Mark Scheme (Results) January 2011
GCE
GCE Further Pure Mathematics FP1 (6667) Paper 1
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January 2011 Publications Code UA026332 All the material in this publication is copyright Š Edexcel Ltd 2011
General Instructions for Marking 1. The total number of marks for the paper is 75. 2. The Edexcel Mathematics mark schemes use the following types of marks: •
M marks: method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated.
•
A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.
•
B marks are unconditional accuracy marks (independent of M marks)
•
Marks should not be subdivided.
3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. •
bod – benefit of doubt
•
ft – follow through
•
the symbol
•
cao – correct answer only
•
cso - correct solution only. There must be no errors in this part of the question to obtain this mark
•
isw – ignore subsequent working
•
awrt – answers which round to
•
SC: special case
•
oe – or equivalent (and appropriate)
•
dep – dependent
•
indep – independent
•
dp decimal places
•
sf significant figures
•
¿ The answer is printed on the paper
•
will be used for correct ft
The second mark is dependent on gaining the first mark
January 2011 Further Pure Mathematics FP1 6667 Mark Scheme Question Number 1. (a)
Scheme
Marks
z = 5 − 3i, w = 2 + 2i z 2 = ( 5 − 3i )( 5 − 3i )
= 25 − 15i − 15i + 9i 2 = 25 − 15i − 15i − 9
An attempt to multiply out the brackets to give four terms (or four M1 terms implied). zw is M0
= 16 − 30i
16 − 30i
A1 (2)
Answer only 2/2 (b)
z ( 5 − 3i ) = w ( 2 + 2i )
=
( 5 − 3i ) ( 2 − 2i ) × ( 2 + 2i ) ( 2 − 2i )
=
10 − 10i − 6i − 6 4+4
=
4 − 16i 8
=
1 − 2i 2
Multiplies
z by w
( 2 − 2i ) ( 2 − 2i )
Simplifies realising that a real number is needed on the denominator and applies i 2 = − 1 on M1 their numerator expression and denominator expression.
1 1 − 2i or a = and b = −2 or 2 2 A1 equivalent Answer as a single fraction A0
GCE Further Pure Mathematics FP1 (6667) January 2011
M1
1
(3) [5]
Question Number 2. (a)
Scheme
Marks
⎛ 2 0⎞ ⎛ −3 −1⎞ A=⎜ ⎟,B=⎜ ⎟ ⎝ 5 3⎠ ⎝5 2⎠ ⎛ 2 0 ⎞⎛ −3 −1⎞ AB = ⎜ ⎟⎜ ⎟ ⎝ 5 3 ⎠⎝ 5 2 ⎠
A correct method to multiply out two matrices. Can be implied by M1 two out of four correct elements.
⎛ 2(−3) + 0(5) 2(−1) + 0(2) ⎞ = ⎜ ⎟ ⎝ 5(−3) + 3(5) 5(−1) + 3(2) ⎠ ⎛ −6 − 2⎞ =⎜ ⎟ 1 ⎠ ⎝ 0
Any three elements correct A1 Correct answer A1 Correct answer only 3/3
(3)
Reflection M1 y-axis (or x = 0 .) A1
(b) Reflection; about the y-axis.
(2) (c)
⎛1 0⎞ C100 = I = ⎜ ⎟ ⎝0 1⎠
⎛1 0⎞ ⎜ ⎟ or I B1 ⎝0 1⎠ (1) [6]
GCE Further Pure Mathematics FP1 (6667) January 2011
2
Question Number 3. (a)
Scheme 3
f ( x) = 5 x 2 − 4 x 2 − 6, x f (1.6) = − 1.29543081... f (1.8) = 0.5401863372...
α − 1.6 "1.29543081..."
=
Marks
0
awrt -1.30 B1 awrt 0.54 B1 Correct linear interpolation method with signs correct. Can be implied M1 by working below.
1.8 − α "0.5401863372..."
⎛
⎞ "1.29543081..." ⎟ 0.2 ⎝ "0.5401863372..." + "1.29543081..." ⎠
α = 1.6 + ⎜
= 1.741143899...
awrt 1.741 A1 (4) Correct answer seen 4/4 At least one of ± a x or ± b x M1 correct. Correct differentiation. A1 1 2
(b)
f ′( x) = 10 x − 6 x 2 1
(2) (c)
f (1.7) = − 0.4161152711... f ′(1.7) = 9.176957114...
f (1.7) = awrt − 0.42 f ′(1.7) = awrt 9.18
B1 B1
Correct application of NewtonRaphson formula using their M1 values.
⎛ "− 0.4161152711..." ⎞ ⎟ ⎝ "9.176957114..." ⎠
α 2 = 1.7 − ⎜
= 1.745343491... = 1.745 (3dp)
1.745 A1 cao (4) Correct answer seen 4/4
[10]
GCE Further Pure Mathematics FP1 (6667) January 2011
1
Question Number 4. (a)
Scheme
Marks
z 2 + p z + q = 0, z1 = 2 − 4i z2 = 2 + 4i
2 + 4i
B1 (1)
(b)
( z − 2 + 4i)(z − 2 − 4i) = 0 ⇒ z 2 − 2 z − 4iz − 2 z + 4 − 8i + 4iz − 8i + 16 = 0
⇒ z 2 − 4 z + 20 = 0
An attempt to multiply out brackets M1 of two complex factors and no i2. Any one of p = − 4, q = 20. A1 Both p = − 4, q = 20. A1 2 (3) ⇒ z − 4 z + 20 = 0 only 3/3 [4]
GCE Further Pure Mathematics FP1 (6667) January 2011
1
Question Number
Scheme
Marks
n
∑ r (r + 1)(r + 5)
5
r =1
(a)
=
n
∑r
3
+ 6 r 2 + 5r
r =1
(b)
Multiplying out brackets and an attempt to use at least one of the M1 standard formulae correctly.
=
1 2 1 1 n (n + 1) 2 + 6. n(n + 1)(2n + 1) + 5. n(n + 1) 4 6 2
=
1 2 5 n (n + 1) 2 + n(n + 1)(2n + 1) + n(n + 1) 4 2
=
1 n( n + 1) ( n(n + 1) + 4(2n + 1) + 10 ) 4
=
1 n(n + 1) ( n 2 + n + 8n + 4 + 10 ) 4
=
1 n(n + 1) ( n 2 + 9n + 14 ) 4
Correct 3 term quadratic factor A1
=
1 n(n + 1)(n + 2)(n + 7) * 4
Correct proof. No errors seen. A1
Sn =
Correct expression. A1
Factorising out at least n(n + 1)
dM1
(5)
50
∑ r (r + 1)(r + 5)
r = 20
= S50 − S19 =
1 4
(50)(51)(52)(57) − 14 (19)(20)(21)(26)
Use of S50 − S19
M1
1837 680
A1
= 1889550 − 51870 = 1837 680
Correct answer only 2/2
GCE Further Pure Mathematics FP1 (6667) January 2011
2
(2) [7]
Question Number 6. (a)
Scheme
C : y 2 = 36 x ⇒ a =
36 4
Marks
=9
S (9, 0)
(9, 0)
B1 (1)
(b)
(c)
x + 9 = 0 or x = − 9
x + 9 = 0 or x = − 9
or ft using their a from part (a).
B1 (1)
Either 25 by itself or PQ = 25 . Do not award if just PS = 25 is B1 seen.
PS = 25 ⇒ QP = 25
x = 16
(d) x-coordinate of P ⇒ x = 25 − 9 = 16
(1)
B1
Substitutes their x-coordinate into M1 equation of C.
y 2 = 36(16)
y = 24
y = 576 = 24
A1 (3)
Therefore P (16, 24) (e) Area OSPQ = 12 (9 + 25)24
1 2
( their a + 25)( their y )
M1
or rectangle and 2 distinct triangles, correct for their values. 408 A1
= 408 (units) 2
(2) [8]
GCE Further Pure Mathematics FP1 (6667) January 2011
3
Question Number 7.
Scheme
Marks
Im
(a)
-24
Re
Correct quadrant with (−24, − 7) B1 indicated.
-7
(1) (b)
arg z = − π + tan −1 ( 247 )
tan −1 ( 247 ) or tan −1 ( 247 )
= − 2.857798544... = − 2.86 (2 dp)
M1
awrt -2.86 or awrt 3.43 A1 (2)
(c)
w = 4 ,arg w =
⇒ r = 4 ,θ =
5π 6
5π 6
w = r cosθ + i r sin θ
Attempt to apply r cosθ + i r sin θ . M1 Correct expression for w. A1
w = 4cos ( 56π ) + 4i sin ( 56π ) =4
( ) + 4i ( − 3 2
1 2
)
= − 2 3 + 2i
either −2 3 + 2i or awrt − 3.5 + 2i
A1 (3)
a = − 2 3, b = 2 z = 25 or (d)
zw = (48 3 + 14) + (14 3 − 48) i or B1 awrt 97.1-23.8i
z = (−24) 2 + (−7) 2 = 25
zw = z × w = (25)(4)
Applies z × w or zw
= 100
M1
100 A1 (3) [9]
GCE Further Pure Mathematics FP1 (6667) January 2011
4
Question Number 8. (a)
Scheme
Marks
⎛ 2 −2 ⎞ A=⎜ ⎟ ⎝ −1 3 ⎠ det A = 2(3) − (−1)(−2) = 6 − 2 = 4
4 B1 (1)
(b)
(c)
A
−1
1 ⎛3 2⎞ ⎜ ⎟ det A ⎝ 1 2 ⎠ 1 ⎛3 2⎞ ⎜ ⎟ 4 ⎝1 2⎠
1 ⎛3 2⎞ = ⎜ ⎟ 4 ⎝1 2⎠
72 or 72 ( their det A ) their det A
72 = 18 (units)2 Area( R ) = 4
M1 A1 (2) M1
18 or ft answer. A1 (2)
(d)
AR = S ⇒ A −1 AR = A −1 S ⇒ R = A −1 S R=
=
1 ⎛ 3 2 ⎞ ⎛ 0 8 12 ⎞ ⎜ ⎟⎜ ⎟ 4 ⎝ 1 2 ⎠ ⎝ 4 16 4 ⎠
At least one attempt to apply A −1 by M1 any of the three vertices in S .
1 ⎛ 8 56 44 ⎞ ⎜ ⎟ 4 ⎝ 8 40 20 ⎠
⎛ 2 14 11⎞ =⎜ ⎟ ⎝ 2 10 5 ⎠
At least one correct column o.e. A1 At least two correct columns o.e.
A1
Vertices are (2, 2), (14, 10) and (11, 5). All three coordinates correct. A1
GCE Further Pure Mathematics FP1 (6667) January 2011
5
(4) [9]
Question Number 9.
Scheme
Marks
un +1 = 4un + 2 , u1 = 2 and un = 23 (4n − 1)
Check that un = 23 (4n − 1)
n = 1; u1 = 23 (41 − 1) = 23 (3) = 2 So un is true when n = 1.
yields 2 when n = 1.
B1
Assume that for n = k that, uk = 23 (4k − 1) is true for k ∈ Z + . Then uk +1 = 4uk + 2 Substituting uk = 23 (4k − 1) into
= 4 ( 23 (4k − 1) ) + 2 =
8 3
( 4)
=
2 3
( 4 )( 4 )
k
−
= 23 4k +1 −
8 3
k
un +1 = 4un + 2.
An attempt to multiply out the M1 brackets by 4 or 83
+2
−
M1
2 3
2 3
= 23 (4k +1 − 1)
2 3
Therefore, the general statement, un = 23 (4n − 1) is true when n = k + 1. (As un is true for n = 1, ) then un is true for all positive integers by mathematical induction
GCE Further Pure Mathematics FP1 (6667) January 2011
6
(4k +1 − 1)
A1
Require ‘True when n=1’, ‘Assume true when n=k’ and ‘True when A1 n = k + 1 ’ then true for all n o.e. (5) [5]
Question Number 10. (a)
Scheme
xy = 36 at ( 6t ,
6 t
Marks
). An attempt at
36 dy 36 = 36 x −1 ⇒ = − 36 x −2 = − 2 y= x x dx At ( 6t ,
6 t
),
So, mT =
T: T: T: T:
dy . dx
dy dx or and dt dt
dy 36 =− dx (6t ) 2
An attempt at
dy 1 =− 2 t dx
M1
dy . in terms of t M1 dx dy 1 =− 2 * A1 t dx
Must see working to award here
6 1 y − = − 2 ( x − 6t ) t t 6 1 6 y− = − 2x+ t t t 1 6 6 y = − 2x+ + t t t 1 12 y = − 2x+ * t t
Applies y −
6 = their mT ( x − 6t ) t
M1
Correct solution . A1 cso (5)
(b) Both T meet at ( −9, 12 ) gives
1 12 ( −9) + 2 t t 9 12 12 = 2 + (× t 2 ) t t 12t 2 = 9 + 12t 12 = −
Substituting (-9,12) into T. M1
An attempt to form a “3 term M1 quadratic”
12t 2 − 12t − 9 = 0
4t 2 − 4t − 3 = 0 (2t − 3)(2t + 1) = 0 t= t=
3 2
3 2
t=−
An attempt to factorise. M1
, − 12
t=
⇒ x = 6 ( 32 ) = 9 , y =
1 2
6
( ) 3 2
= 4 ⇒ ( 9, 4 )
⇒ x = 6 ( − 12 ) = − 3 , y=
6 = − 12 ⇒ ( −3, − 12 ) ( − 12 )
3 2
, − 12
A1
An attempt to substitute either their M1 t = 32 or their t = − 12 into x and y. At least one of A1 ( 9, 4 ) or ( −3, − 12 ) . Both ( 9, 4 ) and ( −3, − 12 ) .
A1 (7) [12]
GCE Further Pure Mathematics FP1 (6667) January 2011
7
Other Possible Solutions Question Number 4. (a) (i) Aliter (ii) Way 2
Scheme
Marks
z 2 + p z + q = 0, z1 = 2 − 4i
2 + 4i
z2 = 2 + 4i
B1
No i2. Attempt Sum and Product M1 of roots or Sum and discriminant
Product of roots = (2 − 4i)(2 + 4i) = 4 + 16 = 20
or b 2 − 4ac = (8 i)2 Sum of roots = (2 − 4i) + (2 + 4i) = 4
= z 2 − 4 z + 20 = 0
Any one of p = − 4, q = 20.
A1
Both p = − 4, q = 20.
A1 (4)
4.
z 2 + p z + q = 0, z1 = 2 − 4i
(a) (i) Aliter (ii) Way 3
z2 = 2 + 4i
2 + 4i
B1
An attempt to substitute either z1 or z2 into z 2 + p z + q = 0 M1 and no i2.
(2 − 4i) + p (2 − 4i) + q = 0 2
−12 − 16i + p (2 − 4i) + q = 0
Imaginary part: −16 − 4 p = 0 Real part: −12 + 2 p + q = 0 4 p = − 16 ⇒ p = − 4 q = 12 − 2 p ⇒ q = 12 − 2(−4) = 20
Any one of p = − 4, q = 20. A1 Both p = − 4, q = 20. A1 (4)
GCE Further Pure Mathematics FP1 (6667) January 2011
8
Question Number Aliter 7. (c) Way 2
Scheme w = 4 ,arg w =
5π 6
Marks
and w = a + i b
Attempts to write down an equation in terms of a and b for either the M1 modulus or the argument of w.
w = 4 ⇒ a 2 + b 2 = 16
arg w =
5π 6
⇒ arctan ( ba ) =
5π 6
⇒
b a
=−
1 3
Either a 2 + b 2 = 16 or
b a
=−
1 3
A1
a = − 3 b ⇒ a 2 = 3b 2
So, 3b 2 + b 2 = 16 ⇒ b 2 = 4 ⇒ b = ± 2 and a = ∓ 2 3
As w is in the second quadrant w = − 2 3 + 2i
either −2 3 + 2i or awrt − 3.5 + 2i
A1 (3)
a = − 2 3, b = 2
GCE Further Pure Mathematics FP1 (6667) January 2011
9
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