1.09a - OCR Spec

15 marks Standard +0.3

The equation $$1 + 5x - x^4 = 0$$ has a positive root $\alpha$.

Show that $\alpha$ lies between 1 and 2. [2]
Use the iterative sequence based on the arrangement $$x = \sqrt[4]{1+5x}$$ with starting value 1.5 to find $\alpha$ correct to two decimal places. [3]
Use the Newton-Raphson method to find $\alpha$ correct to six decimal places. [6]

6 marks Moderate -0.3

The equation $x^3 - 3x + 1 = 0$ has three real roots.

Show that one of the roots lies between $-2$ and $-1$ [2 marks]
Taking $x_1 = -2$ as the first approximation to one of the roots, use the Newton-Raphson method to find $x_2$, the second approximation. [3 marks]
Explain why the Newton-Raphson method fails in the case when the first approximation is $x_1 = -1$ [1 mark]

8 marks Standard +0.3

The curve with equation $y = f(x)$ where $$f(x) = x^2 + \ln(2x^2 - 4x + 5)$$ has a single turning point at $x = \alpha$

Show that $\alpha$ is a solution of the equation $$2x^3 - 4x^2 + 7x - 2 = 0$$ [4]

The iterative formula $$x_{n+1} = \frac{1}{7}(2 + 4x_n^2 - 2x_n^3)$$ is used to find an approximate value for $\alpha$. Starting with $x_1 = 0.3$

calculate, giving each answer to 4 decimal places,
1. the value of $x_2$
2. the value of $x_4$
[2]

Using a suitable interval and a suitable function that should be stated,

6 marks Standard +0.3

You are given that $f(x) = \ln(2x - 5) + 2x^2 - 30$, for $x > 2.5$.

A student takes 4 as the first approximation to $\alpha$. Given $f(4) = 3.099$ and $f'(4) = 16.67$ to 4 significant figures,

apply the Newton-Raphson procedure once to obtain a second approximation for $\alpha$, giving your answer to 3 significant figures. [2]
Show that $\alpha$ is the only root of $f(x) = 0$. [2]

11 marks Challenging +1.2

Fig. 15 shows the graph of $f(x) = 2x + \frac{1}{x} + \ln x - 4$. \includegraphics{figure_11}

Show that the equation $$2x + \frac{1}{x} + \ln x - 4 = 0$$ has a root, $\alpha$, such that $0.1 < \alpha < 0.9$. [2]
Obtain the following Newton-Raphson iteration for the equation in part (i). $$x_{r+1} = x_r - \frac{2x_r^3 + x_r + x_r^2(\ln x_r - 4)}{2x_r^2 - 1 + x_r}$$ [3]
Explain why this iteration fails to find $\alpha$ using each of the following starting values.
1. $x_0 = 0.4$ [2]
2. $x_0 = 0.5$ [2]
3. $x_0 = 0.6$ [2]

9 marks Standard +0.3

The equation $x^3 - x^2 - 5x + 10 = 0$ has exactly one real root $\alpha$.

Show that the Newton-Raphson iterative formula for finding this root can be written as $$x_{n+1} = \frac{2x_n^3 - x_n^2 - 10}{3x_n^2 - 2x_n - 5}.$$ [3]
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. [1]
Use a change of sign method to show that $\alpha = -2.533$ is correct to 4 significant figures. [3]
Explain why the Newton-Raphson method with initial value $x_1 = -1$ would not converge to $\alpha$. [2]

15 marks Challenging +1.3

A curve $C_1$ is defined by the parametric equations $$x = \theta - \sin \theta, \quad y = 1 - \cos \theta,$$ where the parameter $\theta$ is measured in radians.
1. Show that $\frac{dy}{dx} = \cot \frac{1}{2}\theta$, except for certain values of $\theta$, which should be identified. [5]
2. Show that the points of intersection of the curve $C_1$ and the line $y = x$ are determined by an equation of the form $\theta = 1 + A \sin(\theta - \alpha)$, where $A$ and $\alpha$ are constants to be found, such that $A > 0$ and $0 < \alpha < \frac{1}{2}\pi$. [4]
3. Show that the equation found in part (b) has a root between $\frac{1}{4}\pi$ and $\pi$. [2]
A curve $C_2$ is defined by the parametric equations $$x = \theta - \frac{1}{2} \sin \theta, \quad y = 1 - \frac{1}{2} \cos \theta,$$ where the parameter $\theta$ is measured in radians. Find the y-coordinates of all points on $C_2$ for which $\frac{d^2y}{dx^2} = 0$. [4]

11 marks Moderate -0.3

Use a change of sign to verify that the equation $\cos x - x = 0$ has a root $\alpha$ between $x = 0.7$ and $x = 0.8$. [2]
Sketch, on a single diagram, the curve $y = \cos x$ and the line $y = x$ for $0 \leqslant x \leqslant \frac{1}{2}\pi$, giving the coordinates of all points of intersection with the coordinate axes. [2]

An iteration of the form $x_{n+1} = \cos(x_n)$ is to be used to find $\alpha$.

By considering the gradient of $y = \cos x$, show that this iteration will converge. [3]
On a copy of your sketch from part (ii), illustrate how this iteration converges to $\alpha$. [2]
Use a change of sign to verify that $\alpha = 0.7391$ to 4 decimal places. [2]

1.09a Sign change methods: locate roots