General Certificate of Education June 2006 Advanced Level Examination

MATHEMATICS Unit Further Pure 2 Monday 19 June 2006

MFP2 9.00 am to 10.30 am

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 blue or black ink or 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 maximum mark for this paper is 75. The marks for questions are shown in brackets. * *

Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. *

P85475/Jun06/MFP2 6/6/6/

MFP2

2

Answer all questions.

1

(a) Given that  r2 ‡ r 1 1 ˆA‡B r…r ‡ 1† r

1 r‡1

find the values of A and B .



(3 marks)

(b) Hence find the value of 99 2 X r ‡r

1 r…r ‡ 1† rˆ1

(4 marks)

2 A curve has parametric equations xˆt

1 3 t , 3

y ˆ t2

(a) Show that  2  2 dx dy ‡ ˆ …1 ‡ t 2 †2 dt dt

(3 marks)

(b) The arc of the curve between t ˆ 1 and t ˆ 2 is rotated through 2p radians about the x-axis. Show that S, the surface area generated, is given by S ˆ kp, where k is a rational number to be found. (5 marks)

P85475/Jun06/MFP2

3

3 The curve C has equation y ˆ cosh x (a)

(i) The line y ˆ

3 sinh x

1 meets C at the point …k,

1† .

Show that e2k

ek

2ˆ0

(ii) Hence find k, giving your answer in the form ln a . (b)

(4 marks)

(i) Find the x-coordinate of the point where the curve C intersects the x-axis, giving your answer in the form p ln a . (4 marks) (ii) Show that C has no stationary points.

(3 marks)

d2 y (iii) Show that there is exactly one point on C for which 2 ˆ 0 . dx

4

(3 marks)

(1 mark)

(a) On one Argand diagram, sketch the locus of points satisfying: (i) j z

3 ‡ 2ij ˆ 4 ;

(ii) arg…z

1† ˆ

(3 marks)

1 p. 4

(3 marks)

(b) Indicate on your sketch the set of points satisfying both jz and

arg…z

3 ‡ 2ij 4 4 1† ˆ

1 p 4

(1 mark)

Turn over for the next question

P85475/Jun06/MFP2

s

Turn over

4

5 The cubic equation z3

4iz 2 ‡ qz

…4

2i† ˆ 0

where q is a complex number, has roots a, b and g . (a) Write down the value of: (i) a ‡ b ‡ g ;

(1 mark)

(ii) abg .

(1 mark)

(b) Given that a ˆ b ‡ g, show that: (i) a ˆ 2i ; (ii) bg ˆ (iii) q ˆ

(1 mark) (2 marks)

…1 ‡ 2i†; …5 ‡ 2i† .

(3 marks)

(c) Show that b and g are the roots of the equation z2

2iz

…1 ‡ 2i† ˆ 0

(d) Given that b is real, find b and g .

6

(2 marks) (3 marks)

(a) The function f is given by f …n† ˆ 15n

8n 2

Express f …n ‡ 1† in the form k  15n . (b) Prove by induction that 15n

P85475/Jun06/MFP2

8f …n† (4 marks)

8n 2 is a multiple of 7 for all integers n 5 2 . (4 marks)

5

7

(a) Find the six roots of the equation z6 ˆ 1, giving your answers in the form eif , where p 1 4

Responses to this question were usually quite good and it was pleasing to note some quite accurate neat diagrams using a ruler and compasses. Errors in parts (a)(i) and part (a)(ii) were usually errors of sign. For instance in part (a)(i) the centre of the circle was sometimes taken to be the point (–3,2)or even (3,2) and in part (a)(ii) the half line would be drawn from either (0,1)or (–1,0) . Just occasionally the radius of the circle was taken to be 2, or the direction of the line was taken to be

+

π 3π or + 4 4

. In part (b), a substantial number of

candidates thought that the set of points must involve an area and consequently shaded some region in their sketch.

Question 5:

Exam report

0 has roots α , β , γ z − 4iz + qz − (4 − 2i ) = a ) i ) α + β + γ = 4i ii ) αβγ = 4 − 2i 3

2

b) α= β + γ 4i becomes i) α + β + γ = 4i so α = 2i α +α =

Apart from the occasional sign errors, part (a) was answered well. Where sign errors did occur there was some faking to establish the printed answers in part (b).

ii ) αβγ = 4 − 2i 4 − 2i 4 − 2i i 4i + 2 βγ = = × = 2i i α −2 βγ =−2i − 1 =−(1 + 2i ) iii ) q = αβ + αγ + βγ

Part (b) is an example of what was mentioned at the beginning of this report in that with all three answers being printed, sufficient working needed to be shown in order to obtain full credit. Whilst most candidates knew roughly what was required for part (c), few candidates could express their argument succinctly. A number of candidates attempted to divide the cubic equation by z–2i with varying success.

= α ( β + γ ) + βγ = α 2 + βγ = (2i ) 2 − (1 + 2i ) q =−4 − 1 − 2i

Probably the commonest method of approach in part (d) was to substitute

q =−5 − 2i c) β + γ =2i and βγ =−(1 + 2i ) so β and γ are roots of the equations

β 2 = 1 from which a substantial number of candidates assumed that β = 1 instead of considering the imaginary

0 z 2 − 2iz − (1 + 2i ) = d) β

β for z in the quadratic equation in z and

then to equate real parts. Equating real parts led to

parts of the equation as well.

1 is an " obvious "= root (12 − 2i − (1 + 2i ) 0) 0 z 2 − 2iz − (1 + 2i ) = ( z − 1)( z 2 + (1 + 2i )) = roots= are β 1= and γ 1 + 2i

Question 6:

Exam report

a ) f (n + 1) − 8 f (n= ) (15

n +1

n −1

− 8 ) − 8(15 − 8 n

n−2

)

= 15n +1 − 8n −1 − 8 ×15n + 8n −1 = 15n (15 − 8 ) f (n + 1) − 8 f (n)= 7 ×15n b) Propostion, Pn :For all n ≥ 2, 15 − 8 n

n−2

is a multiple of 7.

2− 2

Base case: for n = 2, 15 − 8 = 225 − 1 = 224 = 7 × 32 the proposition is true for n = 2. 2

n k , f (= k ) 15k − 8k − 2 is a multiple of 7. We suppose that Pk is true : for= Let's show that Pk +1 is the true:let's show that f (k + 1)= 15k +1 − 8k −1 is a multiple of 7. −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− According to question a) f (k + 1) =7 ×15k + 8 f (k ) 7 ×15k is a multiple of 7 8f (k ) is a multiple of 7, because f (k ) is (hypothesis) therefore f (k + 1) is a multiple of 7 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Conclusion :If Pk is true then Pk +1 is also true, because P2 is true, according to the induction principle, we can conclude that for all n ≥ 2, Pn is true.

Although there were some good solutions to part(a) of this question it did show in many cases a lack of understanding of the theory of indices. It n was quite common to see 8×15 written n n-2 n-2 as a 120 and 8×8 as 64 . There was also a lack of clarity in part (b). It was not unusual to see the first line of the inductive proof to state “Assume result k k true for n = k i.e. that f (k ) =15 – 8 –2 ” to be followed by “ f (k +1) – 8f (k ) is a multiple of 7 ”, showing a lack of understanding of the proof by induction in the case of multiples of integers. Some candidates tried to establish the result for n = 1 in spite of being told that n was greater than or equal to 2. A substantial minority of candidates ignored the hint in part (a) and in part (b) considered f (k +1) – f (k ) with a measure of success

Question 7:

Exam report iφ

a ) Let note z= re then z = r × e 6

6

i 6φ

and 1 = 1ei 0 The equation z 6 = 1 is equivalent to r 6 × ei 6φ = 1× ei 0 = This gives r 6 1= r 1 6φ = 0 + k 2π = φ k so z = e

−i

2π 3

or e

−i

π

−3< k ≤ 3

−2≤ k ≤3

3

π 3

i0

or e or e

i

π 3

or e

i

2π 3

or eiπ

w2 − 1 ei 2θ − 1 eiθ (eiθ − e − iθ ) eiθ − e − iθ = 2iSinθ = = b)i ) iθ iθ w e e i w i 1 1 − = = × ii ) 2= w − 1 2iSinθ 2iSinθ i 2 Sinθ e − iθ Cosθ − iSinθ 2i 2i 1 = = = iii ) = iθ 2 w − 1 2iwSinθ e × Sinθ Sinθ Sinθ Cosθ Sinθ = −i = Cotθ − i Sinθ Sinθ 2i = z iv) z =− Cotθ i so 2 w −1 = 2i zw2 − z zw 2 z + 2i = z is equivalent to c) i ) ( z + 2i ) = order 5 polynomial=0 6

6

Although part (a) of this question was standard work it was surprising to see many candidates fail to obtain full marks. The 6 commonest errors were either to express the six roots of z =1 in the form a + ib , or to give the roots in the range 0 to 2π . A i

kπ 3

few candidates wrote down the 6 roots as e with k = ± 1,± 2,±3. In part (b), parts (i) and (iv) were often well done, but relatively few candidates spotted part (b)(ii) as the reciprocal of part (b)(i), and it was not unusual to see rewritten as

w w −1 2

w−1 − w .

Part (b)(iii) was beyond all but the most able candidates although quite a number arrived at

1 Sinθ eiθ

at which point

their solutions usually petered out. There was a wide variety of 6 6 reasons why the equation (z + 2i) = z had only 5 roots with about 50% of them spurious.

(the term in z 6 cancel out ) In part (c)(ii) only one or two candidates used the hints given in the earlier parts of the question, but instead, solved the 6 6 equation (z + 2i) = z from first principles by writing

ii ) ( z + 2i )6 = z6  z + 2i    =1  z  ( w2 ) 6 = 1 6

So w2 = e i

i

z + 2i = ze

πk

π 6

− i , cot

π 3

−i

2π 5π − i , cot −i 3 6

−i , 3 −i , z=

2i

z=

π

e 3 −1

w=e 6 This gives

cot

followed by

.

Of the few serious attempts made by candidates at this part of the question, most solutions ended at the point indicated and only the most able candidates found the five roots of the equation in the required form.

(question a )

z= cot 0 − i , cot

kπ 3

i

πk 3

i

3 3 − i, − 3 − i −i , − 3 3

MFP2 – AQA GCE Mark Scheme, 2006 June series

Key To Mark Scheme And Abbreviations Used In Marking M

mark is for method

m or dM A B E

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

or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA

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.

2

AQA GCE Mark Scheme, 2006 June series – MFP2

MFP2 Q 1(a)

Solution

(

2

Marks

)

2

r + r –1 = A r + r + B

M1

A = 1, B = –1

A1 A1F

Total

Comments

Any correct method 3

ft B if incorrect A and vice versa r2 + r −1 1 B1 Or =1− 2 r (r + 1) r +r 1 ⎞ ⎛1 =1− ⎜ − ⎟ M1A1 +1⎠ r r ⎝

(b)

r=1

1–

1 1 + 1 2

r=2

1–

1 1 + 2 3

M1

1 1 + 99 100

A1F

r = 99 1 –

1 100 = 98.01

Sum = 98 +

m1 A1F Total

2(a)

x! = 1 – t 2 , y! = 2t

( ) = (1 + t )

x! 2 + y! 2 = 1 – t 2

2

2

+ 4t 2

(

7

M1 A1

)

S = 2π ∫ 1 +t 2 t 2 dt 1

4

Must have 98 or 99 OE Allow correct answer with no working 4 marks

B1

2 2

(b)

Do not allow M1 if merely 1 1 ∑ r – ∑ r + 1 is summed A1 for suitable (3 at least) number of rows

3

AG; must be intermediate line

M1A1

Must be correct substitutions for M1

m1

Allow if one term integrated correctly

A1F

Any form

2

⎡ t3 t5 ⎤ = 2π ⎢ + ⎥ ⎣⎢ 3 5 ⎥⎦ 1

⎡ 8 32 1 1 ⎤ = 2π ⎢ + – – ⎥ ⎣3 5 3 5⎦ 256π = 15

A1F Total

5 8

3

MFP2 – AQA GCE Mark Scheme, 2006 June series

MFP2 (cont) Q 3(a)(i)

(ii)

ek + e 2

−k

–

Solution

(

3 ek – e – k 2

Marks

) = –1

M1

–2e k + 4e – k = – 2

A1

e 2k – e k – 2 = 0

A1

( ek +1)( ek – 2) = 0

M1

ek ≠ – 1

E1

ek = 2

A1 A1F

k = ln 2 (b)(i)

(ii)

(iii)

Total

cosh x = 3sinh x or in terms of e x 1 tanh x = or 2e x = 4e – x 3 ⎛ 1⎞ 1 ⎜1+ 3 ⎟ or e 2 x = 2 x = ln 2 ⎜⎜ 1 – 1 ⎟⎟ 3⎠ ⎝ 1 x = ln 2 2

Allow if 2’s are missing or if coshx and sinhx interchanged 3

4

M1 A1 A1F

A1

4

CAO

Must give a reason

M1

= 0 when tanh x = 3 or e 2x = – 2

A1

Correct reason

E1

3

B1F

1

(

= y = 0 at 1 ln 2, 0 2 dx ie one point 2

AG Condone x instead of k

Must state something to earn E1. Do not accept ignoring or crossing out.

dy = sinh x – 3cosh x or − e x – 2e – x dx

d2 y

Comments

) Total

4

15

AQA GCE Mark Scheme, 2006 June series – MFP2

MFP2 (cont) Q 4

Solution

Marks

Total

B1 B1 B1

3

(ii) Half line Correct starting point Correct angle

B1 B1 B1

3

(b) Correct part of the line indicated

B1F

(a)(i) Circle Correct centre Enclosing the origin

5(a)(i) (ii) (b)(i) (ii)

B1

1 7 1

αβγ = 4 – 2i

B1

1

α + α = 4i, α = 2i

B1

1

α + β + γ = 4i

βγ =

Total

4 – 2i = – 2i –1 2i

(iii)

(c)

q = αβ + βγ + γα = α ( β + γ ) + βγ

M1 M1

= 2i.2i – 2i – 1 = − 2i – 5

A1

Use of β + γ = 2i and βγ = – 2i –1

M1

z – 2i z – (1 + 2i ) = 0

A1

2

AG Some method must be shown, eg

M1 A1

Comments

2

AG

3

Or α 2 + βγ , ie suitable grouping AG Elimination of say γ to arrive at

2

β 2 – 2iβ – (1 + 2i ) = 0 M1A0 unless also some reference to γ being a root AG

(d)

f ( –1) = 1 + 2i – 1 – 2i = 0

M1

β = –1, γ = 1 + 2i

A1A1 Total

5

2 –1 i

For any correct method 3 13

A1 for each answer

MFP2 – AQA GCE Mark Scheme, 2006 June series

MFP2 (cont) Q Solution 6(a) f ( n + 1) – 8f ( n ) = 15n +1 – 8n –1

Marks

− 8(15n − 8n − 2 )

Total

Comments

M1A1

= 15n +1 – 8.15n = 15n (15 – 8 ) (b)

= 7.15 Assume f ( n ) is M ( 7 )

M1

n

A1

Then f ( n +1) – 8f ( n ) = 7 × 15n

For multiples of powers of 15 only 4

For valid method ie not using 120n etc Or considering f ( n + 1) − f ( n )

M1

f ( n +1) = M ( 7 ) + M ( 7 ) = M (7)

A1

n = 2 : f ( n ) = 15 – 8 = 224 2

0

= 7 × 32

B1

P ( n ) ⇒ P ( n +1) and P ( 2 ) true

E1 Total

6

4 8

n = 1 B0 Must score previous 3 marks to be awarded E1

AQA GCE Mark Scheme, 2006 June series – MFP2

MFP2 (cont) Q 7(a)

(b)(i) (ii)

Solution

z=e

2 kπi 6

w2 – 1 1 = w – = 2isinθ w w w

1 2isinθ w –1 i = – 2sin θ =

2

(iii)

, k = 0, ±1, ± 2, 3

2i 2

w –1

=

(c)(i) (ii)

z=

2i w –1

OE M1A1 only if: (1) range for k is incorrect eg 0,1,2,3,4,5 (2) i is missing

M1A1

2

AG

2

AG

A1

Or z + 2i =

2i 2

w –1

+ 2i

3

M1 2

No coefficient of z 6

E1

1

(w )

=1

w =

kπ z = cot – i , 6

sin θ eiθ

kπi e3

B1 M1 A2,1,0

k = ± 1, ± 2, 3

AG

ie any correct method

A1

2

1

A1

z + 2i = zw2

2 6

Comments

Or for

M1

1 ( cosθ – isinθ ) sin θ

2

3

A1

= cot θ – i (iv)

Total

M1

–2iw –1i 2sinθ

=

Marks M1 A2,1,0

AG

Alternatively: k πi z + 2i = e 3 z

4

z=

B1 2i

k πi e 3

M1 –1

roots A2,1,0 (NB roots are ± 3 – i; ± Total TOTAL

17 75

7

1 – i; – i) 3