June 2024 Paper 1 Q14

AQACurrent spec10 marksLogs & Exponentials Numerical Methods

(a) The equation\[x^3 = \mathrm{e}^{6 - 2x}\]has a single solution, \(x = \alpha\)

By considering a suitable change of sign, show that \(\alpha\) lies between 0 and 4 [2 marks]

(b) Show that the equation \(x^3 = \mathrm{e}^{6 - 2x}\) can be rearranged to give\[x = 3 - \frac{3}{2}\ln x\] [3 marks]

(c)

(i) Use the iterative formula\[x_{n+1} = 3 - \frac{3}{2}\ln x_n\]with \(x_1 = 4\), to find \(x_2\), \(x_3\) and \(x_4\)

Give your answers to three decimal places. [2 marks]

(ii) Figure 1 below shows a sketch of parts of the graphs of\[y = 3 - \frac{3}{2}\ln x \text{ and } y = x\]On Figure 1, draw a staircase or cobweb diagram to show how convergence takes place.

Label, on the \(x\)-axis, the positions of \(x_2\), \(x_3\) and \(x_4\) [2 marks]

Figure 1: graphs of the decreasing curve y = 3 − (3/2) ln x and the line y = x through O, intersecting once, with x = 4 marked on the x-axis — Figure 1

(iii) Explain why the iterative formula\[x_{n+1} = 3 - \frac{3}{2}\ln x_n\]fails to converge to \(\alpha\) when the starting value is \(x_1 = 0\) [1 mark]

Mark scheme (a)

Scheme	Marks	AO
Rearranges the given equation to equal zero and evaluates their non-zero expression in the interval [0,4] at least once. Must have equated to zero.	M1	1.1a
Completes argument with two correct evaluations of their correct expression in the interval [0,4] either side of the solution, with comparison to zero or a comment about change of sign. AND concludes that the solution \(\alpha\) lies between 0 and 4 Evaluations must be correct to at least two significant figures rounded or truncated. Accept exact evaluation at \(x = 0\).	R1	2.1
	(2)

Typical solution

\[x^3 = \mathrm{e}^{6 - 2x} \Rightarrow x^3 - \mathrm{e}^{6 - 2x} = 0\]

Let \(\mathrm{f}(x) = x^3 - \mathrm{e}^{6 - 2x}\)

\[\mathrm{f}(0) = -403 \lt 0\]\[\mathrm{f}(4) = 63.86 \gt 0\]

Hence \(\alpha\) lies between 0 and 4

Mark scheme (b)

Scheme	Marks	AO
Uses natural logs to correctly remove exponential.	M1	1.1a
Uses log rule or cube roots to remove power of 3.	M1	1.1a
Completes reasoned argument to show \(x = 3 - \dfrac{3}{2}\ln x\) AG	R1	2.1
	(3)

Typical solution

\[x^3 = \mathrm{e}^{6 - 2x}\]\[\ln x^3 = 6 - 2x\]\[3\ln x = 6 - 2x\]\[2x = 6 - 3\ln x\]\[x = 3 - \frac{3}{2}\ln x\]

Cobweb diagram starting at x = 4, spiralling inwards between the curve and the line y = x, with x2, x4, x3 and 4 labelled on the x-axis — Mark scheme (c)

Scheme	Marks	AO
(i) Obtains any correct value to at least 3 decimal places, ignoring labels.	M1	1.1a
Obtains \(x_2\), \(x_3\) and \(x_4\) correct to at least 3 decimal places If no labels only accept the three correct answers in the correct order with no extras seen beyond \(x_4\) \(x_2 = 0.92055\ldots\) \(x_3 = 3.12416\ldots\) \(x_4 = 1.29125\ldots\)	A1	1.1b
	(2)
(ii) Draws correct cobweb diagram Condone missing vertical line at \(x = 4\)	M1	1.1a
Shows positions of \(x_2\), \(x_3\) and \(x_4\) on the \(x\)-axis Accept correct values in place of \(x_n\) AWRT 0.92, 3.12 and 1.29 Do not accept labels on \(y = x\) without indication on \(x\)-axis	A1	1.1b
	(2)
(iii) Explains that (it is not possible to evaluate \(x_2\) as) ln 0 has no value or that \(y\) is undefined OE	E1	2.4
	(1)
	(10 marks)