General Certificate of Education Advanced Level Examination January 2011

Mathematics

MFP2

Unit Further Pure 2 Wednesday 19 January 2011

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.

P38405/Jan11/MFP2 6/6/

MFP2

2

Sketch on an Argand diagram the locus of points satisfying the equation

1 (a)

j z  4 þ 3i j ¼ 5 (b) (i)

(3 marks)

Indicate on your diagram the point P representing z1 , where both j z1  4 þ 3i j ¼ 5 and

arg z1 ¼ 0

(ii) Find the value of j z1 j .

(1 mark) (1 mark)

Given that

2 (a)

ur ¼ 1 rðr þ 1Þð4r þ 11Þ 6

show that ur  ur1 ¼ rð2r þ 3Þ

(3 marks)

Hence find the sum of the first hundred terms of the series

(b)

1  5 þ 2  7 þ 3  9 þ ::: þ rð2r þ 3Þ þ :::

Show that ð1 þ iÞ3 ¼ 2i  2 .

3 (a)

(3 marks)

(2 marks)

The cubic equation

(b)

z 3  ð5 þ iÞz 2 þ ð9 þ 4iÞz þ kð1 þ iÞ ¼ 0 where k is a real constant, has roots a, b and g. It is given that a ¼ 1 þ i . (i)

Find the value of k.

(ii) Show that b þ g ¼ 4 . (iii) Find the values of b and g.

(3 marks) (1 mark) (5 marks)

P38405/Jan11/MFP2

3 4 (a)

Prove that the curve y ¼ 12 cosh x  8 sinh x  x has exactly one stationary point.

(b)

5 (a) (b)

(7 marks)

Given that the coordinates of this stationary point are ða, bÞ , show that a þ b ¼ 9 . (4 marks)

Given that u ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi du . 1  x 2 , find dx

(2 marks)

Use integration by parts to show that pffiffi ð 3 2

pffiffiffi sin1 x dx ¼ a 3 p þ b

0

where a and b are rational numbers.

6 (a)

(6 marks)

Given that x ¼ lnðsec t þ tan tÞ  sin t show that dx ¼ sin t tan t dt

(b)

(4 marks)

A curve is given parametrically by the equations x ¼ lnðsec t þ tan tÞ  sin t ,

y ¼ cos t

The length of the arc of the curve between the points where t ¼ 0 and t ¼ denoted by s. Show that s ¼ ln p , where p is an integer.

p is 3 (6 marks)

s

Turn over

P38405/Jan11/MFP2

4

Given that

7 (a)

f ðkÞ ¼ 12k þ 2  5k1 show that f ðk þ 1Þ  5f ðkÞ ¼ a  12k where a is an integer.

(3 marks)

Prove by induction that 12n þ 2  5n1 is divisible by 7 for all integers n 5 1 . (4 marks)

(b)

Express in the form reiy , where r > 0 and p < y 4 p :

8 (a) (i)

pffiffiffi 4ð1 þ i 3Þ ; pffiffiffi

(ii) 4ð1  i 3Þ .

(3 marks)

The complex number z satisfies the equation

(b)

ðz 3  4Þ2 ¼ 48 pffiffiffi Show that z 3 ¼ 4  4 3 i . (c) (i)

(2 marks)

Solve the equation ðz 3  4Þ2 ¼ 48 giving your answers in the form reiy , where r > 0 and p < y 4 p .

(ii) Illustrate the roots on an Argand diagram. (d) (i)

(5 marks) (3 marks)

Explain why the sum of the roots of the equation ðz 3  4Þ2 ¼ 48 is zero.

(ii) Deduce that cos

(1 mark) p 3p 5p 7p 1 þ cos þ cos þ cos ¼ . 9 9 9 9 2

(3 marks)

Copyright ª 2011 AQA and its licensors. All rights reserved.

P38405/Jan11/MFP2

AQA – Further pure 2 – Jan2011 – Answers Question 1:

Exam report

a ) z − 4 + 3i = 5 Let ' s call A( z A ) with z A= 4 − 3i The locus is the circle centre A and radius r = 5 b) i )= z1 − 4 + 3i 5 and = arg( z1 ) 0 The point P( z1 ) is the intersection of the circle and the positive half of the x-axis. ii ) z1 == x1 + 0i x1 and ( x1 − 4 ) + (0 = − 3) 2 25 2

x1 − 4 = ±4 = x1 0= or x1 8 z1 = 8 so z1 = 8

Part (a) was well done and almost universally correct. However part (b) proved to be beyond some of the most able candidates. Candidates were just not able to identify the correct position of P, the commonest positions being either at the origin or at the end of the diameter of the circle parallel to the x-axis. Even when the point P was correctly identified, not all were able to use simple geometrical properties to find the length of OP.

Question 2:

Exam report

1 r (r + 1)(4r + 11) 6 1 1 ur − ur −1= r (r + 1)(4r + 11) − (r − 1)(r )(4r + 7) 6 6 1 = r [ (r + 1)(4r + 11) − (r − 1)(4r + 7) ] 6 1 = r  4r 2 + 15r + 11 − 4r 2 − 3r + 7  6 1 = r (12r + 18 ) = r (2r + 3) 6

a ) ur =

+ ... b)1× 5 + 2 × 7 + 3 × 9 + ... + r (2r + 3)=

100

∑ r (2r + 3) r =1

=

100

∑u r =1

r

− ur −1 = u1 − u0 + u2 − u1 + u3 − u2 + ... + u100 − u99

1 =u100 − u0 =u100 = ×100 × (101) × ( 411) =691850 6

The algebra of part (a) of this question was well done, and many candidates were successful with part (b). In part (b), however, in spite of the question stating ‘Hence’, some candidates chose to use alternative methods, usually using Σr and 2 Σr from the formulae booklet; no credit was given for such attempts.

Question 3:

Exam report

3)a ) (1 + i ) =1 + 3i + 3i + i =1 + 3i − 3 − i =−2 + 2i 3

3

2

3

b) z 3 − (5 + i ) z 2 + (9 + 4i ) z + k (1 + i ) = 0 has root α ,β , γ k is a real number

α = 1+ i i ) α is a root so it satisfies the equation: 0 α 3 − (5 + i )α 2 + (9 + 4i )α + k (1 + i ) = − 2 + 2i − (5 + i )(1 + i ) 2 + (9 + 4i )(1 + i ) + k (1 + i ) = 0 − 2 + 2i + 2 − 10i + 5 + 13i + k + ki = 0 k = −5

(5 + k ) + i (5 + k ) = 0 ii ) α + β + γ = 5 + i

β + γ = 5 + i −1 − i = 4 iii ) αβγ = 5(1 + i )

Again part (a) was well done. Many candidates scored well in part (b) too, obtaining the results to parts (i) and (ii) in a variety of ways, although sign errors, especially in the product of the roots, did lead to a loss of marks. One common error was to think that, because 1 +i was a root, 1 – i also had to be a root, in spite of the fact that the coefficients of the cubic were complex.

5(1 + i ) = 5 (1 + i ) so β and γ are the roots of the equation

βγ =

z2 − 4z + 5 = 0 discriminant:(-4) 2 − 4 × 5 =−4 =(2i ) 2 4 + 2i = β = 2 + i and γ = 2 − i β= 2 Question 4:

a ) y = 12 cosh x − 8sinh x − x dy = 12sinh x − 8cosh= x −1 0 dx 6(e x − e − x ) − 4(e x + e − x ) − 1 =0 −x

2e − 10e −= 1 0 x

(×e ) x

2e 2 x − e x − 10 = 0 (2e x − 5)(e x + 2) = 0 5 or e x = −2 (no solution) 2 5 x = ln   is the x-coordinate of the only one 2 stationay point. ex =

5 5 2 5 2 5 b) a=ln   and b = y (a ) = 6  +  − 4  −  − ln   2 2 5 2 5 2 29 21 90 5 2 5 2 =9 a + b = 6 +  − 4 −  = 6× − 4× = 10 10 10 2 5 2 5

Exam report

There were many good responses to this question. Failure, when it arose, was in the use of methods which did not use the exponential forms for sinh x and cosh x. Such methods usually involved the squaring of the expression for dy , more often than not incorrectly. dx When the squaring was done correctly, there was almost invariably no check that the values obtained were in fact solutions for dy = 0 , an essential step as dx squaring had taken place. Part (b) was often correctly worked and with adequate reasoning.

Question 5:

Exam report

a) u = 1 − x 2 so

−x −2 x du du = = = dx 2 1 − x 2 dx 1 − x2

3 2

3 2

−1 −1 b) ∫ sin −1 x dx = ∫ 1× sin x dx = x × sin x 

3 2 0

−∫

0

0

=

3 −x 3 −1  3  2 − + dx sin  0  ∫ 2 0 2 2 − x 1  

3 2

0

3

x×

π 3  π 3 3 = + 1 − x 2  2= + 1− − 1

∫

3 2

0

x dx sin −1 =

π 3 6

0



6

−

6

4

1 2

Question 6:

1 1 − x2

dx

The differentiation in part (a) was well done, apart from the occasional sign error. In part (b), whilst most candidates were familiar with integration by parts, many candidates stalled at the second integral, even though the hint had been clearly given in part (a). Even when candidates knew how to finish the question, errors of sign spoilt what would otherwise have been a good solution.

Exam report

sin t d ( sec t )= cos 2 t dt

a ) x= ln ( sec t + tan t ) − sin t

sin t 1 + 2 2 sin t + 1 dx t − cos t = cos t cos = − cos t 2 sec t + tan t cos t (sec t + tan t ) dt sin t + 1 1 1 − cos 2 t dx = − cos t = − cos t = cos t cos t dt cos t (1 + sin t ) dx sin 2 t sin t = = × sin t = tan t sin t dt cos t cos t b) x ln ( sec t + tan t ) − sin t and y = = − sin t s=

∫

π

3 0

2

2

 dx   dy    +   dt=  dt   dt 

π

∫

π

3 0

tan 2 t sin 2 t + sin 2 tdt π

1  2  s= s= ∫ sin t (1 + tan t )dt = ∫03 sin t  cos2 t dt 3 0

π

2

2

π sin t 1 3 3 ln cos s == d t t − − ln     ( ) 0 = ∫0 cos t  2 s = ln(2)

Part (a) was extremely badly done, with few candidates scoring more than one out of the four available marks. The reason for this was that candidates failed to realise that the derivative of ln (sect + tan t ) is given in the formulae booklet, and so, in attempting to do the differentiation themselves, they became bogged down with the manipulation of trigonometrical functions. Part (b) was better attempted and there were many correct solutions, but, if a solution did peter out, it was generally at the stage where the candidate obtained sint sect but failed to realise that that was in fact tant.

Question 7:

Exam report

a ) f (k )= 12 + 2 × 5 k

k −1

) 12k +1 + 2 × 5k − 5 × (12k + 2 × 5k −1 ) f (k + 1) − 5 f (k= = 12k +1 + 2 × 5k − 5 ×12k − 2 × 5k = 12 ×12k − 5 ×12k = 7 ×12k a=7 b) The proposition Pn :"for all n ≥ 1,12n + 2 × 5n −1 is divisible by 7" is to be proven by induction. Responses to this question were very mixed. Whilst in part (a) candidates knew what to do, after writing down a correct expression for f (k +1) − f (k ) terms such as k−1 60 appeared quite frequently.

base case: n = 1 121 + 2 × 50 = 12 + 2 = 14 = 7 × 2 P1 is true Let's suppose that for n=k, the proposition Pk is true. Let's show that it is true for n=k+1, i.e Let's show that 12k +1 + 2 × 5k is divisible by 7. −−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 12k +1 + 2 × 5k = f (k + 1) =7 ×12k − 5 f (k ) 7 ×12k is a multiple of 7 5 f (k ) is a multiple of 7 (hypothesis) so,as a difference of two ultiple of , f (k + 1) is also a multiple of 7.

In part (b), there was a lot of muddled logic, with many candidates thinking that the reiteration of part (a) was a sufficient argument to prove part (b). There was also clear evidence of a lack of knowledge regarding how proofs by induction should be set out.

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 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 conlude according to the induction principle that it is true for all n ≥ 1. Question 8:

Exam report

1 i 3 a ) i ) 4(1 + i 3) = 8  + i  = 8e 3 2  2 π

−i

π

8e 3 ii ) 4(1 − i 3) = b) ( z 3 − 4 ) =−48 2

z 3 − 4 =±i 48 z 3= 4 ± 4i 3

c) i ) Let's write z 3 (= reiθ )3 r 3e3θ i = 3 3θ i

4 ± 4i 3 is equivalent to r e z = 3

r =3 8 and 3θ =± r =2 and θ =± i

π 9

solutions :2e , 2e

7π i 9

π 9

π 3

, 2e

π

8e 3 =

+ k × 2π

+k× −i

±i

5π 9

2π 3 , 2e

−i

π 9

, 2e

i

5π 9

, 2e

−i

7π 9

Question 8:continues

Exam report

−48 d ) ( z 3 − 4) 2 = z 6 − 8 z 3 + 16 + 48 = 0 z 6 + 0 z 5 + 0 z 4 − 8 z 3 + 0 z 2 + 0 z + 64 = 0 z5 The coefficient of= i

π

ii ) 2e + 2e 9

−i

π 9

+ 2e

i

5π 9

α ∑= + 2e

−i

0 5π 9

+ 2e

i

7π 9

+ 2e

−i

7π 9

= 0

Parts (a) and (b) of this question were well done; part (c)(i) was reasonably well done especially by the more able candidates. However, the Argand diagrams were very poorly drawn. Many candidates drew a rough circle by hand, with no indication of its radius, and then went on to put crosses on their circle at points which bore little resemblance to the angles in question. Some candidates drew no circle at all but drew lines of differing lengths from the origin. There were few correct responses to part (d) as candidates failed to realise the relevance of part (c), and especially part (d)(i), to this last part.

π  5π   7π   2  2 cos  + 2  2 cos 0 (÷4)  + 2  2 cos = 9  9  9    π 5π 7π + cos = cos + cos 0 9 9 9 π 1 3π = cos = Notice cos 9 3 2 π 3π 5π 7π 1 = + cos + cos hence : cos + cos 9 9 9 9 2 Grade boundaries

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 2 – January 2011

Key to mark scheme abbreviations 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 c sf dp

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 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. 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

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 2 – January 2011

MFP2 Q 1(a)

(b)(i) (ii)

Solution

Marks

Total

B1 B1 B1

3

z1 correctly chosen

B1F

1

ft if circle encloses (0, 0)

z1 = 8

B1F

1

ft if centre misplotted

y

z1 P

x

Circle correct centre through (0, 0)

Total 2(a)

Comments

5

ur − ur −1 = 1 1 r ( r + 1)( 4r + 11) − ( r − 1) r ( 4r + 7 ) 6 6

M1

Correct expansion in any form, eg 1 r 4r 2 + 15r + 11 − 4r 2 − 3r + 7 6 = r ( 2r + 3)

A1

(

)

A1

(b) Attempt to use method of differences S100 = u100 − u 0 = 691850 Total πi 2 3(a) (1 + i ) = 2i or (1+ i ) = 2 e 4

2i (1 + i ) = 2i − 2

M1 A1 A1

AG

3 6

CAO

2

AG Alternative method: 3 (1 + i ) = 1 + 3i + 3i 2 + i3 B1 = 2i − 2 B1

B1 B1

(b)(i) Substitute z = 1 + i Correct expansion

3

M1 A1

allow for correctly picking out either the real or the imaginary parts

k = −5

A1

3

(ii)

β + γ = 5 +i −α = 4

B1

1

(iii)

αβγ = 5 (1 + i ) βγ = 5

M1 A1F

z2 − 4z + 5 = 0 2 Use of formula or ( z − 2 ) = −1 z = 2±i NB allow marks for (b) in whatever order they appear

M1 A1F A1F Total

allow if sign error ft incorrect k

No ft for real roots if error in k 5 11

4

AG

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 2 – January 2011

MFP2 (cont) Q 4(a)

Solution dy = 12sinh x − 8cosh x − 1 dx e x − e− x ) e x + e− x ) ( ( 12 −8 −1 = 0 2 2 2e2 x − e x − 10 = 0 ( 2e x − 5)( e x + 2 ) = 0

Marks

Total

B1

M0 if sinh x and cosh x in terms of e x are interchanged ft slips of sign

M1 A1F M1A1F

x

ft provided quadratic factorises

E1

e ≠ −2 5 x = ln 2

one stationary point

A1F

Comments The B1 and M1 could be in reverse order if put in terms of e first

7

some indication of rejection needed 5 Condone e x = with statement provided 2 quadratic factorises Special Case dy =12 sinh x – 8cosh x If dx For substitution in terms of e x leading to e2 x = 5 Then M0

(b)

⎛5 2⎞ ⎛5 2⎞ ⎜ + ⎟ ⎜ − ⎟ 5 2 5⎠ 2 5⎠ ⎝ b = 12 −8⎝ − ln 2 2 2 174 84 5 = − − ln 10 10 2 =9−a

M1A1F

Total 5(a)

(b)

du 1 = (1 − x dx 2 × ( – 2x)

∫ sin ∫−

−1

1 2 −2

)

x dx = x sin −1 x − ∫ x 2

M1 A1

for substitution into original equation

A1 A1

B0

CAO 4

AG; accept b = 9 − a

11

B1 B1 x 1 − x2

2

M1 A1A1

dx

dx = 1 − x 2 used

1− x 3π 3 + 1− −1 2 3 4 1 1 3π− 6 2

A1 for each part of the integration by parts

du dx

A1F

ft sign error in

m1

substitution of limits

A1 Total

6 8

5

CAO

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 2 – January 2011

MFP2 (cont) Q 6(a)

Solution

Marks

dx = sec t − cos t dt

B1,B1

Use of 1 − cos 2 t = sin 2 t dx = sin t tan t dt (b)

A1

sign error in

dy A0 dt

π

tan t dt = [ ln sec t ] 03

A1F

= ln 2

A1 Total

= 12k +1 + 2 × 5k − 5 (12k + 2 × 5k −1 )

M1

= 12k +1 + 2 × 5k − 5 × 12k − 2 × 5k

A1 A1

= 12 × 12k − 5 × 12k = 7 × 12k

dy dt dy ft sign error in dt CAO ft sign error in

A1F

f ( k + 1) − 5f ( k )

AG

m1

x 2 + y 2 = tan t

7(a)

4

M1A1

Use of 1 + tan 2 t = sec2 t

∫

Comments use of FB for sec t ; if done from first principles, allow B1 when sec t is arrived at

M1

x 2 + y 2 = sin 2 t tan 2 t + sin 2 t

π 3 0

Total

6 10

3

for expansion of bracket 5 × 5k –1 = 5k used clearly shown

(b) Assume f ( k ) = M ( 7 )

Then f ( k + 1) = 5f ( k ) + M ( 7 ) = M (7)

M1

Not merely a repetition of part (a)

A1

clearly shown

f (1) = 12 + 2 = 14 = M ( 7 )

B1

Correct inductive process

E1 Total

4 7

6

(award only if all 3 previous marks earned)

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 2 – January 2011

MFP2 (cont) Q 8(a)(i)

Solution ⎛1 3⎞ 4 1 + i 3 = 8 ⎜⎜ + i ⎟ 2 ⎟⎠ ⎝2

(

Marks

)

Total

(

)

A1

− πi 3

4 1 − i 3 = 8e

(b)

z 3 − 4 = ± − 48

A1

z = 4 ± 4 3i (c)(i)

z = 2e

πi +2kπi 3 3

πi

z = 2e 9 , 2e = 2e

− πi 9

A1

or z = 2e 7 πi 9

, 2e

, 2e

−7 πi 9

− πi +2kπi 3 3

taking square root 2

−5πi 9

A3,2,1F

AG for the 2; ft incorrect 8, but no decimals for either, PI

B1F M1

5πi 9

, 2e

5

Allow A1 for any 2 roots not +/– each other Allow A2 for any 3 roots not +/– each other Allow A3 for all 6 correct roots Deduct A1 for each incorrect root in the interval; ignore roots outside the interval ft incorrect r

(ii)

0

2

Radius 2

B1F

Plotting roots

B2,1

3

E1

1

OE

3

AG

(d)(i) Sum of roots = 0 as coefficient of z 5 = 0 (ii)

Use of, say,

− πi ⎞ 1 ⎛ πi9 π ⎜ e + e 9 ⎟ = cos 2⎝ 9 ⎠

3π 1 = used 9 2 π 3π 5π 7π 1 cos + cos + cos + cos = 9 9 9 9 2

cos

clearly indicated; ft incorrect r allow B1 for 3 correct points condone lines

M1 A1 A1

Total TOTAL

17 75

7

)

3

M1

3

(

If either r or θ is incorrect but the same value in both (i) and (ii) allow A1 π but for θ only if it is given as 6

πi

(ii)

)

for either 4 1 + i 3 or 4 1 − i 3 used

M1

= 8e 3

(

Comments