June 2018 Paper 2 Q5

EdexcelCurrent spec6 marksDifferentiation Numerical Methods

5. The equation \(2x^3 + x^2 - 1 = 0\) has exactly one real root.

(a) Show that, for this equation, the Newton-Raphson formula can be written\[x_{n+1} = \frac{4x_n^3 + x_n^2 + 1}{6x_n^2 + 2x_n}\] (3)

Using the formula given in part (a) with \(x_1 = 1\)

(b) find the values of \(x_2\) and \(x_3\) (2)

Mark scheme (a)

Scheme	Marks	AO
\(\left\{\mathrm{f}(x) = 2x^3 + x^2 - 1 \Rightarrow\right\}\ \mathrm{f}^{\prime}(x) = 6x^2 + 2x\)	B1	1.1b
\(\left\{x_{n+1} = x_n - \dfrac{\mathrm{f}(x_n)}{\mathrm{f}^{\prime}(x_n)} \Rightarrow\right\}\ \{x_{n+1}\} = x_n - \dfrac{2x_n^3 + x_n^2 - 1}{6x_n^2 + 2x_n}\)	M1	1.1b
\(= \dfrac{x_n\left(6x_n^2 + 2x_n\right) - \left(2x_n^3 + x_n^2 - 1\right)}{6x_n^2 + 2x_n} \Rightarrow x_{n+1} = \dfrac{4x_n^3 + x_n^2 + 1}{6x_n^2 + 2x_n}\) *	A1*	2.1
	(3)

Notes

B1: States that \(\mathrm{f}^{\prime}(x) = 6x^2 + 2x\) or states that \(\mathrm{f}^{\prime}(x_n) = 6x_n^2 + 2x_n\) (Condone \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 + 2x\))

M1: Substitutes \(\mathrm{f}(x_n) = 2x_n^3 + x_n^2 - 1\) and their \(\mathrm{f}^{\prime}(x_n)\) into \(x_{n+1} = x_n - \dfrac{\mathrm{f}(x_n)}{\mathrm{f}^{\prime}(x_n)}\)

A1*: A correct intermediate step of making a common denominator which leads to the given answer

Note: Allow B1 if \(\mathrm{f}^{\prime}(x) = 6x^2 + 2x\) is applied as \(\mathrm{f}^{\prime}(x_n)\) (or \(\mathrm{f}^{\prime}(x)\)) in the NR formula \(\{x_{n+1}\} = x_n - \dfrac{\mathrm{f}(x_n)}{\mathrm{f}^{\prime}(x_n)}\)

Note: Allow M1A1 for

\(x_{n+1} = x - \dfrac{2x^3 + x^2 - 1}{6x^2 + 2x} = \dfrac{x\left(6x^2 + 2x\right) - \left(2x^3 + x^2 - 1\right)}{6x^2 + 2x} \Rightarrow x_{n+1} = \dfrac{4x_n^3 + x_n^2 + 1}{6x_n^2 + 2x_n}\)

Note Condone \(x = x - \dfrac{2x^3 + x^2 - 1}{\text{``}6x^2 + 2x\text{''}}\) for M1

Note Condone \(x_n - \dfrac{2x_n^3 + x_n^2 - 1}{\text{``}6x_n^2 + 2x_n\text{''}}\) or \(x - \dfrac{2x^3 + x^2 - 1}{\text{``}6x^2 + 2x\text{''}}\) (i.e. no \(x_{n+1} = \ldots\)) for M1

Note: Give M0 for \(x_{n+1} = x_n - \dfrac{\mathrm{f}(x_n)}{\mathrm{f}^{\prime}(x_n)}\) followed by \(x_{n+1} = 2x_n^3 + x_n^2 - 1 - \dfrac{2x_n^3 + x_n^2 - 1}{6x_n^2 + 2x_n}\)

Note: Correct notation, i.e. \(x_{n+1}\) and \(x_n\) must be seen in their final answer for A1*

Mark scheme (b)

Scheme	Marks	AO
\(\{x_1 = 1 \Rightarrow\}\ x_2 = \dfrac{4(1)^3 + (1)^2 + 1}{6(1)^2 + 2(1)}\) or \(x_2 = 1 - \dfrac{2(1)^3 + (1)^2 - 1}{6(1)^2 + 2(1)}\)	M1	1.1b
\(\Rightarrow x_2 = \dfrac{3}{4},\ x_3 = \dfrac{2}{3}\)	A1	1.1b
	(2)

Notes

M1: An attempt to use the given or their formula once. Can be implied by \(\dfrac{4(1)^3 + (1)^2 + 1}{6(1)^2 + 2(1)}\) or 0.75 o.e.

Note: Allow one slip in substituting \(x_1 = 1\)

A1: \(x_2 = \dfrac{3}{4}\) and \(x_3 = \dfrac{2}{3}\)

Note: Condone \(x_2 = \dfrac{3}{4}\) and \(x_3 =\) awrt 0.667 for A1

Note: Condone \(\dfrac{3}{4}, \dfrac{2}{3}\) listed in a correct order ignoring subscripts

Mark scheme (c)

Scheme	Marks	AO
Accept any reasons why the 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 (or \(y = 2x^3 + x^2 - 1\)) would not meet the \(x\)-axis Tangent to the curve (or \(y = 2x^3 + x^2 - 1\)) is horizontal	B1	2.3
	(1)
	(6 marks)

Notes

B1: See scheme

Note: Give B0 for the following isolated reasons: e.g.

You cannot divide by 0
The fraction (or the NR formula) is undefined at \(x = 0\)
At \(x = 0,\ \mathrm{f}^{\prime}(x_1) = 0\)
\(x_1\) cannot be 0
\(6x^2 + 2x\) cannot be 0
the denominator is 0 which cannot happen
if \(x_1 = 0,\ 6x^2 + 2x = 0\)