General Certificate of Education Advanced Level Examination January 2010
Mathematics
MFP2
Unit Further Pure 2 Friday 15 January 2010
1.30 pm to 3.00 pm
For this paper you must have: * an 8-page answer book * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
Time allowed * 1 hour 30 minutes Instructions * Use black ink or black ball-point pen. Pencil should only be used for drawing. * Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MFP2. * Answer all questions. * Show all necessary working; otherwise marks for method may be lost. 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.
P21940/Jan10/MFP2 6/6/6/
MFP2
2
Answer all questions.
1
1
1
(a) Use the definitions cosh x ¼ 2 ðe x þ ex Þ and sinh x ¼ 2 ðe x ex Þ to show that cosh2 x sinh2 x ¼ 1 (b)
(3 marks)
(i) Express 5 cosh2 x þ 3 sinh2 x in terms of cosh x .
(1 mark)
(ii) Sketch the curve y ¼ cosh x .
(1 mark)
(iii) Hence solve the equation 5 cosh2 x þ 3 sinh2 x ¼ 9:5 giving your answers in logarithmic form.
2
(4 marks)
(a) On the same Argand diagram, draw: (i) the locus of points satisfying j z 4 þ 2i j ¼ 4 ;
(3 marks)
(ii) the locus of points satisfying j z j ¼ j z 2i j .
(3 marks)
(b) Indicate on your sketch the set of points satisfying both j z 4 þ 2i j 4 4 and
P21940/Jan10/MFP2
j z j 5 j z 2i j
(2 marks)
3
3 The cubic equation 2z 3 þ pz 2 þ qz þ 16 ¼ 0 where p and q are real, has roots a, b and g. pffiffiffi It is given that a ¼ 2 þ 2 3 i . (a)
(b)
(i) Write down another root, b, of the equation.
(1 mark)
(ii) Find the third root, g.
(3 marks)
(iii) Find the values of p and q.
(3 marks)
(i) Express a in the form reiy , where r > 0 and p < y 4 p .
(2 marks)
(ii) Show that np pffiffiffi np ð2 þ 2 3 iÞn ¼ 4n cos þ i sin 3 3
(2 marks)
(iii) Show that n
n
n
2nþ1
a þb þg ¼2
n np 1 cos þ 3 2
where n is an integer.
(3 marks)
4 A curve C is given parametrically by the equations 1
x ¼ 2 cosh 2t ,
y ¼ 2 sinh t
(a) Express 2 2 dx dy þ dt dt in terms of cosh t .
(6 marks)
(b) The arc of C from t ¼ 0 to t ¼ 1 is rotated through 2p radians about the x-axis. (i) Show that S, the area of the curved surface generated, is given by S ¼ 8p
ð1
sinh t cosh2 t dt
(2 marks)
0
(ii) Find the exact value of S.
(2 marks)
P21940/Jan10/MFP2
s
Turn over
4
5 The sum to r terms, Sr , of a series is given by Sr ¼ r 2 ðr þ 1Þðr þ 2Þ Given that ur is the rth term of the series whose sum is Sr , show that: (a)
(i) u1 ¼ 6 ;
(1 mark)
(ii) u2 ¼ 42 ;
(1 mark)
(iii) un ¼ nðn þ 1Þð4n 1Þ .
(3 marks)
(b) Show that 2n X
ur ¼ 3n2 ðn þ 1Þð5n þ 2Þ
(3 marks)
r¼nþ1
6
(a) Show that the substitution t ¼ tan y transforms the integral ð dy 9 cos2 y þ sin2 y ð into
dt 9 þ t2
(3 marks)
dy p ¼ 2 þ sin y 18
(3 marks)
(b) Hence show that ðp
3
0
9 cos2 y
7 The sequence u1 , u2 , u3 ,… is defined by u1 ¼ 2 , ukþ1 ¼ 2uk þ 1 (a) Prove by induction that, for all n 5 1 , un ¼ 3 2n1 1
(5 marks)
(b) Show that n X r¼1
P21940/Jan10/MFP2
ur ¼ unþ1 ðn þ 2Þ
(3 marks)
5
8
(a)
(i) Show that o ¼ e
2pi 7
is a root of the equation z7 ¼ 1 .
(ii) Write down the five other non-real roots in terms of o .
(1 mark) (2 marks)
(b) Show that 1 þ o þ o2 þ o3 þ o4 þ o5 þ o6 ¼ 0
(2 marks)
(c) Show that: (i) o 2 þ o 5 ¼ 2 cos (ii) cos
4p ; 7
2p 4p 6p 1 þ cos þ cos ¼ . 7 7 7 2
END OF QUESTIONS
P21940/Jan10/MFP2
(3 marks)
(4 marks)
AQA – Further pure 2 – Jan 2010 – Answers Question 1:
Exam report
1 x −x 2 1 x −x 2 (e + e ) − 4 (e − e ) 4 1 1 = ( e 2 x + e −2 x + 2 ) − ( e 2 x + e −2 x − 2 ) 4 4 2 2 = + =1 4 4 2 2 b) i ) 5cosh x + 3sinh x = 5cosh 2 x + 3(cosh 2 x − 1) a ) cosh 2 x − sinh 2 x =
= 8cosh 2 x − 3 ii ) y = cosh x graph 9.5 iii ) 5cosh 2 x + 3sinh 2 x = 8cosh 2 x − 3 = 9.5 cosh 2 x = 1.5625 cosh x = 1.25 or cosh x = −1.25 (no solution) x = cosh −1 (1.25) or x = − cosh −1 (1.25)
(
)
x = ln 1.25 + 1.252 − 1 = ln 2 or − ln 2 Question 2:
Part (a) was generally well done apart from a few candidates who were unable to square 1 ( e x + e − x ) 2 successfully. Parts (b)(i) and (b)(ii) likewise were well done although 1 ± did appear on the x-axis of some sketches in part 2
(b)(i). However in part (b)(iii), those candidates using the −1 logarithmic formula for cosh x from the formulae booklet arrived at a single value of x, namely ln2, but failed to realise that the sketch in part (b)(ii) was intended to give them a hint that – ln 2 was also a solution of the equation. On the other hand, candidates who used the exponential form of either cosh x or cosh2 x automatically produced both answers provided their working was correct, but some of these candidates were unable to handle the algebra leading to a quadratic x 2x equation in either e or e .
Exam report
a ) i ) z − 4 + 2i = 4 this is the circel centre A(z A ) with z A= 4 − 2i and radius r = 4 ii ) z= z − 2i This is the perpendicular bisector of the line OB = i and zO 0 with z B 2= b) The region is the intersection of the inside of the circle and the half-plane containing B.
This question was well done overall and many candidates scored all of the eight available marks. Those candidates using mathematical instruments, i.e. ruler and compasses, produced superior solutions. Errors, when they did occur, were either the misplotting of the centre of the circle, or more commonly, the misplotting of the line. The commonest mistakes were either to draw the line through the point (0, 2) or more frequently through (0, –1). If serious errors were made in the plotting of the line, loss of marks in the shading were almost inevitable.
Question 3:
Exam report
2 z + pz + qz + 16 = 0 has roots α , β , γ . p and q are REAL numbers 3
2
α= 2 + 2i 3 a )i ) Since the coefficients of the equation are real numbers,
α * is also a root so β= 2 − 2i 3 16 ii ) αβγ = − = −8 2 (2 + 2i 3)(2 − 2i 3)γ = (4 + 12)γ = 16γ αβγ = 1 so γ = − 2 p 1 iii ) − =α + β + γ = 2 + 2i 3 + 2 − 2i 3 − 2 2 1 −2(4 − ) = p= p = −7 2 1 q = αβ + αγ + βγ = αβ + γ (α + β ) = 16 − × 4 2 2 q = 28
This question provided a good source of marks for many candidates. The commonest error in part(a) was to write αβγ as -16, overlooking the fact that 3 the coefficient of x was not unity and, of course, leading to an incorrect value for γ . This error perpetuated in part (a)(ii) with α + β +γ written as p or –p and the same for q. Parts (b)(i) and (b)(ii) were generally well done, although it should be stated that when answers are printed, candidates are expected to provide sufficient detail to show clearly how their answers are arrived at. Part (b)(iii) was also quite well done and it was pleasing to note that in some cases where candidates had not arrived at γ = − 1 , they 2 went back to part (a)(ii) to identify their error. Just occasionally in part (b)(iii), some candidates made blatant errors in their attempt to convert
1 3 π π 2 + 2i 3 = 4 + i 4 cos + i sin b) i ) α = = 2 3 3 2 i
π
α = 4e 3 n
nπ i i π3 nπ nπ n 3 )α 4 4 4n cos ii e e = = = + i sin 3 3 iii ) α n + β n + γ n = n
nπ nπ = 4 cos + i sin 3 3 n
nπ = 4n 2 cos 3
nπ nπ n − i sin + 4 cos 3 3
1 +− 2
4n cos
1 +− 2
n
nπ 1 nπ 1 = 2 × 2 × cos + − = 22 n +1 cos +− 3 2 3 2 n
2n
n
n
nπ nπ nπ into 22 n +1 cos + 4n cos 3 3 3
Question 4:
Exam report
1 = cosh 2t and y 2sinh t 2 dx dy = a) sinh 2t and 2 cosh t dt dt x
2
2
dx dy 2 2 + = (sinh 2t + 4 cosh t ) dt dt = cosh 2 2t − 1 + 4 cosh 2 t Using the identity= cosh t 2 cosh 2 t − 1, we have 2
2
2 dx dy 2 2 += ( 2 cosh t − 1) − 1 + 4 cosh t dt dt = 4 cosh 4 t − 4 cosh 2 t + 1 − 1 + 4 cosh 2 t
= 4 cosh 4 t 2
2
dx dy 4 cosh 4 t + = t d dt 2
2
dx dy 2π ∫ y + dt 0 dt dt 1
b) i ) S
Although a good number of candidates answered part (a) 2 correctly, quite a few stalled at the handling of sinh 2t either by misquoting a formula for the double angle or by using long winded algebraic methods in which they lost direction. Consequently these candidates were unable to score full marks in part (b)(i). Part (b)(ii) proved to be beyond the abilities of the majority of candidates. The usual attempts were either to express 2 2 2 replace cosh t by 1+sinh t or to express cosh t in terms of cosh 2t , thereby making no progress. Few thought of using a simple substitution.
1
= S 2π ∫ 2sinh t × (2 cosh 2 t )dt 0
1
= S 8π ∫ sinh t × cosh 2 t dt 0
1 8π π cosh 3 t cosh 3 1 − 1) 8= ( 3 0 3 1
ii ) S
Question 5:
Exam report
S r = r (r + 1)(r + 2) 2
a ) i ) S r = u1 + u2 + u3 + ... + ur −1 + ur so for r 1,= S1 u1 = u1 =12 × (2) × (3) =u1 = 6 ii ) S 2 =u1 + u2 =6 + u2 and S 2 = 22 × (3) × (4) = 48 u2 = 48 − 6 = u2 = 42 iii ) u= S n − S n −1 n = n 2 (n + 1)(n + 2) − (n − 1) 2 (n)(n + 1) = n(n + 1) n(n + 2) − (n − 1) 2
=n ( n + 1) ( n 2 + 2n − n 2 − 1 + 2n )
un = n(n + 1) ( 4n − 1)
This question proved to be quite discriminating. Either candidates realised what they were asked to do and scored full marks, or the notation puzzled them and they were unable to proceed beyond part (a)(ii), thinking that the answer to this part should be 48 rather than 42. Some of the weaker candidates, whilst realising what to do, failed to take out common factors in their algebraic manipulation in parts (a)(iii) and (b) with the result that correct answers were written down after incorrect algebra.
Question 5:continues 2n
b)
∑u
=
r r= n +1
2n
∑S
Exam report
− S r −1 = S n +1 − S n +
r r= n +1
S n + 2 − S n +1 + S n +3 − S n + 2 + ... + S 2 n − S 2 n −1 2n
∑u
r = n +1
r
= S 2 n − S n = (2n) 2 (2n + 1)(2n + 2) − n 2 (n + 1)(n + 2)
= 8n 2 (2n + 1)(n + 1) − n 2 (n + 1)(n + 2) = n 2 (n + 1) [8(2n + 1) − (n + 2) ] 3n 2 (n + 1)(5n + 2) = n 2 (n + 1) (15n + 6 ) = Question 6:
Exam report
dt 1 + tan 2 θ = 1+ t2 = dθ dt = dθ 1+ t2 1 1 = • cos 2 θ = 2 1 + tan θ 1 + t 2 t2 1 = • sin 2 θ = 1 − cos 2 θ = 1− 1+ t2 1+ t2 dθ dt 1 = × 2 2 ∫ 9 cos2 θ = ∫ t 9 1+ t2 + sin θ + 1+ t2 1+ t2 b) when = t tan= θ 0,= 0 0 a) t = tan θ
θ = π
I
so
π
π
, t tan= = 3 3
dθ 2 ∫= 9 cos θ + sin 2 θ 3 0
dt
∫ 9+t
2
clear how the answer given in the question.
3
∫
0
3
dt 1 t = tan −1 2 9+t 3 0 3
3 1 −1 1 1 π π I = × tan −1 − tan ( 0 ) = × = 3 3 6 18 3 3
In part (a), whilst dt was expressed correctly, the dθ manipulation required to obtain the integral in terms of t was frequently faulty. Also in part (b), whilst many candidates were able to write down the correct definite integral 1 tan −1 t , full marks were not awarded unless it was 3 3
3
π was arrived at, as this answer was
18
Question 7
Exam report
= u1 2 , u= 2uk + 1 k +1 a ) the proposition Pn :for all n ≥ 1, un = 3 × 2n −1 − 1 is to be proven by induction Base case: n = 1 LHS : u1 = 2 RHS :3 × 21−1 − 1 = 3 × 20 − 1 = 3 − 1 = 2 P1 is true.
Candidates generally had difficulty in using the recurrence relationship in their proof by induction, so that responses to this question were rather poor. Proper detail is essential in the proof by induction using a sequence and a sequence relationship so that relatively few candidates scored full marks for part (a).
Let's suppose that for n = k , the propostion is true. Let's show that it is true for n= k + 1. i.e.Let's show that uk +1 =3 × 2k − 1 −−−−−−−−−−−−−−−−−−−−−−−−−−−− uk +1 = 2uk + 1 = 2 ( 3 × 2k −1 − 1) + 1 = 3 × 2k − 2 + 1 uk +1 =3 × 2 − 1 k
Q.E.D
−−−−−−−−−−−−−−−−−−−−−−−−−−−− Conclusion: If the proposition is true for n= k , then it is true for n= k + 1. Because it is true for n = 1, we can conclude, according to the induction principle that it is true for all n ≥ 1. n
n
b) ∑ u = ∑ 3 × 2
r −1
r =r 1 =r 1
n
− 1 = 3∑ 2 =r 1
r −1
remember core 2? : ∑ 2r −1 =1 + 2 + 4 + 8 + ... + 2n −1 =1 r =1
∑u r =1
r
uk +1 = 2uk + 1 as uk +1 − uk = uk + 1 and then
to use the method of differences to sum the series; but this method of solution was extremely rare.
2n − 1 − n = 3× −n 2 −1
n
n
Very few candidates indeed were successful in part (b). Only a handful of candidates recognised that a Geometric Progression was involved. If they did, they usually went on to obtain a correct solution. It should perhaps be added that one method of providing an excellent solution was to rewrite
2n − 1 2 −1
=3 × 2n − 3 − n =(3 × 2n − 1) − (n + 2) = un +1 − (n + 2)
Question 8:
a) i) ω = e
i
Exam report
2π 7 7
i 27π so ω = e 2iπ = cos 2π + i sin 2π = 1 e = ω is a solution of z 7 = 1 2π ii ) 7θ = k × 2π θ= k× 7 the other non − real solutions are 7
i
4π 7
ω = for k 2,= e i
2
8π
= for k 4,= e 7 ω4 i
12π
= for k 6,= e 7 ω6
i
6π 7
ω3 = for k 3,= e i
10π
= for k 5,= e 7 ω5
Whilst part (a) was generally well done, relatively few candidates expressed the other roots in terms of ω , but iθ rather gave them in the form re
Question 8:continues
Exam report
b)1 + ω + ω + ω + ω + ω + ω is a geometric series with common ration ω 2
3
4
5
6
1 − ω 7 1 −1 1 + ω + ω + ω += ω +ω +ω = = 0 1− ω 1− ω 4π 4π i − 4π c) i ) ω 2 + ω 5 = ω 2 + ω −2 = e 7 + e 7 = 2 cos 7 2 3 4 5 6 ii )1 + ω + ω + ω + ω + ω + ω = 0 2
3
4
5
6
ω + ω 2 + ω 3 + ω −3 + ω −2 + ω −1 = −1 ω + ω −1 + ω 2 + ω −2 + ω 3 + ω −3 = −1 2π 4π 6π
2 cos cos
7
+ 2 cos
7
+ 2 cos
7
Few, also, were able to complete part (b). 2 The relation 1+ω +ω = 0 appeared with regularity. Part (c) was poorly answered. Although correct answers were written down as they were given in the question, few responses were convincing and as has already been stated earlier, if answers are given, it is the responsibility of the candidates to supply sufficient working to convince the examiner that they understand the methods involved
= −1
2π 4π 6π 1 + cos + cos = − 7 7 7 2
Grade boundaries
MFP2 - AQA GCE Mark Scheme 2010 January series
Key to mark scheme and abbreviations used in marking M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA
mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach
MC MR RA FW ISW FIW BOD WR FB NOS G c sf dp
mis-copy mis-read required accuracy further work ignore subsequent work from incorrect work given benefit of doubt work replaced by candidate formulae book not on scheme graph candidate significant figure(s) decimal place(s)
No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.
3
MFP2 - AQA GCE Mark Scheme 2010 January series
MFP2 Q
Solution 1(a)
(b)(i)
LHS =
1 4
(ex + e )
–x 2
−
1 4
Marks
(ex − e )
–x 2
Correct expansion of either square Shown equal to 1
A1 A1
3
8cosh 2 x − 3
B1
1
B1
1
cosh x = ( ± )1.25
(
x = ln 1.25 + 1.252 − 1 = ln 2 1 ln by symmetry 2
Comments
M1
(ii) Sketch of y = cosh x (iii)
Total
Must cross y-axis above x-axis OE; ft errors in (b)(i); allow ± missing
B1F
)
AG
M1 A1F A1F
4
Accept − ln 2 written straight down Alternatively, if solved by using e 2 x − 2.5e x + 1 = 0 , allow M1 for ⎛ 2.5 ± 2.52 − 4 ⎞ x = ln ⎜ ⎟ ⎜ ⎟ 2 ⎝ ⎠
Total 2
9
y
x •
(a)(i) Circle
B1
Correct centre
B1
Touching y-axis
B1
(ii) Straight line parallel to x-axis
B1 B1
through (0, 1) (b) Shading: inside circle above line
x-coordinate ≈ − 2 × y-coordinate in correct quadrant; condone ( 4, –2i ) 3
B1
3
B1F B1F
2
Assume (0, 1) if distance up y-axis is half distance to top of circle; no other shading outside circle
Whole question reflected in x-axis loses 2 marks Total
8
4
MFP2 - AQA GCE Mark Scheme 2010 January series
MFP2 (cont) Q 3(a)(i) β = 2 − 2 3 i (ii)
Solution
Marks B1
αβγ = −8 αβ = 16 1 γ =− 2
Total 1
Allow for +8 but not ±16
M1 B1 A1
−p =α + β +γ 2 q = αβ + βγ + γα or 2
3
(iii) Either
A1F, A1F
Alternative to (a)(ii) and (a)(iii): ( z 2 – 4 z + 16 ) ( az + b ) 1 2
(A1)
Equating coefficients p = –7 q = 28
(2 + 2 3 i)
(iii)
(M1) (A1F) (A1F)
π 3 n
B1,B1
2
n
⎛ πi ⎞ = ⎜ 4e 3 ⎟ ⎝ ⎠ nπ nπ ⎞ ⎛ = 4n ⎜ cos + isin ⎟ 3 3 ⎠ ⎝
M1 A1
nπ nπ ⎞ ⎛ = 4n ⎜ cos − isin ⎟ 3 3 ⎠ ⎝ nπ nπ ⎞ ⎛ α n + β n + γ n = 4n ⎜ cos +i sin ⎟ 3 3 ⎠ ⎝
(2 − 2 3 i)
ft incorrect γ
(B1)
a = 2, b = + 1, γ = –
(ii)
3
(M1)
αβ = 16
r = 4, θ =
SC if failure to divide by 2 throughout, allow M1A1 for either p or q correct ft
M1
p = −7, q = 28
(b)(i)
Comments
n
nπ nπ ⎞ ⎛ 1 ⎞ ⎛ + 4n ⎜ cos − i sin ⎟ + ⎜ − ⎟ 3 3 ⎠ ⎝ 2⎠ ⎝ =2
2 n +1
nπ ⎛ 1 ⎞ +⎜− ⎟ cos 3 ⎝ 2⎠
2
AG
3
AG
B1
n
M1
n
A1 Total
14
5
MFP2 - AQA GCE Mark Scheme 2010 January series
MFP2 (cont) Q 4(a)
Solution
2
(b)(i)
Marks
dx = sinh 2t dt dy = 2cosh t dt
Total
Comments
B1 B1 2
⎛ dx ⎞ ⎛ dy ⎞ 2 2 ⎜ ⎟ + ⎜ ⎟ = sinh 2t + 4cosh t ⎝ dt ⎠ ⎝ dt ⎠
M1
Use of sinh 2t = 2sinh t cosh t
m1
Or other correct formula for double angle
= 4cosh 2 t ( sinh 2 t + 1)
A1
For taking out factor
= 4cosh 4 t
A1F
1
S = 2π ∫ 2sinh t.2cosh 2 t dt
6
M1
0
1
= 8π ∫ sinh t.cosh 2 t dt
A1
0
ft errors of sign in
dx dy or dt dt
Using the value obtained in (a) 2
AG
1
⎡ cosh 3 t ⎤ S = 8π ⎢ ⎥ ⎣ 3 ⎦0
M1
8π ⎡cosh 3 1 − 1⎤⎦ 3 ⎣
A1
2
OE eg
u1 = S1 = 1 .2.3 = 6
B1
10 1
AG
(ii)
u2 = S 2 − S1 = 42
B1
1
AG
(iii)
un = S n − S n −1
M1
3
AG
3 8
AG
(ii)
=
Total 5(a)(i)
2
= n 2 ( n + 1)( n + 2 ) − ( n − 1) n ( n + 1)
A1
= n ( n + 1)( 4n − 1)
A1
2
2n
(b)
∑ ur = S 2 n − S n r =n+
M1
= ( 2n ) ( 2n + 1)( 2n + 2 ) − n 2 ( n + 1)( n + 2 )
A1
= 3n 2 ( n + 1)( 5n + 2 )
A1
1
2
Total
6
(
π⎛ e+1 e 3 ⎜⎝
) – 8 ⎞⎟⎠ 3
MFP2 - AQA GCE Mark Scheme 2010 January series
MFP2 (cont) Q 6(a) t = tan θ
I =∫
=∫
(b)
Solution dt = sec2 θ dθ
Marks B1
dt ( 9cos θ + sin 2 θ ) sec2 θ
M1
dt t +9
A1
2
2
t⎤ ⎡1 I = ⎢ tan −1 ⎥ 3⎦ 0 ⎣3
Comments
OE OE
3
AG
3
M1 for tan −1
M1
1 –1 3 1 1 tan or tan –1 3 3 3 3
=
Total
A1
π 18
A1 Total
3
AG
6
7(a) Assume true for n = k
(
)
uk +1 = 2 3 × 2k −1 − 1 + 1 = 3 × 2k − 1
A1
True for n = 1 shown
B1
Method of induction clearly expressed
E1
n
(b)
M1A1 2( k –1) + 1 not necessarily seen
5
Provided all 4 previous marks earned
n
ur = ∑ 3 × 2r −1 − n ∑ r =1 r =1 = 3 ( 2n − 1) − n
M1A1
= un +1 − ( n + 2 )
A1 Total
7
M1 for summation, ie recognition of a GP 3 8
AG
MFP2 - AQA GCE Mark Scheme 2010 January series
MFP2 (cont) Q 8(a)(i)
Solution ⎛ ⎜e ⎝
2πi 7
⎞ 2πi ⎟ = e =1 ⎠
(ii) Roots are ω 2 , ω 3 , ω 4 , ω 5 , ω 6
(b) Sum of roots considered
B1
1
Or z 7 = e 2 kπi z = e
M1A1
2
OE; M1A0 for incomplete set SC B1 for a set of correct roots in terms of eiθ
A1
4πi
10πi 7
M1
4πi
− 4πi 7
A1
ω 2 + ω5 = e 7 + e =e 7 +e = 2cos
(ii)
Total
M1
=0
(c)(i)
Marks
Comments
7
ω + ω 6 = 2cos
4π 7
2π 7
A1
; ω 3 + ω 4 = 2cos
6π
2
Using part (b) Result
M1 A1 Total TOTAL
8
6
ω6 = ∑ r =0
Or cos
3
k =1
ω7 −1 =0 ω −1
4π 4π 4π 4π + isin + cos – isin 7 7 7 7
AG
Allow these marks if seen earlier in the solution
B1,B1
7
⎧ ⎨ or ⎩
2 kπi 7
4 12 75
AG
