Jun 13, 2011 - It is given that x satisfies the equation ... write down the three other roots of this equation in trigonometrical form. ..... Use of quadratic formula.
General Certificate of Education Advanced Level Examination June 2011
Mathematics
MFP2
Unit Further Pure 2 Monday 13 June 2011
9.00 am to 10.30 am
d
For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
e
s
Time allowed * 1 hour 30 minutes
n
Instructions * Use black ink or black ball-point pen. Pencil should only be used for drawing. * Fill in the boxes at the top of this page. * Answer all questions. * Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. * You must answer the questions in the spaces provided. Do not write outside the box around each page. * Show all necessary working; otherwise marks for method may be lost. * Do all rough work in this book. Cross through any work that you do not want to be marked.
e
d
n
o
C
Information * The marks for questions are shown in brackets. * The maximum mark for this paper is 75. Advice * Unless stated otherwise, you may quote formulae, without proof, from the booklet.
P38531/Jun11/MFP2 6/6/
MFP2
2
Draw on the same Argand diagram:
1 (a) (i)
the locus of points for which j z 2 5i j ¼ 5
(3 marks)
p 4
(3 marks)
(ii) the locus of points for which
argðz þ 2iÞ ¼
Indicate on your diagram the set of points satisfying both
(b)
j z 2 5i j 4 5 argðz þ 2iÞ ¼
and
p 4
(2 marks)
Use the definitions of cosh y and sinh y in terms of ey to show that
2 (a)
cosh x cosh y sinh x sinh y ¼ coshðx yÞ
(4 marks)
It is given that x satisfies the equation
(b)
coshðx ln 2Þ ¼ sinh x (i)
5
Show that tanh x ¼ 7 .
(ii) Express x in the form
3 (a)
(b)
(4 marks) 1 ln a . 2
(2 marks)
Show that ðr þ 1Þ! ðr 1Þ! ¼ ðr 2 þ r 1Þðr 1Þ!
(2 marks)
n X ðr 2 þ r 1Þðr 1Þ! ¼ ðn þ 2Þn! 2
(4 marks)
Hence show that
r¼1
(02)
P38531/Jun11/MFP2
3
The cubic equation
4
z 3 2z 2 þ k ¼ 0
ðk 6¼ 0Þ
has roots a, b and g. (a) (i)
Write down the values of a þ b þ g and ab þ bg þ ga .
(ii) Show that a 2 þ b 2 þ g 2 ¼ 4 .
(2 marks) (2 marks)
(iii) Explain why a 3 2a 2 þ k ¼ 0 .
(1 mark)
(iv) Show that a 3 þ b 3 þ g 3 ¼ 8 3k .
(2 marks)
Given that a 4 þ b 4 þ g 4 ¼ 0 :
(b) (i)
show that k ¼ 2 ;
(4 marks)
(ii) find the value of a 5 þ b 5 þ g 5 .
5 (a)
(3 marks)
The arc of the curve y 2 ¼ x 2 þ 8 between the points where x ¼ 0 and x ¼ 6 is rotated through 2p radians about the x-axis. Show that the area S of the curved surface formed is given by pffiffiffi ð 6 pffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ 4 dx S ¼ 2 2p
(5 marks)
0
(b)
6 (a)
(b)
By means of the substitution x ¼ 2 sinh y , show that pffiffiffi pffiffiffi S ¼ pð24 5 þ 4 2 sinh1 3Þ
(8 marks)
ðk þ 1Þð4ðk þ 1Þ2 1Þ ¼ 4k 3 þ 12k 2 þ 11k þ 3
(2 marks)
Show that
Prove by induction that, for all integers n 5 1 , 12 þ 32 þ 52 þ ::: þ ð2n 1Þ2 ¼ 1 nð4n2 1Þ 3
Turn over
s
(03)
(6 marks)
P38531/Jun11/MFP2
4 7 (a) (i)
Use de Moivre’s Theorem to show that cos 5y ¼ cos5 y 10 cos3 y sin2 y þ 5 cos y sin4 y and find a similar expression for sin 5y .
(5 marks)
(ii) Deduce that
tan yð5 10 tan2 y þ tan4 yÞ tan 5y ¼ 1 10 tan2 y þ 5 tan4 y (b)
Explain why t ¼ tan
(3 marks)
p is a root of the equation 5 t 4 10t 2 þ 5 ¼ 0
and write down the three other roots of this equation in trigonometrical form. (3 marks) (c)
Deduce that p 2p pffiffiffi ¼ 5 tan tan 5 5
(5 marks)
END OF QUESTIONS
Copyright Ó 2011 AQA and its licensors. All rights reserved.
(04)
P38531/Jun11/MFP2
MFP2 Q
Solution 1(a)
Marks
Total
Comments
Im
Use average of whole question if 2 diagrams used
Re
(i) Circle correct centre touching x-axis
B1 B1 B1F
(ii) half-line through (0, –2) through point of contact of circle with x-axis
B1 B1 B1
(b) Inside circle On line
B1 B1F Total
2(a)
( e x + e − x ) ( e y + e− y ) − ( e x – e− x ) ( e y – e− y )
2 2 2 2 Correct expansions 1 – x– y = e x – y + e ( ) = cosh ( x − y ) 2
(
(b)(i)
)
3 2 8
ft errors in position of line and circle
M0 if sinh and cosh confused M1 for formula quoted correctly
M1A1
A1
Use of e xy A0 4
AG Alternative:
– sinh x sinh ( ln 2 )
5 cosh ( ln 2 ) = 4 any method 3 sinh ( ln 2 ) = 4 5 7 cosh x = sinh x 4 4 5 tanh x = 7
(ii)
Can be inferred
A1
cosh ( x – ln 2 ) = cosh x cosh ( ln 2 )
1+ 1 x = ln 2 1 – 1 = ln 6 2
3
Circle in any position Must be shown ft incorrect centre
e x − ln 2 + e − x + ln 2 e x − e − x = M1 2 2
M1
Both correct
B1
5 x −x 7 or e − e = 5 5 e x + e− x 7 7
4
M1
A1 Total
ex or e − x + ln 2 = 2e −1 2 used B1
e x = 6 A1
A1F A1
e x− ln 2 =
AG
tanh x =
5 A1 7
Could be embedded in (b)(i)
2 10
MFP2 (cont) Q Solution 3(a) ( r + 1)! = ( r + 1) r ( r – 1)! Result
Marks M1 A1
(b) Attempt to use method of differences n
(r r =1
2
+ r – 1) ( r – 1)! = ( n + 1)! + n! – 1! – 0!
(ii)
α
2
m1 A1
B1
= ( α ) 2 – 2 αβ
(b)(i)
3
AG
2 Used. Watch 2
AG
E1
1
eg α satisfies the cubic equation since it is a root. Accept z = α
M1
= 8 – 3k
A1
Or 2
α
B1
α
M1
Or
A1
ft on
= 2 α – k α 3
= 2 ( 8 – 3k ) – 2k
α
5
A1 = 2 α 4 – k α 2
4
M1
Substitution of values = –8
A1 A1 Total
3 14
3
= ( α ) − 3 α αβ + 3αβγ 3
AG
α 4 = 2α 3 – kα 4
α = − 2 (M1A0)
A1
= 2 α 2 − 3k
k =2 (ii)
4 6
M1
(iii) Clear explanation
α
Must be seen
B1
=4
(iv)
AG
A1
Total
α = 2 αβ = 0
2
Comments
M1
( n + 1)! = ( n + 1) n! ( n + 2 ) n! – 2 4(a)(i)
Total
AG
α
4
= ( α 2 ) − 2 ( αβ ) + 4αβγ α
α = − 2
2
2
MFP2 (cont) Q 5(a)
Solution
2y
Marks
dy = 2x dx
S = 2π
6
0
Total
Or
B1
y 1+
x2 dx y2
Comments –1 dy = x ( x 2 + 8) 2 dx
M1 for use of formula provided
M1 A1F
function of x A1 for substitution for
Eliminating all y
= 2 2π
(b)
6
0
x 2 + 4 dx
A1 dx = 2cosh θ dθ
5
AG
B1
S = 2 2 π 4sinh 2 x + 4 . 2coshθ dθ
)
For eliminating x completely and use of dθ , ie dθ attempted Use of cosh 2 θ – sinh 2 θ = 1 (ignore limits) Use of formula for cosh 2θ ; must be correct
M1
S = 2 2 π 2coshθ . 2coshθ dθ
m1
( cosh2θ + 1) dθ
m1
= 4 2π
dy (one slip) dx
m1
dx = 2cosh θ dθ or
(
dy is a dx
sinh 2θ = 4 2π +θ 2
B1F
Correct integration of a cosh 2θ + b
= 4 2 π [sinhθ coshθ + θ ]
m1
Use of sinh 2θ = 2sinh θ cosh θ Must be seen
M1
Or change limits
x = 4 2π 2
x2 x +1 + sinh –1 4 2
6
0
= π 24 5 + 4 2 sinh –1 3
(
A1 Total
)
6(a) Expansion of ( k + 1) 4 ( k + 1) – 1 2
8 13
M1
= 4k 3 + 12k 2 + 11k + 3
AG
A1
Any valid method – first step correct 2
AG
(b) Assume true for n = k For n = k + 1 : k +1
1 2 k ( 4k 2 – 1) + ( 2k + 1) 3 1 = ( 4k 3 + 12k 2 +11k + 3) 3 1 2 = ( k + 1) 4 ( k + 1) – 1 3 True for n = 1 shown Proof by induction set out properly (if factorised by 3 linear factors, allow A1 at this particular point) Total
( 2r – 1) r =1
2
=
(
)
M1A1
No LHS M1A0
A1F
ft error in (2k + 1)
A1
Using part (a)
B1 E1
6
8
Dependent on all marks correct
MFP2 (cont) Q 7(a)(i)
Solution
cos5θ + isin 5θ = ( cos θ + isin θ ) Expansion in any form Equate real parts:
Marks 5
cos5θ = cos5 θ – 10cos3 θ sin 2 θ + 5cosθ sin 4 θ
Total
M1 A1 m1 A1
Comments
Attempt to expand 3 correct terms Correct simplification AG
Equate imaginary parts: sin 5θ = 5cos 4 θ sin θ – 10cos 2 θ sin 3 θ + sin 5 θ
(ii)
sin 5θ cos 5θ Division by cos5 θ or by cos 4 θ tan 5θ =
tan 5θ =
(b)
(c)
A1
5
M1
CAO Used
m1
tan θ ( 5 – 10 tan 2 θ + tan 4 θ ) 1 –10 tan 2 θ + 5 tan 4 θ
π tan 5θ = 0 5 π ∴ tan satisfies t 4 –10t 2 + 5 = 0 5 kπ k = 2, 3, 4 Other roots tan 5
θ=
A1
AG
M1
Or for tan 4 θ – 10 tan 2 θ + 5 = 0
A1
Or for tan 5θ = 0
B1
Product of roots = 5 π 4π tan = – tan 5 5 π 2π tan 2 tan 2 =5 5 5 π 2π tan tan =+ 5 5 5 – sign rejected with reason
3
3
OE
M1 Or tan
B1
2π 3π = – tan 5 5
A1 A1 E1
5 Alternative (c) Use of quadratic formula t2 = 5 ± 2 5
M1 A1
t = ± 5± 2 5 B1 Correct selection of +ve values E1 Multiplied together to get 5 A1 Total TOTAL
16 75
Scaled mark unit grade boundaries - June 2011 exams A-level
Code
Title
Max. Scaled Mark
Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E
GCE MATHEMATICS UNIT S1B GCE MATHEMATICS UNIT D02 GCE MATHEMATICS UNIT FP2 GCE MATHEMATICS UNIT M2B GCE MATHEMATICS UNIT PC2 GCE MATHEMATICS UNIT S2B GCE MATHEMATICS UNIT XMCA2 GCE MATHEMATICS UNIT FP3 GCE MATHEMATICS UNIT M03 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT S03 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT M04 GCE MATHEMATICS UNIT PC4 GCE MATHEMATICS UNIT S04 GCE MATHEMATICS UNIT M05
75 75 75 75 75 75 125 75 75 75 75 75 75 75 75 75
69 62 68 62 88 69 67 68 68 68 63 58 67 62
59 64 55 62 54 54 76 64 59 59 62 61 57 51 60 55
52 56 48 55 47 46 66 55 51 52 54 53 51 46 52 48
46 49 41 48 41 38 57 46 43 46 46 46 45 41 44 41
40 42 34 41 35 30 48 38 36 40 39 39 39 36 37 34
34 35 28 34 29 23 39 30 29 34 32 32 33 31 30 28
MEST1 MEST2 MEST3 MEST4
GCE MEDIA STUDIES UNIT 1 GCE MEDIA STUDIES UNIT 2 GCE MEDIA STUDIES UNIT 3 GCE MEDIA STUDIES UNIT 4
80 80 80 80
70 74
55 63 61 68
47 54 50 56
40 45 39 45
33 36 28 34
26 28 18 23
MHEB1 MHEB2
GCE MODERN HEBREW UNIT 1 GCE MODERN HEBREW UNIT 2
100 100
80
61 71
54 62
47 54
40 46
34 38
MUSC1 MUS2A MUS2B MUS2C
GCE MUSIC UNIT 1 GCE MUSIC UNIT 2A GCE MUSIC UNIT 2B GCE MUSIC UNIT 2C
Jan 23, 2012 - You must answer the questions in the spaces provided. Do not write outside the box ... The maximum mark for this paper is 75. Advice .... Key to mark scheme abbreviations. M mark is for ..... (ii) Let M=centre of circle then â AMB = 2
Jan 1, 2011 - The curve has equation y ¼ 14 Ð x Ð x4 and the point PÑ1, 12Ю lies on the curve. (a) (i) Find .... or use of quadratic equation formula. 3. â1,. 2.
Jun 2, 2010 - (c) (i) Find the real root a of the cubic equation z3 Ð 2z2 Ð 3z Ñ 10 ¼ 0. (2 marks). (ii) Find the .... squared, obtaining a quadratic in tanh x.
Jun 3, 2010 - Fill in the boxes at the top of this page. â¢. Answer all ... Do all rough work in this book. ..... This was the best answered question on the paper.
(iii) For Salt A and Salt B, compare the effects of water temperature on the maximum weight that will .... proportion of each colour is shown in the table. Colour ... developed and expanded on the basis of students' reactions to a particular paper.
Jan 17, 2012 - Condensed. General Certificate of Education ... Lenny, to select a random sample of 25 of her male patients, aged between 55 and. 65 years ...
The line AB has equation 7x Ñ 3y ¼ 13 . (a) ... The line AB intersects the line with equation 3x Ñ 2y ¼ 12 at the point B. Find ..... within quadratic equation formula.
Jan 18, 2013 - (b) (i) Calculate the equation of the least squares regression line, y ¼ a þ bx. (4 marks) ... (1 mark). (c). Ashok subsequently discovers that students A to F have a more scientific background ..... Correct â within a circle of ra
Jun 4, 2010 - Determine the area of triangle PQR. (4 marks). 2. Let A ¼. 1 x. 2 3. , B ¼. 1 А1. 2. 2 .... or ft or F follow through from previous incorrect result. MC.
Jan 2, 2011 - expression for dy dx .... sin 1 tan sin cos sin. 1 ln cos ln cos. 2 ln(2) t t and y t dx dy s dt t t tdt dt dt ... do, after writing down a correct expression.
Jun 14, 2012 - Use black ink or black ball-point pen. Pencil should only be used for ... The diagram shows a sketch of a curve C, the pole O and the initial line.
Jun 14, 2012 - You must answer each question in the space provided for that .... and Q. Find, in an exact form, the polar coordinates of P and Q. ..... Scaled Mark Grade Boundaries and A* Conversion Points. Code. Title. Scaled Mark. A*. A. B.
Jan 13, 2012 - (ii) Find an equation of the line AB in the form px Ñ qy ¼ r , where p, ... Sketch the graph with equation y ¼ x ..... OR adding 2 to their quadratic.
Jan 23, 2013 - It is given that yÑxЮ satisfies the differential equation dy ... Find the general solution of the differential equation d ..... quadratic formula to find m.
The variables x and y are related by an equation of the form y ¼ ax. 2 Ñ b .... Find a quadratic equation, with integer coefficients, which has roots a Ñ. 2 b and b Ñ.
Jun 3, 2010 - Hence find the general solution of this differential equation. (4 marks) ...... and solving quadratic equations in either sin x, cos x, tan x, sec x , use ...
Jun 22, 2010 - where V m3 is the volume of the liquid in the tank, r kg m ... 3. Three smooth spheres, A, B and C, of equal radii have masses 1 kg, 3 kg and xkg.
Mathematics. MM03. Unit Mechanics 3. Wednesday 22 June 2011 9.00 am to 10.30 am. For this paper you must have: ⢠the blue AQA booklet of formulae and ...
Jan 4, 2011 - Use your answer to part (b) to find the coordinates of the point of intersection of .... Where the permitted calculator has functions which reasonably allow the ...... was the height of a right-angled triangle with base 100, only.
The final answer to questions requiring the use of tables or calculators ... Whilst on his tour, Eduardo noted the prices, x pence, which he paid for 1.5-litre bottles of ... He has two sows, Josie and Rosie, which he feeds at a trough in their field
Jan 2, 2011 - the blue AQA booklet of formulae and statistical tables. ... Given that the coordinates of this stationary point are Ña, bЮ, show that a Ñ b ¼ 9. ... The length of the arc of the curve between the points where t ¼ 0 and t ¼ Ï. 3
During a particular summer holiday, Rick worked in a fish and chip shop at a seaside resort. ... (i) Complete the table of probabilities, printed below, for these events, where J о and W о .... Named. Not affected by unknown/large/small/extreme/.
Jan 4, 2011 - drawing. â¢. Fill in the boxes at the top of this page. â¢. Answer all questions. ... Do all rough work in this book. Cross through any work that you do.
Jun 1, 2012 - The quadratic equation. 5x. 2 Ð 7x Ñ 1 ¼ 0 has roots a and b . (a). Write down the values of a Ñ b and ab . (2 marks). (b). Show that a b Ñ b a. ¼.
