Past Exam Questions

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

(a) Show by calculation that \(\alpha\) lies between 0.5 and 1.

(b) Show that, if a sequence of positive values given by the iterative formula \(x_{n+1} = \frac{1}{3} \left( x_n + 4 \arctan \left( \frac{1}{3x_n} \right) \right)\) converges, then it converges to \(\alpha\).

(c) Use this iterative formula to calculate \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The equation \(x = \frac{10}{e^{2x} - 1}\) 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.

The sequence of values given by the iterative formula

\(x_{n+1} = \frac{x_n(x_n^3 + 100)}{2(x_n^3 + 25)}\),

with initial value \(x_1 = 3.5\), converges to \(\alpha\).

1. Use this formula to calculate \(\alpha\) correct to 4 decimal places, showing the result of each iteration to 6 decimal places.
2. State an equation satisfied by \(\alpha\) and hence find the exact value of \(\alpha\).

The diagram shows the curve \(y = x^4 + 2x^3 + 2x^2 - 4x - 16\), which crosses the x-axis at the points \((\alpha, 0)\) and \((\beta, 0)\) where \(\alpha < \beta\). It is given that \(\alpha\) is an integer.

1. Find the value of \(\alpha\).
2. Show that \(\beta\) satisfies the equation \(x = \sqrt[3]{8 - 2x}\).
3. Use an iteration process based on the equation in part (ii) to find the value of \(\beta\) correct to 2 decimal places. Show the result of each iteration to 4 decimal places.

(i) It is given that \(2 \tan 2x + 5 \tan^2 x = 0\). Denoting \(\tan x\) by \(t\), form an equation in \(t\) and hence show that either \(t = 0\) or \(t = \sqrt[3]{(t + 0.8)}\).

(ii) 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.

(iii) Use the iterative formula \(t_{n+1} = \sqrt[3]{(t_n + 0.8)}\) to find the value of \(t\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

(iv) Using the values of \(t\) found in previous parts of the question, solve the equation \(2 \tan 2x + 5 \tan^2 x = 0\) for \(-\pi \leq x \leq \pi\).

The equation \(x^3 - 8x - 13 = 0\) has one real root.

(i) Find the two consecutive integers between which this root lies.

(ii) Use the iterative formula \(x_{n+1} = (8x_n + 13)^{\frac{1}{3}}\) to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The sequence of values given by the iterative formula \(x_{n+1} = \frac{3x_n}{4} + \frac{15}{x_n^3}\), with initial value \(x_1 = 3\), converges to \(\alpha\).

(i) Use this iterative formula to find \(\alpha\) correct to 2 decimal places, giving the result of each iteration to 4 decimal places.

(ii) State an equation satisfied by \(\alpha\) and hence find the exact value of \(\alpha\).

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

(i) Show by calculation that this root lies between \(x = 1\) and \(x = 2\).

(ii) Prove that, if a sequence of values given by the iterative formula \(x_{n+1} = \frac{2x_n^3 + 2}{3x_n^2 - 2}\) converges, then it converges to this root.

(iii) Use this iterative formula to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The equation \(x^3 - x - 3 = 0\) has one real root, \(\alpha\).

(i) Show that \(\alpha\) lies between 1 and 2.

Two iterative formulae derived from this equation are as follows:

\(x_{n+1} = x_n^3 - 3, \quad (A)\)

\(x_{n+1} = (x_n + 3)^{\frac{1}{3}}, \quad (B)\)

Each formula is used with initial value \(x_1 = 1.5\).

(ii) Show that one of these formulae produces a sequence which fails to converge, and use the other formula to calculate \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

(i) The equation \(x^3 + x + 1 = 0\) has one real root. Show by calculation that this root lies between \(-1\) and \(0\).

(ii) Show that, if a sequence of values given by the iterative formula \(x_{n+1} = \frac{2x_n^3 - 1}{3x_n^2 + 1}\) converges, then it converges to the root of the equation given in part (i).

(iii) 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.

The sequence of values given by the iterative formula \(x_{n+1} = \frac{2}{3} \left( x_n + \frac{1}{x_n^2} \right)\), with initial value \(x_1 = 1\), converges to \(\alpha\).

(i) Use this formula to find \(\alpha\) correct to 2 decimal places, showing the result of each iteration.

(ii) State an equation satisfied by \(\alpha\), and hence find the exact value of \(\alpha\).

The diagram shows the curves \(y = \cos x\) and \(y = \frac{k}{1+x}\), where \(k\) is a constant, for \(0 \leq x \leq \frac{1}{2}\pi\). The curves touch at the point where \(x = p\).

(a) Show that \(p\) satisfies the equation \(\tan p = \frac{1}{1+p}\).

(b) Use the iterative formula \(p_{n+1} = \arctan\left(\frac{1}{1+p_n}\right)\) to determine the value of \(p\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

(c) Hence find the value of \(k\) correct to 2 decimal places.

The diagram shows the curve \(y = x^4 - 2x^3 - 7x - 6\). The curve intersects the \(x\)-axis at the points \((a, 0)\) and \((b, 0)\), where \(a < b\). It is given that \(b\) is an integer.

1. Find the value of \(b\).
2. Hence show that \(a\) satisfies the equation \(a = -\frac{1}{3}(2 + a^2 + a^3)\).
3. Use an iterative formula based on the equation in part (ii) to determine \(a\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

The sequence of values given by the iterative formula

\(x_{n+1} = \frac{2x_n^6 + 12x_n}{3x_n^5 + 8}\),

with initial value \(x_1 = 2\), converges to \(\alpha\).

(i) Use the formula to calculate \(\alpha\) correct to 4 decimal places. Give the result of each iteration to 6 decimal places. [3]

(ii) State an equation satisfied by \(\alpha\) and hence find the exact value of \(\alpha\). [2]

The equation \(x^3 = 3x + 7\) has one real root, denoted by \(\alpha\).

(i) Show by calculation that \(\alpha\) lies between 2 and 3.

Two iterative formulae, \(A\) and \(B\), derived from this equation are as follows:

\(x_{n+1} = (3x_n + 7)^{\frac{1}{3}}\), \quad (A)

\(x_{n+1} = \frac{x_n^3 - 7}{3}\). \quad (B)

Each formula is used with initial value \(x_1 = 2.5\).

(ii) Show that one of these formulae produces a sequence which fails to converge, and use the other formula to calculate \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

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

(i) Show by calculation that \(\alpha\) is greater than 2.5.

(ii) Show that, if a sequence of values in the interval \(0 < x < \pi\) given by the iterative formula \(x_{n+1} = \pi + \arctan \left( \frac{1}{1-x_n} \right)\) converges, then it converges to \(\alpha\).

(iii) Use this iterative formula to determine \(\alpha\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

The diagram shows the curves \(y = x \cos x\) and \(y = \frac{k}{x}\), where \(k\) is a constant, for \(0 < x \leq \frac{1}{2} \pi\). The curves touch at the point where \(x = a\).

1. Show that \(a\) satisfies the equation \(\tan a = \frac{2}{a}\).
2. Use the iterative formula \(a_{n+1} = \arctan \left( \frac{2}{a_n} \right)\) to determine \(a\) 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.

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

1. Verify by calculation that this root lies between 1 and 2. [2]
2. Show that the equation can be rearranged in the form \(x = \sqrt[3]{3x + \frac{4}{x^2} - 1}\). [1]
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]

The equation \(x^3 - x^2 - 6 = 0\) has one real root, denoted by \(\alpha\).

(i) Find by calculation the pair of consecutive integers between which \(\alpha\) lies.

(ii) Show that, if a sequence of values given by the iterative formula \(x_{n+1} = \sqrt{x_n + \frac{6}{x_n}}\) converges, then it converges to \(\alpha\).

(iii) Use this iterative formula to determine \(\alpha\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

The diagram shows the part of the curve \(y = x^2 \cos 3x\) for \(0 \leq x \leq \frac{1}{6}\pi\), and its maximum point \(M\), where \(x = a\).

(a) Show that \(a\) satisfies the equation \(a = \frac{1}{3} \arctan \left( \frac{2}{3a} \right)\).

(b) 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.