Mark Scheme (Results) Summer 2010
GCE
Further Pure Mathematics FP2 (6668)
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Summer 2010 Publications Code UA023928 All the material in this publication is copyright Š Edexcel Ltd 2010
June 2010 Further Pure Mathematics FP2 6668 Mark Scheme Scheme
Question Number 1(a)
(b)
1 1 − 3r − 1 3r + 2
n
3
M1 A1 (2)
1 1 1 1 1
1
1
1
∑ (3r − 1)(3r + 2) = 2 − 5 + 5 − 8 + 8 − 11 + ... 3n − 1 − 3n + 2 r =1
=
(c)
Marks
1 1 3n − = 2 3n + 2 2(3n + 2)
*
Sum = f(1000) – f(99) 3000 297 − = 0.00301 6004 598
M1 A1ft
A1 (3)
or 3.01× 10−3
M1 A1 (2) 7
GCE Further Pure Mathematics FP2 (6668) Summer 2010
Scheme
Question Number 2
f ′′(t ) = − x − cos x,
f ′′(0) = -1
dx , f ′′′(0) = −0.5 dt t2 t3 f (t ) = f (0) + tf ′(0) + f ′′(0) + f ′′′(0) + ... 2 3! 2 1 3 = 0.5t − 0.5t − 12 t + ...
f ′′′(t ) = (−1 + sin x)
GCE Further Pure Mathematics FP2 (6668) Summer 2010
Marks
B1 M1A1
M1 A1 5
Scheme
Question Number 3(a)
Marks
( x + 4)( x + 3) 2 − 2( x + 3) = 0 , ( x + 3)( x 2 + 7 x + 10) = 0 so ( x + 2)( x + 3)( x + 5) = 0 or alternative method including calculator
M1
Finds critical values –2 and -5
A1 A1
Establishes x > -2
A1ft
Finds and uses critical value –3 to give –5 < x < -3
M1A1 (6)
(b)
x > -2
B1ft (1) 7
GCE Further Pure Mathematics FP2 (6668) Summer 2010
Scheme
Question Number 4(a)
Modulus = 16
B1
Argument = arctan(− 3) =
(b)
(c)
Marks
z 3 = 163 (cos(
2π 3
M1A1 (3)
2π 2π ) + i sin( ))3 = 163 (cos 2π + isin2π ) =4096 or 163 3 3
1
w = 16 4 (cos(
2π 2π 1 ⎛π ⎞ ⎛π ⎞ ) + i sin( )) 4 = 2(cos ⎜ ⎟ + isin ⎜ ⎟ ) = 3 + i 3 3 ⎝6⎠ ⎝6⎠
OR −1 + 3i OR − 3 − i OR 1 − 3i
(
M1 A1 (2)
)
M1 A1ft
M1A2(1,0) (5) 10
GCE Further Pure Mathematics FP2 (6668) Summer 2010
Scheme
Question Number 5(a)
1.5 + sin 3θ = 2
and ∴θ =
π 18
or
→
sin 3θ = 0.5 ∴ 3θ =
Marks
π ⎛
5π ⎞ ⎜ or ⎟, 6 ⎝ 6 ⎠
5π 18
M1 A1,
A1 (3)
⎡ 18 ⎤ 1 Area = 12 ⎢ ∫ (1.5 + sin 3θ ) 2 dθ ⎥ , - π × 22 ⎢⎣ 18π ⎥⎦ 9 5π
(b)
M1, M1
⎡ 18 ⎤ 1 ⎢ ∫ (2.25 + 3sin 3θ + 12 (1 − cos 6θ ))dθ ⎥ - π × 22 ⎢⎣ 18π ⎥⎦ 9 5π
=
1 2
5π
1 ⎡ ⎤ 18 1 = 12 ⎢(2.25θ − cos 3θ + 12 (θ − sin 6θ )) ⎥ - π × 22 6 9 ⎣ ⎦π
M1
M1 A1
18
=
13 3 5π − 24 36
M1 A1 (7) 10
GCE Further Pure Mathematics FP2 (6668) Summer 2010
Scheme
Question Number 6(a)
Marks
Imaginary Axis Re(z) = 3
0
6
Real axis Vertical Straight line Through 3 on real axis
B1 B1
(2) (b)
These are points where line x = 3 meets the circle centre (3, 4) with radius 5.
M1
The complex numbers are 3 + 9i and 3 – i.
A1 A1 (3)
(c)
z −6 = z ⇒
30 w
−6 =
30 w
M1
∴ 30 − 6 w = 30 ⇒ ∴ 5 − w = 5
M1 A1
This is a circle with Cartesian equation (u − 5) 2 + v 2 = 25
M1 A1 (5) 10
GCE Further Pure Mathematics FP2 (6668) Summer 2010
Scheme
Question Number 7(a)
dy dy dz dy dy dz = 2 z so = . and = 2 z. dx dz dx dz dx dx
Substituting to get 2 z.
(b)
Marks
I.F. = e ∫ ∴
−2 tan xdx
dz dz − 4 z 2 tan x = 2 z and thus − 2 z tan x = 1 dx dx
= e2ln cos x = cos 2 x
d z cos 2 x ) = cos 2 x ∴ z cos 2 x = ∫ cos 2 xdx ( dx
∴ z cos 2 x
=
∫
1 2
(cos 2 x + 1) dx = 14 sin 2x + 12 x +c
M1 M1 A1
∗
M1 A1 (5)
M1 A1
M1
M1 A1
∴ z = tan x + x sec x + c sec x 1 2
1 2
2
2
A1 (6)
(c)
∴ y = ( 12 tan x + 12 x sec 2 x + c sec 2 x) 2
B1ft (1) 12
GCE Further Pure Mathematics FP2 (6668) Summer 2010
Question Number 8(a)
Scheme
Marks
Differentiate twice and obtaining
dy d2 y = λ sin 5 x + 5λ x cos 5 x and 2 = 10λ cos 5 x − 25λ x sin 5 x dx dx Substitute to give λ =
(b)
3 10
M1 A1 (4)
Complementary function is y = A cos 5 x + B sin 5 x or Pe5ix + Qe −5ix So general solution is y = A cos 5 x + B sin 5 x +
(c)
M1 A1
M1 A1
3 x sin 5 x or in exponential form 10
A1ft (3)
y= 0 when x = 0 means A = 0
B1
dy 3 3 dy = 5B cos 5 x + sin 5 x + x cos 5 x and at x = 0 = 5 and so 5 = 5A dx 10 2 dx
M1 M1
So B = 1
A1
So y = sin 5 x +
3 x sin 5 x 10
A1 (5)
(d) "Sinusoidal" through O amplitude becoming larger
B1
Crosses x axis at
π 2π 3π 4π 5
,
5
,
5
,
B1
5 (2)
14
GCE Further Pure Mathematics FP2 (6668) Summer 2010
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