F2 IAL PAPERS: 2014-6 AND SPECIMEN Further Pure ... · PDF fileEdexcel AS/A level Mathematics Formulae List: Further Pure Mathematics FP2 – Issue 1 – September 2009 9 Further Pure

Edexcel AS/A level Mathematics Formulae List: Further Pure Mathematics FP2 – Issue 1 – September 2009 9

Further Pure Mathematics FP2 Candidates sitting FP2 may also require those formulae listed under Further Pure Mathematics FP1 and Core Mathematics C1–C4. Area of a sector

A = ⎮⌡⌠ θd

21 2r (polar coordinates)

Complex numbers

θθθ sinicosei += )sini(cos)}sini(cos{ θθθθ nnrr nn +=+

The roots of 1=nz are given by nk

zi2

eπ

= , for 1 , ,2 ,1 ,0 −= nk K

Maclaurin’s and Taylor’s Series

KK )0(f!

)0(f!2

)0(f)0f()f( )(2

+++′′+′+= rr

rxxxx

KK )(f!

)( )(f!2

)()(f)()f()f( )(2

+−

++′′−+′−+= a

raxaaxaaxax r

r

KK )(f!

)(f!2

)(f)f()f( )(2

+++′′+′+=+ arxaxaxaxa r

r

xrxxxx

rx allfor

!

!21)exp(e

2

KK +++++==

)11( )1( 32

)1(ln 132

≤<−+−+−+−=+ + xrxxxxx

rr KK

xr

xxxxxr

r allfor )!12(

)1( !5!3

sin1253

KK ++

−+−+−=+

xr

xxxxr

r allfor )!2(

)1( !4!2

1cos242

KK +−+−+−=

)11( 12

)1( 53

arctan1253

≤≤−++

−+−+−=+

xr

xxxxxr

r KK

F2 IAL PAPERS: 2014-6 AND SPECIMEN

IAL F2 SPECIMEN PAPER

JUNE 2014 F2 IAL 1. (a) Show that

1 1 1

1 2 3 2 1 2 2 2 3r r r r r r r

(2)

(b) Hence, or otherwise, find

1

1

1 2 3

n

rr r r

giving your answer as a single fraction in its simplest form.

(4)

2. Use algebra to find the set of values of x for which

62

3x

x

(7)

3. Solve the equation

z5 = 16 – 16i√3 giving your answers in the form reiθ where θ is in terms of π and 0 ≤ θ < 2π.

(5)

4. A transformation from the z-plane to the w-plane is given by

3

zw

z

, z ≠ –3

Under this transformation, the circle |z| = 2 in the z-plane is mapped onto a circle C in the w-plane. Determine the centre and the radius of the circle C.

(7)

5. 2

2

d d2 2 0

d d

y yx y

x x

(a) Show that

4 2

24 2

d d

d d

y yax b

x x

where a and b are constants to be found.

(5)

Given that y = 1 and d

3d

y

x at x = 0,

(b) find a series solution for y in ascending powers of x up to and including the term in x4.

(5)

(c) use your series to estimate the value of y at x = –0.2, giving your answer to four decimal places.

(2)

6. d

1 2d

yx x y x

x , x > 0

Find the general solution of the differential equation, giving your answer in the form y = f(x).

(9)

7. The point P represents a complex number z on an Argand diagram, where

|z + 1| = |2z – 1|

and the point Q represents a complex number w on the Argand diagram, where

|w| = |w − 1 + i| Find the exact coordinates of the points where the locus of P intersects the locus of Q.

(7)

8. (a) Show that the substitution x = et transforms the differential equation

22

2

d d5 13 0

d d

y yx x y

x x , x > 0 (I)

into the differential equation

2

2

d d4 13 0

d d

y yy

t t

(7)

(b) Hence find the general solution of the differential equation (I). (5)

9.

Figure 1

Figure 1 shows the curve C1 with polar equation 2 sin 2r a , 0 ≤ θ ≤ 2

, and the

circle C2 with polar equation r = a, 0 ≤ θ ≤ 2π, where a is a positive constant. (a) Find, in terms of a, the polar coordinates of the points where the curve C1 meets the

circle C2. (3)

The regions enclosed by the curve C1 and the circle C2 overlap and the common region R is shaded in Figure 1. (b) Find the area of the shaded region R, giving your answer in the form

213

12a p q , where p and q are integers to be found.

JUNE 2015 F2 IAL 1. (a) Using algebra, find the set of values of x for which

2

2 5

x

x x

(7) ______________________________________________________________________

_____

2. (a) Express 1

( 6)( 8)r r in partial fractions.

(1)

(b) Hence show that

1

2 ( )

( 6)( 8) 56( 7)( 8)

n

r

n an b

r r n n

where a and b are integers to be found.

(4) ______________________________________________________________________

_____

3. (a) Show that the substitution z = y−2 transforms the differential equation

2 3d

d 2 e xxy x

y

xy (I)

into the differential equation

2dz

de4 2 xxz

xx (II)

(4)

(b) Solve differential equation (II) to find z as a function of x.

(5)

(c) Hence find the general solution of differential equation (I), giving your answer in the form y2 = f(x).

(1) ______________________________________________________________________

_____

4. A transformation T from the z-plane to the w-plane is given by

1

1

zw

z

, z ≠ –1

The line in the z-plane with equation y = 2x is mapped by T onto the curve C in the w-plane. (a) Show that C is a circle and find its centre and radius.

(7) The region y < 2x in the z-plane is mapped by T onto the region R in the w-plane. (b) Show that C is a circle and find its centre and radius.

(2) ______________________________________________________________________

_____

5. Given that y = cot x, (a) Show that

23

2

d2cot 2cot

d

yx x

x

(3)

(b) Hence show that 3

4 23

dcot cot

d

yp x q x r

x

where p, q and r are integers to be found.

(3)

(c) Find the Taylor series expansion of cot x in ascending powers of 3

x

up to and

including the term in 3

3x

.

(3) ______________________________________________________________________

_____

6. (a) Find the general solution of the differential equation

2

2

d d2 3 2sin

d d

y yy x

x x (I)

(8)

Given that y = 0 and d

1d

y

x when x = 0

(b) find the particular solution of differential equation (I).

(5) ______________________________________________________________________

_____

7.

Figure 1 shows the 2 curves given by the polar equations

3sinr , 0 ≤ θ ≤ π

r = 1 + cos θ 0 ≤ θ ≤ π

(a) Verify that the curves intersect at the point P with polar coordinates 3

,2 3

(2) The region R, bounded by the two curves, is shown shaded in Figure 1.

(b) Use calculus to find the exact area of R giving your answer in the form a(π – 3 ), where a is a constant to be found.

(6) ______________________________________________________________________

_____

8. (a) Show that

3 36 2

6 2

1 1 1 1z z z k z

z z z z

where k is a constant to be found.

(3) Given that z = cos θ + i sin θ, where θ is real, (b) show that

(i) 1

2cosnn

z nz

(ii) 1

2isinnn

z nz

(3)

(c) Hence show that

3 3 1cos sin 3sin 2 sin 6

32

(4)

(d) Find the exact value of

3 38

0cos sin d

(4)

TOTAL FOR PAPER: 75 MARKS

END

F2 IAL JUNE 2016 1. (a) Express

2

1

4 1r in partial fractions.

(1)

(b) Hence prove that

2

1

1

4 1 2 1

n

r

n

r n

(3)

(c) Find the exact value of

25

2

9

5

4 1r

r

(2)

(Total 6 marks) ___________________________________________________________________________

2. Use algebra to find the set of values of x for which

2 9 1 2x x

(Total 6 marks)

___________________________________________________________________________

3. Find, in terms of k, where k is a positive integer, the general solution of the differential equation

122

d(1 ) (1 ) , > 0

dky

x ky x x xx

giving your answer in the form y = f(x).

(Total 6 marks) ___________________________________________________________________________

4. 3

f ( ) sin2

x x

(a) Find the Taylor series expansion for f(x) about 3

in ascending powers of

3x

up to and including the term in 4

3x

(6)

(b) Hence obtain an estimate of sin1

2, giving your answer to 4 decimal places.

(2)

(Total 8 marks) ___________________________________________________________________________

5. The transformation T from the z-plane to the w-plane is given by

2 1, z 3

3

zw

z

The circle in the z-plane with equation x2 + y2 = 1, where z = x + iy, is mapped by T onto the circle C in the w-plane. Find the centre and the radius of C.

(Total 7 marks) ___________________________________________________________________________

6. (a) Find the general solution of the differential equation

22

2

d d3 2 3 2 1

d d

y yy x x

x x

(9)

(b) Find the particular solution of this differential equation for which y = 0 and d

d

y

x = 0

when x = 0 (5)

(Total 14 marks) ___________________________________________________________________________

7.

Figure 1 shows a sketch of the curves C1 and C2 with polar equations

C1 : r = 3

cos , 02 2

C2 : r = 3 3 −9

cos , 02 2

The curves intersect at the point P. (a) Find the polar coordinates of P.

(3) The region R, shown shaded in Figure 1, is enclosed by the curves C1 and C2 and the initial line.

(b) Find the exact area of R, giving your answer in the form pπ +q 3 where p and q are rational numbers to be found.

(8)

(Total 11 marks) ___________________________________________________________________________

8. (a) Use de Moivre’s theorem to show that

cos5 θ ≡ p cos 5θ + q cos 3θ + r cos θ

where p, q and r are rational numbers to be found.

(6)

(b) Hence, showing all your working, find the exact value of

3 5

6

cos d

(4)

(Total 10 marks) ___________________________________________________________________________

9. The complex number z is represented by the point P in an Argand diagram.

Given that arg 5

2 4

z

z

.

(a) sketch the locus of P as z varies,

(3)

(b) find the exact maximum value of z .

(4)

(Total 7 marks) ___________________________________________________________________________


Pearson Edexcel International © Pearson Education Limited 2013 Sample Assessment Materials Advanced Level in Mathematics

159

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24

*S45002A02424*

Question 8 continued

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END

Q8

(Total 14 marks)

WFM02/01: Further Pure Mathematics F2

Question Number Scheme Marks

1(a)

1 1

3 1 3 2r r−

− +

M1 A1

(2) (b)

1

3(3 1)(3 2)

n

r r r= − + =1 1 1 1 1 1 1 1...2 5 5 8 8 11 3 1 3 2n n

− + − + − + −− +

M1 A1ft

1 1 32 3 2 2(3 2)

nn n

= − =+ +

* A1

(3) (c)

Sum = f(1000) – f(99)

3000 297 0.003016004 598

− = or 33.01 10−×

M1 A1

(2)

7


160



2

f ( ) cos , f (0) -1t x x′′ ′′= − − =

B1

df ( ) ( 1 sin ) , f (0) 0.5dxt xt

′′′ ′′′= − + = − M1A1

2 3

f ( ) f (0) f (0) f (0) f (0) ...2 3!t tt t ′ ′′ ′′′= + + + +

2 31120.5 0.5 ...t t t= − − + M1 A1

5


161

WFM02/01: Further Pure Mathematics F2 Question Number Scheme Marks

3(a)

2( 4)( 3) 2( 3) 0x x x+ + − + = , 2( 3)( 7 10) 0x x x+ + + = so ( 2)( 3)( 5) 0x x x+ + + =

or alternative method including calculator

M1

Finds critical values –2 and -5

A1 A1

Establishes x > -2 A1ft

Finds and uses critical value –3 to give –5 < x < -3 M1A1 (6)

(b)

x > -2

B1ft

(1)

7


162



4(a)

Modulus = 16

B1

Argument = 2arctan( 3)

3π− =

M1 A1 (3)

(b)

3 3 3 32 216 (cos( ) i sin( )) 16 (cos 2 isin2 )

3 3z π π π π= + = + =4096 or 316 M1 A1

(2)

(c)

1 14 4

2 216 (cos( ) i sin( )) 2(cos isin )3 3 6 6

w π π π π = + = +

( )3 i= +

OR 1 3i− + OR 3 i− − OR 1 3i−

M1 A1ft M1A2 (1,0)

(5)

10


163


5(a)

1.5 sin 3 2θ+ = → sin3 0.5θ = 53 or 6 6π πθ ∴ =

,

M1 A1,

and 5 or

18 18π πθ∴ =

A1 (3)

(b)

Area = 518

18

212 (1.5 sin 3 ) d

π

π

θ θ + , - 21 2

9π ×

M1, M1

=

518

18

1 12 2(2.25 3sin3 (1 cos6 ))d

π

π

θ θ θ + + − - 21 2

9π ×

M1

=

518

1 12 2

18

1(2.25 cos3 ( sin 6 ))6

π

πθ θ θ θ − + −

- 21 29

π × M1 A1

13 3 5

24 36π= −

M1 A1 (7)

10


164


6(a)

Imaginary Axis Real axis Vertical Straight line Through 3 on real axis

B1 B1

(2)

(b) These are points where line x = 3 meets the circle centre (3, 4) with radius 5.

M1

The complex numbers are 3 + 9i and 3 – i. A1 A1 (3)

(c)

30 306 6w wz z− = − =

M1

30 6 30w∴ − = 5 5w∴ − = M1 A1

This is a circle with Cartesian equation 2 2( 5) 25u v− + =

M1 A1 (5)

10

6

Re(z) = 3

0


165

WFM02/01: Further Pure Mathematics F2 Question Number

Scheme

Marks

7(a)

d d d.d d dy y zx z x

= and d 2dy zz

= so d d2 .d dy zzx x

=

M1 M1 A1

Substituting to get 2d2 . 4 tan 2d

zz z x zx

− = and thus d 2 tan 1d

z z xx

− = *

M1 A1

(5)

(b)

2 tan d

I.F. ex x−= = 2 ln cose x = 2cos x

M1 A1

( )2 2d cos cosd

z x xx

∴ = 2 2cos cosz x xdx∴ =

M1

2cosz x∴ = 12 (cos 2 1)x + dx 1 1

4 2sin 2x x= + +c M1 A1

2 21 12 2tan sec secz x x x c x∴ = + +

A1

(6)

(c)

2 2 21 12 2( tan sec sec )y x x x c x∴ = + +

B1ft

(1)

12


166

Further Pure Mathematics F3Advanced/Advanced Subsidiary

You must have:Mathematical Formulae and Statistical Tables (Blue)

Centre Number Candidate Number

Write your name hereSurname Other names

Total Marks

WFM03/01Paper ReferenceSample Assessment Material

Time: 1 hour 30 minutes

S45003A©2013 Pearson Education Ltd.

1/2/2/

*S45003A0128*Turn over

Pearson Edexcel InternationalAdvanced Level

Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.

Instructions• Use black ink or ball-point pen.• If pencil is used for diagrams/sketches/graphs it must be dark (HB or B). Coloured pencils and highlighter pens must not be used.• Fill in the boxes at the top of this page with your name,

centre number and candidate number.• Answer all questions and ensure that your answers to parts of questions are clearly labelled.• Answer the questions in the spaces provided – there may be more space than you need.• You should show sufficient working to make your methods clear. Answers without working may not gain full credit.• When a calculator is used, the answer should be given to an appropriate degree of accuracy.

Information• The total mark for this paper is 75.• The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question.

Advice• Read each question carefully before you start to answer it.• Try to answer every question.• Check your answers if you have time at the end.


Question Number

Scheme

Marks

8(a)

Differentiate twice and obtaining d sin 5 5 cos5d

y x x xx

λ λ= + and 2

2

d 10 cos5 25 sin 5d

y x x xx

λ λ= −

M1 A1

Substitute to give

310

λ = M1 A1

(4)

(b)

Complementary function is cos 5 sin 5y A x B x= + or 5i 5ie ex xP Q −+ M1 A1

So general solution is cos 5 sin 5y A x B x= + +

3 sin 510

x x or in exponential form A1ft

(3)

(c)

y= 0 when x = 0 means A = 0

B1

d 5 cos5d

y B xx

= 3 3sin 5 cos510 2

x x x+ + and at x = 0 d 5dyx

= and so 5 = 5A M1 M1

So B = 1

A1

So sin 5y x= +

3 sin 510

x x

A1

(5)

(d)

B1 B1

(2)

14

"Sinusoidal" through O amplitude becoming larger Crosses x axis at

2 3 4, , ,5 5 5 5π π π π

Question Number F2 IAL JUNE 2014 MARK SCHEME Marks

1.(a)

1 1

2( 1)( 2) 2( 2)( 3)

3 ( 1)

2( 1)( 2)( 3)

r r r r

r r

r r r

Attempt common denominator M1

2

2( 1)( 2)( 3)

1

( 1)( 2)( 3)

r r r

r r r

Correct proof A1

(2)

(a) Way 2

1 1 1 1 1 1

( 1)( 2)( 3) 2 1 3 2 2 1 2 3

1 1

2( 1)( 2) 2( 2)( 3)

r r r r r r r r r

r r r r

M1 A1

M1: Factor of

1

2r and attempt partial fractions

A1: Correct proof

Other methods: Complete method scores M1 All work correct inc final answer reached A1

(b)

1

1

( 1)( 2)( 3)

1 1........

12 241 1

2( 1)( 2) 2( 2)( 3)

n

r r r r

n n n n

Attempt at least the first pair and the last pair of terms as shown. Must start at 1 and end at n

M1

1 1

12 2( 2)( 3)n n

Identifies that the first and last terms do

not cancel. M1

2 5 6 6

12( 2)( 3)

n n

n n

Correctly combined fractions A1

2 5 ( 5)or

12( 2)( 3) 12( 2)( 3)*n n n n

n n n n

Allow either form isw attempts to multiply out the denominatot

A1

(4) Total 6

Question Number

Scheme Marks

2. 6

23

xx

Way 1

6 6

2 2 03 3

x xx x

3 46

2 0 03 3

x xx

x x

Attempt to combine fractions and factorise the numerator

M1

3, 4x x Correct critical values A1, A1 4x Follow through their 4 A1ft 3x Identifies 3 as a critical value B1

3 3x M1: Attempt inside region

M1A1 A1: Correct inequality

(7) Way 2

26 3 2 3

3 4 3 0

x x x

x x x

Multiplies both sides by

23x

and attempt to factorise M1

3, 4x x Correct critical values A1, A1 4x Follow through their 4 A1ft 3x Identifies 3 as a critical value B1

3 3x M1: Attempt inside region

M1A1 A1: Correct inequality

(7)

Way 3

3 0 6 2 3x x x

4 3 0x x

Multiplies both sides by 3x

and attempt to factorise Must state 3 0x

M1

4x Correct critical values A1

4x Follow through their 4 A1ft 3x Identifies 3 as a critical value B1

3 0 6 2 3x x x

3x

Correct critical value

A1

2 3 6 4 3 0

3 3

x x x x

x

M1: Attempt inside region M1A1

A1: Correct inequality

(7)

Question Number

Scheme Marks

3.

5 2 2

1

516 (16 3) 32 32 2r r

Correct value for r B1

5

arg(16 16 3)3

i

Allow 5

or3 3

B1

11 17 23 29

5 , , ,3 3 3 3

52 , 1, 2,3, 4

3n n

At least 2 values which must be positive. May be implied by correct final answers.

M1

11 17 23 29

3 15 15 15 15i i i i i2e ,2e ,2e ,2e ,2ez

515 3

1i i52 or 32 , e or e

B1 A1(all 4 values)

(5)

Total 5

Question Number

Scheme Marks

4 3

zw

z

3

3 1

z ww z

z w

M1: Attempt to make z the subject M1A1

A1: Correct expression for z

32 2

1

wz

w

3 2 1w w

22 2 29 4 1 4u v u v

M1: Uses 2z to obtain an equation in u

and v Pythagoras must be used correctly. No i seen M1A1

A1: Any correct equation in u and v Isw attempts to simplify

2 25 5 8 4 0u v u

2

24 16 40

5 25 5u v

Rearrange to a suitable form for a circle and attempt centre and/or radius. -16/25 (from completing the square) may be omitted. May be implied by centre and radius correct for their previous equation

M1

Centre 4

,05

oe A1

Radius 6

5 oe A1

(7) Total 7

Question Number

Scheme Marks

5 2

2

d d2 2 0

dd

y yx y

xx

(a) 2 2 2 ( 0) 2y y xy y y xy

M1: Attempt to differentiate including use of the product rule on

d2

d

yx

x Equation may have been re-written as y'' = ... before differentiating

M1A1

A1: Correct differentiation

2 2 2 2 ( 0)y y xy y y

M1: Second use of product rule. Dependent on first M1.

dM1A1 A1: Correct differentiation NB A simpler form is obtained if ''' 2 '' 0y xy is used.

2

2 2

2 2 2 4 2

y xy y

x xy y x y

Cao and cso A1

(5)

(b) 0 0 02, 0 4vy y y

B1: 0 2y

B1, M1A1 M1: Attempts 0 0and vy y

A1: All correct and obtained from correct expressions

4

21 36

xy x x

M1: Correct use of Maclaurin series M1A1

A1: Fully correct expansion.

(5)

(c) 4

2 ( 0.2)1 3( 0.2) ( 0.2)

6y

Use of the correct Maclaurin series

and substitution of 0.2x M1

(y =) 0.3597 Allow awrt A1 (2) Total 12

Question Number

Scheme Marks

6. d

(1 2 )d

yx x y x

x

d (1 2 )

( 1)d

y xy

x x

Divides by x - may be implied by subsequent working

M1

1 2 d

Integrating factor ex

x xI

Correct attempt at I, including an attempt at the integration. ln must be seen if not fully correct.

dM1

ln 2e x x Correct expression A1

2e xx No errors in working allowed A1

2 2e e dx xxy x x

Multiply through by their IF and integrate LHS. Can be given if y x their IF is seen

M1

2 21 1

2 2e e dx xx x

M1: Correct application by parts ie differentiate x and attempt to

integrate2e x

M1A1

A1: Correct expression

2 21 12 4e ex xx c

A1: Complete the integration to a correct result. Constant not required.

A1

2e 1 1

4 2

xcy

x x

Oe Must have y = ... Must include a constant. ft their previous line

A1ft

(9) Total 9

Question Number

Scheme Marks

7 : 1 2 1P z z

i i 1 2 i 1z x y x y x y

2 2 221 2 1 2x y x y

M1: Correct use of Pythagoras M1A1

A1: Any correct equation

: 1 iQ w w

i i i 1 iw x y x y x y

2 22 2 1 1 1x y x y y x

M1: Correct use of Pythagoras. Allow with u and v instead of x and y

M1A1 A1: Any correct equation Must have x and y now.

Alternative – M1: identifies perpendicular bisector of (0, 0) and (1, -1)

A1: y = x – 1

2 2 2 21 1 1 or 1x x y y

Attempt to solve simultaneously tie obtain an equation in one variable and get to x = ... or y = ...

M1

1 1 1 1

1 ,1 or ,2 2 2 2

x y Both (oe) A1

1 1 1 1

1 , and 1 ,2 2 2 2

Both (oe) Pairs must be clearly identifiable but coordinate brackets not needed.

A1

(7) Total 7

Question Number

Scheme Marks

8. etx

(a) d d d

d d d

y y t

x t x

Attempt to use an appropriate version of the chain rule

M1

d d 1 1 d

d d e dt

y y y

x t x t

Oe A1

2 2

2 2 2

d 1 d 1 d d,

d d d d

y y y t

x x t x t x

2 2

2 2 2 2

d 1 d 1 d or ,

d d d

y y y

x x t x t

M1: Use of the product rule (penalise chain rule errors by loss of A mark or marks)

d 1Note ln

d

tt x

x x

M1 A1, A1

22

2

22

2 2

2

2

d d5 13 0

d d

1 d d 1 d. 5 13 0

d d d

d d4 13 0*

dt dt

y yx x y

x x

y y yx x y

x t t x t

y yy

M1: Substitutes their first and second derivatives into the given differential equation Depends on both M marks above

ddM1A1

A1: Correct completion to printed answer

(7)

(b)

2 4 13 0

4 16 52

2

m m

m

Attempt to solve the auxiliary equation M1

2 3m i

Correct roots May be implied by a correct GS

A1

2e cos3 sin3ty A t B t

2 3i 2 3ior e et ty A B

Correct GS A1

lnt x B1

2

cos 3ln sin 3lnA x B xy

x

2 3i ln 2 3i lnor e ex xy A B

A1

(5)

Total 12

June 2015 WFM02 Further Pure Mathematics F2

Mark Scheme Question Number Scheme Notes

Marks

1. 2

2 5

x

x x

Critical Values 2 and 5 Seen anywhere in solution Both correct B1B1; one correct B1B0

B1, B1

20

2 5

x

x x

2 3 40

2 5

x x

x x

4 10

2 5

x x

x x

Attempt single fraction and factorise

numerator or use quad formula M1

Critical values 4 and 1 Correct critical values May be seen on a graph or number line.

A1

5 4, 2 1

5, 4 2,1

x x

dM1: Attempt an interval inequality using one of 2 or 5 with another cv

dM1A1,A1

A1, A1: Correct intervals Can be in set notation One correct scores A1A0 Award on basis of the inequalities seen - ignore any and/or between them Set notation answers do not need the union sign.

(7) ALT Critical Values 2 and 5 Seen anywhere in solution B1, B1

2 22

5 2 2 2 52 5

xx x x x x

x x

5 2 5 2 2 0x x x x x

5 2 1 4 0x x x x Multiply by

2 25 2x x

and attempt to factorise a quartic or use quad formula

M1

Critical values 4 and 1 Correct critical values A1

5 4, 2 1

5, 4 2,1

x x

dM1: Attempt an interval inequality using one of 2 or 5 with another cv

dM1A1,A1 A1, A1: Correct intervals Can be in set notation One correct scores A1A0

(7) Any solutions with no algebra (eg sketch graph followed by critical values with no working) scores max B1B1

Question Number Scheme Notes Marks

1

6 8r r

2(a)

1 1

oe2 6 2 8r r

Correct partial fractions, any equivalent form

B1

(1) (b) 1 1 1 1 1 1 1 1 1 1 1

2 ...2 7 9 8 10 9 11 5 7 6 8n n n n

Expands at least 3 terms at start and 2 at end (may be implied) The partial fractions obtained in (a) can be used without multiplying by 2.

Fractions may be 1 1 1 1

2 7 2 9 etc These comments apply to both M1 and A1

M1

1 1 1 1

7 8 7 8n n

Identifies the terms that do not cancel A1

15 7 8 56 2 15

56 7 8

n n n

n n

Attempt common denominator Must have multiplied the fractions from (a) by 2 now

M1

15 113

56 7 8

n n

n n

A1cso

(4) Total 5


3 2 3d2 e

dxy

xy x yx

(a) 122z y y z

321

2

d d

d d

y zz

x x

M1: 32

d d

d d

y zkz

x x

M1A1

A1: Correct differentiation

3 322 2

12

12

d 2e

dxz x

z x zx z

Substitutes for dy/dx M1

2d4 2 e

dxz

xz xx

* Correct completion to printed answer with no errors seen

A1cso

(4) (a) Alternative 1

3d2 oe

d

zy

y

M1: 3d

d

zky

y

M1A1

A1: Correct differentiation 23 31

2

d2 e

dxz

y xy x yx


2dz4 2 e

dxxz x

x * Correct completion to printed answer

with no errors seen A1

(a) Alternative 2

3d d2

d d

z yy

x x

M1: 3d d

d d

z yky

x x inc chain rule

M1A1 A1: Correct differentiation

23 312

d2 e

dxz

y xy x yx


2dz4 2 e

dxxz x

x * Correct completion to printed answer

with no errors seen A1

(b) 24 d 2e e

x x xI

M1: 4 d

ex x

I

M1A1 A1:

22e x 2 22 3e 2 e dx xz x x

2

2 e dxz I x I x dM1

231

e3

x c 2 2

e d eqx qxx x p c M1

2 22 1e e

3x xz c Or equivalent A1

(5) (c) 2 2

2 2

2 22

2

1 1 1e e

13 e e3

x x

x x

c yy c

2 1

( )y

b

2

23

3e

1 e

x

xk

B1ft

(1) Total 10


1

1

zw

z

4(a) 11 ...

1

zw wz w z z

z

Attempt to make z the subject M1

1

1

wz

w

Correct expression in terms of w A1

1 1

1 1

u iv u iv

u iv u iv

Introduces “u + iv” and multiplies

top and bottom by the complex conjugate of the bottom

M1

2 2 1 2,

.... ....

u v vx y

2 22 2 2 2 2y x v u v

Uses real and imaginary parts and y = 2x to obtain an equation connecting “u” and “v” Can have the 2 on the wrong side.

M1

22 1 1

2 4 1u v

Processes their equation to a form that is recognisable as a circle ie coefficients of u2 and v 2 are the same and no uv terms

M1

Centre (0, 12 ), radius

5

2

A1: Correct centre (allow -½i) A1,A1

A1: Correct radius (7)

Special Case:

1 2 i 1 2 ii 1

i 1 1 2 i 1 2 i

x x x xx yw

x y x x x x

M1: rationalise the denominator,

may have 2x or y

2 2

2 2

1 4 2 i 1 1

1 4

x x x x x

x x

A1: Correct result in terms of x only. Must have rational denominator shown, but no other simplification needed

(b) B1ft: Their circle correctly positioned provided their equation does give a circle

B1ft B1

B1: Completely correct sketch and shading

(2)

Total 9

R

Question Number Scheme Notes

Marks

5 coty x

(a) 2dcosec

d

yx

x

2

2

d2cosec cosec cot

d

yx x x

x

M1: Differentiates using the chain rule or product/quotient rule M1A1 A1: Correct derivative

2 32cosec cot 2cot 2cot *x x x x

A1: Correct completion to printed answer 2 21 cot cosecx x or 2 2cos sin 1x x

must be used Full working must be shown

A1cso*

(3) Alternative: 2 2

2 2

cos d sin cos 1

sin d sin sin

x y x xy

x x x x

23

2

d2sin cos ...

d

yx x

x M1A1

A1: Correct completion to printed answer see above A1

(b) 32 2 2

3

d2cosec 6cot cosec

d

yx x x

x Correct third derivative B1

2 2 22 1 cot 6cot 1 cotx x x Uses 2 21 cot cosecx x M1

4 26cot 8cot 2x x cso A1

(3) (c) 1 4 8 16

f , f , f , f3 3 3 3 3 33 3 3

M1: Attempts all 4 values at 3

No working need be shown

M1

2 3

1 4 4 8

3 3 3 9 33 3 3y x x x

M1: Correct application of Taylor using their values. Must be up to and including 3

3x

A1: Correct expression Must start y = .... or cot x f(x) allowed provided defined here or above as f cot or x x y

Decimal equivalents allowed (min 3 sf apart from 0.77), 0.578, 1.33, 0.770, (0.7698.., so accept 0.77) 0.889

M1A1

(3)

Total 9


6(a) 2

2

d d2 3 2sin

d d

y yy x

x x

AE: 2 2 3 0m m

2 2 3 0 ... 1,3m m m Forms Auxiliary Equation and attempts to solve (usual rules)

M1

3e ex xy A B Cao A1

PI: sin cosy p x q x Correct form for PI B1

cos sin

sin cos

y p x q x

y p x q x

sin cos 2 cos sin 3 sin 3 cos 2sinp x q x p x q x p x q x x

Differentiates twice and substitutes M1

2 4 2, 4 2 0q p q p

Correct equations A1

2 1,

5 5p q A1A1 both correct

A1A0 one correct A1A1

1 2cos sin

5 5y x x

3 1 2e e cos sin

5 5x xy A B x x Follow through their p and q and their

CF B1ft

(8) (b) 3 1 2

3 e e sin cos5 5

x xy A B x x Differentiates their GS

M1

1 20 , 1 3

5 5A B A B

M1: Uses the given conditions to give two equations in A and B M1A1 A1: Correct equations

3 1,

10 2A B Solves for A and B Both correct A1

33 1 1 2e e cos sin

10 2 5 5x xy x x Sub their values of A and B in their GS A1ft

(5) Total 13

Question Number Scheme

Notes

Marks

7(a) 33sin

3 3 2r

Attempt to verify coordinates in at least one of the polar equations M1

31 cos

3 3 2r

Coordinates verified in both curves

(Coordinate brackets not needed) A1

(2)

Alternative:

Equate rs: 3sin 1 cos and verify (by substitution) that 3

is a solution

or solve by using tan2

t

or writing 3 1 1 1

sin cos sin2 2 2 6 2 3

Squaring the original equation allowed as is known to be between 0 and

M1

Use

3

in either equation to obtain

3

2r A1

(b)

2 21 1( 3sin ) d , (1 cos ) d

2 2

Correct formula used on at least one curve (1/2 may appear later) Integrals may be separate or added or subtracted.

M1

2 21 13sin d , (1 2cos cos )d

2 2

1 3 1 1

1 cos2 d , (1 2cos 1 cos2 )d2 2 2 2

Attempt to use 2 2 1 1sin or cos cos2

2 2 on either integral

Not dependent 1/2 may be missing

M1

3

03

3 1 1 3 1sin 2 , 2sin sin 2

4 2 2 2 4

Correct integration (ignore limits) A1A1 or A1A0

A1, A1

3 3 1 3 3

0 34 3 4 2 2 2 8

R

Correct use of limits for both integrals Integrals must be added. Dep on both previous M marks

ddM1

3

34 Cao

No equivalents allowed A1

(6) Total 8

Question Number

Scheme

Notes

Marks

8(a) 3 3 32

2

1 1 1z z z

z z z

6 2 62

33z z z

z

M1: Attempt to expand M1A1

A1: Correct expansion

6 26 2

1 13z z

z z

Correct answer with no errors seen A1

(3) (a)

ALT 3 3

3 33 3

1 3 1 1 3 13 , 3z z z z z z

z z z z z z

M1A1

M1: Attempt to expand both cubic brackets A1: Correct expansions

6 26 2

1 13z z

z z

Correct answer with no errors A1

(3) (b)(i)(ii) cos sinnz n i n Correct application of de Moivre B1

cos sin cos sinnz n i n n n

but must be different from their zn

Attempt z-n M1

1 12cos *, 2 sin *n n

n nz n z i n

z z cos sinnz n i n must be seen A1*

(3) (c)

3 3

3 31 12cos 2 sinz z i

z z

B1

6 26 2

1 13 2 sin 6 6 sin 2z z i i

z z

Follow through their k in place of 3 B1ft

3 364 sin cos 2 sin 6 6 sin 2i i i Equating right hand sides and

simplifying 332 2i (B mark

needed for each side to gain M mark)

M1

3 3 1

cos sin 3sin 2 sin 632

*

A1cso

(4) (d)

3 3

0 0

8 8 1cos sin d 3sin 2 sin 6 d

32

0

81 3 1cos 2 cos6

32 2 6

M1: cos2 cos6p q

M1A1 A1: Correct integration Differentiation scores M0A0

1 3 1 3 1 1 4 5 2

32 2 6 32 3 62 2 6 2

dM1: Correct use of limits – lower limit to have non-zero result. Dep on previous M mark dM1A1

A1: Cao (oe) but must be exact

(4) Total 14

Question Number

Scheme Marks

9.

(a) 1

2 sin 2 sin 2 2 ...2

a a C1 = C2 and attempt to solve for 2 M1

1 5

sin 2 2 ,2 6 6

52 or or both

6 6

Decimals

allowed (min 3 sf).

A1

5

, , ,12 12

a a

Both points

Can be written r a , 5

,12 12

Decimals allowed (min 3 sf).

A1

(3)

(b) 21oe

2 3a

Correct expression for the sector B1

221 1

d 2 sin 2 d2 2

r a Use of correct formula Limits not needed (ignore any shown)

M1

2

2

cos 4 1 2sin 2

1sin 2 1 cos 4

2

Uses 2 1 cos4

sin 22

M1

1

1 cos 4 d sin 44

Correct integration Limits not needed (ignore any shown)

A1

122

0

2

1sin 4

4

1sin 4. 0

12 4 12

I a

a

An attempt to find one or both of the regions either side of the sector.

ie uses limits 0,12

and/or

5,

12 2

,

limits to be substituted and subtracted (if non-zero after substitution). Limits to be used the correct way round. If two integrals seen award mark if either correct. Both previous method marks must have been scored.

ddM1

2 2

2 32 2

6 12 8 6

a aR I a

Correct strategy for the complete area (sector + 2I). All areas must be positive.

M1

214 3 3

12R a If decimals seen anywhere (either in rt

3 or the limits) this mark is lost. A1

(7)

Total 10

JUNE 2016 F2 MARK SCHEME


1(a) 2

1

4 1r

1 12 21 1

or2 2 1 2 2 1 2 1 2 1r r r r

or equivalent or

2

1 1 1,

4 1 2 1 2 1 2 2

A BA B

r r r

Correct partial fractions or correct values of ‘A’ and ‘B’. Isw if possible so

if correct values of ‘A’ and ‘B’ are

found, award when seen even if followed by incorrect partial fractions.

B1

(1) (b)

21

1 1 1 1 11 ......

4 1 2 3 2 1 2 1

n

r r n n

Attempt at least first and last terms using their partial fractions.

May be implied by e.g. 1 1

12 2 1n

M1

1 1

or or2 2 1 4 2

1 1 1 11

2 2 1 2 2n nn

Correct expression A1

2 1*

n

n Correct completion with no errors A1*

Allow a different variable to be used in (a) and (b) but final answer in (b) must be as printed i.e. in terms of n.

(3) (c)

25

29

55 f (25) f (8)

4 1r r

f (25) f (8) where f2 1

nn

n

M1

25 8 55

51 17 51

cao A1

Correct answer with no working in (c) scores both marks.

(2) Total 6


Marks

2 2 9 1 2x x

(ignore use of “<” instead of “=” when finding cv’s)

2 29 1 2 2 10 0 ...x x x x x or

2 29 1 2 2 8 0 ...x x x x x

Attempts to solve 2 9 1 2x x OR 2 9 1 2x x

to obtain two non-zero values of x

M1

2 44

2x

OR 2, 4x

One correct pair of values. Allow the irrational roots to be at least as given here or 1 11 or awrt 2.32, -4.32 or truncated 2.3, -4.3

A1

2 29 1 2 2 8 0 ...x x x x x

Attempts to solve 2 9 1 2x x AND 2 9 1 2x x

to obtain four non-zero values of x

M1

2 44

2x

AND 2, 4x

Both pairs of values correct. Allow the irrational roots to be at least as given here or 1 11 or awrt 2.32, -4.32 or truncated 2.3, -4.3

A1

1 11 4x or

1 11 2x

One correct inequality. B1

For 1 11 allow 2 44

2

, for 1 11 allow

2 44

2

but must be exact here.

Allow alternative notation e.g. 1 11, 4 , 1 11, 2

4 1 11, 1 11 4x x x and ,

2 1 11, 1 11 2x x x and

1 11 4x and

1 11 2x

Both inequalities correct. B1

For 1 11 allow 2 44

2

, for 1 11 allow

2 44

2

but must be exact here.

Allow alternative notation e.g. 1 11, 4 , 1 11, 2

4 1 11, 1 11 4x x x and ,

2 1 11, 1 11 2x x x and

(6) Total 6

Q2 Alternative by squaring (ignore use of “<” instead of “=” when finding cv’s)

2 22 4 2 29 1 2 18 81 1 4 4x x x x x x

4 222 4 80 0 ...x x x x Squares and attempts to solve a quartic equation to obtain at least two values of x that are non-zero.

M1

2 44

2x

or 2, 4x

One pair of values correct as defined above

A1

2 44

2x

and 2, 4x

M1: Obtains four non-zero values of x.

M1A1 A1: Both pairs of values correct as defined above

1 11 4x or

1 11 2x

See notes above B1

1 11 4x and

1 11 2x

See notes above B1

In an otherwise fully correct solution, if any extra incorrect regions are given, deduct the final B mark.

Scheme Notes Marks

3

12

2d1 1

dky

x ky x xx

12

21d

d 1 1

kx xy ky

x x x

Divides by (1 + x) including the ky term

M1

d

1I e 1k

x kx x

dM1: Attempt integrating factor. d

1I ek

xx is sufficient for this mark

but must include the k. Condone omission of “dx”.

dM1A1

A1: 1k

x

121 1 d

ky x x x x

Reaches

12 1their I 1 their I dky x x x M1

3 51

2 2 22 2

1 d3 5

x x x x x c

or by parts

3 51

2 2 22 4

1 d 13 15

x x x x x x c

Correct integration A1

3 52 2

2 23 5

1k

x x cy

x

or e.g.

3 52 2

2 41

3 151

k

x x x cy

x

or e.g.

3 52 2

12 41 1 1

3 15

k k ky x x x x c x

or e.g.

3 52 210 1 4

15 1k

x x x cy

x

or e.g.

3 52 2

2 21 1 1

3 5

k k ky x x x x c x

Correct answer with the constant correctly placed. Allow any equivalent correct answer.

A1

(6) Total 6

Question Number

Scheme

Notes

Marks

4(a)

32

' 3 32 2

'' 9 34 2

''' 27 38 2

'''' 81 316 2

f sin

f cos

f sin

f cos

f sin

x x

x x

x x

x x

x x

M1: Attempt first 4 derivatives. Should be sin cos sin cos sin . I.e. ignore signs and coefficients.

M1A2 ' ''3 3 9 32 2 4 2A1:f cos and f sinx x

''' ''''27 3 81 38 2 16 2A1:f cos andf sinx x

Allow un-simplified e.g. '' 3 3 32 2 2f . sin x

' '' ''' ''''9 813 3 3 4 3 3 161, 0, , 0,y y y y y

Attempts at least 1 derivative at 3

x

M1

2 4

9 27f 1

8 3 128 3x x x

dM1: Correct use of Taylor series. I.e.

2 3

3 3 3 3

ff f f ...

2!x x x

Evidence of at least one term of the correct

structure i.e. 3

3

f

!

nn

xn

and not a

Maclaurin series. Dependent on the previous method mark.

dM1A1

A1: Correct expansion. Allow equivalent

single fractions for 9

8and/or

27

128and allow

decimal equivalents i.e. 1.125 and 0.2109375. Ignore any extra terms.

(6) (b)

1f 0.4815

3

M1: Attempts

1f

3

or states 1

3x

M1A1

A1: 0.4815 cao

(2) Total 8

Question Number

Scheme

Notes

Marks

5 3 1

2

wz

w

M1: Attempt to make z the subject as far as z = … M1A1 A1: Correct equation

3 i 13 11 1 1

2 2 i

u vwz

w u v

Uses 1z and introduces w = u + iv M1

2 2 2 23 1 3 2u v u v

M1: Correct use of Pythagoras. Condone missing brackets provided the intention is clear and allow e.g. (3v)2 = 3v2 but there should be no i’s.

M1

2 2 10 30

8 8u v u

225 25 3

8 64 8u v

Attempt to complete the square on the equation of a circle. I.e. an equation where the coefficients of u2 and v2 are the same and the other terms are in u, v or are constant. (Allow slips in completing the square). Dependent on all previous M marks.

ddddM1

225 49 5 7

,0 ,8 64 8 8

u v

A1: Centre 5

,08

A1A1 A1: Radius

7

8

(7) Total 7

Alternative for the first 3 marks 3 1

2

wz

w

M1: Attempt to make z the subject M1A1

A1: Correct equation

2 2

2 22 2

2 22 2

2 22 22 2

5 2 33 i 1 3 1 3i 2 i 7i i

2 i 2 i 2 i 2 2

5 2 3 71 1

2 2

u u vu v u v u v vx y

u v u v u v u v u v

u u v vx y

u v u v

M1

Introduces w = u + iv, multiplies numerator and denominator by the complex conjugate of the denominator and uses 2 2 1x y correctly to obtain an equation in

u and v.


22

2

d d3 2 3 2 1

d d

y yy x x

x x

6(a) 2 3 2 0 1, 2m m m Correct roots (may be implied by their CF)

B1

2e ex xy A B M1: CF of the correct form

M1A1 A1: Correct CF

2y ax bx c Correct form for PI B1

2

2 22

d d2 , 2 2 3 2 2 3 2 1

d d

y yax b a a ax b ax bx c x x

x x M1

M1: Differentiates twice and substitutes into the lhs of the given differential equation and puts equal to 23 2 1x x or substitutes into the lhs of the given differential

equation and compares coefficients with 23 2 1x x .

For the substitution, at least one of , ory y y must be correctly placed.

3

2a A1

7 176 2 2

2 4a b b c

M1: Solves to obtain one of b or c M1A1

A1: Correct b and c

2 23 7 17e e

2 2 4x xy A B x x

Correct ft (their CF + their PI) but must be y = …

B1ft

(9) (b) 17

04

A B Substitutes x = 0 and y = 0 into their GS

M1

2d 7 72 e e 3 0 2

d 2 2x xy

A B x A Bx

Attempts to differentiate and substitutes x = 0 and y’ = 0 M1

17 70 , 0 2 .., ..

4 2A B A B A B

Solves simultaneously to obtain values for A and B

M1

3, 5

4A B Correct values A1

2 23 3 7 17e 5e

4 2 2 4x xy x x

Correct ft (their CF + their PI) but must be y = …

B1ft

(5) Total 14


7. 1 2

3 9: cos , : 3 3 cos

2 2C r C r

(a) 3 9cos 3 3 cos ...

2 2

or 2

cos 3 3 3 ...3

rr r r

Puts C1 = C2 and attempt to solve for or

Eliminates cos θ and solves for r M1

3 3or

6 4r

Correct θ or correct r. Allow θ = awrt 0.524, r = awrt 1.3

A1

3 3and

4 6r

Correct r and θ (isw e.g. 3 3

,6 4

)

Allow θ = awrt 0.524, r = awrt 1.3

A1

(3)

7(b) 2 2

1 9 1 33 3 cos d or cos d

2 2 2 2

M1

Attempts to use correct formula on either curve. The ½ may be implied by later work.

22 cos 2 19 81 81

3 3 cos 27 27 3 cos cos 27 27 3 cos2 4 4 2

M1

Expands to obtain an expression of the form 2cos cosa b c and attempts to use

2 1 cos 2cos

2 2

2

1 9 1 297 813 3 cos d 27 3 sin sin 2

2 2 2 8 16

M1A1

M1: Attempts to integrate to obtain at least two terms from , sin , sin 2

A1: Correct integration with or without the ½ (NB 297 81

278 8

)

6

0

1 297 81 1 297 8127 3 sin sin 2 . 27 3.sin sin 2. 0

2 8 16 2 8 6 6 16 6

M1

M1: Uses the limits 0 and their 6

If the substitution for = 0 evaluates to 0 then the substitution for = 0 does not need to be seen but if it does not evaluate to 0, the substitution for = 0 needs to be seen.

2

6

2

1 3 9 9 1 9 3cos d cos 2 1 d sin 2

2 2 16 16 2 16 3 4

M1

M1: Uses 2 1 cos 2cos

2 2

, integrates to obtain at least sin 2k and uses the limits

of their 6

and

2

to find the other area

NB can be done as a segment : 2 2

1 3 2 1 3sin

2 4 3 2 4 3

Allow 2 2

1 3 1 32 their sin 2 their

2 4 6 2 4 6

297 351 3 9 3 105 453

96 64 16 3 4 32 8

M1A1

M1: Adds their two areas both of which are of the form 3a b

A1: Correct answer (allow equivalent fractions for 105

32and/or

45

8)

(8) Total 11


Marks

8(a) WAY 1

55 3

3 5

1 10 5 15 10z z z z

z z z z

M1: Attempt to expand 51

zz M1A1 A1: Correct expansion with correct

powers of z. 1cos isin 2coszz z May be implied B1

5 35 3

1 1 15 10 2cos5 10cos3 20cosz z z

z z z

Uses at least one of 5 35 3

1 12cos5 or 2cos3z z

z z

M1

5 51 32coszz B1

5 1 5 5cos cos5 cos3 cos

16 16 8 Correct expression A1

(6) WAY 2 (Using ie )

5i i 5i 3i i i 3i 5ie e e 5e 10e 10e 5e e

M1: Attempt to expand

5i ie e M1A1

A1: Correct expansion i i2cos e e May be implied B1

5i 5i 3i 3i i ie e 5 e e 10 e e 2cos5 10cos3 20cos

Uses one of 5i 5i 3i 3ie e 2cos5 or e e 2cos3 M1

5i i 5e e 32cos B1



WAY 3 (Using De Moivre on cos 5θ and identity for cos 3θ)

5 5 4 3 2 2 2 3 3 4 4 5 5cos isin 5i 10 i 10 i 5 i ic c s c s c s c s s

M1: Attempts to expand. NB may only consider real parts here. A1: Correct real terms (may include i’s) (Ignore imaginary parts for this mark)

M1A1

5 3 2 4cos5 cos 10cos sin 5cos sin Correct real terms with no i’s B1

25 3 2 2cos 10cos 1 cos 5cos 1 cos Uses 2 2sin 1 cos to eliminate

sin θ M1

5 316cos cos5 20cos 5cos

3cos3 4cos 3cos Correct identity for cos 3θ B1

516cos cos5 5cos3 10cos



(6)

(b) 1 5 5 1 5 5cos5 cos3 cos d sin5 sin3 sin

16 16 8 80 48 8

M1A1ft

M1: Attempt to integrate – Evidence of 1

cos sin where 5 or 3n n nn

A1ft: Correct integration (ft their p, q, r)

3

6

1 5 5 1 5 5 5 1 5 5 5sin 5 sin 3 sin sin sin sin sin sin sin

80 48 8 80 3 48 8 3 80 6 48 2 8 6

Substitutes the given limits into a changed function and subtracts the right way round.

There should be evidence of the substitution of 3

and

6

into their changed function

for at least 2 of their terms and subtraction. If there is no evidence of substitution and the answer is incorrect, score M0 here.

M1

49 3 203

160 480

Allow exact equivalents e.g. 1 203

4.9 316 30

A1

If they use the letters p, q and r or their values of p, q and r, even from no working, the M marks are available in (b) but not the A marks.

(4) Total 10


5arg

2 4

z

z

9(a) M1

A circle or an arc of a circle anywhere. Allow dotted or dashed.

A1 A circle or an arc of a circle (allow dotted or dashed) passing through or touching at 2 and 5 on the positive real axis. (Imaginary axis

may be missing) A1

Fully correct diagram with 2 and 5 marked correctly with no part of the circle below the

real axis. It must be a major arc and not a semi-circle. The imaginary axis must be

present and the arc must not cross or touch the imaginary axis.

(3)

(b)

Centre ,c cC x y is at 3.5, 1.5

May be implied and may appear on the diagram. Can score anywhere e.g. from finding the equation of the circle in part (a) or as part of the calculation for OC.

B1

2 2 3 3 21.5 1.5

22r

22 cr y or equivalent work e.g.

1.5 1.5,

cos 45 sin 45r r or

3 2

2seen

M1

Max 2 23.5 1.5z OC r r M1

58 3

2 2 Oe e.g

58 3 214.5 4.5,

2

A1

(4)

Special Case – correct work with arc below the real axis: +

Centre ,c cC x y is at 3.5, 1.5 B0

2 2 3 3 2

1.5 1.522

r

22 cr y or equivalent work e.g.

1.5 1.5,

cos 45 sin 45r r or

3 2

2seen

M1

Max 2 23.5 1.5z OC r r M1

58 3

2 2 Oe e.g

58 3 214.5 4.5,

2

A1

Total 7 -

2 5

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F2 IAL PAPERS: 2014-6 AND SPECIMEN Further Pure ... · PDF fileEdexcel AS/A level Mathematics Formulae List: Further Pure Mathematics FP2 – Issue 1 – September 2009 9 Further Pure