June 2024 Paper 3 Q9

OCR MEICurrent spec6 marksNumerical Methods

9 This question is about the equation \(\mathrm{f}(x) = 0\), where \(\mathrm{f}(x) = x^4 - x - \dfrac{1}{3x - 2}\).

Fig. 9.1 shows the curve \(y = \mathrm{f}(x)\).

Fig. 9.1: sketch of y = f(x): a branch above the x-axis crossing the y-axis just above O and rising steeply before an asymptote, and a second branch rising from below to cross the positive x-axis — Fig. 9.1

(a) Show, by calculation, that the equation \(\mathrm{f}(x) = 0\) has a root between \(x = 1\) and \(x = 2\). [2]

(b) Fig. 9.2 shows part of a spreadsheet being used to find a root of the equation.

Fig. 9.2

	A	B
1	x	f(x)
2	1.5	3.1625
3	1.25	0.619977679
4	1.125	-0.250466087
5

Write down a suitable number to use as the next value of \(x\) in the spreadsheet. [1]

(d) Fig. 9.3 shows a similar spreadsheet being used to search for another root of \(\mathrm{f}(x) = 0\).

Fig. 9.3

	A	B
1	x	f(x)
2	0	0.5
3	1	-1
4	0.5	1.5625
5	0.75	-4.4336
6	0.6	4.5296
7	0.7	-10.4599
8	0.65	19.5285
9	0.675	-40.4674
10	0.6625	79.5301
11	0.66875	-160.4687
12

(i) Explain why it looks from rows 2 and 3 of the spreadsheet as if there is a root between 0 and 1. [1]

(ii) Explain why this process will not find a root between 0 and 1. [1]

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{f}(1) = -1\)	B1	1.1
\(\mathrm{f}(2) = 13.75\) or \(\dfrac{55}{4}\) so there is a change of sign	B1	1.1
	[2]

Notes

B1: Finding \(\mathrm{f}(1)\) or \(\mathrm{f}(2)\)

B1: Completion to show change of sign with explanation.

Additional guidance

The first B mark is for finding either f(1) or f(2). The second B mark is for finding the other value AND saying there is a sign change.

Mark scheme (b)

Scheme	Marks	AO
Value of \(x\) in the range \(1.125 \lt x \lt 1.25\)	B1	2.2a
	[1]

Notes

B1: Any value in this range. Candidates may give the range.

Additional guidance

Any value between 1.125 and 1.25 will do for this mark. Some candidates are just halving the range and using 1.0625 which scores B0.

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{f}(1.15) = -0.09\) so \(x \approx 1.2\) (cao)	B1	2.2a
	[1]

Notes

B1: Justified by their calculations (which may not necessarily use 1.15).

Additional guidance

They must give a value for, say f(1.15) = -0.09,(or f(1.15625) = -0.0496, f(1.16) = -0.025, f(1.163) = -0.05, f(1.164) = 0.0015), and a final value of 1.2. The question asks for 1dp so only accept 1.2. 1.2 with no other working scores B0. Other values for f are possible.

Mark scheme (d)

(i)

Scheme	Marks	AO
There is a change of sign;	E1	2.4
	[1]

Notes

E1: Incorrect maths (e.g. it implies a y-intercept) B0

Additional guidance

A change is sign is all that is required here.

(ii)

Scheme	Marks	AO
Clear and correct explanation	E1	2.4
	[1]

Notes

E1: E.g.

The function is undefined for \(x = \frac{2}{3}\) [and it looks as if the spreadsheet is homing in on this value]
Accept ‘discontinuous’ or ‘asymptote’ for ‘undefined’
Fig. 9.1 shows that there is only one root.
Could refer to the table (e.g. \(\mathrm{f}(x)\) values diverging)

isw after a correct answer

Additional guidance

We are allowing various comments. See the MS guidance. ISW after a correct answer has been seen.