Standard +0.3 This is a straightforward fixed point iteration question requiring algebraic rearrangement (routine manipulation to isolate x), simple iterative calculations with a calculator, and recall of when Newton-Raphson fails (f'(x)=0). All parts are standard textbook exercises with no novel problem-solving required, making it slightly easier than average.
5. \(\mathrm { f } ( x ) = \frac { 1 } { 3 } x ^ { 3 } - 4 x - 2\)
a. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form \(x = \pm \sqrt { a + \frac { b } { x } }\), and state the values of the integers \(a\) and \(b\).
\(\mathrm { f } ( x ) = 0\) has one positive root, \(\alpha\).
The iterative formula \(x _ { n + 1 } = \sqrt { a + \frac { b } { x _ { n } } } , \quad x _ { 0 } = 4\) is used to find an approximation value for \(\alpha\).
b. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\) to 4 decimal places.
c. Explain why for this question, the Newton-Raphson method cannot be used with \(x _ { 1 } = 2\).
Part (c): Explain why Newton-Raphson cannot be used with \(x_1=2\)
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
There is a stationary point at \(x=2\); tangent to the curve is horizontal and would not meet the \(x\)-axis
B1
Any reasonable explanation accepted
(1 mark)
## Question 5:
### Part (a): Show $f(x)=0$ can be written as $x = \pm\sqrt{a + \frac{b}{x}}$, state $a$ and $b$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{3}x^3 - 4x - 2 = 0 \Rightarrow x^3 = 12x+6 \Rightarrow x^2 = 12 + \frac{6}{x} \Rightarrow x = \pm\sqrt{12+\frac{6}{x}}$ | M1 | Equates $f(x)=0$ and proceeds to an equation in $x = \pm\sqrt{\ldots}$ |
| $a = 12,\ b = 6$ | A1 | Correct values of $a$ and $b$ |
**(2 marks)**
---
### Part (b): Calculate $x_1, x_2, x_3, x_4$ to 4 d.p. using $x_{n+1}=\sqrt{12+\frac{6}{x_n}},\ x_0=4$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = \sqrt{12+\frac{6}{4}} = 3.6742$ | M1 | Uses formula $x=\sqrt{a+\frac{b}{x}}$ with $x_0=4$ to find $x_1$ correct to 4 d.p. |
| $x_1=3.6742,\ x_2=3.6923,\ x_3=3.6912,\ x_4=3.6913$ | A1 | All four values correct |
**(2 marks)**
---
### Part (c): Explain why Newton-Raphson cannot be used with $x_1=2$
| Answer/Working | Mark | Guidance |
|---|---|---|
| There is a stationary point at $x=2$; tangent to the curve is horizontal and would not meet the $x$-axis | B1 | Any reasonable explanation accepted |
**(1 mark)**
---
5. $\mathrm { f } ( x ) = \frac { 1 } { 3 } x ^ { 3 } - 4 x - 2$\\
a. Show that the equation $\mathrm { f } ( x ) = 0$ can be written in the form $x = \pm \sqrt { a + \frac { b } { x } }$, and state the values of the integers $a$ and $b$.\\
$\mathrm { f } ( x ) = 0$ has one positive root, $\alpha$.\\
The iterative formula $x _ { n + 1 } = \sqrt { a + \frac { b } { x _ { n } } } , \quad x _ { 0 } = 4$ is used to find an approximation value for $\alpha$.\\
b. Calculate the values of $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$ to 4 decimal places.\\
c. Explain why for this question, the Newton-Raphson method cannot be used with $x _ { 1 } = 2$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q5 [5]}}