OCR AS and A Level Mathematics A (H230, H240) · Web viewMathematics A (H230, H240)End of Stage 1/AS Mathematics: Paper 1: Pure Mathematics and Statistics Please note that you may

Mathematics A (H230, H240)

End of Stage 1/AS Mathematics:Paper 1: Pure Mathematics and Statistics

Please note that you may see slight differences between this paper and the original.

Candidates answer on lined paper.

Other materials required:Scientific or graphical calculator

Duration: 1 hour 30 minutes

INSTRUCTIONS

• Use black ink. HB pencil may be used for graphs and diagrams only.• Answer all the questions.• You are permitted to use a scientific or graphical calculator in this paper.• Final answers should be given to a degree of accuracy appropriate to the context.• The acceleration due to gravity is denoted by g m s-2. Unless otherwise instructed, when a

numerical value is needed, use g = 9.8.

INFORMATION

• The total number of marks for this paper is 75.• The marks for each question are shown in brackets [ ].• You are reminded of the need for clear presentation in your answers.• The total number pages is 12.

© OCR 2019. You may photocopy this page. Page 1 Created using ExamBuilder

Formulae A Level Mathematics A (H240)

Arithmetic series

Geometric series

Binomial series

,

where

Differentiation

Quotient rule ,

Differentiation from first principles

Integration

© OCR 2019. You may photocopy this page. Page 2

Integration by parts

Small angle approximations

where θ is measured in radians


Trigonometric identities

Numerical methods

Trapezium rule: … }, where

The Newton-Raphson iteration for solving :

Probability

or

Standard deviation

or

The binomial distribution

If then , Mean of X is np, Variance of X is np(1 – p)

Hypothesis test for the mean of a normal distribution

If then and

Percentage points of the normal distributionIf Z has a normal distribution with mean 0 and variance 1 then, for each value of p, the table gives the value of z

such that

p 0.75 0.90 0.95 0.975 0.99 0.995 0.9975 0.999 0.9995z 0.674 1.282 1.645 1.960 2.326 2.576 2.807 3.090 3.291


KinematicsMotion in a straight line Motion in two dimensions


Section A: Pure MathematicsAnswer all the questions

1 In this question you must show detailed reasoning.

Express in the form , where p, q and r are integers.[4]

2

The diagram shows triangle ABC, with AC = 8 cm and angle CAB = 30°

(a) Given that the area of the triangle is 20 cm2, find the length of AB[2]

(b) Find the length of BC, giving your answer correct to 3 significant figures. [2]


Find the real roots of the equation .[6]

4 (a) Solve [2]

(b) Solve the inequality giving your answer as the intersection of two sets.[3]



You are given that

(a) Sketch the graph of .[3]

(b) State the roots of .[2]

(c) You are also given that .

(i) Show that .[2]

(ii) Verify that [1]

(iii) Hence factorise completely.[4]


6 The diagram shows a sketch of the curve . The point A on the curve has x-coordinate 4. At point B the curve crosses the x-axis.

(a) (i) Use calculus to determine the equation of the normal to the curve at A.[4]

(ii) Verify that the equation of the normal to the curve at A, found in part (a) (i), intersects the x-axis at C (16, 0).

[1]

(b) Find the area of the region ABC bounded by the curve, the normal at A and the x-axis.[5]


7 A fisherman has collected statistics for the number of rod-caught salmon in England and Wales. He obtained the following results.

End of year 2004 2005 2006 2007 2008 2009 2015

Number of rod-caught salmon 28 193 21 418 18 776 17 556 16 243 14 526 11 261

Taking y as the number of rod-caught salmon and t as the time in completed years from 2003,

the fisherman plotted the graph of against . This is shown in the graph below.

The relationship between and is modelled by the straight line with equation

.

(a) Use the graph to write down the value of c.[1]

(b) Show that [3]

(c) Verify that the model works well for the year 2006.[1]

(d) Use the model to estimate the number of rod-caught salmon in

(i) 2012[1]

(ii) 2025[1]


(e) Comment on the reliability of your answers to part (d)[2]


Section B: StatisticsAnswer all the questions

8 The amounts of electricity, x kW h (kilowatt hours), used by 40 households in a three-month period are summarised as follows.

n =40 ∑x = 59 972 ∑x2 = 96 767 028

(a) Calculate the mean of x.[1]

(b) Calculate the standard deviation of x.[1]

9 Sandra makes repeated, independent attempts to hit a target. On each attempt, the probability that she succeeds is 0.1.

(a) Determine the probability that

(i) the first time she succeeds is on her 5th attempt,[2]

(ii) the first time she succeeds is after her 5th attempt.[2]

Jill also makes repeated attempts to hit the target. Each attempt of either Jill or Sandra is independent. Each time that Jill attempts to hit the target, the probability that she succeeds is 0.2. Sandra and Jill take turns attempting to hit the target, with Sandra going first.

(b) Find the probability that the first person to hit the target is Sandra, on her 2nd attempt.[2]


10 A politics student wishes to investigate the proportion of people who are not in employment.The student decides to compare the local authorities in London with the local authorities in Wales and creates two boxplots as shown below.

London WalesMinimum 0.206294 0.243773Lower quartile 0.240326 0.276145Median 0.254547 0.285550Upper quartile 0.277188 0.307418Maximum 0.313341 0.336867

(a) Comment on the differences between these two sets of data.[2]

(b) Determine whether any of the London authorities are outliers.[3]

The student says the number of unemployed people must be higher in Wales than in London.

(c) Give two reasons why this data does not necessarily show this. You should refer to both the information given and your knowledge of the large data set.

[2]


Wales

London

11 (a) André throws a fair six-sided dice 300 times. The number of throws on which the score is six is denoted by X.

(i) State a suitable model for X, including the values of any parameters.[1]

(ii) Find .[1]

(iii) Find [2]

(b) André has another six-sided dice. He suspects that this dice is biased so that it is less likely to show a six than if it were fair. He throws the dice 300 times and it shows a six on 39 throws. Test at the 5% significance level whether André’s suspicion is justified.

[6]

END OF QUESTION PAPER


BLANK PAGE


BLANK PAGE


…day June 20XX – Morning/AfternoonMathematics A (H230, H240)End of Stage 1/AS Mathematics:Paper 1: Pure Mathematics and Statistics

SAMPLE MARK SCHEME

Duration: 1 hour 30 minutes

MAXIMUM MARK 75

This document consists of 19 pages


Question Answer Marks Guidance1 DR Alternate

-[5x2-10x-2]=-5(x2-2x-2/5)=-5[(x-1)2-1-2/5]=-5(x-1)2+7

M1

A1 ,

M1A1

[4]Examiner's Comments

This “completing the square” question was tackled well by the majority of candidates, many securing all four marks. In keeping with previous sessions, almost all earned the first two marks, finding p and q, although q = (+/-)5 was seen relatively often among weaker candidates. Failure to multiply by (+/-) five when working out the constant was the most common error amongst candidates who did not achieve full marks. Those who took out (+/-)5 as a factor of the full expression often made errors with the resulting fractions. Note that the question has been slightly amended (with a negative x squared term) to reflect increase calculator functionality.


Question Answer Marks Guidance2 (a) M1 Equate correct attempt at area of

triangle to 20

Must be using correct formula,

including Allow if subsequently evaluated in radian mode (gives − 3.95AB = 20)

If using then must be valid use of trig to find h

A1 Obtain 10 Must be exactly 10

[2]2 (b) M1 Attempt to use correct cosine rule,

using their AB

Must be using correct cosine ruleAllow M1 if not square rooted, as long as BC2 soiAllow if subsequently evaluated in radian mode (gives 11.8), but 11.8 by itself cannot imply M1Allow if correct formula seen but is then evaluated incorrectly (using (82 + 102 − 2 × 8 × 10) × cos30 gives 1.86)Allow any equiv method as long as valid use of trig

A1[2]

Examiner's Comments

(a) This was a straightforward start to the paper, and nearly all of the candidates were able to find the correct value for the length. The most common and efficient approach was to use the sine rule, but other methods were also employed. As ever, a few candidates worked with their calculator in radian mode, and persisted with their solution despite it resulting in a negative length. As always, candidates should check the reasonableness of their answer and review their method if necessary.

(b) This part of the question was also very well answered by the majority of candidates, and full marks were very common. The cosine rule was usually quoted correctly, but candidates who are unsure should make use of the formula book (this is no longer the case post reform, candidates must learn cosine rule). Some candidates were unable to correctly evaluate the expression with additional, incorrect, brackets being used or square rooting being omitted.


Question Answer Marks Guidance

3 DR

M1 No marks if whole equation square rooted etc.No marks if straight to formula with no evidence of substitution at start and no square rooting/squaring at end.

M1dep If factorising into two brackets:(4x2 − 1)(x2 + 1) = 0 M1 A1

,

A1 (2x + 1) (2x − 1)(x2 + 1) = 0 M1 A1 A1 as before

oe B1 oe explicit discard of non real roots

M1 Attempt to square root

A1 Final answers correct, no extras


This disguised quadratic was well approached by the vast majority of candidates, with just over half achieving all 5 marks of the original tarif. This continued an improving trend over the last few sessions, with fewer candidates going straight to the quadratic formula with no attempt to square root at the end, which has been a problem with similar questions in the past. The most common approach was to perform a substitution and then to factorise, although those who opted for a two-bracket approach using x2 were also often successful. Fewer candidates than in previous sessions used the quadratic formula; those that did were usually successful. Completing the square was rarely seen. Only a small number of candidates opted to square rather than square root, but the main loss of credit was due to lack of accuracy at the end. Although most candidates correctly dismissed any roots from x2 = −1, some did not. More common was the absence of the negative square root.

The mark allocation has been increased to six to reflect the DR requirement and increased emphasis on justification for the discarding of roots.


Question Answer Marks Guidance4 (a) B1 soi, allow −8x > 1 but not just

8x + 1 < 0

Allow ≤ or ≥ for first mark

B1 Correct working only,

allow

Do not ISW if contradictoryDo not allow ≤ or ≥

[2]4 (b) M1* Expand brackets and rearrange to

collect all terms on one side

No more than one incorrect term

boundary values 0 and 5 seen A1 BC 0, 5 seen as roots –could be on sketch graph

Allow (2x + 0)(x − 5)Do not allow (2x − 4)(x − 3), this is the original expression.

A1 oe.Must be the intersection of two sets.


(a) The negative x coefficient increased the difficulty of this linear inequality so that only two-thirds of candidates secured both marks.

(b) Less than half of candidates provided fully correct solutions to this quadratic inequality. Some failed to expand and rearrange initially and thus earned no credit. Most were able to complete both first stages accurately, but on reaching 2x2 − 10x ≤ 0 many “cancelled” x and thus could get no further. Where both roots were found, choosing the correct region still proved difficult, with some choosing the “outside” and other candidates writing x ≤ 0, x ≤ 5. This question has a reduced tariff to take into account the increased functionality of new calculators, however the final requirement to give answer in set notation has been added.


Question Answer Marks Guidance5 (a) DR

sketch of cubic the right way up, with two tps and clearly crossing the x axis in 3 places

B1 no section to be ruled; no curving back; condone slight ‘flicking out’ at ends but not approaching another turning point; condone some doubling (eg erased curves may continue to show); accept min tp on y-axis or in 3rd or 4th quadrant; curve must clearly extend beyond the x axis at both ‘ends’

B1 Intersection at x-axis must be justified and labelled.

B1 Intersection at y-axis must be justified and labelled.

NB to find −24 some are expanding f(x) here, which gains M1 in iiiA. If this is done, put a yellow line here and by (iii)A to alert you; this image appears again there

[3]5 (b) −2, 0 and 7/2 oe isw or ft their intersections B2 B1 for 2 correct or ft

B1 for (−2, 0) (0, 0) and (3.5, 0)or M1 for (x + 2) x (2x − 7) oeor SC1 for −6, −4 and −1/2 oe

[2]


Question Answer Marks Guidance5 (c) (i) DR

correct expansion of product of 2 brackets of M1 need not be simplified; condone lack of brackets for M1

or allow M1 for expansion of all 3 brackets, showing all terms, with at most one error:

e.g. 2x2 + 5x − 12or 2x2 + x − 6or x2 + 6x + 8

may be seen in (a) – allow the M1; the part (a) work appears at the foot of the image for (ci), so mark this rather than in (a)

correct expansion of quadratic and linear and completion to given answer

A1 for correct completion if all 3 brackets already expandedsome reference to show why −24 changes to −9

condone lack of brackets if they have gone on to expand correctly; condone ‘+15’ appearing at some stageNB answer given; mark the whole process

[2]


Question Answer Marks Guidance5 (c) (ii) DR

B1 allow this mark for (x − 1) shown to be a factor and a statement that this means that x = 1 is a root [of g(x) = 0] oe

B0 for just g(1) = 0

[1]5 (c) (iii) attempt at division by (x − 1) M1 or inspection with at least two terms

of quadratic factor correct

M0 for division by x + 1 after g(1) = 0 unless further working such as g(−1) = 0 shown, but this can go on to gain last M1A1

A1 allow B2 for another linear factor found by the factor theorem

NB mixture of methods may be seen in this part – mark equivalentlyeg three uses of factor theorem, or two uses plus inspection to get last factor;

factorising a correct quadratic factor M1 for factors giving two terms correct;e.g. allow M1 for factorising

after division by x + 1

allow M1 for (x + 1)(x + 18/4) oe after −1 and −18/4 oe correctly found by formula

iswA1 allow 2(x + 9/2)(x + 1)(x − 1) oe;

dependent on 2nd M1 only;condone omission of first factor found; ignore ‘= 0’ seen

SC alternative method for last 4 marks: allow first M1A1 for (2x + 9)(x2 − 1) and then second M1A1 for full factorisation

[4]


Examiner's Comments

(a) Most candidates were able to sketch the correct shape for the cubic (the correct way up) and the majority were also able to correctly label the interceptions on the x-axis, although some gave the positive x intercept as ½ or 2/3 or 3. A few candidates failed to label the y-intercept or gave a wrong value such as 12 or -12. Some candidates drew their graph stopping at one of the roots (usually when x = – 4) instead of crossing the x-axis. Only a small number of candidates drew the graph upside-down and a handful drew the wrong shape altogether. The DR statement has been added to encourage candidates to justify the intersections and not penalise those without graphical calculators.

(b) Quite a few errors were seen here, although a minority knew what to do and wrote down the correct values. Some gave factors or coordinates instead of roots, some solved x – 2 = 0 to give x = 2 as the root, and some went back to the equation but made an algebraic error in replacing x with x − 2, reaching 2x − 5 as a factor instead of 2x − 7. Candidates that correctly typed the new function into their calculator have still demonstrated the same understanding of the translation of the graph.

(c) (i) The first part was generally well done; most correctly expanded two brackets and continued to simplify and add 15 to get the required result. Common errors were: not dealing correctly with the 15 such as saying g(x) = −15 to get the result, and errors in expanding or collecting terms. There was some poor ‘mathematical grammar’ with the ‘+15’ often appearing out of nowhere. The increased emphasis on AO2 in the reform means that this type of ‘carelessness’ is less likely to be condoned.

(c) (ii) In this part most candidates correctly showed g(1) = 0 although some failed to show enough working. Candidates were well-versed, in general, with the techniques of long division or inspection so that most achieved the correct quadratic factor and were able to go on and factorise this to gain full marks. Some tried to use the quadratic formula and then only gave (x + 1)(x + 4.5) oe as factors. Note that there are no specific techniques defined, so alternate ‘modern’ approaches are equally valid provided the working can be followed. Reverse engineering from Cubic solve functions should be discouraged, but checking answers is good practice.


Question Answer Marks Guidance6 (a) (i) At A B1 may be embedded

At A M1 BC

oe

Gradient of normal = M1

oe iswA1

[4]6 (a) (ii) substitution of y = 0 and completion to given

result with at least 1 correct interim step www

B1 or substitution of x = 16 to obtain y = 0

correct interim step may occur before substitution

[1]


Question Answer Marks Guidance6 (b) At B, B1 may be embedded

Clear attempt to divide the area B1 could be seen as sketchfind the area of the triangle [= 18] soi B1

soi M1 BC

oe

Area of region = oe isw

A1 19.3 or better


(a) This was done extremely well, with the majority of even the weakest candidates scoring full marks. A few wrote 2x − 4 = 0 to incorrectly obtain m = 2 and made no further progress, and a very small minority tried to answer the question without using calculus and working backwards. Note that in the reform, there is an expectation that candidates can use the numerical calculus functionality on their calculator, hence the slightly reduced tariff.

(b) Nearly all candidates identified the coordinates of B correctly. However, most — as if by rote — subtracted the equation of the line from the equation of the curve and then integrated. Some candidates integrated the equation of the curve correctly, but used the wrong limits (usually 3 to 16) and made no further progress, and of those that did adopt the correct approach, a large number were unable to find the area of the triangle correctly (½ × 12 × 4 was common). As in part (a), the mark scheme adapted to reflect the expectation that the integration could be found using calculator.


Question Answer Marks Guidance7 (a) B1

[1]7 (b)

seen oeB1 may be awarded after combining

logarithmsM1

so AGA1 Clear argument with intermediate

steps seen since answer given.[3]

7 (c)

18781 is close to 18776

B1

[1]7 (d) (i) awrt 12500 B1

[1]7 (d) (ii) awrt 8990 B1

[1]7 (e) Answer to (i) interpolation so more likely to be

reliableB1

Answer to (ii) extrapolation beyond 2015 so unreliable

B1

[2]Examiner's CommentsThis was a question taken from H630 OCR B (MEI) practice paper H630/02 Q9 so no report


Question Answer Marks Guidance8 (a)

3sf

B1 1499.3

[1]8 (b)

3sf

B1 413.8540926411 standard deviation 415

[1]Examiner's CommentsAlmost all candidates found the mean and most also found the standard deviation correctly, although this answer was often over specified to 419.13. It seems that candidates incorrectly think that the number of decimal places is the crucial thing rather than the number of significant figures.

Note that this was not the complete legacy question; it has be adapted to include only those parts appropriate for the reformed specification.


Question Answer Marks Guidance9 (a) (i) M1

A1 awrt 0.066

[2]9 (a) (ii) M1

allow or M1

but , or M0

1 − (0.1 × +0.9 × 0.1 + 0.92 × 0.1 + …0.94 × 0.n

A1 awrt 0.59

[2]9 (b) M1 allow (i.e. girls in wrong order)

(=0.0576)

NOT0.9 × 0.8 × 0.1 × 0.2 = 0.0144: MOAO0.9 × 0.8 × 0.2 = 0.144: MOAO

A1


(a) (i) Most candidates answered this correctly, although a few gave 0.95 × 0.1.

(a) (ii) This type of question involving “before” or “after” often cause problems. Candidates are confused as to whether a “1 −” is needed. Others think that since it is a geometric situation, “× p” must be included. Also sometimes there is confusion over the power. In fact most candidates answered this question correctly, with a few giving 0.94 or 1 − 0.95 or 0.95 × 0.1. Some used the long method (ie the complement method), but (as usual) a few of these omitted a term or added an extra term.

(b) This question was well answered by most candidates. A few misread and thought Jill went first. Others included success for the wrong girl or for both girls.

Note that this was not the complete legacy question; it has be adapted to include only those parts appropriate for the reformed specification.


Question Answer Marks Guidance10 (a) Compare medians – the proportion of people not

in employment is higher in Wales than LondonB1 [0.286>0.254]

Compare IQR – there is slightly more variation in the proportions of people not in employment in London – comparing the ranges, this difference is more marked

B1 [0.037>0.031]

[0.11>0.09]

10 (b) IQR = 0.037 M1

[>0.31…Max data pt]

[<0.21…Min data pt]

A1

There are no outliers E1 FT calculated values10 (c) E.g. data is proportional, rather than a

comparison of actual population numbersE1 Any two distinct reasons

E.g. Data is ‘not in employment’ rather than ‘unemployed’, so includes for example the retired

E1


This question focused on the LDS and has been written for this paper

See page 18 for extra notes on the data used from the LDS

https://www.ocr.org.uk/Images/308727-units-h230-and-h240-large-data-set-lds-sample-assessment-material.xlsx

.


Question Answer Marks Guidance11 (a) (i) B1

[1]11 (a) (ii) 0.0144… B1 BC

0.01444098482[1]

11 (a) (iii) soi M1 BC 0.9453…

0.0546… A1[2]

11 (b) B1

, where

B1 B1B0 for one error, e.g. undefined p or 2 tail

A1 BC

A1 Allow comparison with 0.025 if

There is evidence at the 5% level to reject M1

The evidence suggests that the dice may be biased against six

A1 In context, not definiteFT their 0.04857…, but not comparison with 0.025

[6]Examiner's CommentsThis was amended from the question taken from H230 OCR B (MEI) practice paper H230/01 Q12 so no report. Different numbers and outcome.


Paper 1: Pure Mathematics and Statistics Paper 2: Pure Mathematics and Mechanics1 Competing the square 4 1 Surds 42 Trigonometry 4 2 Log equations 63 Disguised quadratic 6 3 Exponential model 84 Inequalities 5 4 Trig equations 65 Polynomials 12 5 Gradient function 86 Calculus 10 6 Coordinate geometry 147 Reduction to linear 9 7 Vertical projection 68 Mean and standard deviation 2 8 Kinematics 109 Probability 6 9 Force vectors 8

10 LDS representation 7 10 Connected particles 511 Binomial Distribution 10


Q8 LDS Method of Travel by LA 2011

Region local authority: district / unitary

Not in employment

All usual residents

Proportion not in

employment

London Barking and Dagenham 52,351 185,911 0.281591729London Barnet 87,785 356,386 0.246320001London Bexley 55,755 231,997 0.240326384London Brent 86,265 311,215 0.277187796London Bromley 71,171 309,392 0.230035036London Camden 64,867 220,338 0.294397698London City of London 1,560 7,375 0.211525424London Croydon 90,061 363,378 0.247843843London Ealing 87,944 338,449 0.259844171London Enfield 86,402 312,466 0.276516485London Greenwich 68,901 254,557 0.270670223London Hackney 68,867 246,270 0.279640232

LondonHammersmith and Fulham 45,934 182,493 0.251702805

London Haringey 68,766 254,926 0.269748868London Harrow 60,851 239,056 0.254547052London Havering 58,282 237,232 0.24567512London Hillingdon 69,636 273,936 0.254205362London Hounslow 63,125 253,957 0.248565702London Islington 58,384 206,125 0.283245603London Kensington and Chelsea 44,693 158,649 0.281709938London Kingston upon Thames 37,716 160,060 0.235636636London Lambeth 71,236 303,086 0.2350356London Lewisham 70,198 275,885 0.254446599London Merton 44,776 199,693 0.224224184London Newham 96,504 307,984 0.313340953London Redbridge 74,852 278,970 0.268315589London Richmond upon Thames 38,575 186,990 0.206294454London Southwark 77,163 288,283 0.267664066London Sutton 41,052 190,146 0.215897258London Tower Hamlets 75,679 254,096 0.297836251London Waltham Forest 69,957 258,249 0.270889723London Wandsworth 64,951 306,995 0.211570221London Westminster 64,326 219,396 0.293195865


Region local authority: district / unitary

Not in employmen

t

All usual residents

Proportion not in

employment

Wales Anglesey 20,387 69,751 0.292282548Wales Blaenau Gwent 23,518 69,814 0.336866531Wales Bridgend 41,546 139,178 0.298509822Wales Caerphilly 53,462 178,806 0.298994441Wales Cardiff 100,654 346,090 0.290831865Wales Carmarthenshire 52,250 183,777 0.284311965Wales Ceredigion 24,959 75,922 0.328745291Wales Conwy 31,813 115,228 0.276087409Wales Denbighshire 26,520 93,734 0.282928286Wales Flintshire 38,276 152,506 0.250980289Wales Gwynedd 34,952 121,874 0.286787994Wales Merthyr Tydfil 18,278 58,802 0.310839767Wales Monmouthshire 23,290 91,323 0.255028854Wales Neath Port Talbot 44,006 139,812 0.314751237Wales Newport 40,733 145,736 0.279498545Wales Pembrokeshire 33,832 122,439 0.276317187Wales Powys 32,416 132,976 0.243773312Wales Rhondda, Cynon, Taff 72,720 234,410 0.310225673Wales Swansea 74,178 239,023 0.310338336

WalesThe Vale of Glamorgan 32,838 126,336 0.259925912

Wales Torfaen 25,846 91,075 0.283788087Wales Wrexham 35,077 134,844 0.260130225

London Minimum0.2062

9

Lower Quartile0.2403

3 Outliers <0.1850

3

Median0.2545

5 IQR 0.03686

Upper Quartile0.2771

9 Outliers >0.3324

8

Maximum0.3133

4

Wales Minimum0.24377

Lower Quartile0.27614 Outliers <

0.22924

Median0.2855

5 IQR 0.03127Upper Quartile 0.3074 Outliers > 0.3543


2 3

Maximum0.3368

7


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OCR AS and A Level Mathematics A (H230, H240) · Web viewMathematics A (H230, H240)End of Stage 1/AS Mathematics: Paper 1: Pure Mathematics and Statistics Please note that you may