General Certificate of Education January 2009 Advanced Level Examination
MATHEMATICS Unit Further Pure 2 Monday 19 January 2009
MFP2
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 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. *
P10402/Jan09/MFP2 6/6/6/
MFP2
2
Answer all questions.
1
1
1
(a) Use the definitions sinh y ¼ 2ðey ey Þ and cosh y ¼ 2ðey þ ey Þ to show that 1 þ 2 sinh2 y ¼ cosh 2y
(3 marks)
(b) Solve the equation 3 cosh 2y ¼ 2 sinh y þ 11 giving each of your answers in the form ln p .
2
(a) Indicate on an Argand diagram the region for which j z 4i j 4 2 .
(6 marks)
(4 marks)
(b) The complex number z satisfies j z 4i j 4 2 . Find the range of possible values of arg z . (4 marks)
3
1
(a) Given that f ðrÞ ¼ 4 r 2 ðr þ 1Þ2 , show that f ðrÞ f ðr 1Þ ¼ r 3
(3 marks)
(b) Use the method of differences to show that 2n X r¼n
P10402/Jan09/MFP2
3
r 3 ¼ 4 n2 ðn þ 1Þð5n þ 1Þ
(5 marks)
3
4 It is given that a, b and g satisfy the equations a þb þg ¼1 a 2 þ b 2 þ g 2 ¼ 5 a 3 þ b 3 þ g 3 ¼ 23 (a) Show that ab þ bg þ ga ¼ 3 .
(3 marks)
(b) Use the identity ða þ b þ gÞða 2 þ b 2 þ g 2 ab bg gaÞ ¼ a 3 þ b 3 þ g 3 3abg to find the value of abg .
(2 marks)
(c) Write down a cubic equation, with integer coefficients, whose roots are a, b and g. (2 marks)
5
(d) Explain why this cubic equation has two non-real roots.
(2 marks)
(e) Given that a is real, find the values of a, b and g.
(4 marks)
(a) Given that u ¼ cosh2 x , show that
du ¼ sinh 2x . dx
(2 marks)
(b) Hence show that ð1
sinh 2x p dx ¼ tan1 ðcosh2 1Þ 4 4 0 1 þ cosh x
(5 marks)
6 Prove by induction that 2 1 22 2 23 3 2n n 2nþ1 þ þ þ ::: þ ¼ 1 23 34 45 ðn þ 1Þðn þ 2Þ n þ 2 for all integers n 5 1 .
(7 marks)
Turn over for the next question
P10402/Jan09/MFP2
s
Turn over
4
7
(a) Show that d 1 1 1 cosh ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi dx x x 1 x2
(3 marks)
(b) A curve has equation y¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 x 2 cosh1 x
ð0 < x < 1Þ
Show that: dy ¼ (i) dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 ; x
(4 marks) 1
(ii) the length of the arc of the curve from the point where x ¼ 4 to the point where 3
x ¼ 4 is ln 3 .
8
(5 marks)
(a) Show that ðz4 eiy Þðz4 eiy Þ ¼ z8 2z4 cos y þ 1
(2 marks)
(b) Hence solve the equation z8 z4 þ 1 ¼ 0 giving your answers in the form eif , where p < f 4 p . (c) Indicate the roots on an Argand diagram.
END OF QUESTIONS
Copyright Ó 2009 AQA and its licensors. All rights reserved.
P10402/Jan09/MFP2
(6 marks) (3 marks)
AQA – Further pure 2 – Jan 2009 – Answers Question 1:
Exam report
a )1 + 2sinh 2 θ = 1 + 2 ×
2 1 θ 1 e − e −θ ) = 1 + ( e 2θ + e −2θ − 2 ) ( 4 2
1 2θ e + e −2θ ) = cosh 2θ ( 2 b) 3cosh = 2θ 2sinh θ + 11 1 + 2sinh 2 θ =
Part (a) was reasonably well done, although in a number of cases candidates quoted other relationships between coshθ and sinhθ and so were unable to gain credit. In part (b), two points are worthy of note: firstly some candidates thought that sinhθ = –1 had no solutions as a negative sign occurred, and secondly a substantial number of candidates solved sinhθ = 4 and sinhθ = –1 by using 3 the exponential form for sinhθ followed by the θ solving of a quadratic equation in e instead of merely –1 quoting the formula for sinh (x) given on page 5 of the formulae booklet.
3(1 + 2sinh 2 θ ) = 2sinh θ + 11 3 + 6sinh 2 θ − 2sinh θ − 11 = 0 6sinh 2 θ − 2sinh θ − 8 = 0 3sinh 2 θ − sinh θ − 4 = 0 (3sinh θ − 4)(sinh θ + 1) = 0 4 or sinh θ = −1 3 2 4 4 −1 4 θ= sinh = ln + 1 + = ln 3 3 3 3 sinh θ =
) (
(
or θ = sinh −1 ( −1) = ln −1 + 1 + (−1) 2 = ln −1 + 2
)
Question 2: a ) z − 4i ≤ 2 is the region inside the circle
Exam report
centre A(0,4) and radius r = 2. b) Draw the two tangents to the circle from the origin O. We call the points of contact P1 ( z1 ) and P2 ( z2 ). Use trig.properties to work out the argument of z1 and z 2 : In the right-angles triangle OAP1 ,sin α=
opp 2 1 = = hyp 4 2
1 π = = so α sin 2 6 −1
arg( z1 ) =
π 2
−
π 6
π 3
=
π 3
and arg( z2 ) = arg( z1 ) + 2α =
≤ arg( z ) ≤
2π 3
2π 3
Part (a) was well done apart from a few candidates who drew their circle at the mirror image of its correct position in the x-axis. In part (b), although candidates realised that the tangents needed to be drawn from the origin to the circle in order to find the possible value of arg z, few were able to manage the trigonometry involved to reach the correct range, and it was not uncommon to see tan −1 4 appearing, suggesting that 2 candidates thought that these points were in fact the points of intersection of the circle with the line through its centre parallel to the x-axis.
Question 3:
Exam report
1 2 1 r (r + 1) 2 − (r − 1) 2 r 2 4 4 1 2 r (r + 1) 2 − (r − 1) 2 = 4 1 2 2 r ( r + 2r + 1 − r 2 + 2r − 1) = 4 1 = r 2 ( 4r ) 4 f (r ) − f (r − 1) = r 3
1) a ) f (r ) − f (r −=
2n
3 b) ∑ r=
2n
∑ f (r ) − f (r − 1)=
f (n) − f (n − 1) +
r n= r n =
f (n + 1) − f (n) + f (n + 2) − f (n + 1) + .... + f (2n − 1) − f (2n − 2) + f (2n) − f (2n − 1) all the terms cancel except f (2n) − f (n − 1) 2n
= ∑ r 3 f (2n) − f (n − 1) r =n
1 1 2 2 (2n) 2 ( 2n + 1) − ( n − 1) n 2 4 4 1 2 n 4(2n + 1) 2 − (n − 1) 2 = 4 1 2 n (16n 2 + 4 + 16n − n 2 + 2n − 1) = 4 1 2 3 2 n (15n 2 + 18n= n ( 5n 2 + 6n + 1) = + 3) 4 4 2n 3 2 3 r= n ( 5n + 1)( n + 1) ∑ 4 r =n =
There were many good and completely correct solutions to this question. Virtually all candidates completed part (a) correctly, and in part (b), if an error occurred, it was usually in the selection of an incorrect value for r at one end. For instance,
2n
∑r
3
was taken to be f (2n) – f (n) rather than
r =n
f (2n) – f (n – 1) .
Question 4:
Exam report
a ) (α + β + γ ) = α + β + γ + 2 (αβ + αγ + γβ ) 2
2
2
2
12 =−5 + 2 (αβ + αγ + γβ ) 3 so αβ + αγ + γβ = b) (α + β + γ ) (α 2 + β 2 + γ 2 − αβ − αγ − γβ ) = α 3 + β 3 + γ 3 − 3αγβ 1× ( −5 − 3) =−23 − 3αβγ so αβγ = −5 0 c) z 3 − (α + β + γ ) z 2 + (αβ + αγ + γβ ) z − αβγ = z 3 − z 2 + 3z + 5 = 0 d ) α 2 + β 2 + γ 2 =−5 < 0 so at least one of the root is complex; And because the coefficients of the equation are REAL, its conjugate is also a root. e) z 3 − z 2 + 3 z + 5 =0 has an " obvious " root :α =−1 indeed : (−1)3 − (−1) 2 + 3 × (−1) + 5 =−1 − 1 − 3 + 5 =0 Factorise the polynomial ( z + 1)( z 2 − 2 z + 5) = 0 Discriminant of z 2 − 2 z + 5 : (−2) 2 − 4 ×1× 5 =−16 =(4i ) 2 2 + 4i 2 − 4i and γ = β= 2 2 −1, β = 1 + 2i , γ = 1 − 2i α= Question 5:
I
x 1,= u cosh 2 1 = 1 sinh 2 x cosh 2 1 du cosh 2 1 −1 tan d u x = = ∫0 1 + cosh 4 x ∫1 1 + u 2 1 tan −1 ( cosh 2 1) − tan −1 (1)
I = tan −1 ( cosh 2 1) −
Part (e) was poorly answered: it just did not seem to occur to candidates to use the factor theorem to find the real root. Instead they tried to use the symmetric relations between the roots in order to find them. This in turn led to heavy algebra with final abandonment.
Exam report
du 2 2sinh a ) u cosh x, x cosh x sinh 2 x = = = dx du = sinh 2 x dx when= x 0,= u 1
I
Parts (a), (b) and (c) of this question were well done apart from odd sign errors here and there. However, there was much woolly thinking in part (d). For instance, it was not uncommon to see statements such as ‘the cubic equation has real coefficients so it must have one real root and a conjugate pair of non-real roots’ or 2 2 2 ‘since α +β +γ = – 5 , two of α ,β and γ must be nonreal’.
π 4
Part (a) was usually correctly done. However, responses to part (b) were mixed. It was clear that not all candidates du were familiar with ∫ 1 + u 2 , even though it is given on page 8 of the formulae booklet. A significant number of 4 candidates gave this integral as ln(1+ cosh x).
Question 6:
Exam report
The proposition Pn : 2 × 1 2 2 × 2 23 × 3 2n × n 2n +1 for n ≥ 1, + + + ... + = −1 2 × 3 3× 4 4 × 5 (n + 1) × (n + 2) n + 2 is to be proven by induction Base case: n = 1 2 ×1 2 1 21+1 4 1 LHS : = = and RHS : −1 = −1 = 2×3 6 3 1+ 2 3 3 P1 is true Suppose that for n = k the propsotion Pk is true. Let's show that the proposition Pk +1 is true 2 × 1 2 2 × 2 23 × 3 2k +1 × (k + 1) 2k + 2 + + + ... + = −1 2 × 3 3× 4 4 × 5 (k + 2) × (k + 3) k + 3 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
ie let's show that
2 × 1 2 2 × 2 23 × 3 2k +1 × (k + 1) + + + ... + 2 × 3 3× 4 4 × 5 (k + 2) × (k + 3) 2 × 1 2 2 × 2 23 × 3 2k × k 2k +1 × (k + 1) + + + ... + + 2 × 3 3× 4 4 × 5 (k + 1) × (k + 2) (k + 2) × (k + 3) 2k +1 −1 k +2
+
2k +1 × (k + 1) (k + 2) × (k + 3)
2k +1 ( k + 3 + k + 1) 2k +1 (k + 3) + 2k +1 (k + 1) = −1 −1 (k + 2)(k + 3) (k + 2)(k + 3) 2k +1 ( 2k + 4 ) 2k + 2 ( k + 2 ) 2k + 2 = = −1 = −1 −1 (k + 2)(k + 3) (k + 2)(k + 3) (k + 3) −−−−−−−−−−−−−−−−−−−−−−−−−−−− Conclusion :
Q.E.D
If the proposition is true for n= k , then it is true for n= k + 1. Because it is true for n = 1, according to the induction principal we can conclude that is true true for all n ≥ 1.
One thing which became evident in the marking of this question was that although candidates were able to perform the mechanics of proof by induction they did not really understand the theory behind it. In a significant number of solutions not one reference to a series or the use of the Σ symbol occurred. Solutions started ‘assume result true for n = k ’ k +1 k +1 followed by 2 + 1 + 2 (k + 1) (k + 2)(k + 3) n+2 k +2 which was duly shown to be 2 − 1 k +3
Question 7:
Exam report
d 1 1 1 x2 −1 1 cosh = − × = − × 2 2 dx x x2 x2 2 1− x 1 x −1 x2 x x = − 2 x 1 − x2 d 1 −1 1 cosh = − dx x x 1 − x2
a)
1 − x 2 − cosh −1
b) y =
1 x 1
(0 < x < 1)
dy 1 1 i) = × −2 x × + = dx 2 1 − x2 x 1 − x2 1 − x2 dy = = dx x 1 − x 2
−x 1 − x2
+
1 x 1 − x2
1 − x2 x
3 3 3 1 1 − x2 dy 4 4 4 = dx ii ) s = dx 1 1 + = + 1 2 ∫14 ∫ ∫14 dx x x2 4 2
3 4 1 4
s=∫
3 1 3 1 dx = [ ln x ]14 = ln − ln = ln ( 3) x 4 4 4
Few candidates were able to supply a correct
1 proof of part (a). The derivative of cosh −1 x
1
was almost invariably given as
2
1 −1 x followed by the correct answer after a small amount of spurious algebra. Candidates fared better in part (b)(i), although it was a little surprising how many candidates were unable to differentiate 1 − x 2 correctly. Those candidates who continued and attempted part (b)(ii) frequently produced a correct solution. Solutions to this part which went awry were 2 those in which candidates wrote 1 + 1 − x x
2
followed by 1 + 1 − x (which incidentally gave x the correct answer!), or in some cases where candidates arrived at
1 −2 or x but were unable to 2 x
take the square root correctly.
Question 8:
a) ( z − e 4
iθ
)( z
4
−e
− iθ
) =z − z e − z e + 1 = z − z (e + e ) +1 8
4 − iθ
8
4
iθ
− iθ
= z 8 − z 4 × 2 cos θ + 1
(z
4
− eiθ )( z 4 − e − iθ ) = z 8 − 2 z 4 cos θ + 1
1 π ( θ = ) , z 8 − 2 z 4 cos θ + 1 becomes = z8 − z 4 + 1 0 2 3 π π −i i We can factorise as z 4 − e 3 z 4 − e 3 = 0
= b) for cosθ
We need to solve z = e 4
±i
π
4
(re ) = e r 4 e 4iφ = e
φ =± This gives z = e c)
11π ±i 12
,e
π 3
π 12 ±i
5π 12
Part (a) was quite well done, although for some reason the multiplication of the brackets sometimes resulted in 8 iθ –iθ 4 z – 2(e +e )z +1 followed by the printed result. In part (b), a number of candidates lost some marks through attempting to use an ‘otherwise’ method instead of the ‘hence’ method as directed.
±i
±i
π 3
π 3
+ k × 2π +
kπ 2
,e
±i
π 12
,e
i
π
Those candidates arriving at z = e 3 usually went on to solve the given equation correctly. The Argand diagram, however, was poorly drawn. Frequently no circle was indicated and roots appeared at different distances from the origin, and in many cases candidates seemed to think that the eight roots were equally spaced round the origin
3
iφ 4
r =1 or 4φ =±
Exam report
4 iθ
±i
9π 12
MFP2 - AQA GCE Mark Scheme 2009 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 2009 January series
MFP2
Q
Solution
1(a)
LHS = 1 +
Marks Total
1 2θ e – 2 + e –2θ ) ( 2
1 = ( e 2θ + e –2θ ) = cosh 2θ 2
(b)
Comments
M1
Expansion of
A1
Any form
A1
3
1 ( θ −θ )2 correctly e −e 2
AG
3 + 6sinh 2 θ = 2sinh θ + 11
M1
3sinh 2θ – sinh θ – 4 = 0
A1
OE
( 3sinhθ – 4 )( sinh θ + 1) = 0
M1
Attempt to factorise or formula
4 or –1 3
A1F
ft if factorises or real roots found
sinh θ =
θ = ln 3 θ = ln
(
A1F
)
2 –1
A1F Total
2(a)
9
B1
Circle
B1
Correct centre
B1
Correct radius
B1F
α=
2 4
Inside shading
Possibly by tangents drawn ft mirror image of circle in x-axis
M1
π 6
Range is
4
B1F
(b) Correct points P1 and P2 indicated sin α =
6
A1 π 2π - arg z 3 3
A1 Total
4 8
4
Deduct 1 for angles in degrees
MFP2 - AQA GCE Mark Scheme 2009 January series
MFP2 (cont)
Q
Solution
3(a)
Marks Total
f ( r ) − f ( r − 1) 1 1 2 2 = r 2 ( r + 1) – ( r – 1) r 2 4 4 1 2 2 = r ( r + 2r + 1 – r 2 + 2r – 1) 4 = r3
(b)
Comments
M1
A1
1 2 1 2 2 n ( n + 1) – ( n – 1) n 2 4 4
M1 A1
1 1 2 2 2 2 ( 2n ) ( 2n + 1) – ( 2n – 1) ( 2n ) 4 4
A1
r = n : n3 =
Correct expansions of (r + 1)2 and (r – 1)2
A1 3
AG
For either r = n or r = 2n. PI
r = 2n :
( 2n ) 2n
∑r r =n
3
3
=
1 1 2 2 = .4n 2 ( 2n +1) – ( n – 1) n 2 4 4 3 = n 2 ( 5n + 1)( n + 1) 4
M1 A1
5
AG Alternatively 2n
∑r
3
and
r =1
n−1
∑r
3
stated
Difference Answer Total
( ∑α ) = ∑α 1 = – 5 + 2∑ αβ ∑αβ = 3
4(a) Use of
(b)
(c)
(d)
(e)
2
2
+ 2∑ αβ
M1A1A1
r =1
(M1 for either) M1 A1
8
M1 A1
A1
3
1( –5 – 3) = – 23 – 3αβγ αβγ = – 5
M1 A1
2
z 3 – z 2 + 3z + 5 = 0
M1 A1F
2
B1 B1
2
α 2 + β 2 + γ 2 < 0 ⇒ non real roots Coefficients real ∴ conjugate pair f ( –1) = 0 ⇒ z + 1 is a factor
For use of identity
M1A1
( z + 1) ( z 2 – 2 z + 5) = 0
A1
z = –1, 1 ± 2i
A1 Total
5
AG
4 13
For correct signs and “= 0”
MFP2 - AQA GCE Mark Scheme 2009 January series
MFP2 (cont) Q Solution 5(a) du = 2 cosh x sinh x dx = sinh 2 x x =1 du I = ∫ (b) x =0 1 + u 2
Marks
M1 A1
x =1
1
= tan –1
2
AG Ignore limits here
A1
= ⎡⎣ tan –1 ( cosh 2 x ) ⎤⎦ 0 = tan
Comments
Any correct method
M1A1
= ⎡⎣ tan –1u ⎤⎦ x=0
–1
Total
A1
( cosh 1) – tan ( cosh 0) ( cosh 1) – π4 2
–1
Or A1 for change of limits
2
2
A1 Total
5
AG
7
6 Assume result true for n = k
Then =
k +1
2r × r
SC If no series at all indicated on LHS, deduct 1 and give E0 at end
∑ r = 1 ( r + 1)( r + 2 )
2k +1 ( k + 1) 2k + 1 + –1 k + 2 ( k + 2 )( k + 3)
M1A1
=
2k + 1 ( k + 3 + k + 1) –1 ( k + 2 )( k + 3)
M1
=
2k + 1 2 ( k + 2 ) –1 ( k + 2 )( k + 3)
A1
=
2k + 2 –1 k +3
A1
1 22 k = 1: LHS = , RHS = –1 3 3 Pk ⇒ Pk +1 and P1 true
Putting over common denominator (not including the –1, unless separated later)
B1 E1 Total
6
7 7
Must be completely correct
MFP2 - AQA GCE Mark Scheme 2009 January series
MFP2 (cont) Q Solution 7(a) d ⎛ 1 ⎛ 1 ⎞ –1 1 ⎞ ⎜ cosh ⎟= ⎜– 2 ⎟ dx ⎝ x⎠ 1 ⎝ x ⎠ −1 2 x −1 = x 1 – x2 (b)(i)
d dx
)
(
1 – x2 =
Marks
Total
dy = f(y) and no attempt to dx substitute back to x M0 if
M1A1 A1
Comments
3
AG
– 2x
B1
For numerator
2 1 – x2 1
B1
2 For denominator (not (1 − x 2 ) )
1 – x2 x
M1
1
dy –x = + dx x 1 – x2 1 – x2 1 – x2
=
x 1 – x2
(ii)
s=
∫
3 4
1– x 1+ 2 x
1 4
= [ ln x ] = ln
=
2
dx =
A1
∫
3 4 1 4
1 dx x
M1 A1 Total
e
±
πi 12
, e
±
− πi 3
M1
2 kπi 7πi 12
–πi
or e 12 , e
±
5πi 12
, e
+
±
2 kπi
AG
Clearly shown SC If ‘hence’ not used and, say, z 8 − z 4 +1 = 0 is solved by formula, lose π M1A1, but then continue M1m1 etc if 3 is obtained
A1
πi
4
2
M1
z 4 = e 3 or e +
5 12
M1 A1
2cosθ = 1 π θ= 3 πi
2
A1A1
3 4 1 4
3 1 – ln = ln 3 4 4
z = e12
⎛ dy ⎞ For use of 1+ ⎜ ⎟ ⎝ dx ⎠
M1
8(a) Correct multiplication of brackets eiθ + e –iθ = 2cosθ (b)
4
For attempt to put over a common denominator AG
m1
4 11πi 12
A2, 1, 0F
6
A1 if 3 roots correct
(c)
B2,1,0
Indication that r = 1
B1 Total TOTAL
7
B1 for 4 roots indicated correctly on a circle. CAO
3 11 75
