AS/A Level Mathematics A specimen question papers and ...

Specimen Materials - Mathematics © OCR 20001

OCR ADVANCED SUBSIDIARY GCE

IN MATHEMATICS (3840, 3841, 3842, 3843 and 3844)

OCR ADVANCED GCE

IN MATHEMATICS (7840, 7842 and 7844)

Specimen Question Papers and Mark Schemes

These specimen question papers and mark schemes are intended to accompany the OCR AdvancedSubsidiary GCE and Advanced GCE specifications in Mathematics for teaching from September2000.

Centres are permitted to copy material from this booklet for their own internal use.

The GCE awarding bodies have prepared new specifications to incorporate the range of featuresrequired by new GCE and subject criteria. The specimen assessment material accompanying the newspecifications is provided to give centres a reasonable idea of the general shape and character of theplanned question papers in advance of the first operational examination.

www.theallpapers.com


CONTENTS

Advanced Subsidiary GCE

Unit 2631: Pure Mathematics 1 P1

Question Paper Page 5Mark Scheme Page 9

Unit 2637: Mechanics 1 M1


Unit 2641: Probability and Statistics 1 S1


Unit 2645: Discrete Mathematics 1 D1


A2













A2 continued













Unit 2646: Discrete Mathematics 2 D2

Question Paper Page 125Insert Page 133Mark Scheme Page 137

Marking Instructions Page 143


Specimen Materials © OCR 2000Mathematics Oxford, Cambridge and RSA Examinations

4


This question paper consists of 3 printed pages and 1 blank page.

Specimen Materials - Mathematics 5 © OCR 2000

General Certificate of EducationAdvanced Subsidiary (AS) and Advanced Level

MATHEMATICS P1Pure Mathematics 1

Additional materials:Answer paperGraph paperList of Formulae

TIME 1 hour 20 minutes

INSTRUCTIONS TO CANDIDATES

Write your name, Centre number and candidate number in the spaces provided on the answer paper.Answer all the questions.

Give non-exact numerical answers correct to 3 significant figures, unless a different degree ofaccuracy is specified in the question or is clearly appropriate.You are permitted to use only a scientific calculator in this paper.

INFORMATION FOR CANDIDATES

The number of marks is given in brackets [ ] at the end of each question or part question.

The total number of marks for this paper is 60.

Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carryinglarger numbers of marks later in the paper.You are reminded of the need for clear presentation in your answers.



1 (i) Write down the exact value of 27− . [1]

(ii) Simplify 4

3

2)(

xxx . [2]

2 (i) Solve the simultaneous equations

.73 ,232 −=+−= xyxxy [4]

(ii) Interpret your solution to part (i) geometrically. [1]

3 The point A has coordinates )4 ,7( . The straight lines with equations 013 =++ yx and 052 =+ yxintersect at the point B. Show that one of these two lines is perpendicular to AB. [6]

4 Show that the equation

θθ sin13cos15 2 +=

may be written as a quadratic equation in θsin . [2]

Hence solve the equation, giving all values of θ such that °≤≤° 3600 θ . [5]

5 Sketch the graph of °= xy cos , for values of x from 0 to 360. [1]

Sketch, on the same diagram, the graph of °−= )60cos(xy . [2]

Use your diagram to solve the equation

°−=° )60cos(cos xx

for values of x between 0 and 360. Indicate clearly on your diagram how the solutions relate to the graphs.[3]

State how many values of x satisfying the equation

°−=° )6010cos()10cos( xx

lie between 0 and 360. (You should explain your reasoning briefly, but no further detailed working orsketching is necessary.) [2]



6 (i) Evaluate ∫ −4

0d )4( xxx . [3]

(ii)

The diagram shows the curve )4( xxy −= , together with a straight line. This line cuts the curve atthe origin O and at the point P with x-coordinate k , where 40 << k .

(a) Show that the area of the shaded region, bounded by the line and the curve, is 361 k . [4]

(b) Find, correct to 3 decimal places, the value of k for which the area of the shaded region is half ofthe total area under the curve between 0=x and 4=x . [2]

7 A quadratic function is defined by

9)f( 2 ++= kxxx ,

where k is a constant. It is given that the equation 0)f( =x has two distinct real roots. Find the set ofvalues that k can take. [3]

For the case where 34−=k ,

(i) express )f(x in the form bax ++ 2)( , stating the values of a and b, and hence write down the leastvalue taken by )f(x , [4]

(ii) solve the equation 0)f( =x , expressing your answer in terms of surds, simplified as far as possible. [3]

8 The equation of a curve is 326 xxy −= . Find the coordinates of the two stationary points on the curve,and determine the nature of each of these stationary points. [6]

State the set of values of x for which 326 xx − is a decreasing function of x. [2]

The gradient at the point M on the curve is 12. Find the equation of the tangent to the curve at M. [4]



BLANK PAGE


This mark scheme consists of 4 printed pages.




MARK SCHEME

MAXIMUM MARK 60

For live examinations, each Mark Schemeincludes the General Instructions for Markingset out on pages 143 to 145.



1 (i) 491 B1 1 Correct value stated as final answer

--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) 4

4

4

3

22)( 2

1

xx

xxx

= M1 Power 2113× or 2

113 + in numerator

xx 21

21 or 2

1

A1 2 Or any equally simple equivalent

2 (i) EITHER : 2373 2 +−=− xxx M1 Eliminate y to obtain an equation in x only0962 =+− xx A1 Correct 3-term equation in x

3=x only A1 Obtained by any correct solution method2=y only A1 If two values of x are found both y-values

must follow correctly

OR : 23

733

7 2

+

+−

+= yyy M1 Eliminate x to obtain an equation in y only

0442 =+− yy A1 Correct 3-term equation in y2=y only A1 Obtained by any correct solution method3=x only A1 4 If two values of y are found both x-values

must follow correctly--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) The line 73 −= xy is the tangent to the curve

232 +−= xxy at the point )2 ,3( B1 1 For identifying tangency

3 Solve 013 =++ yx and 052 =+ yx simultaneously M1 Attempt soln and obtain at least one answer2 ,5 −== yx at B A1 Identify correct coordinates with B, either

explicitly or implicitly

Gradient of AB is 357

)2(4=

−−−

A1 For simplified follow-through value

Gradients of the lines are 31− and 5

2− B1 For either gradient correctly stated or used

Perpendicular lines require 121 −=mm M1 Any statement or use of the correct relation

AB is perpendicular to 013 =++ yx A1 6 Correct use of 13 31 −=−× , or equivalent

4 θθ sin13)sin1(15 2 +=− M1 Attempted relevant use of 1cossin 22 =+ θθ

02sinsin15 2 =−+ θθ A1 2 Any correct 3-term form--------------------------------------------------------------------------------------------------------------------------------------------------------

0)1sin3)(2sin5( =−+ θθ M1 Any recognisable solution method attempted

31

52 or sin −=θ A1 Both correct values

°°°°= 4.336 ,6.203 ,5.160 ,5.19θ A1 For any one correct valueA1 For a second correct valueA1 5 For both remaining values, and no others



5B1 1 Correct xy cos= over )360 ,0( ; ignore

anything outside this interval---------------------------------------------------------------------------M1 Translation parallel to the x-axis recognisedA1 2 For correct )60cos( −= xy throughout

)360 ,0( ; ignore anything outside this

interval--------------------------------------------------------------------------------------------------------------------------------------------------------Indicate use of points of intersection on diagram B1 For identifying the points, not necessarily

the x-coordinates210 ,30=x B1 For either correct value

B1 3 For second correct value and no others--------------------------------------------------------------------------------------------------------------------------------------------------------Graphs are ‘squashed’ 10× in the x-direction M1 Or any equivalent methodThere are 20 solutions A1 2 ×10 their number of solutions above

6 (i) Expand to 24 xx − and integrate M1 At least one integrated term with correctpower

[ ]4

03

3122 xx − A1 Correct indefinite integral

332

36432 =− A1 3 Follow correct use of limits 4 and 0

--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) (a) 3312

02d )4( kkxxx

k−=−∫ B1 Follow their earlier indefinite integral

)4( triangleof Area 21 kkk −××= M1 To include attempt at y-coordinate of P in

terms of k3

2122 kk −= A1 Correct simplified expression

3613

2123

312 )2(2area Shaded kkkkk =−−−= A1 4 Given answer correctly obtained

--------------------------------------------------------------------------------------------------------------------------------------------------------(b) 3

32213

61 ×=k M1 Using previous results correctly to form an

equation for k175.3=k A1 2 Correct 3dp value

7 9142 ××>k B1 Correct condition stated in any form0)6)(6( >+− kk M1 Factorise or carry out other solution method

6 or 6 ,6 >>−< kkk A1 3 Do not allow 66 −<< k--------------------------------------------------------------------------------------------------------------------------------------------------------(i) EITHER : 22 )32(9)32( −+−x M1 May be implied by correct a and/or b

32−=a A13−=b A1

Least value of )f(x is 3− B1 Their value of b

OR : 9 and 342 2 =+−= baa M1 Expand and equate at least one pair of coeffs

32−=a A13−=b A1

Least value of )f(x is 3− B1 4 Their value of b--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Critical values where 332 ±=−x M1 Or equivalent, via quadratic formula

33or 3=x A1 For either critical value correctly obtained

A1 3 For completely correct answer; allow 27



8 2312dd

xxxy

−= B1 Correct derivative stated or used

0)4(3 =− xx M1 Equate to zero and factorise, or equivalent

4 and 0=x A1Stationary points are )0 ,0( and )32 ,4( A1

=−=+

=−=4when 120when 12

612dd

2

2

xx

xx

yM1 Or other correct alternative method

)0 ,0( is a minimum and )32 ,4( is a maximum A1 6 Correct conclusion from correct working--------------------------------------------------------------------------------------------------------------------------------------------------------

4 ,0 >< xx B1 Either interval statedB1 2 Both intervals, and no others, correct

--------------------------------------------------------------------------------------------------------------------------------------------------------

12312 2 =− xx M1 Equate xy

dd

to 12 and solve for x

2=x A1Tangent passes through )16 ,2( with gradient 12 M1 Use of straight line equation with numerical

gradient and numerical coordinates of M812 −= xy A1 4 Any correct equivalent 3-term form


This question paper consists of 4 printed pages.








Give non-exact numerical answers correct to 3 significant figures, unless a different degree ofaccuracy is specified in the question or is clearly appropriate.

You are permitted to use a graphic calculator in this paper.







1 The cubic polynomial 81873 23 −−− xxx is denoted by )f(x . Use the factor theorem to show that )1( +xis a factor of )f(x . [2]

Hence factorise )f(x completely. [3]

2 Solve the inequality 10100 <−x . [2]

Hence find the set of integers n that satisfy the inequality

1010001.1 <−n . [3]

3 Expand 5)1( x+ in ascending powers of x, simplifying the coefficients. [2]

Hence, by letting 2yyx += , find the coefficient of 4y in the expansion of 52)1( yy ++ in powers of y. [4]

4

The diagram shows the region R which is bounded by the curve 1

2+

=x

y , the x-axis, and the lines 1=x

and 5=x . Use integration

(i) to find the area of R, giving your answer as a single logarithm, [4]

(ii) to show that the volume of the solid formed when R is rotated completely about the x-axis is π34 . [4]

5 At time t minutes after an oven is switched on, its temperature C °θ is given by

t1.0e180200 −−=θ .

(i) State the value which the oven’s temperature approaches after a long time. [1]

(ii) Find the time taken for the oven’s temperature to reach C 150 ° . [3]

(iii) Find the rate at which the temperature is increasing at the instant when the temperature reachesC 150 ° . [4]



6

The diagram shows a semicircle ABC on AC as diameter. The mid-point of AC is O, and angleradians θ=AOB , where πθ 2

10 << . The area of the segment 1S bounded by the chord BC is twice the

area of the segment 2S bounded by the chord AB. Show that

θπθ sin3 += . [4]

Use an iterative method, based on the rearrangement

),sin(31 θπθ +=

together with a suitable starting value, to find θ correct to 3 decimal places. You should show the value ofeach approximation that you calculate. [4]

7 The functions f and g are defined by

. : g,01 : f

2

21

R∈≥+

xxxxxx

aa

(i) Find the domain of the inverse function 1f − . [2]

(ii) Find an expression for )(f 1 x− . [2]

(iii) Find and simplify an expression for )fg(x for the case where 0≥x . [2]

(iv) Explain clearly why the value of )2 [1]fg(− is 3. [1]

(v) Sketch the graph of )fg(xy = , for both positive and negative values of x, and give the equation of thisgraph in a simplified form. [3]



8 (i) A sequence of positive integers K , , , 321 uuu is given by

21 =u and 1for 21 ≥=+ nuu nn .

(a) Write down the first four terms of this sequence. [1]

(b) State what type of sequence this is, and express nu in terms of n. [2]

(ii) A sequence of positive integers K , , , 321 vvv is given by

31 =v and 1for 121 ≥−=+ nvv nn .

(a) Show that the relation between 1+nv and nv may be written in the form

)1(211 −=−+ nn vv . [1]

(b) Hence, by using the results in part (i), show that 12 += nnv for 1≥n . [2]

(iii) The sum of the first N terms of the sequence K , , , 321 vvv is denoted by NS , i.e.

NN vvvvS ++++= K321 .

Express NS in terms of N. [4]


This mark scheme consists of 3 printed pages and 1 blank page.




MARK SCHEME

MAXIMUM MARK 60




1 081873)1f( =−+−−=− M1 Substitute 1−=x and evaluate

Hence )1( +x is a factor A1 2 Zero correctly obtained, and conclusion--------------------------------------------------------------------------------------------------------------------------------------------------------

)8103)(1()f( 2 −−+= xxxx M1 Carry out division (trinomial quotient) or

attempt factorisation by inspection (3 termswith at least 23x and 8± )

A1 Correct quadratic factorAnswer )4)(23)(1( −++ xxx A1 3

2 11090 << x B1 Either end-point obtainedB1 2 Completely correct solution set

--------------------------------------------------------------------------------------------------------------------------------------------------------110ln01.1ln90ln << n M1 Correct use of logs in any equation or

inequality of the form cn =01.14.4722.452 << n A1 Either value, in exact or decimal form

453 to 472 inclusive A1 3 Allow any clear notation

3 543255

45

35

25

151 xxxxx

+

+

+

+

+ M1

5432 5101051 xxxxx +++++ A1 2--------------------------------------------------------------------------------------------------------------------------------------------------------

K+++++++ 32222 )(10)(10)(51 yyyyyy M1 May be implied4y occurs in 22 )( yy + , 32 )( yy + and 42 )( yy + M1 Expand or pick out the relevant terms in at

least two of these cases444 53010 yyy ++ A1 At least two of the three terms correct

Coefficient of 4y is 45 A1 4 Allow answer 445y

4 (i)5

1

5

1)1ln(2d

12

+=⌡

⌠+

xxx

M1 For indefinite integral involving a log

A1 Correct indefinite integral )1ln(2 +x9ln2ln26ln2 =− M1 Use of both limits and at least one law of logs

A1 4 Correct simplified answer 9ln--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) ∫ xyV d 2π= B1 Correct formula stated or used

[ ]51

1)1(4 −+− xπ M1 Integration attempt with negative index result

A1 Correct indefinite integral 1)1(4 −+− x

ππ 34

64 )2( =+− A1 4 Given answer correctly shown

5 (i) C 200 ° B1 1 Allow answer 200 without units--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 18

51.01.0 ee180200150 =⇒−= −− tt M1 Substitute 150=θ and rearrange

( )185ln1.0 =− t M1 Take logs correctly

Answer 12.8 minutes A1 3 Allow answer 8.12=t without units--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) Rate of increase of temperature is td

dθB1 Recognition may be implied

tt

1.0e18dd −=

θM1 Attempt at

tddθ

involving a term tk 1.0e−

A1 Correct differentiationAnswer C 5 ° per minute A1 4 Follow value of t or t1.0e− from (ii)



6 Area of 2S is )sin(221 θθ −r B1

Area of 1S is ))sin((221 θπθπ −−−r B1

Hence )sin(2)sin( 2212

21 θθθθπ −×=−− rr M1 Equate and attempt to simplify )sin( θπ −

i.e. θπθ sin3 += A1 4 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------A suitable initial value is 1=θ (e.g.) B1 State or use any 1θ such that πθ 2

110 ≤≤

...327.1)sin( 131

2 =+= θπθ (e.g.) M1 For one iteration using correct formula

A1 For correct 2θ from their 1θ374.1≈θ A1 4 For correct answer and sufficient iterations

to justify 3dp accuracy

7 (i) Domain of 1f − is the range of f M1 May be impliedi.e. 1≥x A1 2 Allow any intelligible notation

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 2)1(1 −=⇒+= yxxy M1 Attempt to solve for x

21 )1()(f −=− xx A1 2 For answer 12or )1( 22 +−− xxx ; do not

allow final answer in terms of y--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) 21

)(1)fg( 2xx += M1 Attempt composition in the correct orderx+= 1 A1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------(iv) 3)4f( ,4)2g( ==− B1 1 Intermediate value 4 must be seen--------------------------------------------------------------------------------------------------------------------------------------------------------(v)

B1 Sketch of xy += 1 for 0≥x onlyB1 Sketch reflection in the y-axis for 0≤x only

xy 1+= B1 3

8 (i) (a) 2, 4, 8, 16 B1 1 All four terms correctly stated--------------------------------------------------------------------------------------------------------------------------------------------------------

(b) This is a geometric progression B1 No need to state first term or ratio hereHence nn

nu 222 1 =×= − B1 2 For simplified answer n2--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) (a) )1(222112 11 −=−=−⇒−= ++ nnnnn vvvvv B1 1 Given result correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------

(b) 1−nv satisfies same relation as nu M1 For making the connection to part (i)

Hence 121 +=+= nnn uv A1 2 Given result correctly shown; no need for an

explicit check that 123 +=--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) )1()2( Σ+Σ=Σ nnv M1 For considering the two separate sums (use

of Σ notation is not expected in this module)

NN

+−

−=12

)12(2B1 For correct unsimplified GP sum

B1 For correct statement of N=Σ )1(

NN +−= )12(2 or 22 1 −++ NN A1 4


Specimen Materials – M athematics 20 © OCR 2000

BLANK PAGE










Give non-exact numerical answers correct to 3 significant figures, unless a different degree ofaccuracy is specified in the question or is clearly appropriate.You are permitted to use only a scientific calculator in this paper.







1 Expand 21

)21( −− x in ascending powers of x, up to and including the term in 2x . [3]

State the set of values of x for which the expansion is valid. [1]

2 Prove the identity

xxx cos2)30cos(3)30sin( ≡°++°+ ,

where x is measured in degrees. [3]

Hence express °15cos in surd form. [2]

3 By using the substitution xu sin= , or otherwise, find

xxx d 2sinsin 3⌡⌠ ,

giving your answer in terms of x. [6]

4 Find the centre and radius of the circle with equation xyx 622 =+ . [2]

The line kyx =+ is a tangent to this circle. Find the two possible values of the constant k , giving youranswers in surd form. [5]

5 The points A and B have coordinates )4 ,2 ,3( and )3 ,4 ,4( − respectively. The line 1l , which passesthrough A, has equation

+

=

115

423

tr .

Show that AB is perpendicular to 1l . [3]

The line 2l , which passes through B, has equation

−+

−=

212

344

sr .

Show that the lines 1l and 2l intersect, and find the coordinates of their point of intersection. [5]



6 The parametric equations of a curve are

θθθ cos ,sin ayax == ,

where a is a positive constant and πθ 210 << . Find

xy

dd in terms of θ , and hence show that the gradient

of the curve is zero where θ

θ1

tan = . [6]

By sketching a suitable pair of graphs, show that the equation θ

θ1

tan = is satisfied by just one value of θ

in the relevant range. [2]

Determine, with reasons, whether this value of θ is greater or less than π41 . [2]

7 Express )4()1(

413152

2

xxxx

−−+−

in partial fractions. [5]

Hence show that

4ln1d )4()1(

41315 2

23

2

+=−−

+−⌡

⌠x

xxxx . [5]

8

A cylindrical container has a height of 200 cm. The container was initially full of a chemical but there is aleak from a hole in the base. When the leak is noticed, the container is half-full and the level of thechemical is dropping at a rate of 1 cm per minute. It is required to find for how many minutes thecontainer has been leaking. To model the situation it is assumed that, when the depth of the chemicalremaining is x cm, the rate at which the level is dropping is proportional to x . Set up and solve anappropriate differential equation, and hence show that the container has been leaking for about 80 minutes.

[10]



BLANK PAGE






MARK SCHEME

MAXIMUM MARK 60




1 K+−−−

+−−+ 223

21

21 )2(

2))((

)2)((1 xx M1 Either x or 2x term correct (unsimplified)

2231 xx ++ A1 For x+1 correct

A1 3 For 223 x+ correct

--------------------------------------------------------------------------------------------------------------------------------------------------------

21

21

21 or <<−< xx B1 1

2 °−°=°+°+°=°+

30sinsin30coscos)30cos(30sincos30cossin)30sin(

xxxxxx

B1 Both expansions correct

)sincos3(3cossin3 21

21

21

21 xxxx −++ B1 Both exact values substituted throughout

xxx cos2coscos 23

21 =+ B1 3 Given result shown correctly--------------------------------------------------------------------------------------------------------------------------------------------------------

°=°+° 15cos245cos345sin M1 Substitute °= 15x and use exact values for°° 45cos and 45sin

2231

15cos+

=° A1 2 Allow any equivalent surd form

3 xxu

cosdd

= B1 Or equivalent; may be implied

xxx cossin22sin = B1 Stated or used

∫∫∫ == uuuuuxxx d 2d 2 d 2sinsin 433 M1 Substituting for x and dx throughout

A1 Correct simplified integral in terms of u

cxcu +=+ 5525

52 sin M1 Integrate and substitute back

A1 6 Correct answer in terms of x (including c+ )

4 EITHER : 222 3)3( =+− yx M1 Complete the square to obtain standard form

Centre is )0 ,3( and radius is 3 A1 Both correct

OR : Centre ) ,( fg −− is )0 ,3( B1 Formula may be implied

Radius cfg −+ 22 is 3 B1 2 Ditto; allow this mark for a correct radius

even if there appears to be a sign error in g--------------------------------------------------------------------------------------------------------------------------------------------------------EITHER : )(6)( or 6)( 2222 ykyykxxkx −=+−=−+ B1 Eliminate x or y completely

0)62(2 22 =++− kxkx or

0)6()62(2 22 =−+−− kkyky M1 Simplify to quadratic form

0)6(8)62( or 08)62( 2222 =−−−=−+ kkkkk M1 Equate discriminant to zero

0962 =−− kk A1 Or any equivalent 3-term quadratic in k

233 ±=k A1 Allow any equivalent exact form

OR : Gradient of the tangent is 1− B1Perpendicular diameter is )3(10 −+=− xy M1 Using their )0 ,3( and perpendicular gradient

)3(6)3(or 6)3( 2222 +=++=−+ yyyxxx M1 Substitute, and solve for x or y

2 ,23 23

23 ±=±= yx A1 At least one correct value of each

233 ±=k A1 5 For both values, in any equivalent exact form



5

−=

721

AB B1 May be implied

Direction of

115

is 1l B1 May be implied

0171251 =×−×+× , hence result B1 3 Given result correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------

ststst 234 ,42 ,2453 −−=++=++=+ B1 All three equations statedSolve any pair of these equations simultaneously M1

1 ,3 −=−= ts A1 For both values correctCheck the solutions in the remaining equation B1Point of intersection is )3 ,1 ,2(− A1 5 Allow answer as a position vector

6 )sin(cos ,cos θθθθ −== ayax && M1 Differentiate both, with product rule attempt

A1 Both correct, in any form

θθθθ

cossincos

dd −

=xy

M1 Use of xy && /

A1 Any correct form, involving θ only

0cos

sincos=

−θ

θθθM1 Equate

θdd

or dd y

xy

to zero

θθθθ

1tan0tan1 =⇒=− A1 6 Given result correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------

B1 Both curves correctB1 2 Correct conclusion, referring to single point

of intersection--------------------------------------------------------------------------------------------------------------------------------------------------------

EITHER : At 141

and 1tan ,41 >===

πθθπθ M1 Or equivalent calculation(s)

Hence root is greater than π41 A1

OR : Carry out any solution method M1 E.g. )/1(tan 11 nn θθ −

+ =

86.0≈θ , so greater than π41 A1 2

7x

Cx

Bx

A−

+−

+− 4)1(1 2 B1 Correct form stated or implied

2=B B1 May be obtained by the ‘cover-up’ rule3=C B1 Ditto

22 )1()4()4)(1(41315 xCxBxxAxx −+−+−−≡+− M1 Any correct use of this identity to give an

equation involving A1=A A1 5

--------------------------------------------------------------------------------------------------------------------------------------------------------

[ ] 3

21 4ln3)1(21ln xxx −−−+−− − B1 For both log terms, allowing omission of

modulus signs at this stageB1 For 1)1(2 −−+ x , or equivalent

4ln1)2ln321ln()1ln312ln( +=−−−−−−− M1 Use both limits and combine terms

A1 For correct reduction to given form of answerA1 5 Correct use of modulus throughout (not

dependent on correct answer being reached)



8 xktx

xktx

=−=dd

or dd

M1 Use of tx

dd

for rate of change

A1 Correct form of equation (with either sign)

1.0)(1dd

and 100 ±=⇒−== ktx

x M1 Attempted evaluation of k from their DE

xtx

1.0dd

−= A1 Any equivalent correct form

∫∫ −=− txx d1.0d 21

M1 Separate and integrate both sides

Atx +−= 1.02 21

A1 For both 21

2x and kt)(± (the numerical

evaluation of k may be delayed until after theDE is solved)

B1 For one arbitrary constant included, orequivalent statement of both pairs of limits

20020 ,200 =⇒== Atx M1 If limits are used, this mark is combined withthe following one

20021.01002 +−= t M1 Evaluating t8084.82 ≈= K A1 10 Given result correctly obtained


















1 A sequence of positive integers K , , 321 uuu is defined by

1for 23 and 1 11 ≥+== + nuuu nn .

Prove by induction that

1)3(2 1 −= −nnu . [4]

2 Find the general solution of the differential equation

xxy

xy

=−dd ,

giving y in terms of x in your answer. [4]

3 For positive integers r, let

)1(1

)f(+

=rr

r .

Verify that

)2)(1(2

)1f()f(++

=+−rrr

rr . [1]

Hence find the sum of the first n terms of the series

KK )2)(1(

1

4.3.21

3.2.11

+++

+++rrr

. [3]

Show that the series is convergent, and state the sum to infinity. [2]

4 Find the first three terms of the Maclaurin series for )2ln( x+ . [4]

Write down the first three terms of the series for )2ln( x− , and hence show that, if x is small, then

xxx

≈

−+

22

ln . [3]



5

The diagram shows a sketch of the graph of

1332 2

+++

=x

xxy .

Find

(i) the equations of the asymptotes of the curve, [3]

(ii) the values of y between which there are no points on the curve. [4]

6

The diagram shows the curve whose equation, in cartesian coordinates, is

)()( 222222 yxayx −=+ ,

where a is a positive constant. Show that the equation may be expressed, in polar coordinates, in the form

θ2cos22 ar = . [3]

Explain how you can deduce, from the polar form of the equation, that the line πθ 41= is tangential to the

curve at the pole. [2]

Find the area of the region enclosed by one loop of the curve. [4]



7 (i) Find the value of a for which

xx

xx

a

d 1

1d

11

20 0

21

2 +⌡

⌠⌡⌠

=+

∞. [3]

(ii) By means of the substitution θtan=x , or otherwise, find the exact value of

xx

xd

)1( 22

2

0 +⌡⌠

∞. [7]

8 The complex number z satisfies the equation 2 += zz . Show that the real part of z is 1− . [2]

The complex number z also satisfies the equation 2 =z . By sketching two loci in an Argand diagram,find the two possible values of the imaginary part of z, and state the two corresponding values of zarg . [5]

The two possible value of z are denoted by 1z and 2z , where 21 ImIm zz > .

(i) Find a quadratic equation whose roots are 1z and 2z , giving your answer in the form 02 =++ cbzazwhere the coefficients a, b and c are real. [2]

(ii) Determine the square roots of 1z , giving your answers in the form yx i+ . [4]






MARK SCHEME

MAXIMUM MARK 60




1 True for 1=n since 1321 0 −×= B1 Requires at least 121 −=21)3(23 1

1 +−= −+

nnu M1 Substitute given nu into 23 +nu

1)3(2 −= n A1 Correct simplification

1)3(2 1)1( −= +−n , hence result A1 4 Correct conclusion, dependent on all previous

marks and a properly set out proof

2 Integrating factor is ∫ xx d 1e −− M1 Requires integration attempt

i.e. x

x 1e ln =− A1 Simplify to

xx

1or 1−

∫=⇒=

x

xy

xy

xd 11

dd

M1

Axxy += 2 A1 4

3)2)(1(

2)2)(1(

2)2)(1(

1)1(

1++

=++

−+=

++−

+ rrrrrrrr

rrrrB1 1 Given result correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------

++−

+++−+−

)2)(1(1

)1(1

4.31

3.21

3.21

2.11

21

nnnnK M1 Using differences

)2)(1(21

41

++−

nnM1 Cancelling pairs of terms

A1 3 Correct answer; allow any equivalent form--------------------------------------------------------------------------------------------------------------------------------------------------------

∞→→++

nnn

as 0)2)(1(2

1; convergent B1 Or equivalent argument

Sum to infinity is 41 B1 2

4 EITHER : If x

xxx+

=′+=2

1)(f then )2ln()f( , M1 At least one differentiation attempt

and 2)2(1

)(fx

x+

−=′′ A1 Correct first and second derivatives

41

21 )0(f ,)0(f ,2ln)0f( −=′′=′= A1 All three correct

281

212ln)2ln( xxx −+≈+ A1 Three correct terms

OR : )1(2ln)2ln( 21 xx +=+ M1

)1ln(2ln 21 x++= A1

2)(

2ln2

21

21 x

x −+≈ M1 Use of standard series for )1ln( kx+

281

212ln xx −+= A1 4 Three correct terms

--------------------------------------------------------------------------------------------------------------------------------------------------------2

81

212ln)2ln( xxx −−≈− B1 Replacing x by x−

)2(ln)2(ln 281

212

81

21 xxxx −−−−+ M1 Subtracting series

xxx ≈

−+

22ln A1 3 Given result correctly shown



5 (i)1

212

+++=

xxy M1 Algebraic division, or equivalent

12 += xy is an asymptote A11−=x is an asymptote B1 3

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) EITHER : Consider discriminant of

)3()3(2 2 yxyx −+−+ M1 Form quadratic in x and refer to ∆

)3(8)3( 2 yy −=− A1 Allow equation or inequality

0)5)(3( =+− yy M1 Or equivalent solution method

5 and 3 −=y A1

OR : 2)1(2

2dd

+−=

xxy

or

2

2

)1()332()24)(1(

+++−++

xxxxx

M1 Differentiate and equate to zero

042or 1)1( 22 =+=+ xxx A1 Correct simplification

5 and 30 and 2 −=⇒−= yx M1 Solve for x and substitute to find y

A1 4 Both values of y correct

6 θθ sin ,cos ryrx == B1 For both; may be implied

)sincos()sincos( 2222222222 θθθθ rrarr −=+ M1 Substitute, and use at least one trig identity

θ2cos22 ar = A1 3 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------

02cos0 =⇒= θr M1 Using 0=r for form at pole

πθ 41= is a solution A1 2 Given result correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------θθθ 2sind 2cos 2

412

21 aa =∫ M1 Using correct formula ∫ θd 2

21 r

B1 Indefinite integral of the form θ2sink

[ ] [ ] πππ

θθ 41

41

41 0

2412

41 2sin2or 2sin aa

−M1 Using correct limits

221Area a= A1 4

7 (i) a1tanLHS −= B1

π21

21RHS ×= B1 Recognising π2

11tan =∞−

1tan 411 =⇒=− aa π B1 3 Correct answer for a

--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) θθ

2secdd

=x

B1 Or equivalent; may be implied

⌡⌠

+=⌡

⌠+

∞ π

θθθ

θ21

0

222

2

022

2d sec

)tan1(tan

d )1(

xx

xM1 Substitute for x and dx throughout

∫=π

θθ21

02 d sin M1 Use trig identities to simplify integrand

A1 Correct simplification; ignore limits so far

∫ −=π

θθ21

0 21 d )2cos1( M1 Use relevant double-angle formula

[ ] πθθ 21

041

21 2sin−= A1 For correct indefinite integral

π41= A1 7 Correct answer and no previous error



8 EITHER : Locus 2 += zz is a perp bisector M1 For recognis ing linear locusHence 1Re −=z A1 Needs mention of points 2 ,0 −== zz , or

equivalent

OR : 2222 )2( yxyx ++=+ M1Hence 1−=x , i.e. 1Re −=z A1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------

B1 Both loci correct

22 21 =+ y M1 Using Pythagoras or equivalent

3Im ±=z A1

( ))3/1(tanarg 1−−±= πz M1 Or equivalent correct method for either case

π32±= A1 5 Both correct

--------------------------------------------------------------------------------------------------------------------------------------------------------(i) 0)3i1)(3i1( =−+++ zz M1 Form equation and expand LHS; allow any

equivalent complete method0422 =++ zz A1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) EITHER : 2 2 11 =⇒= zz B1 For 2

πππ 32

31

132

1 or )arg(arg −=⇒= zz B1 For either possibility

)sini(cos2 31

31 ππ +± M1 Convert either case to cartesian form

)3i1(221 +± A1 Both correct; allow any equivalent exact

yx i+ expression

OR : If yxz i1 += then

xyyx 23 and 1 22 =−=− B1 Both equations correct

0344or 0344 2424 =−−=−+ yyxx M1 Form and solve quadratic in 2x or 2y

232

212 or == yx A1 Correct single value for 2x or 2y

( )23

21

1 i+±=z A1 4 Both correct; allow any equivalent exact

yx i+ expression


















1 The cubic equation 03 =−+ baxx has roots γβα , , . Given that αβγ = , express each of a and b in terms

of γ only, and hence show that bba =+ 2)( . [5]

2 The part of the curve 2xy = between 0=y and 2=y is rotated completely about the y-axis. Show that

the area of the curved surface formed is π313 . [5]

3 Starting from the definitions of xsinh and xcosh in terms of exponentials, show that

xxx 22 sinhcosh2cosh +≡ . [2]

Given that kx =2cosh , where 1>k , express xtanh in terms of k . [4]

4 The differential equation

xyxy

+= 1dd ,

with 1=y when 0=x , is to be solved numerically by a step-by-step method.

(i) Use two steps of Euler’s method, with step-length 0.1, to find an approximation for the value of ywhen 2.0=x . [3]

(ii) Use one step of the modified Euler method, with step-length 0.2, to find an alternative approximationfor the value of y when 2.0=x . [3]

5 Given that xxI nn d )(ln

e

1⌡⌠= , show that, for 1≥n ,

1e −−= nn nII . [4]

Hence find the exact value of 4I . [4]

6 The curve with equation

xx

ycosh

=

has one maximum point for 0>x . Show that the x-coordinate of this maximum point satisfies theequation 01tanh =−xx . [2]

The positive root of the equation 01tanh =−xx is denoted by α . Use the Newton-Raphson method,taking first approximation 11 =x , to find further approximations 2x and 3x for α . [4]

By considering the approximate errors in 1x and 2x , estimate the error in 3x . [3]



7

The diagram shows the curve 1

1+

=x

y together with four rectangles of unit width and heights 51

41

31

21 , , ,

respectively. Explain how the diagram shows that

xx

d 1

14

051

41

31

21

+⌡

⌠<+++ . [2]

The curve 2

1+

=x

y passes through the top left-hand corner of each of the four rectangles shown. By

considering the rectangles in relation to this curve, write down a second inequality involving 51

41

31

21 +++

and a definite integral. [2]

By considering a suitable range of integration and corresponding rectangles, show that

∑=

<<1000

2

)1000ln(1

)5.500ln(

rr

. [4]

Explain briefly how you can deduce that a reasonable estimate for the value of ∑=

1000

2

1

rr

is 6.5. [2]

8 The phenomenon of ‘resonance’ in a simple electrical circuit can be modelled by the differential equation

tVtV

10cos2100dd

2

2=+ ,

where V represents the voltage in the circuit and t represents time.

(i) Verify that tkt 10sin is a particular integral for this differential equation, where k is a constant whosenumerical value is to be found. [4]

(ii) Find the general solution of the differential equation. [3]

(iii) Find the particular solution for which both V and tV

dd are zero when 0=t . [3]

(iv) By considering the values of V when t becomes large, explain briefly why the mathematical modelcannot give an entirely satisfactory representation of the voltage in the circuit. [1]



BLANK PAGE






MARK SCHEME

MAXIMUM MARK 60




1 ba ==++=++ αβγγαβγαβγβα , ,0 B1 All three correct, at any stage2γαβγ =⇒= b B1 Correct answer for b

2)( γγβαγγ −=++=a B1 Correct answer for a

bbaba ==+⇒−= 22)( γγ M1 Eliminating 2γ , or equivalent

A1 5 Given result correctly shown

2 EITHER :yy

x2

1dd


⌡⌠ +=

2

0d

41

12Area yy

yπ M1 Stating ∫ sx d 2π all in terms of y

⌡⌠ +=

2

041 d

2 yyπ A1 Or equivalent simplification

( )[ ]2

041

34 2

3

+= yπ A1 Correct indefinite integral

π313= A1 Given result correctly shown

OR : xxy

2dd


⌡⌠ +=

2

0

2 d 41 2Area xxxπ M1 Stating ∫ sx d 2π all in terms of x

A1 Correct limits for x

( )2

0

261 2

3

41

+= xπ A1 Correct indefinite integral

π313= A1 5 Given result correctly shown

3 )e2e()e2(eRHS 224122

41 xxxx −− +−+++= M1 Squaring and adding

LHS)e(e 2221 =+= − xx A1 2 Correct simplification and conclusion

--------------------------------------------------------------------------------------------------------------------------------------------------------1sinhcosh 22 =− xx B1 May be implied

)1(sinh ,)1(cosh 212

212 −=+= kxkx B1 Both correct

11

coshsinhtanh

+−±==

kk

xxx M1 Using

xxx

coshsinhtanh =

A1 4 Both values needed

4 (i) 11.011 ×+=y M1 Use of correct Euler formula for 1st step

1.1= A1211.1)1.11.01(1.01.12 =×+×+=y A1 3 Allow 3sf value 1.21 correctly obtained

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 2.112.01temp =×+=y B1

)2.12.01(12.01 21

1 ×++××+=y M1 Use of correct modified Euler method

224.1= A1 3 Allow 3sf value 1.22 correctly obtained



5 [ ] ∫ −−−=e

111e

1 d )(ln )(ln xxxnxxxI nnn M1 Using relevant integration by parts

A1 Correct unsimplified result1e −−= nnI A1 First term shown to simplify to e

A1 4 Given result fully justified--------------------------------------------------------------------------------------------------------------------------------------------------------

)2e(12e3)3e(4e4e 1234 IIII −+−=−−=−= M1 Reduction formula used at least twice

00 24e15)e(24e9 II +−=−−= A1 For 04 24e15 II +−= or 024e9 I−1e0 −=I or 11 =I B1 For either correct

24e94 −=I A1 4

6 0cosh

sinhcoshdd

2 =−

=x

xxxxy

M1 Differentiate and equate to zero

Maximum when xxx sinhcosh = , i.e. 1tanh =xx A1 2 Given result correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------

nnn

nnnn xxx

xxxx 21 sechtanh1tanh

+−−=+ B1

11 =x gives K20177.12 =x M1 Newton-Raphson used at least onceA1 2x correct to at least 3sf

K1996785.13 =x A1 4 3x correct to at least 4sf--------------------------------------------------------------------------------------------------------------------------------------------------------

002.0 ,2.0 21 −≈≈ ee B1 For both; ignore signs of errors

732

1

222

3 102 −×−≈⇒≈ eee

ee

M1 Use of quadratic convergence property

A1 3 Ignore sign of answer

7 LHS is the total area of the four rectangles B1RHS is the corresponding area under the curve; henceresult B1 2--------------------------------------------------------------------------------------------------------------------------------------------------------

⌡⌠

+>+++

4

051

41

31

21 d

21 x

xM1 Attempt at relevant new inequality

A1 2 Correct statement--------------------------------------------------------------------------------------------------------------------------------------------------------Σ is the area of 999 rectangles M1 May be implied

Bounds are ⌡⌠

+

999

0d

21 x

x and ⌡

⌠+

999

0d

11 x

xM1 Accuracy of 999 not essential here

Lower limit is )5.500ln( A1 Given value correct shown

Upper limit is )1000ln( A1 4 Ditto--------------------------------------------------------------------------------------------------------------------------------------------------------

K56.6)1000ln()5.500ln(21 =+ M1 For considering the average

Round down to 6.5 as it’s an overestimate A1 2 Some reason for given answer 6.5 rather than6.6 needed



8 (i) tkttkV 10cos1010sin +=& and

tkttkV 10sin10010cos20 −=&& M1 Differentiate twice

A1 Correct VV &&& and ttk 10cos210cos20 ≡ M1 Substitute throughout the DE

1.0=k A1 4--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) i1001002 ±=⇒=+ mm M1 State and solve auxiliary equation

CF is tBtA 10cos10sin + A1 Or equivalentGS is tttBtAV 10sin1.010cos10sin ++= A1 3 Follow their CF and k

--------------------------------------------------------------------------------------------------------------------------------------------------------(iii) BtV =⇒== 00 ,0 M1

AtV 1000 ,0 =⇒==& M1 Differentiate and substituteHence ttV 10sin1.0= A1 3

--------------------------------------------------------------------------------------------------------------------------------------------------------(iv) V is unbounded as t increases; hence unrealistic B1 1 Follow errors if the result remains true


















1 The matrices A and B are given by

−

=

−=

αααα

cossinsincos

, 1001

BA ,

respectively. Find the matrix 1−BAB , simplifying your answer. [5]

2 The set 5 ,4 ,3 ,2 ,1 ,0=G is a cyclic group under addition modulo 6, and the set , , , , , tsrqpeH = is amultiplicative group whose group table is shown below.

eqpsrttpeqrtssqpetsrrrtspeqqsrteqpptsrqpeetsrqpe

(i) Find an element of G of order 6. [1]

(ii) Show that G and H are not isomorphic. [2]

The set 6 ,5 ,4 ,3 ,2 ,1=K is a group under multiplication modulo 7. Determine the order of the element 3in K, and hence or otherwise determine which of G and H is isomorphic to K. [3]

3 G is a multiplicative group with identity element e. The group is not commutative, but the elements a andb in G each commute with every element in G, i.e. for any x in G, xaax = and xbbx = .

(i) Prove that the element ab commutes with every element in G. [2]

(ii) Prove that the element 1−a commutes with every element in G. [2]

(iii) Deduce that the set of all those elements in G which commute with every element in G forms asubgroup of G. [4]

4 Use de Moivre’s theorem to prove that

)sin16sin205(sin5sin 42 θθθθ +−= . [4]

By letting πθ 51= , show that the exact value of )(sin 5

12 π is )55(81 − . [4]



5 Write down the six 6th roots of unity. [1]

By writing the equation 66)1( zz =+ in the form

11 6

=

+

zz ,

show that the values of z satisfying the equation are given by

1e1

i31

−= πkz ,

where k is an integer, and state a set of values of k that gives all the roots of the equation. [4]

Express each of these roots of 66)1( zz =+ in the form yx i+ , where x and y are real. [5]

6 Find the value of a for which the simultaneous equations

,5,1745,1023

bazyxzyxzyx

=++=−−=−+

do not have a unique solution for x, y and z. [4]

Show that, for this value of a, the equations are inconsistent unless 3=b . [2]

For the case where the equations represent three planes having a common line of intersection, L, find

equations for L, giving your answer in the form n

rzm

qyl

px −=

−=

− . [5]

7

The diagram shows a cuboid ABCDA'B'C'D' in which the lengths of AB, AD, and AA' are 3a, 2a and a

respectively. The point A is taken as origin, with unit vectors i, j, k in the directions of 'AAADAB , ,respectively.

(i) Find a normal vector for the plane through A', B and C', and find also the perpendicular distance fromD to this plane. [6]

(ii) Find the shortest distance between the skew lines A'B and AD'. [6]



BLANK PAGE






MARK SCHEME

MAXIMUM MARK 60




1

−=−

αααα

cossinsincos1B B1

−=−

αααα

cossinsincosor 1ABBA M1 Correct multiplication process

A1 Correct product of either pair

−

−=−αααα

αααα22

221

cossincossin2cossin2sincosBAB A1 Correct unsimplified product

−= αα

αα2cos2sin2sin2cos A1 5

2 (i) 1 or 5 B1 1 For either answer--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) EITHER : 222331 :Order: ofElement tsrqpeH B1

No element of order 6, hence result B1

OR : H is not commutative,e.g. srptpr == , B1 Counter-example is required

G is commutative, hence result B1 2--------------------------------------------------------------------------------------------------------------------------------------------------------

13 ,53 ,43 ,63 ,23 65432 ===== M1 At least 3 calculationsThe order of 3 in K is 6 A1K is isomorphic to G B1 3 If not deduced from orders of elements, a

reason is required (e.g. that K iscommutative)

3 (i) )()()()()()( abxbxabaxxbabxaxab ===== M1 At least one correct interchange

A1 2 Completely correct proof--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 1111 )()( −−−− =⇒= axaaaaxaxaax M1 Pre- and/or post-multiply and simplify at

least onceHence xaxa 11 −− = A1 2 Completely correct proof

--------------------------------------------------------------------------------------------------------------------------------------------------------(iii) Closure follows from (i) B1

Associativity is true within G B1Identity is e, since xeex = B1Inverses follow from (ii) B1 4

4 5)siniIm(cos5sin θθθ += M1 Expand and take imaginary part

θθθθθ 5324 sinsincos10sincos5 +−= A1 All three terms correct53222 )1(10)1(5 sssss +−−−= M1 Use of θθ 22 sin1cos −= throughout

)sin16sin205(sin 42 θθθ +−= A1 4 Given answer shown correctly--------------------------------------------------------------------------------------------------------------------------------------------------------

05sin51 =⇒= θπθ B1

Hence )sin(51 π is a root of 052016 24 =+− ss B1

)55(812 ±=s M1 For solving the quadratic in an exact form

)55()(sin 81

512 −=π A1 4 Given answer fully justified, e.g. negative

sign chosen since )(sin 512 π is certainly 2

1<



5 i31

e πk for 6 ,5 ,4 ,3 ,2 ,1=k B1 1 Or equivalent values for k ; allow exponentialtrigonometrical or cartesian form for roots

--------------------------------------------------------------------------------------------------------------------------------------------------------

ω=+z

z 1, where ω is a 6th root of unity M1 1≠ω not required at this stage

11

1−

=⇒=+ω

ω zzz M1 Solving for z

1e

1i

31

−=

πkA1 Given answer correctly shown

5 ,4 ,3 ,2 ,1=k B1 4 Or any correct set of 5 values--------------------------------------------------------------------------------------------------------------------------------------------------------

3=k gives real root 21− B1

)1)(e1(e

1eii

i

31

31

31

−−

−−

−

ππ

π

kk

k

M1 Use of conjugate for general, or for any one

specific, case

π

ππ

k

kk

31

31

31

cos22

1sinicos

−

−−A1 Correct trigonometric or numerical form

i)3( i,)3( 61

21

21

21 ±−±− A1 Any one complex root correct

A1 5 All four correct

6 051

415123

det =

−−−

aB1

0)125()45(2)20(3 =+−+−+− aa M1 Correct expansion method

A1 Correct unsimplified equation2=a A1 4

--------------------------------------------------------------------------------------------------------------------------------------------------------EITHER : 213 2 rrr −= M1 Complete method for relation between rows

3=b A1 Given answer correctly shown

OR : 051

17151023

det =

−

bM1

3=b A1 2 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------

Direction of L is

−−×

− 415

123

M1 For relevant vector product, or complete

equivalent method

i.e.

−

−

1379

A1 Or any multiple

449131745 ,1023 =−⇒=−−=−+ zxzyxzyx (e.g.) M1 Solve two equations simultaneously

So )2 ,1 ,2( − lies on L A1 Or any other correct point

L is 13

27

192

−+

=−

=−− zyx

A1 5 Or correct equivalent



7 (i) EITHER :

=′′

−=′

023

,03

aa

CAa

aBA M1 Or equivalent pair of vectors

Normal n is

×

− 023

03

aa

a

aM1 Or equivalent complete method

i.e.

−=

632

n A1 Or any multiple

aaa

DB 12632

. 0

23

. =

−

−=n M1 Or other relevant scalar product

Perpendicular distance is n

a12M1 Calculate n and divide

a712= A1

OR : Equation srzqypx =++ givessaraqapsapsar =++== 23 ,3 , M1 Substitute all three points

sarsaqsap =−== , , 21

31 M1 Solve for p, q, r or p : q : r

Normal vector is

−

632

A1 Or any multiple

Plane is azyx 6632 =+− M1

Perp distance from )0 ,2 ,0( a is

222 632

66

++

−− aaM1 Relevant use of formula

a712= A1 6

--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii)

=′

−=′

aaDA

a

aBA 2

0 ,0

3B1 Or equivalent

Direction of common perpendicular is

×

− aa

a

a2

00

3M1

i.e.

−=

632

n is perpendicular to both A1 or any multiple

aa

AA 6632

. 00

. =

−

=′ n M1 Or equivalent scalar product

Shortest distance is aa

766

=n

M1 Divide by n

A1 6 Correct answer





MATHEMATICS M1Mechanics 1






Where a numerical value for the acceleration due to gravity is needed, use 9.8 m s–2.








1

Two forces, of magnitudes 1 N and 3 N, act on a particle in the directions shown in the diagram. Calculatethe magnitude of the resultant force on the particle and the angle between this resultant force and the forceof magnitude 3 N. [5]

2

The diagram shows a railway engine of mass 50 tonnes pulling two trucks horizontally along a straighttrack. The trucks are coupled together behind the engine and have masses 8 tonnes and 4 tonnesrespectively, starting with the truck nearer to the engine. The acceleration of the train is 2s m 5.0 − .Assuming that there are no resistances to motion, find

(i) the driving force of the engine, [2]

(ii) the tensions in the two couplings. [4]

3

Two particles, of masses x kg and 0.1 kg, are moving towards each other in the same straight line andcollide directly. Immediately before the impact, the speeds of the particles are 1s m 2 − and 1s m 3 −

respectively (see diagram).

(i) Given that both particles are brought to rest by the impact, find x. [2]

(ii) Given instead that the particles move with equal speeds of 1s m 1 − after the impact, find the threepossible values of x. [6]

4 A moving particle P travels in a straight line. At time t seconds after starting from the point O on the line,the velocity of P is 1s m −v , where

)6(2 ttv −= .

Show that the acceleration of P is zero when 4=t . [3]

After a certain time, P comes instantaneously to rest at the point A on the line. State the time taken for themotion from O to A, and find the distance OA. [5]



5

A heavy ring of mass 5 kg is threaded on a fixed rough horizontal rod. The coefficient of friction betweenthe ring and the rod is

21 . A light string is attached to the ring and is pulled downwards with a force of

magnitude T newtons acting at an angle of 30° to the horizontal (see diagram). Given that the ring is aboutto slip along the rod, find the value of T. [9]

6

The diagram shows an approximate ) ,( vt graph for the motion of a parachutist falling vertically; 1s m −vis the parachutist’s downwards velocity at time t seconds after he jumps out of the plane. Use theinformation in the diagram

(i) to give a brief description of the parachutist’s motion throughout the descent, [4]

(ii) to calculate the height from which the jump was made. [2]

The mass of the parachutist is 90 kg. Calculate the upwards force acting on the parachutist, due to theparachute, when 7=t . [5]



7 (i)

Particles A, of mass 5m, and B, of mass 3m, are attached to the ends of a light inextensible string. Thestring passes over a fixed peg, and the system is released from rest with both parts of the string tautand vertical, and each particle a distance d above a fixed horizontal plane (see diagram). Neglectingall resistances to motion,

(a) find the acceleration of A in terms of g and show that the tension in the string is mg415 , [6]

(b) find an expression in terms of d and g for the time after release at which A hits the plane. [2]

(ii) The results in part (i) are based on a mathematical model in which resistances to motion areneglected. Describe briefly one resisting force, other than air resistance, which would be present in areal system in which objects of unequal mass, hanging from a string passing over a fixed support, arein motion. [1]

When this force is taken into account, state with brief reasons whether each of the following would besmaller or larger than the value calculated in part (i):

(a) the acceleration of A;

(b) the tension in the string acting on A;

(c) the tension in the string acting on B.

What can you conclude about the tension in the string in this case? [4]






MARK SCHEME

MAXIMUM MARK 60




1 EITHER : ‘Vertical’ component of resultant is o40sin1 B1‘Horizontal’ component is °+ 40cos13 B1

Magnitude is 22 6428.0766.3 + M1 Allow M mark for either Pythagoras or trigi.e. 3.82 N A1 For correct magnitude

Angle is °=

− 69.9

766.36428.0tan 1 A1 For correct angle

OR : Vector triangle with sides 1, 3 and includedangle of 140º (not 40º) B1 May be implied

°×××−+= 140cos13213 222R M1 For use of cosine formula with °140 ,1 ,3Hence magnitude is 3.82 N A1

82.3140sin

1sin °=θ

M1 For sin formula, or other complete method

Hence angle is °69.9 A1 5

2 (i) 5.0)4000800050000(Force ×++= M1 For use of NII applied to whole systemkN 31or N 31000= A1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) For back t ruck: 5.040001 ×=C M1 Use of NII for the rear truck only

i.e. Force in rear coupling is 2000 N or 2 kN A1For both trucks: 5.0120002 ×=C M1 Use of NII for the pair of trucks with one

force, or equivalent, i.e. 400012 =− CC or2500031000 2 =− C

i.e. Force in front coupling is 6000 N or 6 kN A1 4 Follow through if earlier answer is used

3 (i) 01.032 =×−×x M1 For relevant use of momentum conservationHence 15.0=x A1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) xxxx −+−+=− 1.0or )1.0(or )1.0(3.02 M1 For any one relevant momentum equation

A1 For any one correct (unsimplified) equationM1 For appreciating at least 2 correct cases

Hence 133.0or 0.0667or 4.0=x A1 For any one correct valueA1 For a second correct valueA1 6 For all three correct answers

4 0312dd 2 =−= tt

tv

M1 For expanding v and differentiating

A1 For correct derivative equated to zero4 when 0 so ,0)4(3 ===− tatt A1 3 Given answer correctly found or verified

--------------------------------------------------------------------------------------------------------------------------------------------------------P reaches A when 6=t B1

[ ]60

4413

6

032 2d )6( ttttts −=−= ∫ M1 For integrating v

A1 For correct indefinite integral324432 −= M1 Use of limits or evaluation of arbitrary const

Distance m 108=OA A1 5



5

B1 Correct forces identified, by diagram orotherwise

Resolving horizontally: M1 For attempting one resolution equationFT =°30cos A1

Resolving vertically: M1 For attempting a second resolutionRTg =°+ 30sin5 A1 Other correct equations are possible

For limiting equilibrium RF 21= B1 Available at any stage

)5(3. 21

21

21 TgT += M1 For eliminating F and R

A1 Correct unsimplified equation in T only8.39=T A1 9

6 (i) Initially the parachutist falls with constant acc B1 Allow ‘free-fall’ etc hereThen decelerates at a constant rate B1Then falls with constant speed B1And finally hits the ground and comes to rest B1 4

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Area is 15106)1040(404 2

121 ×+×++×× M1 For sensible attempt at total area under graph

Height is 380 m A1 2--------------------------------------------------------------------------------------------------------------------------------------------------------

Acceleration when 7=t is 54104010 −=

−−

(downwards) M1 For use of gradient to find acceleration

A1 For value 5)(± even if sign/direction muddleHence )5(9090 −×=− Tg M1 For use of NII with three relevant terms

B1 For consistent signs in T and ma termsForce from parachute is 1330 N A1 5

7 (i) (a) Equations of motion for the particles are: M1 For use of NII for either particle separatelymaTmg 55 =− A1 The ‘system’ equation mamg 28 = is anmamgT 33 =− A1 alternative for one of these A marks

Hence acceleration is g41 and M1 For finding T or a from sufficient equation(s)

Tension is 415 mg A1 For correct acceleration

A1 6 For obtaining given tension correctly--------------------------------------------------------------------------------------------------------------------------------------------------------

(b) 241

21 tgd ××= M1 Use appropriate uvast equation and solve for t

Time is gd8

A1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Friction between the string and the support B1 1--------------------------------------------------------------------------------------------------------------------------------------------------------

(a) Acceleration is smaller, as the resistanceopposes the motion B1

(b) Tension at A is larger, becausemamgTA 55 −= , and a is less than before B1

(c) Tension at B is smaller, becausemgmaTB 33 += and a is less than before B1

The tensions in the two parts of the string are nowunequal B1 4


Specimen Materials – Mathematics 60 © OCR 2000

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1

The diagram shows the cross-section of a uniform solid rectangular block. This cross-section hasdimensions 20 cm by 10 cm and lies in a vertical plane. The block rests in equilibrium on a rough planewhose inclination α to the horizontal can be varied. The coefficient of friction between the block and theplane is 0.7. Given that α is slowly increased from zero, determine whether equilibrium is broken bytoppling or sliding. [5]

2 A small ball of mass 0.2 kg is dropped from rest at a height of 1.2 m above a horizontal floor. The ballrebounds vertically from the floor, reaching a height of 0.8 m. Assuming that air resistance can beneglected, calculate

(i) the coefficient of restitution between the ball and the floor, [4]

(ii) the impulse exerted by the floor on the ball when the ball bounces. [2]

If air resistance were taken into account, would the value calculated for the coefficient of restitution belarger or smaller than the value calculated in part (i)? Justify your conclusion. [1]

3

A uniform lamina ABCD has the shape of a square of side a adjoining a right-angled isosceles trianglewhose equal sides are also of length a. The weight of the lamina is W. The lamina rests, in a verticalplane, on smooth supports at A and D, with AD horizontal (see diagram).

(i) Show that the centre of mass of the lamina is at a horizontal distance of a911 from A. [4]

(ii) Find, in terms of W, the magnitudes of the forces on the supports at A and D. [4]



4

Fig. 1 shows the cross-section of a hollow container. The base of the container is circular, and ishorizontal. The sloping part of the side makes an angle of 15° with the horizontal, and the vertical part ofthe side forms a circular cylinder of radius 0.4 m. A small steel ball of mass 0.1 kg moves in a horizontalcircle inside the container, in contact with the vertical and sloping parts of the side at A and B respectively,as shown in Fig. 2.

It is assumed that all contacts are smooth and that the radius of the ball is negligible compared to 0.4 m.

(i) Given that the ball is moving with constant speed 1s m 3 − , find the magnitudes of the contact forcesacting on the ball at A and at B. [5]

(ii) Calculate the least speed that the ball can have while remaining in contact with the vertical part of theside of the container. [3]

5 A car of mass 650 kg is travelling on a straight road which is inclined to the horizontal at 5°. At a certainpoint P on the road the car’s speed is 1s m 15 − . The point Q is 400 m down the hill from P, and at Q thecar’s speed is 1s m 35 − .

(i) Assume that the car’s engine produces a constant driving force on the car as it moves down the hillfrom P to Q, and that any resistances to the car’s motion may be neglected. By considering thechange in energy of the car, or otherwise, calculate the magnitude of the driving force of the car’sengine. [4]

(ii) Assume instead that resistance to the car’s motion between P and Q may be represented by a constantforce of magnitude 900 N. Given that the acceleration of the car at Q is zero, show that the power ofthe car’s engine at this instant is approximately 12.1 kW. [4]

Given that the power of the car’s engine is the same when the car is at P as it is when the car is at Q,calculate the car’s acceleration at P. [2]



6

A uniform rectangular box of weight W stands on a horizontal floor and leans against a vertical wall. Thediagram shows the vertical cross-section ABCD containing the centre of mass G of the box. AD makes anangle θ with the horizontal, and the lengths of AB and AD are 2a and 8a respectively.

(i) By splitting the weight into components parallel and perpendicular to AD, or otherwise, show that theanticlockwise moment of the weight about the point D is )sincos4( θθ −Wa . [3]

(ii) The contact at A between the box and the wall is smooth. Find, in terms of W and θ , the magnitudeof the force acting on the box at A. [3]

(iii) The contact at D between the box and the ground is rough, with coefficient of friction µ . Given that

the box is about to slip, show that 18

4tan

+=

µθ . [4]

7

A shell is fired from a stationary ship O which is at a distance of 1000 m from the foot of a vertical cliff ABof height 100 m. The shell passes vertically above B and lands at a point C on horizontal ground, levelwith the top of the cliff (see Fig. 1). The shell is fired with speed 1s m 300 − at angle of elevation θ , andair resistance to the motion of the shell may be neglected.

(i) Given that °= 30θ , find the time of flight of the shell and the distance BC. [6]

(ii)

Given instead that the shell just passes over B, as shown in Fig. 2, find the value of θ , correct to thenearest degree. [6]






MARK SCHEME

MAXIMUM MARK 60




1 Topples when CG is above lowest corner B1 May be impliedi.e. when 2

1tan =α B1

Slides when αα sin7.0 and cos mgRmgR == M1 Both equations attemptedi.e. when 7.0tan =α A1 Allow B2 in place of M1 A1 if αµ tan= is

quotedHence it topples, since 7.02

1 < B1 5 Conclusion and reason needed

2 (i) Speed before impact is 8.02 and 2.12 ×× gg M1 For one relevant use energy or const acc

A1 For both correct (unsimplified) values

Hence 0.8164.26.1

32 ≈==

gge M1 For use of evv =′ to calculate e

A1 4 For correct exact or decimal answer--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) sN 76.1)6.14.2(2.0Impulse ≈+= gg M1 Allow M mark even if there is a sign error

A1 2--------------------------------------------------------------------------------------------------------------------------------------------------------With air resistance, speed before impact is smaller, and

speed after impact is larger; hence e is larger B1 1 For correct conclusion with correct reasons

3 (i) CG of triangle is a32 horizontally from A B1

Moments: xWaWaW ×=×+× 23

32

32

31 M1 For equating moments about A, or equivalent

A1 For a correct unsimplified equationHence ax 9

11= A1 4 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) WRaWaR AA 18

7972 =⇒×=× M1 For one moments equation

A1 For one correct answerWRWRR DDA 18

11=⇒=+ M1 For resolving, or a second moments equation

A1 4 For a second correct answer

4 (i) gRB 1.015cos =° B1

Hence N 01.1=BR B1

4.03

1.015sin2

×=°+ BA RR M1 For using NII horizontally (3 terms needed)

A1 Correct equationHence N 99.1=AR A1 5

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 01.1 and ,0 == BA RR as before B1 May be implied

4.01.015sin

2vRB ×=° M1

1s m 02.1 −=v A1 3



5 (i) EITHER : )1535(650KE 2221 −××=∆ B1 For correct unsimplified expression

°×=∆ 5sin400650PE g B1 Sign of change not required073222000325400 −=T M1 For equating work and energy change

N 257=T A1

OR : 25.14002

1535 22=

×−

=a B1 For correct unsimplified expression for a

agT 6505sin650 =°+ M1 For 3-term NII equation || slope

A1 Correct equationN 257=T A1 4

--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) Driving force at Q is 35H

B1 For any correct statement of ‘ FvP = ’

°+= 5sin65035

900 gH

M1 For 3-term force equation || slope at Q

A1 Correct equationHence kW 1.12 i.e.W 12069≈H A1 4 Given answer correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------

agH

6509005sin65015

=−°+ M1 For 4-term NII equation

2s m 707.0 −=a A1 2

6 (i) EITHER : Moments of weight components are:aW 4cos ×θ (anticlockwise) B1

aW ×θsin (clockwise) B1Hence total anticlockwise is

)sincos4( θθ −Wa B1 Given answer correctly shown

OR : Horizontal distance G – D is M1 For using horizontal projectionsθθ cos4sin aa +− A1 Correct expression

Hence anticlockwise moment is)sincos4( θθ −Wa A1 Given answer correctly shown

OR : Horizontal distance G – D is

4122 tan where)cos()4( =++ ααθaa B1

Anticlockwise moment is

( )θθ sincos17171

1742 −× aW M1

i.e. )sincos4( θθ −Wa A1 3 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) θθθ sin8)sincos4( aRWa A ×=− M1 For moments equation about D, using (i)

A1 Correct equation)1cot4(8

1 −= θWRA A1 3 For correct answer, in any form--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) WRD = B1

AD RF = B1

81cot4 −= θµ M1 For use of RF µ= in equation involving θ

Hence 18

4tan+

=µ

θ A1 4 For showing given result correctly



7 (i) EITHER : 29.4150100 tt −= M1 For use of 221 atuts += vertically

A1 For correct equation

8.920540150 ±=t M1 For any solution method

Time to C is 29.9 s A19.2930cos300 ×°=x M1

m 67801000 ≈−= xBC A1

OR :432

2

30028.9

3100

××−= xx

M1 For trajectory equation with 100=y ,

300=V , °= 30θA1 For correct unsimplified equation

7776≈x M1 For any solution methodHence m 6780≈BC A1

°=

30cos300x

t M1

s 9.29= A1 6--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) )tan1(300210008.9

tan1000100 22

2θθ +

××

−= M1 For trajectory equation with 100=y ,

300 ,1000 == VxM1 For use of θθ 22 tan1sec +=

0139tan900tan49 2 =+− θθ A1 Correct simplified quadratic

9827244900900tan

2 −±=θ M1 For any solution method

M1 For taking arctan of the smaller root°≈ 9θ A1 6



















1

A ball of mass 0.2 kg falls vertically onto a sloping grass bank, and rebounds horizontally (see diagram).Immediately before the bounce the speed of the ball is 1s m 8 − , and immediately after the bounce the speedis 1s m 3 − . Calculate the magnitude and direction of the impulse on the ball due to the impact. [4]

2 A light elastic string of modulus 28 N and natural length 0.8 m has one end attached to a fixed point O. Aparticle of mass 0.5 kg is attached to the other end.

(i) The particle hangs in equilibrium at the point E. Calculate the distance OE. [2]

(ii) The particle is held at O and is released from rest. Calculate the speed of the particle as it passes thepoint E. [4]

3

Two uniform smooth spheres A and B, of equal radius, are free to move on a smooth horizontal table. Themass of B is twice the mass of A. Initially B is at rest and A is moving with speed 1s m 5 − . The spherescollide, and immediately before impact the direction of motion of A makes an angle of 30° with the line ofcentres. After the collision A moves at right angles to its original direction (see diagram). Show that

(i) the speed of A immediately after the collision is 135 s m 3 − , [2]

(ii) the speed of B immediately after the collision is also 135 s m 3 − , [3]

(iii) the collision is perfectly elastic. [3]



4 A particle of mass 0.2 kg is connected by two equal light elastic springs, each of natural length 0.5 m andmodulus of elasticity 5 N, to two points A and B on a smooth horizontal table. The mid-point of AB is Oand the length of AB is 1 m. The particle is displaced from O, towards B, through a distance of 0.3 m tothe point C and released from rest. In the subsequent motion air resistance may be neglected. Aftert seconds the displacement of the particle from O is x metres. Show that

xtx

100dd

2

2−= . [3]

The particle moves a distance 0.1 m from C to D. Find

(i) the speed of the particle at D, [3]

(ii) the time taken to reach D. [3]

5 A particle of mass m is attached to one end of a light inextensible string of length 10a. The other end ofthe string is attached to a fixed point O. The particle is released from rest with the string taut andhorizontal. Assuming there is no air resistance, find

(i) the speed of the particle when the string has turned through 30°, [2]

(ii) the tension in the string at this instant. [3]

When the string reaches the vertical position, it comes into contact with a small fixed peg A which is adistance 7a below O. The particle begins to move in a vertical circle of radius 3a with centre A (seediagram). Determine, showing your working, whether the particle describes a complete circle about A. [5]



6

Two uniform beams AB and BC, each of length 5a, have masses 3m and 2m respectively. The beams arefreely jointed to fixed points at A and C, and to each other at B. The points A and C are on the samehorizontal level at a distance 8a apart, and the beams are in equilibrium with B vertically below the mid-point of AC, as shown in the diagram.

(i) Find the vertical component of the force acting on BC at C, and show that the horizontal componentof this force is mg3

5 . [6]

(ii) Find the magnitude and direction of the force acting on AB at B. [5]

7 A body falls vertically, the forces acting being gravity and air resistance. The air resistance is proportionalto v, where v is the body’s speed at time t. The value of v for which the acceleration is zero is known as the‘terminal velocity’ for the motion, and is denoted by U. Show that the equation of motion of the body maybe expressed as

)(dd

vUUg

tv

−= . [3]

A parachutist jumps from a helicopter which is hovering at a height of several hundred metres, and fallsvertically. Assume that, before the parachute is opened, the terminal velocity for the motion is 1s m 50 − .The parachutist opens the parachute 10 s after jumping. Find the speed at which the parachutist is fallingjust before the parachute opens. [5]

The diagram shows a ) ,( vt graph for the parachutist’s motion, as modelled using the above differentialequation.

(i) Explain the significance of the speed of 1s m 10 − in relation to the differential equation. [1]

(ii) What has been assumed about the opening of the parachute? [1]

(iii) Find the decele ration of the parachutist just after the parachute opens. [2]






MARK SCHEME

MAXIMUM MARK 60




1

B1 For identifying impulse vector with thechange of momentum, by means of a triangleor otherwise

=

+=−

6.06.1

tanhorizontal aboveDirection

6.16.0Magnitude1

22

M1 For either Pythagoras or trig calculation

Impulse is 1.71 N s at °4.69 to horizontal A1 For magnitudeA1 4 For angle to horizontal, or equivalent

2 (i)8.0

285.0 xg = M1 For equilibrium equation and Hooke

m 94.014.0 =⇒= OEx A1 2 Correct answer for OE--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Conservation of energy: M1 For equation involving KE, PE and EE

94.05.08.0214.0285.0

22

21 ×=

××+× gv B1 For correct EE term

B1 For PE term, correct apart possibly from sign606.4343.025.0 2 =+v , hence 1s m 13.4 −=v A1 4

3 (i) °=° 60cos530cosAv M1 Equating components ⊥ line of centres

( ) 33 35

21

25 =÷=Av A1 2 Given answer correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) °−=°× 60cos230cos5 AB mvmvm M1 Using momentum || line of centres

A1 Correct equation ( Av need not be numerical)

3332 35

65

25 =⇒+= BB vv A1 3 Given answer correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------(iii) °×=+° 30cos560cos evv BA M1 Using restitution || line of centres

°=+° 30cos5360cos3 35

35 e A1 Correct equation

Hence 1=e , as required A1 3 Given result correctly shown

4 Force in each spring is 5.0

5xB1 Correct expression for a general position

Equation of motion is xxx &&2.01010 −=+ M1 For relevant use of NIIi.e. xx 100−=&& A1 3 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------(i) Motion is SHM with amplitude 0.3 m B1 Allow at any stage in the question

)2.03.0(100 222 −=Dv M1

Speed at D is 1s m 24.2 − A1 3--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) )10cos(3.02.0 Dt= B1 For correct SHM equation involving Dt

( )321cos1.0 −=Dt M1 Or equivalent complete solution method

Time to reach D is 0.0841 s A1 3



5 (i) °×= 30sin10221 amgmv M1 For relevant use of conservation of energy

Hence gav 10= A1 2--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii)agammgT

101060cos ×=°− M1 3-term NII equation || string

A1 Correct unsimplified equationmgT 2

3= A1 3--------------------------------------------------------------------------------------------------------------------------------------------------------Critical case is 0=T at highest point B1 May be implied

amgmvH 4221 ×= M1 Use of energy to find Hv at highest point

gavH 82 = A1

amvTmg3

2=+ M1 Resolving to find T ( mg3

5= ) when Hvv = or

to find critical )3 ( 2 gav = for 0=THence it does make a complete circle A1 5 Correct result and reason

6 (i) Moments about A for the system: M1 Equation with 3 terms neededaYamgamg C 86223 ×=×+× A1 For correct unsimplified equation

mgYC 49= A1 Correct answer for the vertical component

Moments about B for BC : M1 Equation with 3 terms needed

amgaXamg C 4322 49 ×=×+× A1 For correct unsimplified equation

mgXC 35= A1 6 Given answer correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) mgX B 3

5= B1

mgYB 41= B1 Follow the answer for CY in (i)

=

+=− tanhorizontal aboveDir

Magnitude1

22

B

BBB

XY

YXM1 For numerical Pythagoras or trig calculation

mgmg 69.1409Magnitude 121 ≈= A1 Correct exact or approximate value

( ) °≈= − 5.8tanhorizontal aboveDir 2031 A1 5 Correct exact or approximate angle

7 Equation of motion is kvmgtv

m −=dd

B1

At terminal velocity kUmg = B1

Hence )(dd

vUUg

tv

−= B1 3 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------

⌡⌠=⌡

⌠−

tgvv

d 50

d 50

1M1 For separation and attempt at integration

ctv +=−− 196.0)50ln( A1 For both indefinite integrals correct

ctv =−⇒== 50ln0 ,0 M1 Evaluation of constant or equiv use of limits50ln96.1)50ln( 10 −=−− v A1 Correct equation for 10v

-196.110 s m 043)e1(50 .v ≈−= − A1 5 For correct exact or approximate answer

--------------------------------------------------------------------------------------------------------------------------------------------------------(i) 1s m 10 − is the terminal velocity (value of U) after

the parachute opens B1 1--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) The parachute is assumed to open instantaneously B1 1--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) )0.4310(10

8.9dd −=

tv

M1

Hence deceleration is 2s m 3.32 − A1 2



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74


1 The region bounded by the part of the curve xy = from 0=x to ax = , the x-axis and the line ax = isrotated completely about the x-axis to form a uniform solid of revolution. Find by integration thex-coordinate of the centre of mass of the solid. [6]

2

The diagram shows a uniform circular disc, of mass m and radius a, which is free to rotate in a verticalplane about a smooth fixed horizontal axis through its centre O. A light inextensible string is wrappedround the circumference of the disc, and has one end attached to the circumference. A particle of mass mhangs freely at the other end of the string. The system is released from rest. Show that the angular

acceleration of the disc is ag

32 , and find the tension in the string. [6]

3 A rigid square frame consists of four uniform rods, each of mass m and length 2a, joined at their ends toform a square. Show that the moment of inertia of the frame, about an axis through one of its corners andperpendicular to its plane, is 2

340 ma . [3]

The frame is suspended from one corner, and can rotate in a vertical plane about a smooth horizontal axisthrough that corner. Show that the motion in which the frame makes small oscillations about itsequilibrium position is approximately simple harmonic, and find the period of this simple harmonicmotion. [5]

4 A uniform circular disc, of mass m and radius a, can rotate in a vertical plane about a fixed horizontal axispassing through its centre O. When the disc is at rest, a particle of mass 2m is released, from rest, at aheight 2a vertically above one end A of the horizontal diameter of the disc. The particle falls freely, strikesthe disc at A, and adheres to the disc. Find the angular speed with which the disc starts to rotate. [5]

While the disc (with the attached particle) rotates, a constant frictional couple C acts on the disc. The disccomes to rest after one quarter of a revolution, when the particle is at the lowest point of the disc. Find C.

[3]


75


5 A uniform rod AB, of mass m and length 2a, is free to rotate in a vertical plane about a smooth horizontalaxis through A. When the rod is hanging at rest in equilibrium with B vertically below A, it is given anangular speed Ω, where

ag

432 =Ω .

Show that the rod comes instantaneously to rest when it has turned through an angle π31 . [2]

At this position of instantaneous rest, the force acting on the rod at A has horizontal and verticalcomponents X and Y respectively. Find X and Y in terms of m and g. [8]

6

A light aircraft flies flies from A due east to B and then flies directly back to A. The distance AB is d, andthe speed of the aircraft relative to the air is 4V. During the flight, there is a steady wind, of speed V in adirection making an angle θ with AB (see diagram). Show that the total flying time for the journey from Ato B and back is

Vd

15cos152 2 θ+ . [10]


76


7

A small smooth bead B, of mass m, is threaded on a circular wire with centre O and radius a. The wire isfixed in a vertical plane. A light elastic string, of natural length a and modulus of elasticity λ , has one endfixed at O. The string passes through a small smooth ring A fixed at the highest point of the wire, and theother end of the string is attached to B. The diagram shows the system at an instant when OB makes anangle θ with the downward vertical at O.

(i) Taking the horizontal level of O as the reference level for gravitational potential energy, show that thetotal potential energy of the system in the position shown is

θλλ cos)( mgaa −+ . [4]

(ii) Hence show that there is a position of stable equilibrium with 0=θ so long as mg<λ . [3]

(iii) Given that mg21=λ , show that the approximate period of small oscillations about the equilibrium

position is

ga2

2π . [5]






MARK SCHEME

MAXIMUM MARK 60




1 ∫=a

xx0

d mass Total πρ M1 For relevant use of ∫ xy d 2

221 aπρ= A1 Introduction of ρ is required

∫=a

xx0

2 d moment Total πρ M1 For relevant use of ∫ xxy d 2

331 aπρ= A1 No further penalty if ρ omitted

Hence axaxa 323

312

21 =⇒= πρπρ M1 For equating and solving for x

A1 6 For correct final answer

2 For particle: mfTmg =− B1

For disc: α221 maTa = B1

String moves with disc, so αaf = B1 Stated or used at any stage

Hence αα mamamfmg 23

21 =+= M1 For obtaining an equation in α (or T)

ag

32

=α A1 Given answer shown correctly

mgT 31= A1 6

3 M.I. for one of the rods opposite the axis is: M1 Relevant use of parallel axes theorem2

316222

31 )4( maaamma =++ A1 For correct unsimplified expression

23402

3162

34 )(2frame of M.I. mamama =+= A1 3 Given answer shown correctly

--------------------------------------------------------------------------------------------------------------------------------------------------------Weight 4mg acts at 2a from the axis B1

Equation of motion is: θθ &&23

40sin24 mamga −= M1 For relevant use of θ&&IC = at general posn

A1 Correct equation

i.e. θθa

g10

23−≈&& A1 Reduction to standard SHM form

Hence SHM with period 23

102

ga

π A1 5

4 Speed of particle before impact is ga4 B1

( )ω2221 242 mamaagam +=× M1 Equating ang mom before and after

B1 For total M.I. ( )2221 2mama + , or equivalent

A1 Correct equation

Hence ag

58

=ω A1 5

--------------------------------------------------------------------------------------------------------------------------------------------------------Work done against friction is π2

1×C B1

Hence mgamaC 22245

21 += ωπ M1 Equating energy (KE and PE) to work

π552mga

C = A1 3



5 )cos1(43

.. 234

21 θ−= mga

ag

ma M1 For relevant use of conservation of energy

πθ 31

21 so ,LHS == mga A1 2 Given answer obtained or verified correctly

--------------------------------------------------------------------------------------------------------------------------------------------------------

They may have X and/or Y reversed; anyconsistent work is acceptable

( ) απ 234

31sin mamga = M1 Taking moments about A to find ang acc

i.e. a

g8

33=α A1

Acceleration component of CG || rod is zero B1 May be implied

and perpendicular to rod is 833 g

A1

EITHER : Res hor: ( )π31cos

833

.g

mX = M1 To include attempt at resolving transverse acc

i.e. mgX16

33= A1

Res vert: ( )π31sin

833

.g

mYmg =− M1

Hence mgY 167= A1

OR : Res || rod: mgYX 21

21

21 3 =+ B1

Res ⊥ rod: M1 4 terms requiredmgmgYX 333 8

321

21

21 =+− A1 Correct equation

mgYmgX 167 ,

1633

== A1 8 For both answers

6B1 For correct triangle; may be implied

θθα cos216or sinsin 22241 RVVRV −+== B1 For appropriate use of sine or cosine rule

θθ 241 sin14cos −+= VVR or M1 Any method for R in terms of V and θ

15coscos 2 ++= θθ VVR or equivalent A1 For any correct (unsimplified) expression

B1 For correct return triangle; may be implied

15coscos 2 ++−= θθ VVS or equivalent M1 Any method for S in terms of V and θA1 For any correct (unsimplified) expression

Total time is Sd

Rd + M1 For expressing this in terms of d, V , θ

i.e. θθ

θ2222

2

cos)15(cos15cos2)(

VVdV

RSSRd

−++=+

M1 For combining the terms algebraically

Vd

15cos152 2θ+= A1 10 Given answer correctly shown



7 (i) θθ cos22or cos2 22221 aaABaAB +== M1

aa

2)cos1(2

energy Elastic2 θλ +

= A1 Or equivalent

PE of B is θcosmga− B1

Total energy θλλ cos)( mgaaV −+= A1 4 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) 0sin)(dd

=−−= θλθ

mgaV

M1 Differentiate and equate to zero, or equivalent

argument from properties of cos graphHence 0=θ A1 Given answer correctly shown

mgmgaV <>−==

λλθ θ

if 0)(dd

02

2A1 3 Stability explained via identification of min V

--------------------------------------------------------------------------------------------------------------------------------------------------------(iii) constant)( 2

21 =+ Vam θ& B1

0sin212 =+ θθ mgama && M1 Differentiate with respect to t

A1 Correct simplified equation

i.e. θθag2

−≈&& , so SHM M1 Reduction to standard form

Period is ga2

2π A1 5 Given answer correctly shown





MATHEMATICS S1Probability & Statistics 1













1 Janet and John wanted to compare their daily journey times to work, so they each kept a record of theirjourney times for a few weeks. Janet’s daily journey times, x minutes, for a period of 25 days, weresummarised by 2120=Σx and 0441802 =Σx . Calculate the mean and standard deviation of Janet’sjourney times. [3]

John’s journey times had a mean of 79.7 minutes and a standard deviation of 6.22 minutes. Describebriefly, in everyday terms, how Janet and John’s journey times compare. [2]

2 Two independent assessors awarded marks to each of 5 projects. The results were as shown in the table.

Project A B C D E

First assessor 38 91 62 83 61Second assessor 56 84 41 85 62

Calculate Spearman’s rank correlation coefficient for the data. [4]

Show, by sketching a suitable scatter diagram, how two assessors might have assessed 5 projects in such away that Spearman’s rank correlation coefficient for their marks was 1+ while the product momentcorrelation coefficient for their marks was not 1+ . (Your scatter diagram need not be drawn accurately toscale.) [2]

3 Each packet of the breakfast cereal Fizz contains one plastic toy animal. There are five different animals inthe set, and the cereal manufacturers use equal numbers of each. Without opening a packet it is impossibleto tell which animal it contains. A family has already collected four different animals at the start of a yearand they now need to collect an elephant to complete their set. The family is interested in how manypackets they will need to buy before they complete their set.

(i) Stating any necessary assumption, name an appropriate distribution with which to mode l thissituation. What is the expected number of packets that they family will need to buy? [3]

(ii) Find the probability that the family will complete their set with the third packet they buy after the startof the year. [2]

(iii) Find the probability that, in order to complete their collection, the family will need to buy more than 4packets after the start of the year. [3]

4 A sixth-form class consists of 7 girls and 5 boys. Three students from the class are chosen at random. Thenumber of boys chosen is denoted by the random variable X. Show that

(i) 447)0P( ==X , [2]

(ii) 227)2P( ==X . [3]

The complete probability distribution of X is shown in the following table.

221

227

4421

447)P(

3210xX

x=

Calculate )E(X and )(Var X . [5]



5

The diagram shows the cumulative frequency graphs for the marks scored by the candidates in anexamination. The 2000 candidates each took two papers; the upper curve shows the distribution of markson paper 1 and the lower curve shows the distribution on paper 2. The maximum mark on each paper was100.

(i) Use the diagram to estimate the median mark for each of paper 1 and paper 2, and the interquartilerange for paper 1. [6]

(ii) State with a reason which of the two papers you think was the easier one. [2]

(iii) The candidates’ marks for the two papers could also be illustrated by means of a pair of box-andwhisker plots. Give two brief comments on any advantages or disadvantages in using cumulativefrequency graphs and box-and-whisker plots to represent the data. [2]

6 Items from a production line are examined for any defects. The probability that any item will be found tobe defective is 0.15, independently of all other items.

(i) A batch of 16 items is inspected. Using tables of cumulative binomial probabilities, or otherwise, findthe probability that

(a) at least 4 items in the batch are defective, [2]

(b) exactly 4 items in the batch are defective. [2]

(ii) Five batches, each containing 16 items, are taken.

(a) Find the probability that at most 2 of these 5 batches contain at least 4 defective items. [4]

(b) Find the expected number of batches that contain at least 4 defective items. [2]



7 An experiment was conducted to see whether there was any relationship between the maximum tidalcurrent, 1s cm −y , and the tidal range, x metres, at a particular marine location. [The tidal range is thedifference between the height of high tide and the height of low tide.] Readings were taken over a periodof 12 days, and the results are shown in the following table.

x 2.0 2.4 3.0 3.1 3.4 3.7 3.8 3.9 4.0 4.5 4.6 4.9

y 15.2 22.0 25.2 33.0 33.1 34.2 51.0 42.3 45.0 50.7 61.0 59.2

.]78.1837 ,75.91520 ,69.164 ,9.471 ,3.43[ 22 =Σ=Σ=Σ=Σ=Σ xyyxyx

The scatter diagram below illustrates the data.

(i) Calculate the product moment correlation coefficient for the data, and comment briefly on youranswer with reference to the appearance of the scatter diagram. [4]

(ii) Calculate the equation of the regression line of maximum tidal current on tidal range. [3]

(iii) Estimate the maximum tidal current on a day when the tidal range is 4.2 m, and indicate briefly howreliable an estimate you consider your answer to be. [3]

(iv) It is suggested that the equation found in part (ii) could be used to predict the maximum tidal currenton a day when the tidal range is 15 m. Comment briefly on the validity of this suggestion. [1]






MARK SCHEME

MAXIMUM MARK 60




1 Mean is 84.8 minutes B1

28.8425

180044deviation Standard −= M1 May be implied

27.3= minutes A1 3--------------------------------------------------------------------------------------------------------------------------------------------------------John’s average time is about 5 mins less than Janet’s B1John’s times are more variable than Janet’s B1 2

2 Ranks are: 3514224351 B1 Or with ranks reversed

Values of d are 1 ,1 ,2 ,1 ,1 −−− M1 Or reversed, or values of 2d

6.024586

1 =××

−=sr M1 Correct formula for Spearman used

A1 4 Correct answer (fraction or decimal)--------------------------------------------------------------------------------------------------------------------------------------------------------

(e.g.) B2 2 For 5 points, showing any non-linear‘increasing’ relationship

3 (i) Each packet is equally likely to contain any of the5 animals, independently of other packets B1 Allow either ‘equally likely’ or ‘independent’Geometric distribution B1 No need to state 5

1=p here

Expected number is 5 B1 3--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) ( ) ( ) 128.0or 12516

512

54 =× M1 Any numerical ‘ pqn ’ calculation

A1 2 Correct answer--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) ( ) ( )( ) ( ) ( ) ( ) ( ) 513

54

512

54

51

54

514

54 1or +++− M1 Allow M mark even if there is an error of 1 in

the number of termsA1 For correct expression for the answer

410.0or 4096.0or 625256 A1 3 Correct fraction or decimal

4 (i) 447

22035

312

37)0P( ==

÷

==X M1 For ratio of relevant

rn

terms, or equivalent

A1 2 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) )girl 1 and boys 2P()2P( ==X B1 May be implied

227

220107

312

25

17 =×=

÷

×

= M1 For use of relevant

rn

terms, or complete

alternative multiplication/addition of probsA1 3 Given answer correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------

45

221

227

4421

447 3210)E( =×+×+×+×=X M1 For correct calculation process

A1 Correct answer (fraction or decimal form)

4495

221

227

4421

4472 9410)E( =×+×+×+×=X B1 For correct numerical expression for px2Σ

( ) 597.0or )(Var 1761052

45

4495 =−=X (to 3dp) M1 For correct overall method for variance

A1 5 For correct answer



5 (i) Medians correspond to 1000 candidates M1 Reading off at 1000; may be implied63 ,38 21 == mm A1 Correct value for either median

A1 For both correctQuartiles correspond to 1500 and 500 candidates M1 Reading off at either; may be implied

26 ,56 13 == qq A1 Both correct30IQR = A1 6

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Paper 2 was easier B1

Marks were higher on paper 2 B1 2 Or similar statement, e.g. ‘higher median’--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) Possible valid comments include:Box plots give quick direct comparisons of

medians and IQRsBox plots don’t include all the information that

CF graphs doCF graphs can be used to read off values both

ways round B1 For any one valid commentetc B1 2 Any other valid comment

6 (i) (a) )1(210.07899.01 =− M1 Complement of relevant tabular value

A1 2 Correct answer--------------------------------------------------------------------------------------------------------------------------------------------------------

(b) 131.07899.09209.0 −− M1 Subtracting relevant tabular valuesA1 2 Correct answer

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) (a) 2345 210.0790.010210.0790.05790.0 ××+××+ M1 Any use of B(5, 0.210)

M1 Correct 3 cases identifiedA1 Correct numerical expression for required

probability, with their value from (i)(a)0.934= A1 4

--------------------------------------------------------------------------------------------------------------------------------------------------------(b) 05.1210.05 =× M1 For relevant use of np

A1 2 For correct answer

7 (i)

−

−

×−

=

129.471

75.2091512

3.4369.164

129.4713.43

78.1837

22r M1 Or equivalent; may be implied

956.0= A1The value is close to 1+ , B1 For relating the value to 1

and the points in the diagram lie (fairly) close toa straight line with positive gradient B1 4 For a reasonable comment about linearity

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Gradient of regression line is

9789.15

123.43

69.164

129.4713.43

78.18372

=−

×−

M1 May be implied if calculator routine used

−=−

123.43

9789.1512

9.471xy M1 May similarly be implied

3.180.16 −= xy A1 3--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) 3.182.40.16 −×=y M1

Current is 1s cm 8.48 − A1 Units required in answerDiagram indicates some uncertainty, e.g. 1s cm 5 −± B1 3 Allow any reasonable comment

--------------------------------------------------------------------------------------------------------------------------------------------------------(v) The prediction would be (very) unreliable because

of the extrapolation involved B1 1 For conclusion and idea of extrapolation



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90


1 Before being packed in boxes, apples in a fruit-packing plant have to be checked for bruising. The applespass along a conveyor belt, and an inspector removes any of the apples that are badly bruised. Badlybruised apples arrive at random times, but at a constant average rate of 1.8 per minute.

(i) Find the probability that at least one badly bruised apple arrives in a one-minute period. [3]

(ii) In a period of a minutes, the probability of at least one badly bruised apple arriving is 0.995. Find thevalue of a. [3]

2 A student answers a test consisting of 16 multiple-choice questions, in each of which the correct responsehas to be selected from the four possible answers given. The student only gets 2 of the questions correct,and the teacher remarks that this ‘shows that the student did worse than anyone would do just by guessingthe answers’. The probability of the student answering a question correctly is denoted by p, assumed to bethe same for all the questions in the test.

(i) State suitable null and alternative hypotheses, in terms of p, for a test to examine whether theteacher’s remark is justified. [2]

(ii) Carry out the test, using a 10% significance level, and state your conclusion clearly. [4]

3 Lessons in a school are supposed to last for 40 minutes. However, a Mathematics teacher finds that pupilsare usually late in arriving for his lessons, and that the actual length of teaching time available can bemodelled by a normal distribution with mean 34.8 minutes and standard deviation 1.6 minutes.

(i) Find the probability that the length of teaching time available will be less than 37.0 minutes. [2]

(ii) The probability that the length of teaching time available exceeds m minutes is 0.75. Find m. [3]

The teacher has a weekly allocation of 5 lessons with a particular class. Assuming that these 5 le ssons canbe regarded as a random sample, find the probability that the mean length of teaching time available inthese 5 lessons will lie between 34.0 and 36.0 minutes. [4]

4 It is given that 93% of children in the UK have been immunised against whooping cough. The number ofchildren in a random sample of 60 UK children who have been immunised is X, and the number who havenot been immunised is Y. State, with reasons, which of X or Y has a distribution which can beapproximated by a Poisson distribution. [3]

Using a Poisson approximation, find the probability that at least 58 children in the sample of 60 have beenimmunised against whooping cough. [3]

Three random samples, each of 60 UK children, are taken. Find the probability that in one of thesesamples exactly 59 children have been immunised while in each of the other two samples exactly 58children have been immunised. Give your answer correct to 1 significant figure. [3]



5 The continuous random variable X has probability density function f given by

≤≤−=

otherwise,0,30)3()f(

2 xxkxx

where k is a constant.

(i) Show that 274=k , and find )E(X . [5]

(ii) Find )2P( <X . [2]

(iii) Use your answer to part (ii) to state, with a reason, whether the median of X is less than 2, equal to 2or greater than 2. [2]

6 The ‘reading age’ of children about to start secondary school is a measure of how good they are at readingand understanding printed text. A child’s reading age, measured in years, is denoted by the randomvariable X. The distribution of X is assumed to be ) ,N( 2σµ . The reading ages of a random sample of 80

children were measured, and the data obtained is summarised by 7.892=Σx , 82.266 102 =Σx .

(i) Calculate unbiased estimates of µ and 2σ , giving your answers correct to 2 decimal places. [3]

(ii) Previous research has suggested that the value of µ was 10.75. Determine whether the evidence ofthis sample indicates that the value of µ is now different from 10.75. Use a 10% significance levelfor your test. [5]

(iii) State, giving a brief reason, whether your conclusion in part (ii) would remain valid if

(a) the distribution of X could not be assumed to be normal, [1]

(b) the 80 children were all chosen from those starting at one particular secondary school. [1]

7 The breaking strength of a certain type of fishing line has a normal distribution with standard deviation0.24 kN. A random sample of 10 lines is tested. The mean breaking strengths of the sample and of thepopulation are kN x and kN µ respectively. The null hypothesis 75.8=µ is tested against the

alternative hypothesis 75.8<µ at the %2 21 significance level.

(i) Show that the range of values of x for which the null hypothesis is rejected is given by 60.8<x ,correct to 2 decimal places. [4]

(ii) Explain briefly what is meant, in the context of this question, by a Type I error, and state theprobability of making a Type I error. [2]

(iii) Explain briefly what is meant, in the context of this question, by a Type II error, and find theprobability of making a Type II error when 50.8=µ . [5]



BLANK PAGE






MARK SCHEME

MAXIMUM MARK 60




1 (i) 8.1e1)oneleast at P( −−= B1 For any use of Po(1.8)

M1 For use of complementary probability835.0= A1 3

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 995.0e1 8.1 =− − a B1

005.0ln8.1 =− a M1 For correct use of logs in solving for a94.2=a A1 3

2 (i) Null hypothesis: 41=p B1

Alternative hypothesis: 41<p B1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Under the NH, ) ,16B(~correctnumber 4

1 M1 May be implied

Hence 1971.0)correctfewer or 2P( = A1 Using tables or direct calculation

This is greater than 0.1 (not significant) M1 For comparing with the significance levelThere is insufficient evidence to justify the

teacher’s suggestion that the score was worsethan would be produced by pure guesswork A1 4

3 (i) 915.0)375.1(6.1

8.340.37)0.37P( =Φ=

−Φ=<T M1 Standardising and using tables

A1 2 Correct answer 0.915--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) 674.06.1

8.34 −=−mM1 Equating standardised m to a z value

B1 For use of 674.0)(± in an equation

Hence 7.33=m A1 3--------------------------------------------------------------------------------------------------------------------------------------------------------

−Φ−

−Φ=<<

5/6.18.340.34

5/6.18.340.36)0.360.34P( T M1 For using )5/6.1 ,8.34N( 2

A1 For both end-points standardised correctly)118.1(1)677.1( Φ−−Φ= M1 Correct process for prob between end-points

821.0= A1 4

4 )07.0 ,60B(~ ),93.0 ,60B(~ YX M1 For either binomial distribution identified

Hence Y is suitable for a Poisson approximation, A1 For correct conclusionsince n is large and p is small A1 3 For correct justification

--------------------------------------------------------------------------------------------------------------------------------------------------------)2.4(Po~Y B1 May be implied

Required probability is )2P( ≤Y M1 For attempted evaluation of relevant Po prob

i.e. 0.210 A1 3 Using tables or direct calculation--------------------------------------------------------------------------------------------------------------------------------------------------------

Required probability is 3e22.4

e2.42

2.42

2.4 ×

× −− B1 For 221 ppp ×× , using Poisson or binomial

B1 For correct factor of 3i.e. 0.003 to 1 sf B1 3 Follow through wrong value of 4.2 only



5 (i) 1d )3(3

032 =−∫ xxxk M1 Equating to 1 and attempting to integrate

[ ] 130

4413 =− xxk A1 For correct integration

2741

48127 =⇒=

− kk A1 Given answer correctly obtained

∫ −=3

043

274 d )3()E( xxxX M1 For correct application of ∫ xxx d )f(

59= A1 5

--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) [ ]20

4413

274)2P( xxX −=< M1

2716= A1 2 Or equivalent decimal answer

--------------------------------------------------------------------------------------------------------------------------------------------------------(iii) 27

16 is greater than 21 M1 For comparison of answer (ii) with 2

1

Hence the median is less than 2 A1 2

6 (i) 16.1180

7.892 ==x to 2dp B1 For correct value 11.16

−=

807.892

82.662 10791 2

2s M1 For this expression, or equivalent

87.3= to 2dp A1 3--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 75.10:H0 =µ , 75.10:H1 ≠µ B1 Both hypotheses

Test statistic is 86.180/87.3

75.1016.11=

−=z M1 Standardising attempt using 80/2s

A1 Correct value; follow their sThis is greater than critical (2 tail) value 645.1=z M1 Or comparing )86.1(Φ with 5%There is evidence to suggest that the value of µ is

now different A1 5--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) (a) Still valid, since the sample size (80) is large Allow any reasoned conclusion mentioningenough to appeal to the CLT B1 1 the CLT

--------------------------------------------------------------------------------------------------------------------------------------------------------(b) Not valid, since the children starting at one

school may not be representative of allchildren of this age. B1 1 For conclusion and reason

7 (i) Critical value is 1024.0

96.175.8 ×− M1 Calculation of correct form S.E.75.8 ×− z

B1 Relevant use of 96.1−A1 Relevant use of 10/24.0

i.e. reject null hypothesis when 60.8<x A1 4--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) NH 75.8=µ would be rejected when the mean

breaking strength is in fact 8.75 kN B1025.0)error I Type(P = B1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------(iii) NH 75.8=µ would be accepted when the mean

breaking strength is in fact less than 8.75 kN B1Type II error occurs when 60.8>x B1 May be implied

Probability is

−Φ−

10/24.050.860.81 M1 Using normal distribution with mean 8.50

A1 Correct standardising, and use of tables0938.0= A1 5



BLANK PAGE


















1 A test of the null hypothesis of independence of two characteristics is required for the followingcontingency table.

Characteristic 1 Totals

12 9 19 40Characteristic 2

28 61 71 160

Totals 40 70 90 200

(i) Find the expected frequency corresponding to the cell with observed frequency 12. [1]

(ii) Given that the value of the 2χ test statistic is 4.80, correct to 3 significant figures, carry out the test atthe 10% significance level. [3]

2 The ‘customer services’ section in a department store deals both with enquiries by telephone and also withenquiries in person by shoppers in the store. Telephone enquiries occur at random times at an average rateof 4 per half hour. Shoppers in the store arrive to make enquiries at random times at an average rate of 5per half hour. Assuming that the two types of enquiries occur independently of each other, find theprobability that a total of between 10 and 20 enquiries (inclusive) have to be dealt with in a randomlychosen half hour period. [5]

3 A supermarket sells 2 kg bags of new potatoes, in which the potatoes have been selected to be all roughlythe same size. The potatoes used to fill the bags may be assumed to be randomly chosen items from apopulation in which the mass, in grams, of an individual potato is normally distributed with mean 90 andstandard deviation 4.

(i) Show that the probability that the total mass of 21 of these potatoes exceeds 2 kg is very small. [4]

(ii) Find the probability that the total mass of 22 of these potatoes exceeds 2 kg. [2]

(iii) The machine filling the bags delivers potatoes one by one until a total mass of at least 2 kg is reached.Show that the bags are almost certain to contain either 22 or 23 potatoes. [2]

4 A random sample of six observations of the random variable X gave the following values:

.7.1,1.6,1.2,0.4,8.1,2.3 −.]99.73,7.14[ 2 =Σ=Σ xx

The population mean of X is µ . Calculate a 90% confidence interval for µ ,

(i) assuming that X has a normal distribution with variance 8.5,

(ii) assuming instead that X has a normal distribution with unknown variance.[9]



5 State what distributional assumptions are necessary for it to be valid to use

(i) the two-sample t-test,

(ii) the paired-sample t-test,

to test for a difference in population means. [3]

Two different types of nylon fibre were tested for the amount of stretching under tension. Ten randomsamples of each fibre, of the same length and diameter, were stretched by applying a standard load. ForFibre 1 the increases in length, x mm, were as follows.

12.84 14.26 13.23 14.75 15.13 14.15 13.37 12.96 15.02 14.38.]0513.1969 ,09.140[ 2 =Σ=Σ xx

For Fibre 2 the increases in length, y mm, were as follows.

14.27 13.25 14.17 13.11 14.92 12.12 14.21 13.68 15.14 14.81

.]9794.1958 ,68.139[ 2 =Σ=Σ yy

Assuming that any necessary conditions for the validity of your test hold, test whether the mean increase inlength of the two types of fibre is different. Use a 10% significance level. [7]

6 Six hens are observed over a period of 20 days and the number of eggs laid each day is summarised in thefollowing table.

Number of eggs 3 4 5 6

Number of days 2 2 10 6

Show that the mean number of eggs per day is 5. [2]

It may be assumed that a hen never lays more than one egg in any day. State one other assumption thatneeds to be made in order to consider a binomial model, with 6=n , for the total number of eggs laid in aday. [1]

Calculate the expected frequencies using a binomial model for the above data and carry out a 2χ goodnessof fit test, using a 10% significance level. [9]



7 The continuous random variable X has a triangular distribution with probability density function given by

≤≤−

≤≤−+=

otherwise.0,101

,011)f( xx

xxx

Show that, for 10 ≤≤ a ,

22)P( aaaX −=≤ . [3]

The random variable Y is given by 2XY = . Express )P( yY ≤ in terms of y, for 10 ≤≤ y , and henceshow that the probability density function of Y is given by

11

)g( −=y

y , for 10 ≤< y . [3]

Use the density function of Y to find )E(Y , and show how the value of )E(Y may also be obtained directlyusing the probability density function of X. [4]

Determine )E( Y . [2]






MARK SCHEME

MAXIMUM MARK 60




1 (i) 8200

4040=

×B1 1 For correct answer; working not needed

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 4.80 is greater than critical value 4.605 M1 Comparing with a tabular value

A1 Using correct figure 4.605There is evidence to suggest that the characteristics

are not independent A1 3

2 Model for all enquiries is Po(9) B1 For any mention of a Poisson distributionM1 Summing two Poisson distributionsA1 For statement of correct parameter 9

5874.09996.0yProbabilit −= M1 Subtracting relevant tabular values412.0= A1 5

3 (i) )336 ,1890N(~21T M1 For normal distribution with correct mean

A1 For variance 2421 × (both these first twomarks may be implied by later working)

)6(1336

189020001)2000P( 21 Φ−=

−Φ−=>T M1 For correct processes for relevant tail

Hence prob is very small since 1)6( ≈Φ A1 4 Correct conclusion, based on suff large z--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii)

××−Φ−=>

1622902220001)2000P( 22T M1 For relevant new normal calculation

143.0)066.1(1 =Φ−= A1 2 For correct probability--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) 1)649.3(368

207020001)2000P( 23 ≈Φ=

−Φ−=>T M1 For relevant calculation for 23 potatoes

Hence 23 potatoes is almost always enough A1 2 For correct conclusion based on correct figs

4 (i) 45.26

7.14 ==x B1 At any stage; may be implied

Interval is 9579.145.265.8645.145.2 ±=×± M1 For calculation of the form nzx /2σ±

A1 For relevant use of 645.1=zi.e. 41.449.0 << µ A1 4 For correct interval

--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) 595.767.14

99.7351 2

2 =

−=s M1 For correct unsimplified expression for 2s

A1 For correct value (may be implied)

Interval is 267.245.26595.7015.245.2 ±=×± M1 For calculation of the form nstx /2±

A1 For relevant use of 015.2=ti.e. 72.418.0 << µ A1 5 For correct interval



5 (i) The distributions must be normal B1The distributions must have equal variances B1

(ii) The differences must be normally distributed B1 3--------------------------------------------------------------------------------------------------------------------------------------------------------

0:H ;0:H 10 ≠−=− yxyx µµµµ B1 For both

1810

68.1399794.1958

1009.140

0513.196922

2

−+

−=pS M1 Use of correct formula

8033.0= A1 May be implied

Test statistic ( ) 1023.08033.0

968.13009.14

101

101

=+

−=t M1 Use of correct formula

A1 Correct value of tThis is less than the critical value 1.734 M1 Comparing to tabular value of tThere is not significant evidence of a difference in mean

increase A1 7

6 520

661052423Mean =

×+×+×+×= M1

A1 2 Obtain given answer correctly--------------------------------------------------------------------------------------------------------------------------------------------------------Assume that the probability of any hen laying an egg on

any day is constant and that hens lay eggsindependently of each other B1 1 For either constant prob or independence

--------------------------------------------------------------------------------------------------------------------------------------------------------Distribution to be fitted is ( )6

5 ,6B B1 May be implied

Expected frequencies are ( ) ( ) iii if

−××

×=

661

65620 M1 For any one calculation using correct method

02.4 ,04.8 ,70.6 456 === fff A1 All three correct to at least 2dp

00.0 ,02.0 ,16.0 ,07.1 0123 ==== ffff A1 For all four correct, or 25.13 =≤f stated

Combining cells: 70.604.827.56104654

e

off

≤M1 Use of criterion 5<ef for combining

857.070.670.0

04.896.1

27.527.1 222

2 =++=χ M1

A1 Allow anything between 0.85 and 0.86 (inc)This is less than the critical value 2.706 M1 Compare with tabular valueHence there is a satisfactory fit A1 9



7 )P()P( aXaaX <<−=< M1 For consideration of two areas, or equiv

∫∫ −++=−

a

axxxx

0

0d )1(d )1( A1 Or equivalent trapezium areas

[ ] [ ] 20

22102

21 2 aaxxxx a

a−=−++=

−A1 3 Given answer correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------( ) yyyXyXyY 2P)P()P( 2 −=≤=≤=≤ B1 For correct expression

Hence the pgf of Y is ( ) 11 2dd −=−

yyy

yM1 For differentiation of previous expression

A1 3 Given answer correctly shown--------------------------------------------------------------------------------------------------------------------------------------------------------

[ ]61

1

021

321

021

23

21

d )E( =−=−= ∫ yyyyyY M1 For correct integral

A1 For answer 61

∫∫ −++=−

1

032

0

1322 d )(d )()E( xxxxxxX M1

[ ] [ ] 61

121

1211

04

413

310

14

413

31 =+=−++=

−xxxx A1 4

--------------------------------------------------------------------------------------------------------------------------------------------------------

[ ]31

1

0321

0

1

023

21

21

d )1(d )g( )E( =−=−== ∫∫ yyyyyyyY M1 For forming the correct integral

A1 2 Correct answer


















1 To compare the effect of a new drug on men and women, a small-scale trial was conducted, involving 6randomly chosen men and 4 randomly chosen women. For each person, the time, in minutes, before thedrug began to take effect was recorded, with the results shown in the table below.

Men 19 32 24 33 31 28

Women 14 13 18 27

Use a suitable non-parametric test to determine if there is evidence, at the 5% significance level, that thedrug acts more quickly in women than it does in men. [5]

2 The continuous random variable X has probability density function given by

<≥=

−

,00,0e)f(

xxx

xλλ

where λ is a positive constant. Show that the moment generating function of X is given by

ttX −

=λ

λ)(M . [3]

Use this moment generating function to find the mean and variance of X. [4]

3 Events A and B are such that

21)P( =A , 3

1)P( =B , 43)P( =∪ BA .

(i) Determine, giving your reasons clearly, whether A and B are

(a) mutually exclusive, [1]

(b) independent. [3]

(ii) Find the values of

(a) )|P( BA , [2]

(b) )|P( BA ′ , where B′ denotes the complement of B. [3]



4 A bin contains a large number of seeds of which 20% will produce a plant with a red flower and 30% willproduce a plant with a yellow flower. The remaining 50% will fail to germinate. Two seeds are chosen atrandom from the bin and planted in a pot. The random variables R and Y denote the number of red andyellow flowers, respectively, that will be produced from the seeds. The table below shows the jointprobability distribution of R and Y.

R0 1 2

0 0.25 0.20 0.04Y 1 0.30 0.12 0

2 0.09 0 0

(i) Find the marginal distributions of R and Y. [2]

(ii) Show that 4.0)(E =R and find )(E Y . [2]

(iii) Find ) ,(Cov YR . [3]

(iv) Find the distribution of Y conditional on 1=R , and hence state the expected value of the number ofyellow flowers in a pot in which there is one red flower. [2]

5 The discrete random variable X denotes the score obtained in a single throw of an ordinary fair die. Showthat the probability generating function of X may be expressed as

)1(6)1( 6

ttt

−− . [2]

Write down the probability generating function for the total score obtained when three fair dice are thrown.[1]

Hence show that the probability of obtaining a total score of 10 when three fair dice are thrown is 81 . [6]

6 Explain briefly the circumstances under which a non-parametric test of significance should be used inpreference to a parametric test. [1]

The acidity of soil can be measured by its pH value. As a part of a Geography project a student measuredthe pH values of 14 randomly chosen samples of soil in a certain area, with the following results.

37.568.589.571.610.516.552.680.541.510.676.664.673.567.5

Use a suitable non-parametric test to test whether the average pH value for soil in this area is 6.24. Use a10% level of significance. [4]

Some time later, the pH values of soil samples taken at exactly the same locations as before were againmeasured. It was found that, for 3 of the 14 locations, the new pH value was higher than the previousvalue, while for the other 11 locations the new value was lower. Test, at the 5% significance level,whether there is evidence that the average pH value of soil in this area is lower than previously. [4]



7 The continuous random variable X has a uniform distribution on the interval ax ≤≤0 , where the value ofthe parameter a is unknown. Three independent observations, 321 , , XXX , of X are taken.

(i) An estimator θ is defined by )( 32132 XXX ++=θ . Show that θ is an unbiased estimator of a, and

find ) [6](Var θ in terms of a. [6]

(ii) Another estimator φ is based on the greatest of the three values 321 , , XXX . Denoting the greatest ofthe three values by the variable G, use the fact that, for any value x between 0 and a,

) and and ( 321 xXxXxXxG <<<⇔<

to write down the cumulative distribution function of G, and hence to obtain the probability densityfunction of G. [2]

Hence show that, if G34=φ , then φ is an unbiased estimator of a, and determine which of θ and φ

is the more efficient estimator. [4]






MARK SCHEME

MAXIMUM MARK 60




1 0H : same distributions of times for men and women,

1H : lower average for women B1 For both NH and AH

Ranks are: 6 312W

7810594M M1 For ranking all 10 values

Test statistic is 126321 =+++ A1 For sum of women’s ranksCritical Wilcoxon rank-sum value is 13 M1 For comparing with correct tabular valueHence reject 0H and conclude that there is evidence to

suggest that drug acts more quickly in women than itdoes in men A1 5

2 ∫∞ −=0

d ee)(M xt xtxX

λλ B1 Correct integral stated

∞−

−=

0

)(e xtt

λλ

λM1 For correct integration method

t−=

λλ

A1 3 Given answer correctly shown

--------------------------------------------------------------------------------------------------------------------------------------------------------

K+

++=

2

1)(MλλtttX B1 Three correct terms

λ1

)E( =X B1

2

2

211!2)Var(λλλ

=

−=X M1 For using correct variance formula

A1 4 Correct answer for variance

3 (i) (a) Not exclusive, since 43

31

21 ≠+ B1 1 Conclusion and reason both required

--------------------------------------------------------------------------------------------------------------------------------------------------------(b) 12

131

21

43)P( =−−=∩ BA M1 Use of correct formula

A1 Correct value 121

Not independent, as 121

31

21 ≠× A1 3 Conclusion and reason both required

--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) (a) 41

)P()P(

)|P( =∩

=B

BABA M1 Use of correct formula

A1 2 Follow answer to (i)(b)--------------------------------------------------------------------------------------------------------------------------------------------------------

(b) )P()P()P( BAABA ∩−=′∩ M1 Or equivalent, e.g. from Venn diagram

125

121

21 =−= A1

85

32

125)|P( =÷=′BA A1 3



4 (i) R : 04.032.064.0210 M1 Adding rows (or columns)

Y : 09.042.049.0210 A1 2 Both distributions correct

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 4.004.0232.0164.00)E( =×+×+×=R M1 Correct process for either mean

6.009.0242.0149.00)E( =×+×+×=Y A1 2 Both means correct--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) 12.0)E( =RY B1

6.04.012.0) ,Cov( ×−=YR M1

12.0−= A1 3--------------------------------------------------------------------------------------------------------------------------------------------------------

(iv) Conditional distribution is: 0210

83

85 B1

Expected value is 83 B1 2

5 PGF is )( 6543261 tttttt +++++ B1

i.e. )1(6)1( 6

ttt

−−

B1 2 Use of GP sum to deduce given answer

--------------------------------------------------------------------------------------------------------------------------------------------------------

3

363

)1(216)1(t

tt−

−B1 1 For cube of answer (i), in any form

--------------------------------------------------------------------------------------------------------------------------------------------------------)631)(31( 263

2161 KK ++++− tttt M1 For using both binomial expansions

A1 Both correct at least as far as shownTerms in 7t from product of brackets are: M1 Picking out relevant term(s)

tt 33 6 ×− A17361 t× A1

81

2161 )936(yProbabilit =−= A1 6 Given answer correctly shown

6 A non-parametric test is needed when there is noinformation (or reasonable assumption) availableabout an underlying distribution B1 1

--------------------------------------------------------------------------------------------------------------------------------------------------------Deviations from NH value 6.24 are:

87.056.035.047.014.108.128.044.083.014.052.040.051.057.0

−−−−−−−−−− M1 For calculating signed differences from 6.24

Signed ranks are:1293614132511184710

−−−−−−−−−− A1 For calculating correct signed ranks

Test statistic is 25208642 <=+++ M1 For calculating T and comparingConclude that there is evidence to suggest that the

average pH value is not 6.24 A1 4 For correct conclusion based on correct work--------------------------------------------------------------------------------------------------------------------------------------------------------

0H : same average pH as before; 1H : lower value B1 For both hypotheses

Under 0H , tail probability for 3 or fewer out of 14 is M1 For use of relevant binomial distribution

( ) 05.00287.0)36491141(1421 <=+++× A1 For correct value tail probability 0.0287

Hence reject 0H and accept that average pH is lower A1 4 For correct conclusion based on correct work



7 (i) aX 21)E( = B1 May be implied

)E(3)E( 32 X×=θ M1

a= , as required A1 For conclusion correctly shown and stated

( )221

0

2d )Var( ax

axX

a−⌡

⌠= M1

2121 a= A1

2912

123

94)Var( aa =×=θ A1 6

--------------------------------------------------------------------------------------------------------------------------------------------------------

(ii)3

)P(

=<

axxG B1

Hence pdf is 3

23ax

B1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------

⌡⌠=

ax

ax

03

3d 3

34)E(φ M1

a= , as required A1 For conclusion correctly shown and stated

2

03

4d 3

916)Var( ax

axa

−⌡⌠=φ M1

2912

151 aa <=

Hence φ is more efficient A1 4 Correct )Var(φ and conclusion





MATHEMATICS D1Discrete Mathematics 1













1 Andy wants to record the following twelve TV programmes onto video tape. Each video tape has space forup to three hours of programmes.

Programme

Length (hours) 221111111 43

21

21

43

21

21

LKJIHGFEDCBA

(i) Suppose that Andy records the programmes in the order A to L using the first fit algorithm. Find thenumber of tapes needed, and show which programmes are recorded onto which tape. [3]

(ii) Suppose instead that Andy is transferring the programmes from previously recorded tapes, so thatthey can be copied in any order, and that Andy uses the first fit decreasing algorithm. Find thenumber of tapes needed, and show which programmes are recorded onto which tape. [3]

2

(a) (b) (c)

(i) By considering the order of each node, classify each of the graphs (a), (b) and (c) shown in thediagram as Eulerian, semi-Eulerian or neither. [4]

(ii) Explain briefly how your classification in part (i) relates to the problem of finding a route through thegraph that includes each arc exactly once. Are there any restrictions on where such a route can startand finish? [3]

3 A company has offices is six towns, A, B, C, D, E and F. The costs, in £, of travelling between these townsare shown in the table.

Town A B C D E FA – 15 26 13 14 25B 15 – 16 16 25 13C 26 16 – 38 16 15D 13 16 38 – 15 19E 14 25 16 15 – 14F 25 13 15 19 14 –

Use Prim’s algorithm, starting by deleting row A, to find the cheapest way of connecting the six towns.You should show all your working and indicate the order in which the towns were included. [7]



4 An express delivery pizza company promises to deliver pizzas within 30 minutes of an order beingtelephoned in. Four customers telephone in orders at the same time. While the pizzas are cooking, whichtakes 10 minutes, a delivery route must be planned for the person who will deliver all four pizzas. Thediagram below shows the pizza company P and the four customers A, B, C and D, and the accompanyingtable shows the travel times for each possible leg of a journey between them.

The travel times shown exclude the time taken to stop at a customer’s house and deliver the pizza; thisstopping and delivery time is 2

12 minutes.

(i) Explain why a travelling salesperson solution taking less than 13 minutes guarantees that all fourcustomers will get their pizzas (including stopping and delivery times) within 30 minutes of theirtelephone call. [4]

(ii) The pizzas are delivered using the nearest neighbour algorithm, as follows.

The first delivery is to the customer who is nearest to P (in the sense of having the shortest traveltime).

The second delivery is to the customer who is nearest to the first one.

The third delivery is to the customer nearest to the second who is still waiting for a pizza.

The fourth delivery is to the remaining customer.

Write down the order in which the pizzas are delivered using this algorithm, and calculate how longthe fourth customer has to wait for their pizza (including the stopping and delivery time). [4]

P A B C DP – 5 3 4 2A 5 – 1 4 4B 3 1 – 3 5C 4 4 3 – 7D 2 4 5 7 –



5 Lou Zitt has a budget of £2000 to spend on storage units for his office. The storage units must not covermore than 2m 50 of floor space, and Lou wants to maximise the storage capacity. The three types ofstorage unit that he can choose from are shown in the following table.

Type Storage capacity (m3) Floor space covered (m2) Cost (£)

Antique pine units 3 1 100Beech wood units 8 4 500Cedar wood units 6 3 200

Suppose that Lou buys a antique pine units, b beech wood units and c cedar wood units.

(i) Write down two constraints that must be satisfied by a, b and c, other than 0≥a , 0≥b , 0≥c . [2]

(ii) Write down the objective function for this problem. [1]

(iii) Set up the problem as an LP formulation. (You are not expected to solve the problem.) [4]

(iv) Identify which aspect of the original problem has been overlooked in the LP formulation. [1]

6 A graph has five vertices, P, Q, R, S, T, and each vertex is directly connected to every other vertex.Describe how to apply Dijkstra’s algorithm to find the shortest path from P to T, and explain why thisrequires 6 addition calculations in the worst case. [6]

Show that when Dijkstra’s algorithm is used on a graph with six vertices it requires 10 additioncalculations in the worst case. [2]

The number of additions affects the amount of time that Dijkstra’s algorithm takes to run on a computer.

(i) Assuming that the problem has already been put into a suitable format, what is the other main factorthat would affect the time that Dijkstra’s algorithm takes to run on a computer? [1]

(ii) Dijkstra’s algorithm is of quadratic order (order 2n ). Explain what this means. [2]

7 A linear programming problem gives the following LP formulation.

Maximise zyxP 543 ++=

subject to26393582

≤++≤++

zyxzyx

and .0 ,0 ,0 ≥≥≥ zyx

(i) Set up an initial simplex tableau for this problem. Perform two iterations, choosing first to pivot onan element chosen from the z column. [9]

(ii) State the values of x, y, z and P that result from each of the two iterations carried out in part (i). [2]

(iii) Explain how you know whether or not the optimal solution has been achieved. [2]






MARK SCHEME

MAXIMUM MARK 60




1 (i) A, B, C, D all fit onto one tape but E doesn’t M1 Applying the algorithm with first tapeTapes hold ABCD, EFG, HI, J, K, L A1So 6 tapes are needed A1 3

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) First tape contains K and D (or other ‘2 + 1’ pair) M1 Applying the algorithm with first tape

Tapes hold KD, LE, JF, HI, GCAB (or equiv) A1So 5 tapes are needed A1 3

2 (i) Orders are: (a) 55455EDCBA

(b) 34445EDCBA M1 For recognisable attempt for any one case

(c) 44444EDCBA

Hence (a) is neither, since there are 4 odd nodes A1(b) is semi-Eulerian, since there are 2 odd nodes A1(c) is Eulerian, since all the nodes are even A1 4

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) No such route exists for graph (a) B1

There are routes for graph (b), but they have tostart and finish at odd nodes, i.e. A and E B1

There are routes for graph (c), where all routesstart and finish at the same node, which can beany one of the five B1 3

3

B1 Deleting row A and choosing AD

M1 Deleting row D and choose from cols A, D

A1 Correct choice at this stage

M1 For carrying out the complete algorithm

Arcs: 15 ,13 ,14 ,14 ,13 ===== FCFBEFAEAD A1 All arcs correctOrder of adding is A, D, E, F, B, C B1Total cost is 69£1513141413 =++++ B1 7

A1 B C D E FA — 15 26 13 14 25B 15 — 16 16 25 13C 26 16 — 38 16 15D 13 16 38 — 15 19E 14 25 16 15 — 14F 25 13 15 19 14 —

A1 B C D2 E FA — 15 26 13 14 25B 15 — 16 16 25 13C 26 16 — 38 16 15D 13 16 38 — 15 19E 14 25 16 15 — 14F 25 13 15 19 14 —

A1 B C D2 E3 FA — 15 26 13 14 25B 15 — 16 16 25 13C 26 16 — 38 16 15D 13 16 38 — 15 19E 14 25 16 15 — 14F 25 13 15 19 14 —

A1 B C D2 E3 F4

A — 15 26 13 14 25B 15 — 16 16 25 13C 26 16 — 38 16 15D 13 16 38 — 15 19E 14 25 16 15 — 14F 25 13 15 19 14 —

A1 B5 C D2 E3 F4

A — 15 26 13 14 25B 15 — 16 16 25 13C 26 16 — 38 16 15D 13 16 38 — 15 19E 14 25 16 15 — 14F 25 13 15 19 14 —



4 (i) Travel time available is mins 10241030 21 =×−− B1

Return time from last delivery is not relevant M1This can be at least 3 mins (since any route can befollowed in either direction) A1Hence TSP solution of 13310 =+ or less suffices A1 4 Given answer correctly explained

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) The order is DABC B2 Allow B1 for DACB

Total time is 2124314210 ×+++++ M1 Cooking and stopping times required

i.e. 30 minutes A1 4

5 (i) 5034 ≤++ cba B12025 ≤++ cba B1 2 Allow any correct unsimplified version

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) cba 683 ++ B1 1 Allow in the form of an equation--------------------------------------------------------------------------------------------------------------------------------------------------------

(iii) Maximise cbaP 683 ++= B1Subject to 5034 ≤++ cba B1

and 2025 ≤++ cba B1Together with 0 ,0 ,0 ≥≥≥ cba B1 4

--------------------------------------------------------------------------------------------------------------------------------------------------------(iv) The variables a, b, c must be integers B1 1

6 Label Q, R, S, T with direct distances from P , and selectsmallest (say Q) B1 Correct description of labelling and selecting

Update distances to R, S, T if route from P via Q isshorter B1 Correct description of first stage of Dijkstra

The initial 3 possible updates require 3 additions so far B1Select the smallest of R , S, T and repeat the updating

process; this stage requires another 2 additions B1 Explanation of 2 additions at this stageThe final stage similarly requires 1 addition, so the total

is 6123 =++ B1 Given total of 6 correctly explainedTrace back to find the shortest path, including an arc XY

whenever the distance to X plus the length of XY isthe same as the distance to Y B1 6 For explanation of tracing back

--------------------------------------------------------------------------------------------------------------------------------------------------------Worst case number is 1234 +++ M1i.e. 10 A1 2--------------------------------------------------------------------------------------------------------------------------------------------------------(i) The number of comparisons B1 1

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) The number of operations needed to complete the

algorithm is (approximately) proportional to thesquare of the number of nodes B1 Allow e.g. ‘time taken prop to square of size’

So, for example, doubling the number of nodeswill (roughly) quadruple the amount to be done B1 2



7 (i) Simplex tableau is: M1 Table with correct numbers of rows and cols

210639030158200005431 −−−

zyxP

A1 Correct initial values throughout

Select 6 in the z column as pivot B1Result of first iteration is: M1 For correct row operations

31

61

21

21

31

65

21

21

32

65

21

21

0110110550100141

−−−

zyxP

A1 Correct values throughout

Select 215 in y column for second pivot B1

Result of second iteration is: M1 For correct row operations

337

338

111

338

335

112

3367

3320

113

102001100031

−−−

zyxP

A2 9 All values correct; allow A1 if one error

--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) After 1 iteration: 3

231 1 , ,0 ,0 ==== Pzyx B1

After 2 iterations: 3367

337

338 , , ,0 ==== Pzyx B1 2

--------------------------------------------------------------------------------------------------------------------------------------------------------(iii) There are no negative numbers in the top row

of the tableau B1Hence the solution is optimal B1 2


This question paper consists of 5 printed pages, 3 blank pages and an insert.

Specimen Materials – Mathematics © OCR 2000125















1

The diagram represents a system of pipes. The weights show the (directed) maximum capacity for eachpipe in litres per minute.

(i) Calculate the capacity of the cut C marked in the diagram. [2]

(ii) Draw a diagram showing a flow from S to T of 230 litres per minute in which each of the pipes JL,KM and MT is carrying its full capacity. [1]

(iii) Explain carefully what can be deduced from parts (i) and (ii) about flows through this network. [2]

2 Granny has bought Christmas presents for her five grandchildren.

The teddy bear is suitable for Cathy, Daniel or Elvis;the book is suitable for Annie or Ben;the football is suitable for Daniel or Elvis;the money box is suitable for Annie or Daniel;the drum is suitable for Cathy or Elvis.

Draw a bipartite graph, G, to show which present is suitable for which grandchild. [2]

Granny decides to give Annie the book, Cathy the teddy bear, Daniel the money box and Elvis the drum.This leaves Ben without a present, since the football is not suitable for him.

(i) Show the incomplete matching, M, that describes which present Granny had decided to give to eachchild. [1]

(ii) Use a matching algorithm to construct an alternating path for M in G, and hence find a maximalmatching between the presents and the grandchildren. [2]



3 Richard and Carol play a two-person zero-sum simultaneous play game. The table shows Richard’s pay-off matrix for the game.

CarolStrategy X Strategy Y Strategy Z

Strategy A 2 3 –2Richard Strategy B –4 –1 –1

Strategy C –5 0 1

(i) Find the play-safe strategy for each player, and hence show that this game does not have a stablesolution. [4]

(ii) Richard’s optimal mixed strategy can be found by using linear programming. Set up a suitable LPformulation, defining the symbols you use. (You are not required to solve the LP problem.) [4]

4 A relay team consists of four runners, A, B, C and D, each of whom runs one leg of the race. The besttraining times, in seconds, for each of the runners over each of the legs are given in the table.

1st leg 2nd leg 3rd leg 4th legA 47 45 45 43B 48 44 45 44C 46 45 44 43D 50 47 46 44

Use the Hungarian algorithm to decide which runner should be allocated to which leg of the race. [8]

5 The Rolling Pebbles have been playing as a band for many years. When they tour they sometimes play oldsongs, they sometimes play new songs and they sometimes play a mixture of old and new songs. Theirchoice depends on the age of the audience. The table shows the audience reaction (as a score out of 10) foreach of the possible combinations. High scores are good.

AudienceYoung Mixed Older

Old 1 3 8Songs played Mixture 3 1 2

New 8 5 3

Explain why, according to this data, the band should never choose to play a mixture of old and new songs.[1]

The band do not know whether their audience will be young, older or of mixed ages. Suppose that theychoose to play old songs with probability p and new songs with probability p−1 .

(i) Calculate, in terms of p, the expected reaction from each of the three types of audience. [4]

(ii) Use a graphical method to decide what value p should take to maximise the minimum expectedreaction from part (i). Mark clearly on your graph the vertex where the optimal value occurs. [5]



6 [Answer both parts of this question on the Insert Sheet provided.]

Jim is playing a space adventure game. He has found the alien headquarters and now has four turns toescape from the planet, or else perish. On each turn, Jim must play one of three tactics. He can attack, runaway or dodge the aliens. The number of energy pods used up during the turn is shown in Table 1.

Table 1

Tactic Energy pods used

Attack 2Run away 1Dodge 0

The number of squares that Jim travels with each of these tactics is shown in Table 2.

Table 2

Energy pods remaining at start of turn5 4 3 2 1

Attack 6 7 6 4 –

Run away 5 4 4 3 1

Dodge 1 2 1 1 0

Jim currently has five energy pods, and needs to maximise the number of squares that he travels in the fourturns that he has left. He should finish the game with no energy pods remaining, but he needs at least oneenergy pod at the start of each of his four turns.

(i) The diagram on the Insert Sheet shows the action (A, R or D) for each possible transition. Mark onthe diagram the cost associated with each action, i.e. the number of squares travelled. [3]

(ii) The dynamic programming tabulation on the Insert Sheet shows stages (turns), states (numbers ofenergy pods remaining), actions (A, R or D) and a column for costs (numbers of squares travelled).Complete this tabulation to find Jim’s best strategy, and the number of squares that Jim travels usingthis strategy. [8]



7 The table shows the six activities in a project, together with their durations, precedences and the number ofpeople required for each activity.

Activity Duration (days) Preceded by People requiredA 3 — 2B 2 — 1C 1 A, B 3D 5 B 2E 4 B, C 1F 5 D, E 2

Draw an activity network to represent these activities and their precedences. [2]

In addition to the precedences shown in the table, activity E must not start until at least 2 days after activityB has finished.

(i) Determine the earliest and latest starting times for each activity, for completion of the project in theminimum time. [4]

(ii) Identify the critical activities, and state the minimum time for completion of the project, assuming thatthere are sufficient people available. [2]

(iii) Find the least number of people required for the project to be completed in the minimum time,explaining your reasoning. [3]

(iv) By how many days must the project over-run if the least possible number of people are used? Justifyyour answer. [2]



BLANK PAGE



BLANK PAGE



BLANK PAGE


This insert consists of 3 printed pages and 1 blank page.


Candidate Name ___________________________________________


INSERT


This insert contains a diagram for use in Question 6 (i) and a table for use in Question 6 (ii).

Write your name, Centre number and candidate number in the spaces at the top of this page andattach it to your answers.

Centre NumberCandidateNumber



6 (i) Write your answer on this sheet and hand it in with your answer booklet.



6 (ii) Write your answer on this sheet and hand it in with your answer booklet.

Stage State Action Cost Current maximum

1 R4 2 A1 D

R2

DA

3R

3

4 AAR3DAR4DA

2

5RAR1 5D

Best strategy is: ..............................................................................................................................

......................................................................................................................................................

......................................................................................................................................................

Number of squares travelled: ...........................................................................................................



BLANK PAGE






MARK SCHEME

MAXIMUM MARK 60




1 (i) Capacity of 80150 is +C M1 For adding appropriate weightsi.e. 230 litres per minute A1 2 For correct answer

---------------------------------------------------------------------------------------------------------------------------------------------------------(ii)

B1 1 For all the flows correct

---------------------------------------------------------------------------------------------------------------------------------------------------------(iii) The maximum flow is 230 litres per minute B1 For stating the correct maximum flow

Max flow is at least 230 from (ii) and min cut isat most 230 from (i), and result follows frommax flow = min cut B1 2 For explanation using max flow = min cut

2

M1 For labels and appropriate connectionsA1 2 For completely correct graph

---------------------------------------------------------------------------------------------------------------------------------------------------------(i)

B1 1 For correct incomplete matching

---------------------------------------------------------------------------------------------------------------------------------------------------------(ii)

M1 Correct labelling procedure

Alternating path is B–bk–A–mb–D–fb, and so amaximal matching is (Annie, Money Box),(Ben, Book), (Cathy, Teddy Bear),(Daniel, Football), (Elvis, Drum) A1 2 Alternating path and maximal matching



3 (i) The row minima for A , B, C are 5 ,4 ,2 −−−and the negatives of the column maxima for X,Y, Z are 1 ,3 ,2 −−− M1 For row minima and/or (neg) column maxima

Hence Richard’s play-safe strategy is A A1Carol’s play-safe strategy is Z A1Not stable, since 0)1()2( <−+− A1 4 Or equivalent demonstration of given result

---------------------------------------------------------------------------------------------------------------------------------------------------------

(ii) Convert to positive values, e.g. 761552498

M1 For adding 6 to each entry

Maximise P subject to 1≤++ cba and028 ≤−−− cbaP ,0659 ≤−−− cbaP , A1 For at least two correct constraints0754 ≤−−− cbaP . A1 For all four correct

a, b, c are the probabilities with which Richardplays strategy A, B, C, and P represents thevalue of the game. B1 4

4 Row reduction (or column reduction) gives: M1 For correct row (column) process

1234001011020111

4th3rd2nd1st

or

0236012301040224

4th3rd2nd1st

DCBA

DCBA

A1 For correct reduced matrix

Column reduction (or row reduction) now gives: M1 For correct subsequent column (row) process

0123001011020111

4th3rd2nd1st

or

0133002000010121

4th3rd2nd1st

DCBA

DCBA

A1 For correct reduced matrix

Matching is incomplete (3 lines will cover zeros) B1 May be impliedAugmenting the matrix (there are 2 possible results) M1 For correct augmentation process

0012101021020000

4th3rd2nd1st

or

0022102010010010

4th3rd2nd1st

DCBA

DCBA

A1 Their reduced matrix correctly augmented

Possible complete matchings are:1A, 2B, 3C, 4D or 1C, 2B, 3A, 4D or 1C, 2B, 3D, 4A B1 8 For any one correct solution

5 Because ‘mixture’ is dominated by ‘new’ B1 1---------------------------------------------------------------------------------------------------------------------------------------------------------

(i) For ‘young’ expectation is : )1(81 pp −×+× M1 Correct calculation method for any one casei.e. p78 − A1

For ‘mixed’: ppp 25)1(53 −=−×+× A1

For ‘older’: ppp 53)1(38 +=−×+× A1 4---------------------------------------------------------------------------------------------------------------------------------------------------------(ii)

B2 For correct lines plotted/sketchedB1 For identification of ‘optimal’ vertex

Optimum occurs where pp 5325 +=− M1 For equating relevant expectations

i.e. where 72=p A1 5



6 (i)

M1 For showing costs appropriatelyA1 For all Turn 1 and Turn 2 costs correctA1 3 For all remaining costs correct

---------------------------------------------------------------------------------------------------------------------------------------------------------(ii)

B1 Correct details for stage 4

M1 For method of updating costsM1 For sub-optimizationA1 Correct details for stage 3

A1 Correct details for stage 2

A1 Correct details for stage 1

The best strategy is 1:Run, 2:Dodge, 3:Attack,4:Attack A1The number of squares is 18 A1 8

Stage State Action Cost1 R 1 ←42 A 4 ← *1 D 1 + 0 = 1 ←

R 1 + 3 = 42 D 4 + 1 = 5 ←A 1 + 6 = 73 R 4 + 4 = 8 ←

3

4 A 4 + 7 = 11 ← *A 1 + 6 = 7R 5 + 4 = 9 ←3D 8 + 1 = 9 ←A 5 + 7 = 12R 8 + 4 = 124D 11 + 2 = 13 ← *A 8 + 6 = 14

2

5 R 11 + 5 = 16 ←A 9 + 6 = 15R 13 + 5 = 18 ← *1 5D 16 + 1 = 17



7

M1 For correct structure and labels (diagram mayalternatively be activity on arc)

A1 2 For all arrows correctly shown (start andfinish nodes may be omitted)

---------------------------------------------------------------------------------------------------------------------------------------------------------(i) Earliest starting times are: M1 Carrying out forward pass

842300FEDCBA A1

Latest starting times are: M1 Carrying out reverse pass

843300FEDCBA A1 4

---------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Critical events are A, B , C, E and F B1

Minimum completion time is 1358 =+ days B1 2---------------------------------------------------------------------------------------------------------------------------------------------------------(iii)

M1 For scheduling attempt, either diagrammaticor otherwise

A1 For a correct schedule, or equivalent

Since C and D overlap, 5 people are needed A1 3---------------------------------------------------------------------------------------------------------------------------------------------------------(iv)

Least number of people is 3 (for activity C) M1One extra day is required A1 2 A correct schedule or argument is needed



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MARKING INSTRUCTIONS

1. Mark in red. Correct answers should be ticked, errors which determine marks should be indicated byringing or by a cross or by underlining, and omissions by . Do not cross out or obliterate any work. Incases of particular difficulty, brief comments written on the script may be helpful should the script bereviewed at a later stage. Each page of the script (including unused pages in an answer booklet) must havesome indication that it has been seen, e.g. a tick in the margin.

2. Marks are of the following three types.

M Method mark, awarded for a valid method applied to the problem. Method marks are not lost fornumerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidatejust to indicate an intention of using some method or just to quote a formula; the formula or idea mustbe applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy markscannot be given unless the associated Method mark is earned (or implied).

B Mark for a correct result or statement independent of Method marks.

The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidatehas earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks oncegained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored.

3. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unlessthe scheme specifically says otherwise; and similarly where there are several B marks allocated. (Thenotation ‘dep *’ is used to indicate that a particular M or B mark is dependent on an earlier, asterisked,mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in apart of a question, the work from there on is worthless so that no more marks can sensibly be given. On theother hand, when two or more steps are run together by the candidate, the earlier marks are implied and fullcredit must be given.

4. The symbol implies that the A or B mark indicated is allowed for work correctly following on frompreviously incorrect results. Otherwise, A and B marks are given for correct work only — differences innotation are of course permitted. A and B marks are not given for ‘correct’ answers or results obtainedfrom incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution,there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable willbe agreed at the standardisation meeting.

5. Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specificallyindicates otherwise. Candidates are expected to give numerical answers to an appropriate degree ofaccuracy, with 3 significant figures being the norm. Small variations in the degree of accuracy to which ananswer is given (e.g. 2 or 4 significant figures where 3 is expected) should not be penalised, while answerswhich are grossly over- or under-specified should result in the loss of a mark. The situation regarding anyparticular cases where the accuracy of the answer may be a marking issue will be decided at thestandardisation meeting.

6. If work is deleted and replaced, mark the replacement. If work is deleted without replacement, mark thedeleted work provided that it is legible. When two solutions are offered (neither crossed out), count whatappears the more serious attempt or the more complete attempt at the question. If attempts areindistinguishable in these respects, count the better.

7. For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of thequestion remain unaltered, mark according to the scheme but following through from the candidate’s data.(Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.) All M marksare available for a ‘misread’ solution, and A or B marks are initially given, as per the scheme, but for



results as modified by the misread. At the end of each part of the question affected, deduct 0, 1 or 2according as the number of ‘misread’ A and B marks earned in that part is 0, 1–4 or >4. If the misreadmakes the question easier, a further deduction of 1 or more marks may be made at your discretion; thisdeduction can include M marks.

8. For a partially correct part of a question, exhibit the detailed marks, e.g. M1 A0, in the margin at the pointwhere the marks have been first earned. Please give sufficient detail to allow your marking to beunderstood. For a completely correct part of a question, only the total mark for that part need be given, inthe margin . Do NOT use subtotals (underlined or otherwise). The question total should be ringed andplaced in the margin at the end of the question. This total MUST equal the sum of all the marks in themargin for that question and should be entered against the question number in the question grid on thefront of the script. (N.B. Addendum to the booklet ‘Instructions for Examiners’: please use the left handmargin of left hand pages.)

If a candidate’s answer is in two instalments, indicate the carried forward total at the end of the first partby, for example, and the brought forward total at the start of the second instalment by, for example, .

The total mark for the paper should be obtained (a) by adding all the unringed marks through the script(checking at the same time that all pages have been marked) and (b) by adding the question marks in thegrid in reverse order. The two totals must, of course, tally, and the resulting figure should be written,ringed, on the front of the script.

9. The following abbreviations may be used in a mark scheme, or may be found useful for notating a script.

AEF Any Equivalent Form (of answer or result is equally acceptable).

AG Answer Given on the question paper (so extra care is needed in checking that the detailed workingleading to the result is valid).

BOD Benefit Of Doubt (allowed for work whose validity may not be absolutely plain).

CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed).

ISW Ignore Subsequent Working.

MR Misread.

PA Premature Approximation (resulting in basically correct work that is numerically insufficientlyaccurate).

SOS See Other Solution (the candidate makes a better attempt at the same question).

SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where somestandard marking practice is to be varied in the light of a particular circumstance).

10. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions getfull marks. Be alert for correct but unfamiliar or unexpected methods — often signalled by a correct resultfollowing an apparently incorrect method. Such work must be carefully assessed. On the other hand, workmust not be judged on the answer alone, and answers that are given in the question, especially, must bevalidly obtained. Key steps in the working must always be looked at and anything unfamiliar must beinvestigated thoroughly. If a method is not catered for in the scheme, mark at discretion, imitating thescheme as closely as possible. If a number of candidates are involved, or you are not sure what to do,telephone your Team Leader.

11. For papers in which graphic (and programmable) calculators are allowed, some answers may be obtainedwith little or no working visible. Allow full marks for correct answers (provided, of course, that there isnothing in the wording of the question specifying that analytical methods are required). Where an answer iswrong but there is some evidence of method, allow appropriate method marks. Wrong answers with nosupporting method score zero. If in doubt, consult your Team Leader.



12. If in any case the scheme operates with considerable unfairness, mark at discretion but please give a briefreason and initial the mark. This discretion should only be used very rarely.

13. If there is any suspicion of cheating or copying, mark according to the scheme and enter the marks on themarksheet as usual. Send the script(s) to your Team Leader, as per OCR instructions. Notes concerningillness etc should be sent to OCR with the marksheets. Scripts should be marked as per the scheme.

14. Examiners are reminded of the VITAL importance of checking the accuracy of the addition of marks and ofthe transcriptions onto the marksheets; in particular that the marks are entered against the right candidates.Do not assume that the scripts are in the same order as the names on the marksheet. As detailed in §8above, each Examiner must check the paper total, obtaining the same figure twice by different methods.The transcription to the marksheet should also be checked; ideally, the Checker should read out thecandidate’s name and mark from the marksheet, while the Examiner checks with the front of the script.

The Examiner has final responsibility for the accuracy of the mark recorded on the marksheet.



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AS/A Level Mathematics A specimen question papers and ...

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