OCR AS Level Mathematics A H230/01 Pure Mathematics ... · Web viewOCR Created Date 05/03/2019 05:10:00 Title OCR AS Level Mathematics A H230/01 Pure Mathematics and Statistics Sample

Mathematics A (H230, H240)

End of Stage 1/AS Mathematics:Paper 2: Pure Mathematics and Mechanics

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 8.

© 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 each of the following in the form , where a is an integer

(a) [2]

(b) [2]

2 Solve the equation , giving your answer in the form .[6]

3 The value £V of a car t years after it is new is modelled by the equation , where A and k are positive constants which depend on the make and model of the car.

(a) Brian buys a new sports car. Its value is modelled by the equation . Calculate how much value, to the nearest £100, this car has lost after 1 year.

[2]

At the same time as Brian buys his car, Kate buys a new hatchback for £15 000. Her car loses £2000 of its value in the first year.

(b) (i) Show that, for Kate’s car, correct to 3 significant figures.[3]

In this question you must show detailed reasoning

(ii) Find how long it is before Brian’s and Kate’s cars have the same value.[3]


4 In this question you must show detailed reasoning

(a) Show that the equation can be expressed in the form .

[2]

(b) Hence solve the equation for [4]

5 The sketch shows the curve with equation

(a) Show that [3]

(b) (i) Hence determine the coordinates of the stationary point on the curve.[3]

(ii) Verify that the stationary point is a maximum.[2]


6 The sketch shows the circle with equation

(a) Find the centre and radius of the circle[3]

(b) The circle crosses the positive x-axis at the point A.

(i) Find the equation of the tangent to the circle at A.[6]

(ii) A second tangent to the circle is parallel to the tangent at A. Find the equation of this second tangent.

[3]

(c) Another circle has centre at the origin O and radius r. This circle lies wholly inside the first circle. Find the set of possible values of r.

[2]


Section B: MechanicsAnswer all the questions

7 A particle is projected vertically upwards with speed 7 ms−1 from a point on the ground.

(a) (i) Find the speed of the particle 0.4 s after projection.[2]

(ii) Find the distance above ground of the particle 0.4 s after projection[2]

(b) Find the maximum height of the particle above the ground.[2]

8

The sketch shows the velocity-time graph of an athlete running in a straight line on a horizontal track in a 100 m race. He starts from rest and has constant acceleration until he reaches a speed of 15 ms−1 when t = T. He maintains this constant speed until he decelerates at a constant rate of 1.75 ms−2 for the final 4 s of the race. He completes the race in 10 s. 15 ms−1

(a) Calculate T.[5]

The athlete races against a robot which has a displacement from the starting line of

m, at time t s after the start of the race.

(b) (i) Show that the speed of the robot is 15 ms−1 when t=5.[3]

(ii) Verify that the athlete and the robot reach the finish line simultaneously.[2]


9 The unit vectors i and j shown below are the horizontal and vertically upwards directions

Forces p and q are given, in newtons, by and .

(a) Show that the force p+q is parallel to 8i-j[3]

(b) Show that the force 3p + 10q acts in the horizontal direction.[2]

(c) A particle is in equilibrium under forces kp, 3q and its weight w.

(i) Verify that the value of k must be -4.[1]

(ii) Find the mass of the particle.[2]

10

Particles P and Q, of masses m kg and 0.05 kg respectively, are attached to the ends of a light inextensible string which passes over a smooth pulley. Q is attached to a particle R of mass 0.45 kg by a light inextensible string. The strings are taut, and the portions of the strings not in contact with the pulley are vertical. P is in contact with a horizontal surface when the particles are released from rest. The tension of the string QR is 2.52 N during the descent of R.

(a) Find the acceleration of R during its descent.[2]

(b) Determine the tension in the string PQ during the descent of R.[3]

END OF QUESTION PAPER


…day June 20XX – Morning/AfternoonMathematics A (H230, H240)End of Stage 1/AS Mathematics:Paper 2: Pure Mathematics and Mechanics

SAMPLE MARK SCHEME

Duration: 1 hour 30 minutes

MAXIMUM MARK 75

This document consists of 13 pages


Question Answer Marks Guidance1 (a) M1 or

(not just )

For method mark, makes a correct start to manipulate the expression i.e. at least combines two parts correctly or splits one part correctly

or or or betterA1 Correctly simplified answer[2]

1 (b) or

M1 do not allow without a clear intermediate step. Do not allow for decimals

A1 cao[2]

Examiner's Comments

(a) Most candidates were successful with this easy starter, but a significant minority found it quite challenging. Most earned at least a method mark for correct surd manipulation of some kind, but the accuracy was more of a problem, with

some arithmetic errors and also conceptual ones such as .

(b) This was generally less successful than parts (a), with just under three-quarters of candidates earning the mark. Many of those who did not give the answer in the required form did at least understand the notation as was often

seen, but then simplified to .

The DR statement has been added here to reflect that the majority of modern calculators will automatically simplify surds. Whilst using a calculator to obtain the answer will not gain full credit, students should be encouraged to check their written work by calculator to eliminate careless errors.


Question Answer Marks Guidance2 M1* Introduce logs throughout and drop

power(s)Allow no base or base other than 10 as long as consistent, including log3 on LHS or log2 on RHSDrop single power if log3 or log2 or both powers if any other base

A1 Any correct linear equationie 4x − 1 = (5 − 2x) log23or (4x − 1)log32 = 5 − 2x

Brackets must be seen, or implied by later working Allow no base, or base other than 10 if consistent

M1* Attempt to make x the subject Expand bracket(s) and collect like terms - as far as their 4xlog102 + 2xlog103 = log102 + 5log103Expressions could include log23 or log32Must be working exactly, so M0 if log(s) now decimal equivs

x(4log102 + 2log103) = log102 + 5log103 A1 Obtain a correct equation in which x only appears once

LHS could be x(4log102 + 2log103), xlog10144 or even log10144x

Expressions could include log23 or log32RHS may be two terms or single term

xlog10144 = log10486 M1d* Attempt correct processes to combine logs

Use b log a = log ab, then log a + log b = log ab correctly on at least one side of equation (or log a − log b)Dependent on previous M1 but not the A1 so log10486 will get this M1 irrespective of the LHS

A1 Obtain correct final expression Base 10 required in final answer - allow A1 if no base earlier, or if base 10 omitted at times, but A0 if different base seen previously (unless legitimate working to change base seen)Do not isw subsequent incorrect log work eg x = log 27/log 8

[6]Examiner's Comments

Most candidates were able to gain the first two marks for taking logarithms of both sides and using the power rule, though a number of candidates failed to use brackets. This lack of precision was penalised unless subsequent working clearly showed the correct intention. In order to make further progress candidates had to then expand the brackets and gather like terms which, only the better candidates realised the need to do. Even fewer managed the next step of making x the subject of the equation although some did manage to get a method mark for correctly combining two relevant logarithms. Recent examination sessions have shown candidates becoming more proficient in using logarithms to solve equations when a decimal answer is required, but it appears that algebraic manipulation of logarithms is still a challenge for many. Nevertheless, a pleasing number of fully correct solutions were still seen.


Question Answer Marks Guidance3 (a)

when , B1 (soi) art 16 400

so car loses (£)3 600 B1 condone no £, must be to nearest £100 or B2 for correct answer[2]

3 (b) (i) when , If k = 0.143 verified, e.g.

M1 15000 e−0.143 = 13001[.31 …], SCB1

M1 taking lns correctlyoe e.g. ln 13000 = ln15000 –k [lne]

need not have substituted for V and A

(3sf) A1 cao NB AG must show some working if 4th d.p. not shown

e.g. k = − ln (13000/15000) = 0.143

[3]3 (b) (ii) 15000e−0.143t = 20000e−0.2t M1* must be correct, but could use a more

accurate value for k⇒ (15000/20000) = e(0.143 − 0.2)t M1dep dep* o.e. e.g. ln15000–0.143t = ln 20000 − 0.2t⇒ t = ln 0.75 / − 0.057 = 5.05 years so

after 5 years A1 cao accept answers in the range 5 − 5.1

If M0, SCB1 for 5 − 5.1 years from correct calculations for each car, e.g. t = 5, £7358 (Brian), £7338 (Kate) or (£7334 with more accurate k)


(a) This was a very accessible two marks, provided candidates answered question – the loss in value rounded to the nearest £100.(b)(i) Again, this was very well answered. Occasionally the final ‘A1’ was missed by skipping straight from –k = ln(13/15) to k = 0.143: as this is a given answer, some additional working was required. Occasionally, the result was verified by substituting k = 0.143 and evaluating 15000e-0.143. This was treated as a special case and got 1 mark only. It is important candidates know the difference between ‘show’ and ‘verify’.(b)(ii) This was a little more demanding, requiring candidates to combine the e-0.2 and e-0.143 terms. This defeated quite a few candidates. Some listed the values of each car for t = 1, 2, 3, etc years, and if successful picked out t = 5 as when they were closest. This was judged to be worth 1 mark only.


Question Answer Marks Guidance4 (a) M1

Use correctly once

Must be used clearly at least once – either explicitly of by writing e.g. ‘divide by cos x at side of solution

Allow M1 for any equivalente.g. sin x = cos x tan x

AG A1 Obtain Correct equation in given form, including ‘=0’Correct notation throughout so A0 if tan rather than tan x seen in solution

[2]4 (b) DR

M1 Attempt to solve quadratic in tan x This M mark is for explicit attempt to solve disguised quadratic. Condone any substitution used, inc x = tan x

tan x = 3, tan x = -2

, M1 Attempt to solve tan x = k at least

onceNot dependent on previous mark so M0M1 possible

x = 71.6°, 117°, 252°, 297° A1 Obtain two correct solutions Allow 3sf of betterMust come from a correct method to solve the quadratic (as far as correct factorisation or substitution into formula)Allow radian equivs i.e. 1.25/4.39/2.03/5.18

A1 Obtain all 4 correct solutions, and no others in range

Must now be in degreesAllow 3sf or betterA0 if other incorrect solutionsA0 if solutions outside range


(a) A variety of methods were seen for this proof, some more efficient than others. Most candidates did get there in the end, but full credit was only given if the correct notation had been used throughout. Candidates must also ensure that each step is clearly and convincingly detailed when a ‘proof’ has been requested.

(b) This question was generally very well done, and many candidates gained full marks on this question. The most common error was to completely discount the solution resulting from tan −1(−2) as it resulted in a negative angle rather than appreciating it would still generate other angles within the given range. It was also disappointing to see candidates with a correct method failing to gain full marks due to rounding errors. As in previous questions involving trigonometry, some candidates did not ensure their calculator was in the correct mode before proceeding. Angles given in radians could gain some credit, but candidates did not actually consider which measure they were using so the typical error was tan−1(3) = 1.25 and hence 189.25


Question Answer Marks Guidance5 (a) M2 M1 for just 8x-3 or 1-8x-3

A1

But not just as AG[3]

5 (b) (i) Their M1

A1 A0 if more than one x-value x=-2must have been correctly obtained for all marks after first M1

A1 A0 if more than one y-value[3]

5 (b) (ii)

Substitution of :

M1 Or considering signs of gradients either side of -2 with negative x-values

Condone any bracket error

<0 or = -1.5 oe correctly obtained isw A1 Signs for gradients identified to verify maximum


(a) Most knew what to do here, but 8x−3 and 1 - 8x−3 were often seen. Only a few candidates failed to show sufficient detail of their working to earn the third mark following a fully correct dy/dx

(b) There were many correct solutions, although even some of the better candidates neglected to find the corresponding value of y, or evaluated the second derivative as + 1.5, and concluded the stationary value must be a local maximum. A few obtained x = 2 following correct differentiation, but never bothered to look at the graph to realise that this must be wrong. It was surprising just how many candidates solved 8x−3 = 0 to obtain x = 2, without realising that something must have gone wrong.


Question Answer Marks Guidance6 (a) Centre of circle (4,3) B1

M1and seen

(or implied by correct answer)or

A1 or better www[3]

6 (b) (i) At A, y=0 so M1 Valid method to find A e.g. put y=0 and attempt to solve quadratic (allow slips) or Pythagoras’ theorem

A=(10,0) A1 BC

Gradient of radius

M1 Attempts to find gradient of radius (3 out of 4 terms correct for their centre, their A)

Gradient of tangent = 2 B1M1 Equation of line through their A, any

non-zero gradient

oe A1 Correct answer in any three term form

[6]6 (b) (ii) B1 Finds the opposite end of the diameter

M1 Line through their A' parallel to their line in (ii)

Not through centre of circle

A1 Correct answer in any three-term form[3]


Question Answer Marks Guidance6 (c)

OC= M1Attempts to find the distance from O to their centre and subtract from their radius

ISW incorrect simplification

A1 Correct inequality, condone ≤[2]

Examiner's Comments

(a) This proved to be a very successfully answered question, with around nine in ten candidates securing all three marks.

(b)(i) Just over half of candidates obtained full marks in this part, with errors appearing at all stages. Some put x rather than y equal to 0 when trying to find A and the alternative method of using Pythagoras’ theorem often led to slips. There were a significant number of problems finding the gradient and errors such as -3/6=-1/3 were commonly seen.

(b)(ii) Many candidates did not realise that the point required for the parallel line was the opposite end of the diameter. Most did use the same gradient as in (i), but some used the negative reciprocal. An interesting method sometimes seen was consideration of translation of the original line.

(c) This proved very demanding, with many candidates unable to start; those who drew a diagram were generally more successful but less than a quarter of candidates secured both marks. Even amongst those who found the maximum

length of the radius to be it was quite rare to see the correct inequality.


Question Answer Marks Guidance7 (a) (i) M1 v=7 +/-0.4g

ms-1 A1 Exact, or correct to 3sf from g=9.81 (3.076..) or g=10 (3)

[2]7 (a) (ii) M1 s=7 x 0.4 +/- g0.42/2 or

(their v)2=72-2 x 9.8 x s

m A1 Exact but accept 2.02,g=9.81 (2.0152…) or g=10 (2)

[2]7 (b) M1 oe when v=0

m A1 g=9.81 (2.497…), g=10 (2.45)[2]

Examiner's Comments

(a) Nearly all candidates found both values correctly. Marks were lost either through finding only one quantity, or using an incorrect sign with the value of g.

(b) No examiners report for this amended part.


Question Answer Marks Guidance8 (a) M1 Or

and A1 May be implied

M1 Any attempt at combined 3 stage distances being 100

A1ft Simplification not essential ft cv( , numerical)

s A1[5]

8 (b) (i) M1 Attempt at differentiating

A1Accept

=15 ms-1 AGA1 Must show explicit substitution

[3]8 (b) (ii) M1 Substitute 10 into formula or sets up and solves

equation for robot

Athlete and robot both finish race in 10 s

A1 Needs comment about athlete or both finishing race in 10 s


(a) Most candidates gained full marks. The commonest source of error lay in the calculation of the distance the athlete travelled while decelerating. The setting up of the equation in T was in general done well.

(b)(i) Nearly all candidates obtained full marks.

(b)(ii) Though most candidates followed the route indicated in the question (“Verify...”), some tried to solve the equation for the time taken by the robot to reach the 100 m line. Lacking the knowledge to solve this, they usually lost one, or both, marks.


Question Answer Marks Guidance9 (a) , B1 p+q=28i-3.5j

tan θ=-3.5/28, θ=-7.13°

M1 Or equivalent. k may be implied by going straight to 3.5 8i-jtan θ=-1/8, θ=-7.13°

k=3.5, (so they are parallel) A1 So they are parallel

[3]9 (b)

Therefore no force acting in vertical directionM1A1

oe

[2]9 (c) (i) The horizontal component must be zero

so B1 Substituting and showing i component is zero is acceptable

AG[1]

9 (c) (ii) B1 Award for 24.5 seen

The mass is 2.5 kg B1 Award for 2.5 seen. FT from their weight


This question was about vectors defined using i, j notation. It was well answered with many candidates obtaining full marks. Most of the errors that did occur were in part (c) where candidates were required to interpret the vectors as forces on a particle in equilibrium, with its weight now introduced. Some did not interpret the equilibrium conditions sufficiently rigorously, failing to recognise that both the horizontal and vertical components of the resultant had to be zero. The question ended with a request for the mass of the particle and some candidates confused this with its weight.


Question Answer Marks Guidance10 (a) M1 N2L for R. 2 vertical forces. Accept

+/-0.45a = 0.45g +/– 2.52 ms-2 A1 Accept –4.2

[2]10 (b) M1 N2L for Q, 3 vertical forces, 0.05 ×

4.2 = 0.05g +/− 2.52 +/− T accn not 9.8;

ACCEPT A COMBINED Q AND R METHOD(0.45 + 0.05) × 4.2 = 0.45g + 0.05g +/− T M1(0.45 + 0.05) × 4.2 =0.45g + 0.05g − T A1ftT = 2.8 A1

A1ft ft cv(a). Any equivalent form of equation

N A1[3]

Examiner's Comments

(a) Often this was answered correctly; erroneous methods usually entailed the inclusion of an irrelevant mass or weight. Mis-reading the tension as 2.25 N was also seen.

(b) The motion of the particle Q is determined by three forces. The equation for its motion frequently omitted one of these.


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


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OCR AS Level Mathematics A H230/01 Pure Mathematics ... · Web viewOCR Created Date 05/03/2019 05:10:00 Title OCR AS Level Mathematics A H230/01 Pure Mathematics and Statistics Sample