Monday 14 May 2018 – Afternoon - Mathematics in Education … · 2018. 8. 28. · OC 2018 Further...

INSTRUCTIONS• Use black ink. HB pencil may be used for graphs and diagrams only.• Complete the boxes provided on the Printed Answer Booklet with your name, centre

number and candidate number.• Answer all the questions.• Write your answer to each question in the space provided in the Printed Answer

Booklet. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s).

• Do not write in the barcodes.• You are permitted to use a scientific or graphical calculator in this paper.• Final answers should be given to a degree of accuracy appropriate to the context.

INFORMATION• The total number of marks for this paper is 60.• The marks for each question are shown in brackets [ ].• You are advised that an answer may receive no marks unless you show sufficient detail

of the working to indicate that a correct method is used. You should communicate your method with correct reasoning.

• The Printed Answer Booklet consists of 12 pages. The Question Paper consists of 4 pages.

Turn over

Oxford Cambridge and RSA

AS Level Further Mathematics B (MEI)Y410/01 Core PureQuestion Paper

Monday 14 May 2018 – AfternoonTime allowed: 1 hour 15 minutes

You must have:• Printed Answer Booklet• Formulae Further Mathematics B (MEI)

You may use:• a scientific or graphical calculator

OCR is an exempt Charity

*7063170595*

© OCR 2018 [T/508/5549]DC (SC/SW) 162697/2

2

Y410/01 Jun18© OCR 2018

Answer all the questions.

1 The matrices A, B and C are defined as follows:

, ,123

21

0133 1 3A B C= =

-=

J

L

KKK

c ^N

P

OOO

m h.

Calculate all possible products formed from two of these three matrices. [4]

2 Find, to the nearest degree, the angle between the vectors 102-

J

L

KKK

N

P

OOO and

233

-

-

J

L

KKK

N

P

OOO. [3]

3 Find real numbers a and b such that ( ) ( )a b3 5 17i i i- - = - . [5]

4 Find a cubic equation with real coefficients, two of whose roots are 2 - i and 3. [5]

5 A transformation of the x-y plane is represented by the matrix cossin

sincos22i

i

i

i-c m, where i is a positive acute

angle.

(i) Write down the image of the point (2, 3) under this transformation. [2]

(ii) You are given that this image is the point (a, 0). Find the value of a. [5]

6 Find the invariant line of the transformation of the x-y plane represented by the matrix 24

01-

c m. [4]

7 (i) Express r r2 11

2 11

--+

as a single fraction. [2]

(ii) Find how many terms of the series

( ) ( )r r1 32

3 52

5 72

2 1 2 12

# # #f f+ + + +

- ++

are needed for the sum to exceed 0.999 999. [7]

8 Prove by induction that 1012

102 12

n n

n=-c cm m for all positive integers n. [6]

3

Y410/01 Jun18© OCR 2018

9 Fig. 9 shows a sketch of the region OPQ of the Argand diagram defined by

z z 4 2| kG% / argz z41

31| G Gr r% /.

O Re

P

Im

Q

Fig. 9

(i) Find, in modulus-argument form, the complex number represented by the point P. [2]

(ii) Find, in the form a + ib, where a and b are exact real numbers, the complex number represented by the point Q. [3]

(iii) In this question you must show detailed reasoning.

Determine whether the points representing the complex numbers

• 3 + 5i

• . ( . . )cos sin5 5 0 8 0 8i+ lie within this region. [4]

10 Three planes have equations

-x + 2y + z = 0 2x - y - z = 0 x + y = a

where a is a constant.

(i) Investigate the arrangement of the planes:

• when a 0= ;

• when a 0=Y . [6]

(ii) Chris claims that the position vectors ,2 2i j k i j k- + + - - and i + j lie in a plane. Determine whether or not Chris is correct. [2]

END OF QUESTION PAPER

4

Y410/01 Jun18© OCR 2018


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*3369126002*

Further Mathematics B (MEI)© OCR 2018 [603/1364/X]DC (LK/TP) 159339/3


The information in this booklet is for the use of candidates following the Advanced Subsidiary in Further Mathematics B (MEI)(H635) or the Advanced GCE in Further Mathematics B (MEI)(H645) course.The formulae booklet will be printed for distribution with the examination papers.Copies of this booklet may be used for teaching.This document consists of 16 pages.

Advanced Subsidiary GCE (H635)Advanced GCE (H645)

Further Mathematics B (MEI)

Formulae Booklet

Instructions to Exams Officer/Invigilator• Do not send this Data Sheet for marking; it should be retained in the centre or

destroyed.

CST323

AS and A Level Further M

athematics B

(MEI)

2

Further Mathematics B (MEI)© OCR 2018

Contents

A level MathematicsCore PureMechanicsFurther Pure with TechnologyExtra PureNumerical MethodsStatisticsStatistical tables

A level Mathematics

Arithmetic series

( ) { ( ) }S n a l n a n d2 1n 21

21= + = + -

Geometric series

( )S r

a r11

n

n

=--

forS ra r1 11=-3

Binomial series

( ) ( ),C C Ca b a a b a b a b b n Nn n n n n n nr

n r r n1

12

2 2 f f !+ = + + + + + +- - -

where ! ( ) !!C C

nr r n r

nnr n r= = =

-

J

LKKN

POO

( ) ,!( )

!( ) ( )

x nx x nn n

x rn n n r

x1 1 121 1 1

Rn r2 ff

f 1 !+ = + +-

+ +- - +

+ ^ h

Differentiation

( )f x ( )f xltan kx seck kx2

sec x sec tanx xcot x cosec x2-

cosec x cosec cotx x-

Quotient Rule , dd d

ddd

y vu

xy

v

v xu u x

v

2= =-

Differentiation from first principles

( )( ) ( )

f limf f

x hx h x

h 0=

+ -

"l

3

Further Mathematics B (MEI) Turn over© OCR 2018

Integration

( )( )

( )ff d ln fx

xx x c= +

ly

( ) ( ) ( )f f d fx x x n x c11n n 1

=+

++l ^ ^h hy

Integration by parts dd d d

d du vx x uv v x

u x= -y y

Small Angle Approximations

, ,sin cos tan1 21 2. . .i i i i i i- where i is measured in radians

Trigonometric identities

( )sin sin cos cos sinA B A B A B! !=

( )cos cos cos sin sinA B A B A B! "=

( ) ( ( ) )tan tan tantan tanA B A B

A B A B k1 21

!"!

! ! r= +

Numerical methods

Trapezium rule: {( ) ( )}dy x h y y y y y2a

bn n2

10 1 2 1f. + + + + +

-y , where h nb a

=-

The Newton-Raphson iteration for solving ( ) : ( )( )

f ff

x x x xx

0 n nn

n1= = -+ l

Probability

( ) ( ) ( ) ( )P P P PA B A B A B, += + -

( ) ( ) ( ) ( ) ( )P P P P PA B A B A B A B+ = = or ( ) ( )( )

P PP

A B BA B+

=

Sample Variance

s n S11

xx2 =

- where ( )S x x x n

xx nxxx i i

ii

2 22

2= - = - = - 2^ h/ /

//

Standard deviation, variances =

4


The Binomial Distribution

If ( , )BX n p+ then ( ) CP X r p qnr

r n r= = - where q p1= -Mean of X is np

Hypothesis test for the mean of a Normal distribution

If ( , )NX 2+ vn then ,NXn

2+ n

vJ

LKK

N

POO and

/( , )N

nX

0 1+v

n-

Percentage points of the normal distribution

z

%p21 %p2

1

p 10 5 2 1

z 1.645 1.960 2.326 2.576

Kinematics

Motion in a straight line Motion in two and three dimensions

v u at= + v u a t= +

s ut at21 2= + s u at t2

1 2= +

( )s u v t21= + ( )s u v t2

1= +

v u as22 2= + -

s vt at21 2= - s v at t2

1 2= -

5


Core Pure

Complex Numbers

De Moivre’s theorem:{ ( )} ( )cos isin cos isinr r n nn ni i i i+ = +

Roots of unity:

The roots of z 1n = are given by exp izn

k2r=

J

LKK

N

POO for , , , ,k n0 1 2 1f= -

Vectors and 3-D geometry

Cartesian equation of a plane isn x n y n z d 01 2 3+ + + =

Cartesian equation of a line in 3-D is

dx a

dy a

dz a

1

1

2

2

3

3-=-

=-

Vector product a ba b a ba b a ba b a b

2 3 3 2

3 1 1 3

1 2 2 1

# =

-

-

-

J

L

KKKK

N

P

OOOO

sina bijk

a b na b a ba b a ba b a b

aaa

bbb

2 3 3 2

3 1 1 3

1 2 2 1

1

2

3

1

2

3

# i=

-

-

-

= = t

J

L

KKKK

N

P

OOOO

where , ,a b nt , in that order, form a right-handed triple.

Distance between skew lines is . ( )d d

d da a1 2

1 21 2

#

#- where a1 is the position vector of a point on the first line and

d1 is parallel to the first line, similarly for the second line.

Distance between point (x1, y1) and line ax by c 0+ + = is a b

ax by c2 2

1 1

+

+ +

Distance between point (x1, y1, z1) and plane n x n y n z d 01 2 3+ + + = is n n n

n x n y n z d

12

22

32

1 1 2 1 3 1

+ +

+ + +

Hyperbolic functions

cosh sinhx x 12 2- =

[ ( )]arsinh lnx x x 12= + +

[ ( ],)arcosh lnx x x x1 12 $= + -

,artanh lnxxx x2

111 1 11 1=-

+-

J

LKK

N

POO

6


Calculus

( )f x ( )f xlarcsinx

x11

2-arccosx

x11

2-

-arctanx

x11

2+

( )f x ( )f dx xy

a x1

2 2-arcsin

ax x a1J

LKK ^N

POO h

a x1

2 2+ arctana ax1 J

LKKN

POO

a x1

2 2+( )arsinh or ln

ax x x a2 2+ +J

LKKN

POO

x a1

2 2-( ) ( )arcosh or ln

ax x x a x a2 2 2+ -J

LKKN

POO

The mean value of f(x) on the interval [a,b] is ( )f db a x x1a

b

-y

Area of sector enclosed by polar curve is dr21 2 iy

Series

( ) ( )r n n n61 1 2 1

r

n2

1= + +

=

/ ( )r n n41 1

r

n3

1

2 2= +=

/

( ) ( ) ( ) !( )

!( )

f f ff f

x x x r x0 0 20 0( )r

r2 f f= + + + + +lm

( ) ! !e exp x x xrx1 2

xr2

f f= = + + + + + for all x

( ) ( ) ( )ln x x x xrx x1 2 3 1 1 1r

r2 31f f 1 #+ = - + - + - + -+

! ! ( ) ( ) !sin x x x xrx

3 5 1 2 1r

r3 5 2 1f f= - + - + -

++

+

for all x

! ! ( ) ( ) !cos x x x xr1 2 4 1 2

rr2 4 2

f f= - + - + - + for all x

( ) !( )

!( ) ( )

,x nxn n

x rn n n r

x x n1 1 21 1 1

1 Rn r2 ff

f 1 !+ = + +-

+ +- - +

+ ^ h

7


Mechanics

Motion in a circle

For motion in a circle,

tangential velocity is v ri= o

radial acceleration is orrv r

22io towards the centre

tangential acceleration is rip

Further Pure with Technology

Numerical solution of differential equations

For ( , ):dd

fxy

x y=

Euler’s method: ( , )fx x h y y h x yn n n n n n1 1= + = ++ +

Modified Euler method (A Runge-Kutta method of order 2):

( , )fk h x yn n1 =

( , )fk h x h y kn n2 1= + +

, ( )x x h y y k k21

n n n n1 1 1 2= + = + ++ +

Runge-Kutta method of order 4:

( , )fk h x yn n1 =

( , )fk h x h yk

2 2n n21= + +

( , )fk h x h yk

2 2n n32= + +

( , )fk h x h y kn n4 3= + +

( )k ky y k k61 2 2n n1 1 2 3 4= + + + +

+

Gradient of tangent to a polar curve

For a curve ( ),f dd

dd cos sindd sin cos

r xy

r r

r ri

ii i

ii i

= =+

-

8


Extra Pure

Multivariable calculus

g g

g

g

g

gradx

y

z

d

2

2

2

2

2

2

= =

J

L

KKKKKKK

N

P

OOOOOOO

. If g(x, y, z) can be written as , )(fz x y= then g

f

fgradx

y1

22

22=

-

J

L

KKKKK

N

P

OOOOO

9


Numerical methods

Solution of equations

The Newton-Raphson iteration for solving ( ) : ( )( )

f ff

x x x xx

0 n nn

n1= = -+ l

For the iteration ( )gx xn n1 =+ the relaxed iteration is ( ) ( )gx x x1n n n1 m m= - +

+.

Numerical integration

To estimate ( )f dx xa

by :

The midpoint rule:

( )M h y y y yn n n21

23

23

21f= + + + +

- - where h n

b a=-

The trapezium rule:

{( ) ( )}T h y y y y y21 2n n n0 1 2 1f= + + + + +

- where h n

b a=-

Simpson’s rule

{( ) ( ) ( )}S h y y y y y y y y31 4 2n n n n2 0 2 1 3 2 1 2 4 2 2f f= + + + + + + + + +

- -

where h nb a2=-

These are related as follows:

( )T M T21

n n n2 = +

( ) ( )S M T T T31 2 3

1 4n n n n n2 2= + = -

Interpolation

Newton’s forward difference interpolation formula:

( ) ( )( )

( )!

( ) ( )( )f f f fx x h

x xx

hx x x x

x20

00 2

0 1 20 fT T= +

-+- -

+

Lagrange’s polynomial:

( ) ( ) ( )P L fx x xn r r=/ where ( )L x x xx x

rr i

i

i ri

n

0=

-

-

!=

%

10


Statistics

Discrete distributions

X is a random variable taking values xi in a discrete distribution with ( )P X x pi i= =

Expectation: ( )E X x pi in = =/Variance: ( ) ( )Var X x p x pi i i i

2 2 2 2v n n= = - = -//

Probability E(X) Var(X)

Uniform distribution over 1, 2, …, n ( )P X r n1

= =n 1

2+ ( )n12

1 12 -

Geometric distribution ( )P X r q pr 1= = -

q p1= -

p1

pp1

2-

Poisson distribution( ) !eP X r r

rm= = m-

Correlation and regression

For a sample of n pairs of observations ( , )x yi i

, ,S x nx

S y ny

S x y nx y

xx ii

yy ii

xy i ii i2

2

2

2

= - = - = -a ak k/

//

/ ///

product moment correlation coefficient: rS S

S

x nx

y ny

x y nx y

xx yy

xy

ii

ii

i ii i

2

2

2

2= =

- -

-

J

LKK

J

LKK

a aN

POO

N

POO

k kR

T

SSS

V

X

WWW

//

//

///

least squares regression line of y on x is ( )y y b x x- = - where b SS

x nx

x y nx y

xx

xy

ii

i ii i

2

2= =

-

-

a k//

///

least squares regression line of x on y is ( )bx x y y- = -l where b SS

n

x y nx y

yy

xy

ii

i ii i

yy2

2= =

-

-l

a k//

///

Spearman’s coefficient of rank correlation:

( )r

n nd

11

6s

i2

2

= --

/

11


Confidence intervals

To calculate a confidence interval for a mean or difference in mean in different circumstances, use the given distribution to calculate the critical value, k.

To estimate… Confidence interval Distribution

a mean x kn

!v N(0, 1)

a mean x kn

s! tn–1

difference in mean of paired populations treat differences as a single distribution

Hypothesis tests

Description Test statistic Distribution

Pearson’s product moment correlation test r

S S

S

xx yy

xy=

xn

xy

n

y

x y nx y

ii

ii

i ii i

2

2

2

2=

- -

-

J

L

KKK

J

L

KKK

a aN

P

OOO

N

P

OOO

k kR

T

SSS

V

X

WWW

//

//

///

Spearman’s rank correlation test

( )r

n nd

11

6s

i2

2

= --

/

2| testf

f fo

e

e2-^ h/

v2|

Normal test for a mean

n

xv

n-J

LKK

N

POO

N(0, 1)

t-test for a mean

sn

x n-J

LKK

N

POO

tn–1

Wilcoxon single sample test A statistic T is calculated from the ranked data

Continuous distributionsX is a continuous random variable with probability density function (pdf) ( )f x

Expectation: ( ) ( )E f dX x x xn = = yVariance: ( ) ( ) ( ) ( )Var f d f dX x x x x x x2 2 2 2v n n= = - = -yyCumulative distribution function ( ) ( ) ( )F P f dx X x t tx

#= =3-y

E(X) Var(X)

Continuous uniform distribution over [a,b] a b2+ ( )b a12

1 2-

12


Cri

tical

val

ues f

or th

e pr

oduc

t mom

ent c

orre

latio

n co

effic

ient

, rC

ritic

al v

alue

s for

Spe

arm

an’s

ran

k co

rrel

atio

n co

effic

ient

, rs

5%2½

%1%

½%

1-Ta

il Te

st5%

2½%

1%½

%5%

2½%

1%½

%1-

Tail

Test

5%2½

%1%

½%

10%

5%2%

1%2-

Tail

Test

10%

5%2%

1%10

%5%

2%1%

2-Ta

il Te

st10

%5%

2%1%

nn

nn

1-

--

-31

0.30

090.

3550

0.41

580.

4556

1-

--

-31

0.30

120.

3560

0.41

850.

4593

2-

--

-32

0.29

600.

3494

0.40

930.

4487

2-

--

-32

0.29

620.

3504

0.41

170.

4523

30.

9877

0.99

690.

9995

0.99

9933

0.29

130.

3440

0.40

320.

4421

3-

--

-33

0.29

140.

3449

0.40

540.

4455

40.

9000

0.95

000.

9800

0.99

0034

0.28

690.

3388

0.39

720.

4357

41.

0000

--

-34

0.28

710.

3396

0.39

950.

4390

50.

8054

0.87

830.

9343

0.95

8735

0.28

260.

3338

0.39

160.

4296

50.

9000

1.00

001.

0000

-35

0.28

290.

3347

0.39

360.

4328

60.

7293

0.81

140.

8822

0.91

7236

0.27

850.

3291

0.38

620.

4238

60.

8286

0.88

570.

9429

1.00

0036

0.27

880.

3300

0.38

820.

4268

70.

6694

0.75

450.

8329

0.87

4537

0.27

460.

3246

0.38

100.

4182

70.

7143

0.78

570.

8929

0.92

8637

0.27

480.

3253

0.38

290.

4211

80.

6215

0.70

670.

7887

0.83

4338

0.27

090.

3202

0.37

600.

4128

80.

6429

0.73

810.

8333

0.88

1038

0.27

100.

3209

0.37

780.

4155

90.

5822

0.66

640.

7498

0.79

7739

0.26

730.

3160

0.37

120.

4076

90.

6000

0.70

000.

7833

0.83

3339

0.26

740.

3168

0.37

290.

4103

100.

5494

0.63

190.

7155

0.76

4640

0.26

380.

3120

0.36

650.

4026

100.

5636

0.64

850.

7455

0.79

3940

0.26

400.

3128

0.36

810.

4051

110.

5214

0.60

210.

6851

0.73

4841

0.26

050.

3081

0.36

210.

3978

110.

5364

0.61

820.

7091

0.75

4541

0.26

060.

3087

0.36

360.

4002

120.

4973

0.57

600.

6581

0.70

7942

0.25

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3044

0.35

780.

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120.

5035

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6783

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940.

3955

130.

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3008

0.35

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130.

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040.

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430.

3014

0.35

500.

3908

140.

4575

0.53

240.

6120

0.66

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0.25

120.

2973

0.34

960.

3843

140.

4637

0.53

850.

6264

0.67

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0.25

130.

2978

0.35

110.

3865

150.

4409

0.51

400.

5923

0.64

1145

0.24

830.

2940

0.34

570.

3801

150.

4464

0.52

140.

6036

0.65

3645

0.24

840.

2945

0.34

700.

3822

160.

4259

0.49

730.

5742

0.62

2646

0.24

550.

2907

0.34

200.

3761

160.

4294

0.50

290.

5824

0.63

5346

0.24

560.

2913

0.34

330.

3781

170.

4124

0.48

210.

5577

0.60

5547

0.24

290.

2876

0.33

840.

3721

170.

4142

0.48

770.

5662

0.61

7647

0.24

290.

2880

0.33

960.

3741

180.

4000

0.46

830.

5425

0.58

9748

0.24

030.

2845

0.33

480.

3683

180.

4014

0.47

160.

5501

0.59

9648

0.24

030.

2850

0.33

610.

3702

190.

3887

0.45

550.

5285

0.57

5149

0.23

770.

2816

0.33

140.

3646

190.

3912

0.45

960.

5351

0.58

4249

0.23

780.

2820

0.33

260.

3664

200.

3783

0.44

380.

5155

0.56

1450

0.23

530.

2787

0.32

810.

3610

200.

3805

0.44

660.

5218

0.56

9950

0.23

530.

2791

0.32

930.

3628

210.

3687

0.43

290.

5034

0.54

8751

0.23

290.

2759

0.32

490.

3575

210.

3701

0.43

640.

5091

0.55

5851

0.23

290.

2764

0.32

600.

3592

220.

3598

0.42

270.

4921

0.53

6852

0.23

060.

2732

0.32

180.

3542

220.

3608

0.42

520.

4975

0.54

3852

0.23

070.

2736

0.32

280.

3558

230.

3515

0.41

320.

4815

0.52

5653

0.22

840.

2706

0.31

880.

3509

230.

3528

0.41

600.

4862

0.53

1653

0.22

840.

2710

0.31

980.

3524

240.

3438

0.40

440.

4716

0.51

5154

0.22

620.

2681

0.31

580.

3477

240.

3443

0.40

700.

4757

0.52

0954

0.22

620.

2685

0.31

680.

3492

250.

3365

0.39

610.

4622

0.50

5255

0.22

410.

2656

0.31

290.

3445

250.

3369

0.39

770.

4662

0.51

0855

0.22

420.

2659

0.31

390.

3460

260.

3297

0.38

820.

4534

0.49

5856

0.22

210.

2632

0.31

020.

3415

260.

3306

0.39

010.

4571

0.50

0956

0.22

210.

2636

0.31

110.

3429

270.

3233

0.38

090.

4451

0.48

6957

0.22

010.

2609

0.30

740.

3385

270.

3242

0.38

280.

4487

0.49

1557

0.22

010.

2612

0.30

830.

3400

280.

3172

0.37

390.

4372

0.47

8558

0.21

810.

2586

0.30

480.

3357

280.

3180

0.37

550.

4401

0.48

2858

0.21

810.

2589

0.30

570.

3370

290.

3115

0.36

730.

4297

0.47

0559

0.21

620.

2564

0.30

220.

3328

290.

3118

0.36

850.

4325

0.47

4959

0.21

620.

2567

0.30

300.

3342

300.

3061

0.36

100.

4226

0.46

2960

0.21

440.

2542

0.29

970.

3301

300.

3063

0.36

240.

4251

0.46

7060

0.21

440.

2545

0.30

050.

3314

13


Percentage points of the 2| (chi-squared) distribution

p%

2|

p%

2|

p% 99 97.5 95 90 10 5 2.5 1 0.5

v = 1 .0001 .0010 .0039 .0158 2.706 3.841 5.024 6.635 7.8792 .0201 .0506 0.103 0.211 4.605 5.991 7.378 9.210 10.603 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.34 12.844 0.297 0.484 0.711 1.064 7.779 9.488 11.14 13.28 14.865 0.554 0.831 1.145 1.610 9.236 11.07 12.83 15.09 16.756 0.872 1.237 1.635 2.204 10.64 12.59 14.45 16.81 18.557 1.239 1.690 2.167 2.833 12.02 14.07 16.01 18.48 20.288 1.646 2.180 2.733 3.490 13.36 15.51 17.53 20.09 21.959 2.088 2.700 3.325 4.168 14.68 16.92 19.02 21.67 23.5910 2.558 3.247 3.940 4.865 15.99 18.31 20.48 23.21 25.1911 3.053 3.816 4.575 5.578 17.28 19.68 21.92 24.72 26.7612 3.571 4.404 5.226 6.304 18.55 21.03 23.34 26.22 28.3013 4.107 5.009 5.892 7.042 19.81 22.36 24.74 27.69 29.8214 4.660 5.629 6.571 7.790 21.06 23.68 26.12 29.14 31.3215 5.229 6.262 7.261 8.547 22.31 25.00 27.49 30.58 32.8016 5.812 6.908 7.962 9.312 23.54 26.30 28.85 32.00 34.2717 6.408 7.564 8.672 10.09 24.77 27.59 30.19 33.41 35.7218 7.015 8.231 9.390 10.86 25.99 28.87 31.53 34.81 37.1619 7.633 8.907 10.12 11.65 27.20 30.14 32.85 36.19 38.5820 8.260 9.591 10.85 12.44 28.41 31.41 34.17 37.57 40.0021 8.897 10.28 11.59 13.24 29.62 32.67 35.48 38.93 41.4022 9.542 10.98 12.34 14.04 30.81 33.92 36.78 40.29 42.8023 10.20 11.69 13.09 14.85 32.01 35.17 38.08 41.64 44.1824 10.86 12.40 13.85 15.66 33.20 36.42 39.36 42.98 45.5625 11.52 13.12 14.61 16.47 34.38 37.65 40.65 44.31 46.9326 12.20 13.84 15.38 17.29 35.56 38.89 41.92 45.64 48.2927 12.88 14.57 16.15 18.11 36.74 40.11 43.19 46.96 49.6428 13.56 15.31 16.93 18.94 37.92 41.34 44.46 48.28 50.9929 14.26 16.05 17.71 19.77 39.09 42.56 45.72 49.59 52.3430 14.95 16.79 18.49 20.60 40.26 43.77 46.98 50.89 53.6735 18.51 20.57 22.47 24.80 46.06 49.80 53.20 57.34 60.2740 22.16 24.43 26.51 29.05 51.81 55.76 59.34 63.69 66.7750 29.71 32.36 34.76 37.69 63.17 67.50 71.42 76.15 79.49100 70.06 74.22 77.93 82.36 118.5 124.3 129.6 135.8 140.2

14


Percentage points of the t distribution

t

%p21 %p2

1

p%v

10 5 2 1

1 6.314 12.71 31.82 63.662 2.920 4.303 6.965 9.9253 2.353 3.182 4.541 5.8414 2.132 2.776 3.747 4.6045 2.015 2.571 3.365 4.0326 1.943 2.447 3.143 3.7077 1.895 2.365 2.998 3.4998 1.860 2.306 2.896 3.3559 1.833 2.262 2.821 3.250

10 1.812 2.228 2.764 3.16911 1.796 2.201 2.718 3.10612 1.782 2.179 2.681 3.05513 1.771 2.160 2.650 3.01214 1.761 2.145 2.624 2.97715 1.753 2.131 2.602 2.94720 1.725 2.086 2.528 2.84530 1.697 2.042 2.457 2.75050 1.676 2.009 2.403 2.678

100 1.660 1.984 2.364 2.626∞ 1.645 1.960 2.326 2.576 = percentage points of the Normal distribution N(0, 1)

Percentage points of the normal distribution

z

%p21 %p2

1

p 10 5 2 1

z 1.645 1.960 2.326 2.576

15


Critical values for the Wilcoxon Single Sample test

1-tail 5% 2½% 1% ½% 1-tail 5% 2½% 1% ½%2-tail 10% 5% 2% 1% 2-tail 10% 5% 2% 1%n n

26 110 98 84 752 – – – – 27 119 107 92 833 – – – – 28 130 116 101 914 – – – – 29 140 126 110 1005 0 – – – 30 151 137 120 1096 2 0 – – 31 163 147 130 1187 3 2 0 – 32 175 159 140 1288 5 3 1 0 33 187 170 151 1389 8 5 3 1 34 200 182 162 14810 10 8 5 3 35 213 195 173 15911 13 10 7 5 36 227 208 185 17112 17 13 9 7 37 241 221 198 18213 21 17 12 9 38 256 235 211 19414 25 21 15 12 39 271 249 224 20715 30 25 19 15 40 286 264 238 22016 35 29 23 19 41 302 279 252 23317 41 34 27 23 42 319 294 266 24718 47 40 32 27 43 336 310 281 26119 53 46 37 32 44 353 327 296 27620 60 52 43 37 45 371 343 312 29121 67 58 49 42 46 389 361 328 30722 75 65 55 48 47 407 378 345 32223 83 73 62 54 48 426 396 362 33924 91 81 69 61 49 446 415 379 35525 100 89 76 68 50 466 434 397 373

16








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Oxford Cambridge and RSA Examinations

GCE

Further Mathematics B (MEI)

Unit Y410: Core Pure

Advanced Subsidiary GCE

Mark Scheme for June 2018

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OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. © OCR 2018

Y410/01 Mark Scheme June 2018

3

Annotations and abbreviations

Annotation in scoris Meaning

and

BOD Benefit of doubt

FT Follow through

ISW Ignore subsequent working

M0, M1 Method mark awarded 0, 1

A0, A1 Accuracy mark awarded 0, 1

B0, B1 Independent mark awarded 0, 1

SC Special case

^ Omission sign

MR Misread

Highlighting

Other abbreviations in mark scheme

Meaning

E1 Mark for explaining a result or establishing a given result

dep* Mark dependent on a previous mark, indicated by *

cao Correct answer only

oe Or equivalent

rot Rounded or truncated

soi Seen or implied

www Without wrong working

AG Answer given

awrt Anything which rounds to

BC By Calculator

DR This indicates that the instruction In this question you must show detailed reasoning appears in the question.


4

Subject-specific Marking Instructions for AS Level Further Mathematics B (MEI)

a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.

b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. If you are in any doubt whatsoever you should contact your Team Leader.

c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.


5

d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular 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 a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.

e The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from 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, what is acceptable will be detailed in the mark scheme. If this is not the case please, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.

f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.) We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so. When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value. This rule should be applied to each case. When a value is not given in the paper accept any answer that agrees with the correct value to 2 s.f. Follow through should be used so that only one mark is lost for each distinct accuracy error, except for errors due to premature approximation which should be penalised only once in the examination. There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question.

g Rules for replaced work: if a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests; if there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ will not affect an earlier decision about a misread. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.


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i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader.

j If in any case the scheme operates with considerable unfairness consult your Team Leader.


7

Question Answer Marks AOs Guidance

1

BA = 1

2 0 3 112

1 1 3 83

AC =

1 1 3

2 1 3 2 6

3 3 9

CB = 2 0 3

1 3 5 3 121 1 3

M1 A1

A1

A1 [4]

1.1a 1.1

1.1

1.1

BA, AC or CB calculated with correct shape deduct A1 for any incorrect products pursued

allow one arith error for M1

2 1 ( 2) 0 3 ( 2) ( 3)cos 0.381

5 22

K

= 68 to nearest degree

M1

A1

A1 [3]

1.1a,

1.1

1.1

Use of .

cos a b

a b

0.381… or 4

5 22

must be nearest degree

allow unsupported correct answers

3 (a 3i)(5 i) = 5a 3 15i ai

5a 3 = b 15 + a = 17

a = 2, b = 7

M1 M1 A1 A1 A1 [5]

1.1a 1.1 3.1a 1.1 1.1

expanding and i2 = 1 equating Re and Im parts soi o.e. eg 15i + ai = 17i both correct

allow 1 sign error (e.g. +3)


8


4 third root is 2 + i B1 1.2 soi

2 i + 2 + i + 3 = 7

(2 i)(2 + i) + (2 + i)3 + 3(2 i) = 17

(2 i)(2 + i)3 = 15

so eqn is z3 7z2 + 17z 15 = 0 OR

(z – 2 + i)(z 2 – i) = z2 – 4z + 5

(z2 4z + 5)(z 3) = z3 7z2 + 17z 15

so eqn is z3 7z2 + 17z 15 = 0

B1ft B1ft B1ft

B1cao

M1 A1 M1 A1 [5]

3.1a 1.1 1.1 1.1

ft their 2 + i must include ‘ = 0’

(z – 2 + i)(z 2 – i) = z2 – 4z + 5

their (z2 4z + 5) (z 3)

any variable , e.g. x

5 (i) (2cos 6sin , 4sin 3cos )

B1B1 [2]

1.1,1.1

Accept in vector form

5 (ii) 4sin 3cos 0

34

tan

= 36.9 or 0.644 rad 2cos 6sin 5.2a

M1 M1 A1

M1

A1 [5]

3.1a 1.1 1.1

1.1

1.1cao

their 4sin 3cos 0

tan = sin / cos used

= 36.9 or 0.644 rad or better

substituting their into their

2cos + 6 sin 5.2

or sin2 + cos2 = 1 used

or 3 45 5

sin , cos

or sin and cos

6 '2 0

4 1 '

xx

y y

x’ = 2x, y’ = 4x y y = mx + c and y’ = mx’ + c

4x mx c = m.2x + c

43

m , c = 0

so invariant line is 43

y x

M1 M1 A1

A1 [4]

2.1 1.1b 2.4

1.1b

soi [condone x = 2x, y=4x y] condition for invariance forming identity in x (may assume c = 0)

assuming c = 0: y = mx and y’ = mx’

4x mx = m.2x 43

m

7 (i) 1 1 2 1 (2 1) 2

2 1 2 1 (2 1)(2 1) (2 1)(2 1)

r r

r r r r r r

M1

A1

[2]

1.1a

1.1

combining fractions

or 2

2

4 1r

mark final answer


9


7 (ii)

1

2 2 2 2

1 3 3 5 5 7 (2 1)(2 1)

2

(2 1)(2 1)

n

r

n n

r r

K

1

1 1

2 1 2 1

n

r r r

1 1 1 1 1 1 11

3 3 5 5 7 2 1 2 1n n

K

11

2 1n

so 1

1 0.999 9992 1n

2n + 1 > 1 000 000 n > 499 999.5

Number of terms is 500 000

M1

M1

M1

A1

M1

M1 A1cao

[7]

3.1a

3.1a

2.1

2.1

1.1a

1.1 3.2a

rth term is 2

(2 1)(2 1)r r soi

splitting into partial fractions soi

must end with 1

...2 1n

11

2 1r

is A0

or 2

0.999 9992 1

n

n

allow

equality, condone r re-arranging (correctly) for n

not 1

...2 1r

dep use of difference method [e.g. not by inspection] trial and error: must show both n = 499999 and n = 500000 for final A1

8 When n = 1,

11 1

0 2

1

1

1 2 1

0 2

[Assume] n = k: 1 1

0 2

k

1 2 1

0 2

k

k

Then 1

1 1

0 2

k

1 2 1

0 2

k

k

1 1

0 2

1 1

1 1

1 1 2 2 1 2 1

0 2 0 2

k k

k k

so if true for n = k then true for n = k + 1 [as true for n = 1] therefore true for all n.

B1

M1

M1

A1*

M1dep* A1dep*

[6]

2.1

2.2a

1.1

1.1

2.2a 2.4

or 1 1

0 2 1 2 1

0 2

k

k

or 1

1 2 1 2

0 2

k k

k

must clearly ‘assume’ n = k must have established truth for n = 1 for final mark

‘true for n = k and k + 1’ is M0 but need not re-state it at the end


10


9 (i) 14

arg , | | 4 2z z

so 1 14 4

4 2 cos isinz

B1

B1 [2]

1.1

2.5

allow 45 for both marks

9 (ii) 13

arg , | | 4 2z z

so 1 13 3

4 2 cos isinz

2 2 2 6i

M1

A1

A1 [3]

1.1a

1.1

1.1

o.e., must be exact

9 (iii) DR

| 3 5i| = 9 25 34

34 32 so not in the region

5.5(cos0.8 isin0.8) has mod 5.5 and arg 0.8

5.5 32 5.656 K

0.785 < 0.8 < 1.047 so is in the region

M1

A1

M1 A1 [4]

3.1a

2.1

3.1a 2.1

finding modulus (correct method) comparing mod or argument checking both conditions

10 (i) det 0M [so no unique solution]

no planes parallel [so prism or sheaf] when a = 0, they form a sheaf as the system has solutions when a ≠ 0, they form a prismatic intersection as there are no solutions

B1 B1 B1 B1 B1 B1 [6]

3.1a

1.1

2.2a

2.2a

2.2a

3.1a

or M is singular

allow ‘intersect in a line’ o.e. e.g. finding solutions allow ‘prism’

or state no unique solution by direct solution of equations

10 (ii) These are the normals to the three planes In either of the above cases, they must lie in the same plane

M1

A1dep

1.1a

2.3

dep previous part correct

OR

(e.g) i + j = i + 2j + k + 2i j k

coplanar

M1

A1

[2]

showing linear dependence

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QualificationAccredited

AS LEVEL

FURTHER MATHEMATICS B (MEI)

Examiners’ report

Y410/01 Summer 2018 seriesVersion 1

H635For first teaching in 2017

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AS Level Further Mathematics B (MEI) - Y410/01 - Summer 2018 Examiners’ report

2 © OCR 2018

Contents Introduction .............................................................................................................................................. 3

Paper Y410/01 series overview ................................................................................................................ 4

Question 1 ............................................................................................................................................ 4

Question 2 ............................................................................................................................................ 4

Question 3 ............................................................................................................................................ 5

Question 4 ............................................................................................................................................ 5

Question 5(i) ......................................................................................................................................... 5

Question 5(ii) ........................................................................................................................................ 5

Question 6 ............................................................................................................................................ 6

Question 7(i) ......................................................................................................................................... 6

Question 7(ii) ........................................................................................................................................ 7

Question 8 ............................................................................................................................................ 7

Question 9(i) ......................................................................................................................................... 9

Question 9(ii) ........................................................................................................................................ 9

Question 9(iii) ........................................................................................................................................ 9

Question 10(i) ..................................................................................................................................... 10

Question 10(ii) .................................................................................................................................... 10


3 © OCR 2018

Introduction Our examiners’ reports are produced to offer constructive feedback on candidates’ performance in the examinations. They provide useful guidance for future candidates. The reports will include a general commentary on candidates’ performance, identify technical aspects examined in the questions and highlight good performance and where performance could be improved. The reports will also explain aspects which caused difficulty and why the difficulties arose, whether through a lack of knowledge, poor examination technique, or any other identifiable and explainable reason.

Where overall performance on a question/question part was considered good, with no particular areas to highlight, these questions have not been included in the report. A full copy of the question paper can be downloaded from OCR.


4 © OCR 2018

Paper Y410/01 series overview The overall standard of work achieved by the relatively small candidature for the first run of this paper was impressive. Nearly all candidates scored over half marks, and over a quarter achieved over 50 out of 60 marks. For candidates with experience of only one year’s work in advanced mathematics, this was highly creditable. Even though 75 minutes is quite a short time to complete an examination, there was no evidence of candidates running out of time. The performance on many of the questions was excellent, with accurate, well presented work.

Most successful questions Least successful questions

• Q1 (matrices)• Q2 (angle between vectors)• Q3 (complex numbers)• Q4 (symmetric properties of roots)• Q8 (mathematical induction)

• Q6 (invariant line of a transformation)• Q7 (difference method for summing a series)• Q10 (arrangement of planes)

Question 1

The product BA and the product CB were generally calculated accurately. The product AC was omitted by a significant number of candidates.

Question 2

A misconception that there should be a second value of 112° (180 – 68) was seen in the responses from a few candidates. The angle between vectors is unique (unlike the angle between lines).

The number of candidates who misread one of the vectors – usually giving the z entry in the first vector as +2 - was well into double figures. As a result, the answer 140 was quite common. The other common error was to forget to round the final answer to the nearest degree, as requested in the question.


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Question 3

Virtually all candidates expanded and equated real and imaginary parts. The few who lost marks did so through slips in the expansion of the brackets, most commonly getting +3 instead of −3.

Question 4

There were no significant issues with candidates identifying the conjugate pair to find the third root of 2 + i. Thereafter, candidates divided fairly evenly between either using symmetric properties of roots or formulating the product of factors from the roots, with neither method proving more successful. The only blemish, which was quite common, was to omit the ‘= 0’ in stating the final equation, at the expense of a mark.

Question 5(i)

A few candidates attempted to find a numerical value for the image, but the majority gained full marks, albeit expressing the coordinates in vector form (which was condoned).

Question 5(ii)

A regulatory requirement of the reformed qualifications is that some questions require deployment of knowledge from across a number of different topic areas. Here, candidates were required to use trigonometry from AS Mathematics to solve the equation 4sinθ − 3cosθ = 0, and a significant minority struggled with the manipulation. Nevertheless, the majority gained full marks here.


6 © OCR 2018

Question 6

The main problem that candidates had here was an inability to differentiate between invariant points and invariant lines.

Many candidates gained a method mark for calculating the image of (x, y) as (2x, 4x − y). For an invariant point, this implies that 2x = x, so x = 0 and y = −y, so y = 0. So, the only invariant point is the origin. However, for an invariant line, y = mx + c and 4x − y = m.2x + c, leading to 4x − mx − 2c = 2mx . This is an identity which needs to be true for all x, so c = 0 and m = 4/3, giving an invariant line y = 4x/3. There is in fact another solution to the identity, namely x = 0, which is indeed another invariant line. However, very few spotted this, and there was no penalty for omitting this. In fact, with linear transformations such as this, it suffices to use the line y = mx, as the origin is necessarily invariant.

Exemplar 1

Some candidates, as in this response, incorrectly cancelled c’s rather than finding 2c = 0 and hence c = 0.

Question 7(i)

This straightforward combining of fractions was negotiated successfully by virtually the whole candidature.


7 © OCR 2018

Question 7(ii)

The difference method for summing the series was attempted by most candidates, but a significant minority used invalid methods based on the results for standard series.

Those who used the valid difference method usually got to a sum of 1 − 1/(2n + 1); if they did not, it was due to muddling n’s and r’s and getting 1 − 1/(2r + 1), which lost a couple of marks.

The final three marks were then open to candidates, and many gained all three, though those who solved the inequality (rather than the equality) sometimes lost the final ‘A’ mark through errors in handling inequality signs.

Question 8

The induction step in this question was a relatively straightforward one. However, it is important that candidates show some working to justify this step. Errors and omissions here proved quite costly, as the final two marks were dependent on this. These errors, though rare, were of two types: one was to add M to Mk (where M is the given matrix); another involved manipulating handling indices, such as writing 2k−1 instead of 2k − 1, or 2× 2k = 4k.

In order to gain full marks for this question, it was important that candidates used precise language that showed a clear mathematical justification of the induction process. For example, in their concluding statement, if they wrote ‘so, as it is true for n = 1, n = k and n = k + 1, it is true for all n’, they missed out on the final two marks; whereas if they wrote ‘so, as it is true for n = 1, and if true for n = k, then true for n = k + 1, it is true for all n’, they got full marks (assuming the induction step was negotiated without errors). Although many candidates are well schooled in the logic of induction, the use of precise language to convey this logic is required, and it is important that candidates understand that the truth for n = k is an assumption, not a given.


8 © OCR 2018

Exemplar 2

In this response, the candidate loses a mark in summarising the induction step for the above reason.


9 © OCR 2018

Question 9(i)

This was well answered. Most correctly identified the modulus and argument of P as 4√2 and π/4, but a few did not express the point in modulus-argument form as 4√2(cosπ/4 + isinπ/4).

Question 9(ii)

Again, the majority of candidates found the correct modulus and argument, but there were occasional errors in finding a and b.

Question 9(iii)

This question was, on the whole, well answered. Most compared the modulus of 3+5i to 4√2 and concluded it was outside the region. Some, instead, compared the imaginary part of 3+5i with that of Q to reach the same conclusion. The second point was handled appropriately by most, though occasionally they went round in a circle by first expressing the point in x + iy form and then finding the modulus and argument of their x + iy to get 5.5 and 0.8 (or approximations to these).


10 © OCR 2018

Question 10(i)

This question proved to be more demanding, and even the best candidates lost marks here. While some concluded correctly that in the case a = 0 the planes formed a sheaf, and in the case a ≠ 0 they formed a prismatic intersection, complete solutions demanded a systematic investigation to eliminate the other possible arrangements.

The following strategy might help for future questions on this topic:

1. Is the determinant of the matrix of coefficients zero? If yes, we eliminate a unique point of intersection.

2. Are the coefficients of x, y and z for any two equations multiples of each other? If no, this eliminates parallel planes.

3. Are the equations consistent, so that there are solutions? In the case a = 0, they are, so the planes form a sheaf.

4. In the case a ≠ 0, the equations are inconsistent and there are no solutions, so they form a prismatic intersection.

In the case a = 0, some candidates attempted to give solutions to the equations, but quite often gave the solutions in the form of a plane rather than a line.

Question 10(ii)

Candidates who recognised the three vectors as normal to the planes gained a mark. However, many wrongly concluded that Chris was incorrect. If the correct arrangement of the planes was given, then it suffices to state that the normal must be coplanar for a sheaf or a prism. The other approach was to show that the given vectors are linearly dependent, as the third is the sum of the first two. Some tried to find a normal to each of the three planes, with some success on occasions.

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Published: 15 August 2018 Version 1.0

Qualification and notional component raw mark grade boundaries June 2018 series

New AS Levels

The following qualifications are linear and therefore do not use UMS. For an explanation of how the new linear qualifications work, check out our blog: www.ocr.org.uk/blog/view/how-linear-qualifications-and-grade-boundaries-work

AS GCE Mathematics B (MEI)Max Mark a b c d e u

H630 Raw 70 44 38 33 28 23 0H630

01 Pure Mathematics and Mechanics02 Pure Mathematics and Statistics Raw 70 50 45 39 33 28 0

Overall 140 94 83 72 61 51 0

AS GCE Further Mathematics B (MEI) (H635)Max Mark a b c d e u

Y410 Raw 60 46 41 36 32 28 0Y411 Raw 60 37 32 27 22 18 0Y412 Raw 60 42 38 34 30 26 0Y413 Raw 60 37 33 29 25 22 0Y414 Raw 60 35 29 24 19 14 0Y415 RawY416

01 Core Pure01 Mechanics a01 Statistics a01 Modelling with Algorithms01 Numerical Methods01 Mechanics b01 Statistics b Raw 60 43 38 33 28 24 0

H635 Overall 180 125 111 98 85 72 0H635 Overall 180 120 107 94 81 68 0H635 Overall 180 118 103 88 74 60 0H635 Overall 180 125 112 100 88 76 0H635 Overall 180 123 109 95 81 68 0H635 Overall 180 131 117 104 91 78 0H635

Option Y410+Y411+Y412Option Y410+Y411+Y413Option Y410+Y411+Y414Option Y410+Y412+Y413Option Y410+Y412+Y414Option Y410+Y412+Y416Option Y410+Y413+Y414 Overall 180 118 104 90 77 64 0

No entry in June 2018

Published: 15 August 2018 Version 1.0

New A Levels

Qualification and notional component raw mark grade boundaries June 2018 series

The following qualifications are linear and therefore do not use UMS. For an explanation of how the new linear qualifications work, check out our blog: www.ocr.org.uk/blog/view/how-linear-qualifications-and-grade-boundaries-work

A Level Mathematics B (MEI)Max Mark a* a b c d e u

H640 Raw 100 81 74 67 59 52 45 0H640 Raw 100 75 68 61 54 47 40 0H640

01 Pure Mathematics and Mechanics02 Pure Mathematics and Statistics03 Pure Mathematics and Comprehension Raw 75 62 55 48 42 36 30 0

Overall 275 218 197 176 155 135 115 0

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