Question 5 - A-Level Maths

Edexcel Paper 2 2018 June — Question 5 6 marks

Exam Board	Edexcel
Module	Paper 2 (Paper 2)
Year	2018
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Newton-Raphson method
Type	Derive Newton-Raphson formula
Difficulty	Moderate -0.5 This is a straightforward application of the Newton-Raphson formula requiring differentiation of a polynomial and algebraic manipulation to reach the given form. Part (b) involves simple substitution, and part (c) tests understanding of when the method fails (division by zero). All steps are routine for A-level Further Maths students with no novel problem-solving required.
Spec	1.09d Newton-Raphson method 1.09e Iterative method failure: convergence conditions

The equation $2 x ^ { 3 } + x ^ { 2 } - 1 = 0$ has exactly one real root.
1. Show that, for this equation, the Newton-Raphson formula can be written
$$x _ { n + 1 } = \frac { 4 x _ { n } ^ { 3 } + x _ { n } ^ { 2 } + 1 } { 6 x _ { n } ^ { 2 } + 2 x _ { n } }$$ Using the formula given in part (a) with $x _ { 1 } = 1$
find the values of $x _ { 2 }$ and $x _ { 3 }$
Explain why, for this question, the Newton-Raphson method cannot be used with $x _ { 1 } = 0$

Show mark scheme Show mark scheme source

Question 5:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\{f(x)=2x^3+x^2-1\Rightarrow\}\ f'(x)=6x^2+2x$	B1	1.1b
$\left\{x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\Rightarrow\right\}\ \{x_{n+1}\}=x_n-\frac{2x_n^3+x_n^2-1}{6x_n^2+2x_n}$	M1	1.1b
$=\frac{x_n(6x_n^2+2x_n)-(2x_n^3+x_n^2-1)}{6x_n^2+2x_n} \Rightarrow x_{n+1}=\frac{4x_n^3+x_n^2+1}{6x_n^2+2x_n}$	A1*	2.1

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\{x_1=1\Rightarrow\}\ x_2=\frac{4(1)^3+(1)^2+1}{6(1)^2+2(1)}$ or $x_2=1-\frac{2(1)^3+(1)^2-1}{6(1)^2+2(1)}$	M1	1.1b
$\Rightarrow x_2=\frac{3}{4}$, $x_3=\frac{2}{3}$	A1	1.1b

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Accept any reasons why Newton-Raphson method cannot be used with $x_1=0$ which refer or allude to either the stationary point or the tangent. E.g.: there is a stationary point at $x=0$; tangent to the curve $y=2x^3+x^2-1$ would not meet the $x$-axis; tangent to the curve is horizontal	B1	2.3

Question 5 (continued):

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Attempt to use formula once, e.g. $\frac{4(1)^3+(1)^2+1}{6(1)^2+2(1)}$ or 0.75	M1	Allow one slip in substituting $x_1=1$
$x_2=\frac{3}{4}$ and $x_3=\frac{2}{3}$	A1	Condone $x_2=\frac{3}{4}$ and $x_3=$ awrt 0.667; condone $\frac{3}{4}, \frac{2}{3}$ listed in correct order ignoring subscripts

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
See scheme	B1	Give B0 for isolated reasons only, e.g. "you cannot divide by 0", "fraction undefined at $x=0$", "at $x=0$, $f'(x_1)=0$", "$x_1$ cannot be 0", "$6x^2+2x$ cannot be 0", "denominator is 0", "if $x_1=0$, $6x^2+2x=0$"

# Question 5:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{f(x)=2x^3+x^2-1\Rightarrow\}\ f'(x)=6x^2+2x$ | B1 | 1.1b |
| $\left\{x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\Rightarrow\right\}\ \{x_{n+1}\}=x_n-\frac{2x_n^3+x_n^2-1}{6x_n^2+2x_n}$ | M1 | 1.1b |
| $=\frac{x_n(6x_n^2+2x_n)-(2x_n^3+x_n^2-1)}{6x_n^2+2x_n} \Rightarrow x_{n+1}=\frac{4x_n^3+x_n^2+1}{6x_n^2+2x_n}$ | A1* | 2.1 |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{x_1=1\Rightarrow\}\ x_2=\frac{4(1)^3+(1)^2+1}{6(1)^2+2(1)}$ or $x_2=1-\frac{2(1)^3+(1)^2-1}{6(1)^2+2(1)}$ | M1 | 1.1b |
| $\Rightarrow x_2=\frac{3}{4}$, $x_3=\frac{2}{3}$ | A1 | 1.1b |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Accept any reasons why Newton-Raphson method cannot be used with $x_1=0$ which refer or allude to either the stationary point or the tangent. E.g.: there is a stationary point at $x=0$; tangent to the curve $y=2x^3+x^2-1$ would not meet the $x$-axis; tangent to the curve is horizontal | B1 | 2.3 |

## Question 5 (continued):

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to use formula once, e.g. $\frac{4(1)^3+(1)^2+1}{6(1)^2+2(1)}$ or 0.75 | M1 | Allow one slip in substituting $x_1=1$ |
| $x_2=\frac{3}{4}$ and $x_3=\frac{2}{3}$ | A1 | Condone $x_2=\frac{3}{4}$ and $x_3=$ awrt 0.667; condone $\frac{3}{4}, \frac{2}{3}$ listed in correct order ignoring subscripts |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| See scheme | B1 | Give B0 for isolated reasons only, e.g. "you cannot divide by 0", "fraction undefined at $x=0$", "at $x=0$, $f'(x_1)=0$", "$x_1$ cannot be 0", "$6x^2+2x$ cannot be 0", "denominator is 0", "if $x_1=0$, $6x^2+2x=0$" |

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Show LaTeX source

\begin{enumerate}
  \item The equation $2 x ^ { 3 } + x ^ { 2 } - 1 = 0$ has exactly one real root.\\
(a) Show that, for this equation, the Newton-Raphson formula can be written
\end{enumerate}

$$x _ { n + 1 } = \frac { 4 x _ { n } ^ { 3 } + x _ { n } ^ { 2 } + 1 } { 6 x _ { n } ^ { 2 } + 2 x _ { n } }$$

Using the formula given in part (a) with $x _ { 1 } = 1$\\
(b) find the values of $x _ { 2 }$ and $x _ { 3 }$\\
(c) Explain why, for this question, the Newton-Raphson method cannot be used with $x _ { 1 } = 0$

\hfill \mbox{\textit{Edexcel Paper 2 2018 Q5 [6]}}

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