \(x\)-axis at points \(A\) and \(B\) as shown in Figure 2 .

1. Calculate the \(x\) coordinate of \(A\) and the \(x\) coordinate of \(B\), giving your answers to 3 decimal places.
2. Find \(\mathrm { f } ^ { \prime } ( x )\). The curve has a minimum turning point at the point \(P\) as shown in Figure 2.
3. Show that the \(x\) coordinate of \(P\) is the solution of $$x = - \frac { 3 \left( 2 x ^ { 2 } + 1 \right) } { 2 \left( x ^ { 2 } + 2 \right) }$$
4. Use the iteration formula $$x _ { n + 1 } = - \frac { 3 \left( 2 x _ { n } ^ { 2 } + 1 \right) } { 2 \left( x _ { n } ^ { 2 } + 2 \right) } , \quad \text { with } x _ { 0 } = - 2.4$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places. The \(x\) coordinate of \(P\) is \(\alpha\).
5. By choosing a suitable interval, prove that \(\alpha = - 2.43\) to 2 decimal places.

1. $$\mathrm { f } ( x ) = 6 \sqrt { x } - x ^ { 2 } - \frac { 1 } { 2 x } , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 3,4 ]\).
2. Taking 3 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.
   [0pt]
3. Use linear interpolation once on the interval [3,4] to find another approximation to \(\alpha\). Give your answer to 3 decimal places.

2. $$f ( x ) = x ^ { 3 } - 3 x ^ { 2 } + \frac { 1 } { 2 \sqrt { x ^ { 5 } } } + 2 , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 2,3 ]\).
2. Taking 3 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\). Give your answer to 3 decimal places.

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

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [1.6,1.7]
2. Taking 1.6 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\). Give your answer to 3 decimal places.

6. $$f ( x ) = x ^ { 3 } - \frac { 1 } { 2 x } + x ^ { \frac { 3 } { 2 } } , \quad x > 0$$ The root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval [0.6, 0.7].

1. Taking 0.6 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.
2. Show that your answer to part (a) is correct to 3 decimal places.

1. $$f ( x ) = 3 x ^ { 2 } - \frac { 5 } { 3 \sqrt { x } } - 6 , \quad x > 0$$ The single root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval [1.5, 1.6].

1. Taking 1.5 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.
   [0pt]
2. Use linear interpolation once on the interval [1.5, 1.6] to find another approximation to \(\alpha\). Give your answer to 3 decimal places.

4. $$f ( x ) = 1 - \frac { 1 } { 8 x ^ { 4 } } + \frac { 2 } { 7 \sqrt { x ^ { 7 } } } \quad x > 0$$ The equation \(\mathrm { f } ( x ) = 0\) has a single root, \(\alpha\), that lies in the interval \([ 0.15,0.25 ]\)

1. 1. Determine \(\mathrm { f } ^ { \prime } ( x )\)
   2. Explain why 0.25 cannot be used as an initial approximation for \(\alpha\) in the Newton-Raphson process.
   3. Taking 0.15 as a first approximation to \(\alpha\) apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\) Give your answer to 3 decimal places.
2. Use linear interpolation once on the interval \([ 0.15,0.25 ]\) to find another approximation to \(\alpha\) Give your answer to 3 decimal places.