General Certificate of Education Advanced Level Examination June 2010

Mathematics

MFP2

Unit Further Pure 2 Wednesday 9 June 2010 1.30 pm to 3.00 pm

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For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.

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Time allowed * 1 hour 30 minutes

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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.

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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.

P27888/Jun10/MFP2 6/6/

MFP2

2 1 (a)

Show that 9 sinh x
cosh x ¼ 4ex
5e
x

(b)

(2 marks)

Given that 9 sinh x
cosh x ¼ 8 find the exact value of tanh x .

2 (a) (b)

Express

(7 marks)

1 in partial fractions. rðr þ 2Þ

(3 marks)

Use the method of differences to find 48 X

1 rðr þ 2Þ r¼1

giving your answer as a rational number.

(5 marks)

Two loci, L1 and L2 , in an Argand diagram are given by

3

L1 : j z þ 1 þ 3i j ¼ j z
5
7i j L2 : arg z ¼

p 4

(a)

Verify that the point represented by the complex number 2 þ 2i is a point of (2 marks) intersection of L1 and L2 .

(b)

Sketch L1 and L2 on one Argand diagram.

(c)

Shade on your Argand diagram the region satisfying both

j z þ 1 þ 3i j 4 j z
5
7i j

and

p p 4 arg z 4 4 2

(5 marks)

(2 marks)

P27888/Jun10/MFP2

3

The roots of the cubic equation

4

z 3
2z 2 þ pz þ 10 ¼ 0 are a, b and g. It is given that a 3 þ b 3 þ g 3 ¼
4 . (a)

Write down the value of a þ b þ g .

(1 mark)

(b) (i)

Explain why a 3
2a 2 þ pa þ 10 ¼ 0 .

(1 mark)

(ii) Hence show that

a 2 þ b 2 þ g 2 ¼ p þ 13

(4 marks)

(iii) Deduce that p ¼
3 . (c) (i)

(2 marks)

Find the real root a of the cubic equation z 3
2z 2
3z þ 10 ¼ 0 .

(2 marks)

(ii) Find the values of b and g.

(3 marks)

Using the identities

5 (a)

cosh2 t
sinh2 t ¼ 1 , tanh t ¼

sinh t 1 and sech t ¼ cosh t cosh t

show that: (i)

tanh2 t þ sech2 t ¼ 1 ;

(2 marks)

(ii)

d ðtanh tÞ ¼ sech2 t ; dt

(3 marks)

(iii)

d ðsech tÞ ¼
sech t tanh t . dt

(3 marks)

A curve C is given parametrically by

(b)

x ¼ sech t , y ¼ 4
tanh t (i)

1

Show that the arc length, s, of C between the points where t ¼ 0 and t ¼ 2 ln 3 is given by ð 1 ln 3 2 sech t dt (4 marks) s¼ 0

(ii) Using the substitution u ¼ et , find the exact value of s.

(6 marks) s

Turn over

P27888/Jun10/MFP2

4

Show that

6 (a)

1 kþ1 2
¼ . ðk þ 2Þ! ðk þ 3Þ! ðk þ 3Þ!

(2 marks)

Prove by induction that, for all positive integers n,

(b)

n X r  2r 2nþ1 ¼1
ðr þ 2Þ! ðn þ 2Þ! r¼1

7 (a) (i)

Express each of the numbers 1 þ

(6 marks)

pffiffiffi 3 i and 1
i in the form reiy , where r > 0 . (3 marks)

(ii) Hence express

ð1 þ

pffiffiffi 8 3 iÞ ð1
iÞ5

in the form reiy , where r > 0 . (b)

(3 marks)

Solve the equation z 3 ¼ ð1 þ

pffiffiffi 8 3 iÞ ð1
iÞ5

pffiffiffi giving your answers in the form a 2 eiy , where a is a positive integer and
p < y 4 p . (4 marks)

END OF QUESTIONS

Copyright Ó 2010 AQA and its licensors. All rights reserved.

P27888/Jun10/MFP2

AQA – Further pure 2 – Jun 2010 – Answers Question 1:

Exam report

9 x −x 1 x −x (e − e ) − 2 (e + e ) 2 8 x 10 − x e − e = 2 2 x = 4e − 5e − x 8 is equivalent to b) 9sinh x − cosh x = a ) 9sinh x − cosh x =

x 4e x − 5e −= 8

(×e x )

4e 2 x − 5 − 8e x = 0 4 ( e x ) − 8e x − 5 = 0 2

(2e x − 5) ( 2e x + 1) = 0 5 1 or e x = − (no solution) 2 2 5 2 − x −x − 4 21 e −e 2 5 25 = = = = tanh x x −x 5 2 25 + 4 29 e +e + 2 5 21 tanh x = 29 ex =

Almost all candidates scored the two available marks in part (a). However in part (b) a number of candidates did not draw on the hint of part (a) but instead tried to manipulate the equation given in part (b). A few of these candidates expressed the given equation in tanh x and sech x and then squared, obtaining a quadratic in tanh x. When factorised these candidates obtained two values for tanh x, only one of which was the correct one. However, virtually no one rejected the incorrect solution so that it was almost impossible to award full marks when this method was used.

Question 2:

a)

Exam report

1 A B = + r (r + 2) r (r + 2) 1= A(r + 2) + Br 0= gives 2 A 1

r

× r (r + 2)

= A

r= −2 gives − 2 B = 1

1 2

B= −

1 2

1 1 1 = − r (r + 2) 2r 2(r + 2) 48

48 1 1 1 = ∑ − r (r + 2) r 1 2r 2(r + 2) 1=

b) ∑ r

=

sum, notably

1 1 1 1 1 1 − + − + − + ... 2 6 4 8 6 10 1 1 1 1 1 1 − + − + − 92 96 94 98 96 100

48

1

1

1

1

1

894

∑ r (r + 2) = 2 + 4 − 98 − 100 = 1225 r =1

This question was generally well done, with many candidates scoring full marks. When errors did occur they were usually in the omission of one of the four fractions that made up the

1 98

Question 3:

Exam report

z= 2 + 2i and M ( z ) Does M belong to L1 ? z + 1 + 3i = 2 + 2i + 1 + 3i = 3 + 5i =

9 + 25 =

34

z − 5 − 7i =2 + 2i − 5 − 7i =−3 − 5i = 9 + 25 = 34 M ( z= 2 + 2i ) belongs to L1 Does M belong to L 2 ? 2 π arg( z )= arg(2 + 2i )= tan −1  = 2 4 M ( z = 2 + 2i ) belongs to L 2 M ( z ) is a point of the intersection between L1 and L2 b) L1 is the perpendicular bisector of the line AB with A(z A =−1 − 3i ) and B( z B =5 + 7i ) L2 is the half line from O with gradient tan c)

π 4

= 1.

The verifications in part (a) were not always convincing, especially the verification that the point representing the complex number 2 + 2i lay on the line L 1 . The sketches in part (b) varied considerably. Those candidates who made a reasonably careful drawing generally scored higher marks as they were able to clearly show that the point representing 2 + 2i lay on both L 1 and L 2 . Careful sketches also improved a candidate’s chance of scoring full marks in part (c).

Question 4:

Exam report

a ) z − 2 z + pz + 10 = 0 has roots α , β , γ 3

2

−4 α3 + β3 +γ 3 = a) α + β + γ = 2 b) i ) α is a root so it satifies the equation

α 3 − 2α 2 + pα + 10 = 0 ii ) The same applies to β and γ α 3 − 2α 2 + pα + 10 = 0 β 3 − 2β 2 + pβ + 10 = 0 γ 3 − 2γ 2 + pγ + 10 = 0 by adding 0 ( α 3 + β 3 + γ 3 ) − 2 ( α 2 + β 2 + γ 2 ) + p (α + β + γ ) + 30 = Whilst parts (a) and (b)(i) were well done, few

candidates were able to complete part (b)(ii) correctly through not taking note of the hint given in part (b)(i). Those candidates attempting to work 3 out ( α + β + γ ) were inevitably doomed to failure.

− 4 − 2 ( α 2 + β 2 + γ 2 ) + 2 p + 30 = 0 − 2( α 2 + β 2 + γ 2 ) = −2 p − 26

α 2 + β 2 + γ 2 =p + 13

Part (b)(iii) was usually attempted by assuming the result of part (b)(ii). There were many correct solutions to part(c) although slips of sign often led to a solution with three real roots, contrary to the statement of part (c)(i).

iii ) α + β + γ = (α + β + γ ) − 2 (αβ + αγ + γβ ) 2

2

2

2

=22 − 2 p =4 − 2 p 4 − 2 p =p + 13

so

−3p = 9 p = −3 c)i ) Test the values − 2, − 1,0,1,or 2 for α . here, α =−2

indeed (−2)3 − 2 × ( −2 ) − 3 × (−2) + 10 = 2

0 − 8 − 8 + 6 + 10 = ii ) z 3 − 2 z 2 − 3 z + 10 = ( z + 2 ) ( z 2 − 4 z + 5 ) z 2 − 4 z + 5 discriminant =(−4) 2 − 4 ×1× 5 =−4 =(2i ) 2 4 + 2i = β = 2 + i and γ = 2 − i β= 2 Question 5:

Exam report

sinh t 1 sinh t + 1 cosh t + = = = 1 2 2 cosh t cosh t cosh 2 t cosh 2 t d d  sinh t u  u ' v − uv ' cosh t × cosh t − sinh t × sinh t ii ) ( tanh= t) ) = = =   ( dt dt  cosh t v  v2 cosh 2 t cosh 2 t − sinh 2 t 1 sech 2t = = = 2 2 cosh t cosh t d d 1 1  u'  sinh t iii ) = − ( sech t ) =  = − 2  = dt dt  cosh t u   u  cosh 2 t sinh t 1 = − × = −sech t tanh t cosh t cosh t i ) tanh 2 t + sech 2t =

2

2

2

Question 5:continues

Exam report

b) x= sech t and y= 4 − tanh t i)

dx dy = −sech t × tanh t and = −sech 2t dt dx 2

2

 dx   dy  2 2 4   +   = sech t × tanh t + sech t dt dx    

= sech 2t ( tanh 2 t + sec= h 2t ) sech 2t 2

Part (a) was a source of good marks for almost all candidates. If errors did occur they were usually errors of sign. Part (b) was also generally well done although it was disappointing to see the square root 2 2 4 of sech t tanh t + sech t written as sech t tanh t + 2 sech t a significant number of times. Responses to part (b)(ii) were mixed. Poor algebraic manipulation in the handling of sech t when expressed in terms of u let many candidates down badly so that they ended up with a polynomial in u to integrate.

2

1 ln 3  dx   dy  2 = + = s ∫ dt sech t dt     ∫ 0  dt   dx  du du t ii ) u= et = e= u = dt dt u whe= n t 0,= u 1 1 ln 3 2 0

1 = = u ln 3 ln 3, 3 2 1 ln 3 3 2 s ∫ 2 sech t dt ∫ x = dx = = 0 1 e + e− x = t

∫

1

3

2 u+

1 u

×

du u

3 2 −1   du u 2 tan 2 tan −1 3 − 2 tan −1 1 = = ∫1 u 2 + 1  1 π π 2π π π s = 2× − 2× = − = 3 4 3 2 6

s =

3

Question 6:

a)

1 k +1 k +3 k +1 2 − = − = (k + 2)! (k + 3)! (k + 3)! (k + 3)! (k + 3)!

Exam report

r × 2r 2n +1 = 1− (n + 2)! r =1 ( r + 2)! n

b) the proposition Pn :for all n ≥ 1, ∑ is to be proven by induction Base case: n = 1 1× 21 2 1 r × 2r = = = (1 + 2)! 3! 3 r =1 ( r + 2) ! 1

LHS : ∑ RHS :1 −

21+1 4 2 1 =1 − =1 − = (1 + 2)! 3! 3 3 P1 is true

Let's suppose that for n = k , the proposition is true. Let's show that it is true for n= k + 1 r × 2r 2k + 2 1 = − ∑ (k + 3)! r =1 ( r + 2)! −−−−−−−−−−−−−−−−−−−−−−−−−−− k +1

i.e Let's show that

k r × 2r r × 2r (k + 1)2k +1 2k +1 (k + 1)2k +1 = + = 1 − + ∑ ∑ (k + 3)! (k + 2)! (k + 3)! r =1 ( r + 2)! r =1 ( r + 2)! k +1

 1 (k + 1)  2 − = 1 − 2k +1  1 − 2k +1 × from Qa ) = (k + 3)!  (k + 2) ! (k + 3)!  2k + 2 = 1− Q.E.D (k + 3)! −−−−−−−−−−−−−−−−−−−−−−−−−−−− 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 principal that it is true for all n ≥ 1.

Again responses to this question were mixed. It was evident that some candidates thought that (k + 2)! started at k, and wrote it as k(k + 1)(k + 2). Others wrote down the result after some rather dubious algebra. Although there has been considerable improvement in the way that solutions by induction have been expressed, in this case what would otherwise have been acceptable solutions were spoilt by errors of sign. The same error occurred frequently. It occurred when a candidate tried to combine, in one bracket, a negative expression followed by a positive expression by placing a negative sign outside the combining bracket and then by forgetting to alter the sign before the positive term to compensate.

Question 7:

Exam report

1 i 3 a ) i )1 + i 3 = 2  + i  = 2e 3 2  2 π

= 1− i

1   1 − i= 2  2  2

2e

−i

π 4

8

π −i   i π3   ii ) 1 + i 3 (1 −= i )  2e  ×  2e 4     

(

)

8

= 28 e

i

8π 3

5 −i

× 2e

5π 4

 2π 3π  +  i 3 4 

5

21

iθ 3 i 3θ e 22 e ( re= ) r= 3

= 28 e 21

= 28 × 2 2 × e  = b) z 3

Part (a)(i) was generally well done. The less successful candidates usually wrote the argument of 1 – i as 3π/4 instead of – π/4. In part (a)(ii) there was some poor handling of fractions in the argument of the product of the two complex numbers, and also some omission of raising the moduli of the two complex numbers to their respective powers. Many of the candidates who had been successful in part (a) often went on to complete part (b) correctly, although some candidates lost marks either through not giving z in the form asked for or by giving values for θ outside the specified range.

5

5

i

=22 e i

2π 3

i

5 i

× 2e

17π 12

3π 4

= 1024 2 e

17π 12

17 π 12

17π + k × 2π 12 7 17π 2π +k× r =2 2 =8 2 and θ = 36 3 21

i

and 3θ=

3 so r= 22

i

17π

−i

7π

k =−2, −1, 0

= solutions : z 8 = 2e 36 or z 8= 2e 36 or z 8 2e

−i

31π 36

Grade boundaries

MFP2 - AQA GCE Mark Scheme 2010 June 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 June series

MFP2 Q 1(a)

9 (e – e x

–x

)

2 = 4e x – 5e – x

Solution x

–

e +e 2

Marks

–x

A1

AG

M1

x

A1

( 2e x – 5)( 2e x + 1) = 0

M1

ft provided quadratic factorises (or use of formula)

E1F

PI but not ignored

ex ≠ –

1 2

5 2

ex =

tanh x =

A1F 5 2 5 2

– 52 21 = + 52 29

M1 A1F Total

M1

1 1 , B=– 2 2

A1, A1F

1 1 ⎛1 1 ⎞ = ⎜ – ⎟ 1.3 2 ⎝1 3 ⎠ 1 1 ⎛1 1⎞ r =2 = ⎜ – ⎟ 2.4 2 ⎝2 4⎠ 1 1 ⎛1 1⎞ r=3 = ⎜ – ⎟ 3.5 2 ⎝3 5⎠ 1 1⎛ 1 1 ⎞ r = 48 = ⎜ – ⎟ 48.50 2 ⎝ 48 50 ⎠ Cancelling appropriate pairs 1 ⎛1 1 1 1 ⎞ Sum = ⎜ + – – ⎟ 2 ⎝ 1 2 49 50 ⎠ 894 = 1225

7

M1 PI for attempt to use tanh x = or equivalent fraction

9

A B = + r r+2 r ( r + 2) 1

A=

(b)

2

– 8e – 5 = 0

4e

2(a)

Comments M0 if cosh x mixed up with sinh x

M1

(b) Attempt to multiply by e x 2x

Total

3

ft incorrect A

r =1

M1

3 rows (PI) numerical values only

A1F

Last row – could be implied

M1 Allow if the

A1F A1 Total

5 8

4

1 is missing only 2

CAO (or equivalent fraction)

sinh x cosh x

MFP2 - AQA GCE Mark Scheme 2010 June series

MFP2 (cont) Q 3

Solution

Marks

Total

Comments

Im

(2,2) Re

(a)

2 + 2i +1 + 3i = 2 + 2i –5 –7i arg ( 2+2i ) =

(b)

π 4

B1

L1 : straight line with negative gradient perpendicular to line joining

B1

through ( 2, 2 )

B1

L2 : half line through O

B1

through ( 2, 2 )

B1

π π and 4 2

B1

(b)(i) (ii)

α is a root and so satisfies the equation

∑α

3

– 2 ∑ α 2 + p ∑ α + 30 = 0

Substitution for

∑α (iii)

(c)(i)

(ii)

2

∑α

3

and

2

5 No marks for shading if circles drawn in (b)

E1

1

∑α

ml

+ 2 ∑αβ used

4

A1

f ( –2 ) = 0 α =–2

M1 A1 – 4 z + 5) = 0

2

4 ± –4 2 =2 ± i

For attempting to find quadratic factor

m1 A1 Total

5

AG

2

M1

z=

AG do not allow this M mark if used in (b)(ii)

M1

p= –3

( z + 2) ( z 2

Clearly shown

The point (2,2) must be shown either by (2,2) or 2+2i or with numbered axes

B1

A1 2

without comment

M1A1

= p + 13

( ∑α ) = ∑α

2

2 9 1

Total

4(a) α + β + γ = 2

3 + 5i = –3 – 5i

B1

Below L1

do not allow

B1

( –1, –3) to ( 5,7 )

(c) Shading between

Clearly shown

B1

3 13

Use of formula or completing the square m0 if roots are not complex CAO

MFP2 - AQA GCE Mark Scheme 2010 June series

MFP2 (cont) Q

Solution Divide cosh t – sinh 2 t = 1 by cosh 2 t

Marks

Total

5(a)(i)

Rearrange

(ii)

(iii)

d ⎛ sinh t ⎞ cosh 2 t – sinh 2 t ⎜ ⎟= dt ⎝ cosh t ⎠ cosh 2 t

M1A1

= sech 2t

A1

2

∴s =

∫

A1

AG If solved back to front with no conclusion ending cosh 2 t – sinh 2 t =1 B1 only

3

AG Allow A1 if negative sign missing

3

2

1 ln 3 2 0

u = et du = et dt 2 ∫ sech t dt = ∫ u 2 + 1 du

AG Allow slips of sign before squaring for this M1

M1 m1 A1 A1

sech t dt

Correct formula only for m1 4

AG (including limits)

B1 CAO M1 for putting integrand in terms of u (no sech (lnu))

M1A1

⎡⎣ 2 tan –1u ⎤⎦

Change limits correctly or change back to t 2π 2π π = − = 3 4 6

A1

Or 2 tan –1et

m1

At some stage

A1 Total

6(a)

2

M1A1

⎛ dx ⎞ ⎛ dy ⎞ (b)(i) ⎜ ⎟ + ⎜ ⎟ = sech 4t + sech 2 t tanh 2 t ⎝ dt ⎠ ⎝ dt ⎠ Use of tanh 2t + sech 2t = 1 = sech 2t

(ii)

M1 A1

d –2 ( sech t ) = – ( cosh t ) sinh t dt = − sech t tanh t

Comments Or sinh 2 t + 1 2 cosh t cosh t 2

2

1 k +3 = ( k + 2 )! ( k + 3) ! Result

6

CAO

18

M1 A1

(b) Assume true for n = k For n = k + 1 k +1 k + 1) 2k +1 ( r × 2r 2k +1 = 1– + ∑ ( k + 2 ) ! ( k + 3) ! r =1 ( r + 2 )! ⎛ 1 k +1 ⎞ – = 1 – 2 k +1 ⎜ ⎜ ( k + 2 )! ( k + 3 )! ⎟⎟ ⎝ ⎠

2k + 2 ( k + 3) ! True for n = 1 Method of induction set out properly = 1–

2

M1A1

If no LHS of equation,

M1A0

m1

m1 for a suitable combination clearly shown

A1

clearly shown or stated true for n = k + 1

B1 E1 Total

6

6 8

Shown Provided previous 5 marks all earned

MFP2 - AQA GCE Mark Scheme 2010 June series

MFP2 (cont) Q 7(a)(i)

Solution 1 + 3 i = 2e 1– i = 2e

(ii)

(b)

Marks

πi 3

B1

πi − 4

B1B1

21

2 2 or equivalent single expression

B1F

Raising and adding powers of e 17π or equivalent angle 12

M1

z = 3 210 2 e

Total

AIF

17 πi 2 kπi + 36 3

Comments

B1 both correct 3

OE No decimals; must include fractional powers

3

Denominators of angles must be different

M1 B1

3

210 2 = 8 2 17π 7π 31π θ= ,− ,− 36 36 36

A2,1F Total TOTAL

CAO 4 10 75

7

Correct answers outside range: deduct 1 mark only