1.09a Sign change methods: locate roots

4

1. \includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-05_753_944_278_630} The diagram shows the graph of \(y = 3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x }\).
   On the diagram, sketch the graph of \(y = | 5 x - 4 |\), and show that the equation \(3 - e ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) has exactly two real roots. It is given that the two roots of \(3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) are denoted by \(\alpha\) and \(\beta\), where \(\alpha < \beta\).
2. Show by calculation that \(\alpha\) lies between 0.36 and 0.37 .
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 5 } \left( 7 - \mathrm { e } ^ { - \frac { 1 } { 2 } x _ { n } } \right)\) to find \(\beta\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

5 A curve has equation \(\mathrm { y } = \frac { 1 + \mathrm { e } ^ { 2 \mathrm { x } } } { 1 + 3 \mathrm { x } }\). The curve has exactly one stationary point \(P\).

1. Find \(\frac { \mathrm { dy } } { \mathrm { dx } }\) and hence show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 1 } { 6 } + \frac { 1 } { 2 } \mathrm { e } ^ { - 2 x }\).
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.35 and 0.45 .
3. Use an iterative formula based on the equation in part (a) to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{971a1d8d-a82e-4a3a-b72d-3c147e4f30bb-10_451_647_258_699} The diagram shows the curve with equation \(\mathrm { y } = \sqrt { \sin 2 \mathrm { x } + \sin ^ { 2 } 2 \mathrm { x } }\) for \(0 \leqslant x \leqslant \frac { 1 } { 6 } \pi\). The shaded region is bounded by the curve and the straight lines \(x = \frac { 1 } { 6 } \pi\) and \(y = 0\).

6 \includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-10_417_700_310_685} The diagram shows the curve with equation \(y = \frac { \ln ( 2 x + 1 ) } { x + 3 }\). The curve has a maximum point \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \frac { x + 3 } { \ln ( 2 x + 1 ) } - 0.5\).
3. Show by calculation that the \(x\)-coordinate of \(M\) lies between 2.5 and 3.0 .
4. Use an iterative formula based on the equation in part (b) to find the \(x\)-coordinate of \(M\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

5

1. By sketching a suitable pair of graphs, show that the equation $$\ln x = 2 - x ^ { 2 }$$ has exactly one root.
2. Verify by calculation that the root lies between 1.0 and 1.4 .
3. Use the iterative formula $$x _ { n + 1 } = \sqrt { } \left( 2 - \ln x _ { n } \right)$$ to determine the root correct to 2 decimal places, showing the result of each iteration.

6

1. By sketching a suitable pair of graphs, show that there is only one value of \(x\) that is a root of the equation \(x = 9 \mathrm { e } ^ { - 2 x }\).
2. Verify, by calculation, that this root lies between 1 and 2 .
3. Show that, if a sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left( \ln 9 - \ln x _ { n } \right)$$ converges, then it converges to the root of the equation given in part (i).
4. Use the iterative formula, with \(x _ { 1 } = 1\), to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

7

1. By sketching a suitable pair of graphs, show that the equation $$\mathrm { e } ^ { 2 x } = 2 - x$$ has only one root.
2. Verify by calculation that this root lies between \(x = 0\) and \(x = 0.5\).
3. Show that, if a sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( 2 - x _ { n } \right)$$ converges, then it converges to the root of the equation in part (i).
4. Use this iterative formula, with initial value \(x _ { 1 } = 0.25\), to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
5. By differentiating \(\frac { \cos x } { \sin x }\), show that if \(y = \cot x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
6. By expressing \(\cot ^ { 2 } x\) in terms of \(\operatorname { cosec } ^ { 2 } x\) and using the result of part (i), show that $$\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } \cot ^ { 2 } x \mathrm {~d} x = 1 - \frac { 1 } { 4 } \pi$$
7. Express \(\cos 2 x\) in terms of \(\sin ^ { 2 } x\) and hence show that \(\frac { 1 } { 1 - \cos 2 x }\) can be expressed as \(\frac { 1 } { 2 } \operatorname { cosec } ^ { 2 } x\). Hence, using the result of part (i), find $$\int \frac { 1 } { 1 - \cos 2 x } \mathrm {~d} x$$

7

1. By sketching a suitable pair of graphs, show that the equation $$\mathrm { e } ^ { 2 x } = 14 - x ^ { 2 }$$ has exactly two real roots.
2. Show by calculation that the positive root lies between 1.2 and 1.3.
3. Show that this root also satisfies the equation $$x = \frac { 1 } { 2 } \ln \left( 14 - x ^ { 2 } \right) .$$
4. Use an iteration process based on the equation in part (iii), with a suitable starting value, to find the root correct to 2 decimal places. Give the result of each step of the process to 4 decimal places.
5. Express \(4 \sin \theta - 6 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places.
6. Solve the equation \(4 \sin \theta - 6 \cos \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
7. Find the greatest and least possible values of \(( 4 \sin \theta - 6 \cos \theta ) ^ { 2 } + 8\) as \(\theta\) varies.

6

1. By sketching a suitable pair of graphs, show that the equation $$\cot x = 4 x - 2$$ where \(x\) is in radians, has only one root for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
2. Verify by calculation that this root lies between \(x = 0.7\) and \(x = 0.9\).
3. Show that this root also satisfies the equation $$x = \frac { 1 + 2 \tan x } { 4 \tan x }$$
4. Use the iterative formula \(x _ { n + 1 } = \frac { 1 + 2 \tan x _ { n } } { 4 \tan x _ { n } }\) to determine this 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 $$3 \mathrm { e } ^ { x } = 8 - 2 x$$ has only one root.
2. Verify by calculation that this root lies between \(x = 0.7\) and \(x = 0.8\).
3. Show that this root also satisfies the equation $$x = \ln \left( \frac { 8 - 2 x } { 3 } \right)$$
4. Use the iterative formula \(x _ { n + 1 } = \ln \left( \frac { 8 - 2 x _ { n } } { 3 } \right)\) to determine this root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

4

1. By sketching a suitable pair of graphs, show that the equation $$3 \ln x = 15 - x ^ { 3 }$$ has exactly one real root.
2. Show by calculation that the root lies between 2.0 and 2.5.
3. Use the iterative formula \(x _ { n + 1 } = \sqrt [ 3 ] { } \left( 15 - 3 \ln x _ { n } \right)\) to find the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

7

1. The equation \(x ^ { 3 } + x + 1 = 0\) has one real root. Show by calculation that this root lies between - 1 and 0 .
2. Show that, if a sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } - 1 } { 3 x _ { n } ^ { 2 } + 1 }$$ converges, then it converges to the root of the equation given in part (i).
3. Use this iterative formula, with initial value \(x _ { 1 } = - 0.5\), to determine the root correct to 2 decimal places, showing the result of each iteration.

7

1. By sketching a suitable pair of graphs, show that the equation $$\operatorname { cosec } x = \frac { 1 } { 2 } x + 1$$ where \(x\) is in radians, has a root in the interval \(0 < x < \frac { 1 } { 2 } \pi\).
2. Verify, by calculation, that this root lies between 0.5 and 1 .
3. Show that this root also satisfies the equation $$x = \sin ^ { - 1 } \left( \frac { 2 } { x + 2 } \right)$$
4. Use the iterative formula $$x _ { n + 1 } = \sin ^ { - 1 } \left( \frac { 2 } { x _ { n } + 2 } \right)$$ with initial value \(x _ { 1 } = 0.75\), to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6 \includegraphics[max width=\textwidth, alt={}, center]{8580dddb-cc72-4745-9e0f-1ac641c6506d-2_355_601_1562_772} The diagram shows a sector \(A O B\) of a circle with centre \(O\) and radius \(r\). The angle \(A O B\) is \(\alpha\) radians, where \(0 < \alpha < \pi\). The area of triangle \(A O B\) is half the area of the sector.

1. Show that \(\alpha\) satisfies the equation $$x = 2 \sin x$$
2. Verify by calculation that \(\alpha\) lies between \(\frac { 1 } { 2 } \pi\) and \(\frac { 2 } { 3 } \pi\).
3. Show that, if a sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( x _ { n } + 4 \sin x _ { n } \right)$$ converges, then it converges to a root of the equation in part (i).
4. Use this iterative formula, with initial value \(x _ { 1 } = 1.8\), to find \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

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

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

4 The equation \(x = \frac { 10 } { \mathrm { e } ^ { 2 x } - 1 }\) has one positive real root, denoted by \(\alpha\).

1. Show that \(\alpha\) lies between \(x = 1\) and \(x = 2\).
2. Show that if a sequence of positive values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( 1 + \frac { 10 } { x _ { n } } \right)$$ converges, then it converges to \(\alpha\).
3. Use this iterative formula to determine \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

5 \includegraphics[max width=\textwidth, alt={}, center]{d1377d66-73c8-4d97-9cae-d784b41fb0a8-2_519_800_1359_669} The diagram shows a circle with centre \(O\) and radius \(r\). The tangents to the circle at the points \(A\) and \(B\) meet at \(T\), and the angle \(A O B\) is \(2 x\) radians. The shaded region is bounded by the tangents \(A T\) and \(B T\), and by the minor \(\operatorname { arc } A B\). The perimeter of the shaded region is equal to the circumference of the circle.

1. Show that \(x\) satisfies the equation $$\tan x = \pi - x .$$
2. This equation has one root in the interval \(0 < x < \frac { 1 } { 2 } \pi\). Verify by calculation that this root lies between 1 and 1.3.
3. Use the iterative formula $$x _ { n + 1 } = \tan ^ { - 1 } \left( \pi - x _ { n } \right)$$ to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

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

1. Show by calculation that \(\alpha\) is greater than 2.5.
2. Show that, if a sequence of values in the interval \(0 < x < \pi\) given by the iterative formula \(x _ { n + 1 } = \pi + \tan ^ { - 1 } \left( \frac { 1 } { 1 - x _ { n } } \right)\) converges, then it converges to \(\alpha\).
3. Use this iterative formula to determine \(\alpha\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

3 The equation \(x ^ { 5 } - 3 x ^ { 3 } + x ^ { 2 } - 4 = 0\) has one positive root.

1. Verify by calculation that this root lies between 1 and 2 .
2. Show that the equation can be rearranged in the form $$\left. x = \sqrt [ 3 ] { ( } 3 x + \frac { 4 } { x ^ { 2 } } - 1 \right)$$
3. Use an iterative formula based on this rearrangement to determine the positive root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

3

1. By sketching suitable graphs, show that the equation \(\mathrm { e } ^ { - \frac { 1 } { 2 } x } = 4 - x ^ { 2 }\) has one positive root and one negative root.
2. Verify by calculation that the negative root lies between - 1 and - 1.5 .
3. Use the iterative formula \(x _ { n + 1 } = - \sqrt { } \left( 4 - e ^ { - \frac { 1 } { 2 } x _ { n } } \right)\) to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

7 \includegraphics[max width=\textwidth, alt={}, center]{b89c016e-dc56-48f4-b4c4-b432418e1b28-3_435_672_273_684} The diagram shows a curved rod \(A B\) of length 100 cm which forms an arc of a circle. The end points \(A\) and \(B\) of the rod are 99 cm apart. The circle has radius \(r \mathrm {~cm}\) and the arc \(A B\) subtends an angle of \(2 \alpha\) radians at \(O\), the centre of the circle.

1. Show that \(\alpha\) satisfies the equation \(\frac { 99 } { 100 } x = \sin x\).
2. Given that this equation has exactly one root in the interval \(0 < x < \frac { 1 } { 2 } \pi\), verify by calculation that this root lies between 0.1 and 0.5.
3. Show that if the sequence of values given by the iterative formula $$x _ { n + 1 } = 50 \sin x _ { n } - 48.5 x _ { n }$$ converges, then it converges to a root of the equation in part (i).
4. Use this iterative formula, with initial value \(x _ { 1 } = 0.25\), to find \(\alpha\) correct to 3 decimal places, showing the result of each iteration.

5

1. By sketching suitable graphs, show that the equation $$\sec x = 3 - x ^ { 2 }$$ has exactly one root in the interval \(0 < x < \frac { 1 } { 2 } \pi\).
2. Show that, if a sequence of values given by the iterative formula $$x _ { n + 1 } = \cos ^ { - 1 } \left( \frac { 1 } { 3 - x _ { n } ^ { 2 } } \right)$$ converges, then it converges to a root of the equation given in part (i).
3. Use this iterative formula, with initial value \(x _ { 1 } = 1\), to determine the root in the interval \(0 < x < \frac { 1 } { 2 } \pi\) correct to 2 decimal places, showing the result of each iteration.

5 \includegraphics[max width=\textwidth, alt={}, center]{8c533469-393c-4e4c-a6ec-eab1303741e7-2_385_476_1653_836} The diagram shows a sector \(O A B\) of a circle with centre \(O\) and radius \(r\). The angle \(A O B\) is \(\alpha\) radians, where \(0 < \alpha < \frac { 1 } { 2 } \pi\). The point \(N\) on \(O A\) is such that \(B N\) is perpendicular to \(O A\). The area of the triangle \(O N B\) is half the area of the sector \(O A B\).

1. Show that \(\alpha\) satisfies the equation \(\sin 2 x = x\).
2. By sketching a suitable pair of graphs, show that this equation has exactly one root in the interval \(0 < x < \frac { 1 } { 2 } \pi\).
3. Use the iterative formula $$x _ { n + 1 } = \sin \left( 2 x _ { n } \right)$$ with initial value \(x _ { 1 } = 1\), to find \(\alpha\) correct to 2 decimal places, showing the result of each iteration.

5

1. Show that the equation $$\tan \left( 45 ^ { \circ } + x \right) - \tan x = 2$$ can be written in the form $$\tan ^ { 2 } x + 2 \tan x - 1 = 0$$
2. Hence solve the equation $$\tan \left( 45 ^ { \circ } + x \right) - \tan x = 2$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

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.

4

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