June 2009 6668 Further Pure Mathematics FP2 (new) Mark Scheme Question Number Q1
(a)
Scheme
1 1 1 = − r (r + 2) 2r 2( r + 2)
Marks
1 1 − 2r 2(r + 2)
B1 aef (1)
n
(b)
∑
4 = r (r + 2) r =1
n
⎛2
2 ⎞
∑ ⎜⎝ r − r + 2 ⎟⎠ r =1
⎛2 2⎞ ⎛2 2⎞ = ⎜ − ⎟ + ⎜ − ⎟ + ...... ⎝1 3⎠ ⎝2 4⎠ ⎛ 2 2 ⎞ ⎛2 2 ⎞ .............. + ⎜ − ⎟ + ⎜ − ⎟ ⎜ n −1 n +1 ⎟ ⎜n n + 2 ⎟ ⎝ ⎠ ⎝ ⎠
=
Includes the first two underlined terms and includes the final two M1 underlined terms. 2 2 2 2 + − − A1 1 2 n +1 n + 2
2 2 2 2 + ;− − 1 2 n +1 n + 2
= 3−
List the first two terms M1 and the last two terms
2 2 − n +1 n + 2
=
3( n + 1)(n + 2) − 2(n + 2) − 2(n + 1) (n + 1)(n + 2)
=
3n 2 + 9n + 6 − 2n − 4 − 2n − 2 (n + 1)(n + 2)
=
3n 2 + 5n (n + 1)(n + 2)
=
n (3n + 5) (n + 1)(n + 2)
Attempt to combine to an at least 3 term fraction to a single fraction and an attempt to take M1 out the brackets from their numerator.
Correct Result A1 cso AG (5) [6]
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Question Number Q2
(a)
Scheme
Marks
z3 = 4 2 − 4 2 i , − π < θ „ π y
O α
4 2
x
arg z 4 2
( 4 2 , − 4 2)
r=
(4 2 ) + (− 4 2 ) 2
θ = − tan −1
( )=− 4 2
π
4 2
4
2
=
32 + 32 =
64 = 8
A valid attempt to find the modulus and argument of M1 4 2 − 4 2 i.
z 3 = 8 ( cos ( − π4 ) + isin ( − π4 ) ) 1 ⎛ ⎛−π So, z = ( 8 ) 3 ⎜ cos ⎜ 4 ⎝ 3 ⎝
Taking the cube root of the modulus and dividing the M1 argument by 3.
⎞ ⎛ − π4 ⎞ ⎞ isin + ⎟ ⎜ 3 ⎟⎟ ⎠ ⎝ ⎠⎠
⇒ z = 2 ( cos ( − 12π ) + i sin ( − 12π ) )
2 ( cos ( − 12π ) + isin ( − 12π ) )
Also, z 3 = 8 ( cos ( 74π ) + isin ( 74π ) )
Adding or subtracting 2π to the argument for z 3 in order to find M1 other roots.
or
z 3 = 8 ( cos ( − 94π ) + isin ( − 94π ) )
⇒ z = 2 ( cos 712π + isin 712π )
Any one of the final two roots A1
( ( ) + isin ( ))
and z = 2 cos
− 3π 4
− 3π 4
Both of the final two roots. A1
Special Case 1: Award SC: M1M1A1M1A0A0 for ALL three of 2 ( cos 12 + isin 12 ) , π
π
( ( ) + isin ( )) .
2 ( cos 34π + isin 34π ) and 2 cos
− 7π 12
− 7π 12
Special Case 2: If r is incorrect (and not equal to 8) and candidate states the brackets ( ) correctly then give the first accuracy mark ONLY where this is applicable.
8371_9374 GCE Mathematics January 2009
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[6]
Question Number Q3
Scheme
sin x
Marks
dy − y cos x = sin 2 x sin x dx An attempt to divide every term in the differential equation by M1 sin x . Can be implied.
dy y cos x sin 2 x sin x − = dx sin x sin x dy y cos x − = sin 2 x dx sin x
cos x
Integrating factor = e
=
cos x
∫ − sin x dx
= e
e
− ln sin x
1 sin x
∫ ± sin x ( dx )
± their P( x ) ( dx ) or e ∫ e − ln sin x or eln cosec x
1 or (sin x) −1 or cosec x sin x
dM1 A1 aef A1 aef
sin 2 x ⎛ 1 ⎞ dy y cos x − = ⎜ ⎟ 2 sin x ⎝ sin x ⎠ dx sin x
d dx
d ( y × their I.F.) = sin 2 x × their I.F M1 dx
⎛ y ⎞ 1 ⎜ ⎟ = sin 2 x × sin x ⎝ sin x ⎠
d dx
d dx
⎛ y ⎞ ⎜ ⎟ = 2cos x ⎝ sin x ⎠
y = sin x
⎛ y ⎞ ⎜ ⎟ = 2cos x or ⎝ sin x ⎠ A1 y = 2cos x ( dx ) sin x
∫
∫ 2cos x dx
y = 2sin x + K sin x
A credible attempt to integrate dddM1 the RHS with/without + K
y = 2sin 2 x + K sin x
y = 2sin 2 x + K sin x
A1 cao [8]
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Question Number
Scheme
Q4
1 A= 2
Marks
1 Applies 2
2π
∫ ( a + 3cosθ ) dθ 2
2π
∫ r ( d θ ) with 2
0
B1
correct limits. Ignore dθ .
0
(a + 3cosθ ) 2 = a 2 + 6a cosθ + 9cos 2 θ ± 1 ± cos 2θ M1 2 Correct underlined expression. A1
cos 2 θ =
⎛ 1 + cos 2θ ⎞ = a + 6a cosθ + 9 ⎜ ⎟ 2 ⎝ ⎠ 2
1 A= 2
2π
∫ 0
9 9 ⎛ 2 ⎞ ⎜ a + 6a cos θ + + cos 2θ ⎟ dθ 2 2 ⎝ ⎠ Integrated expression with at least 3 out of 4 terms of the form ± Aθ ± B sin θ ± Cθ ± D sin 2θ . M1* Ignore the 12 . Ignore limits.
2π
9 9 ⎛1⎞ ⎡ ⎤ = ⎜ ⎟ ⎢ a 2θ + 6a sin θ + θ + sin 2θ ⎥ 2 2 4 ⎝ ⎠⎣ ⎦0
=
a 2θ + 6a sin θ + correct ft integration. A1 ft 1 Ignore the 2 . Ignore limits.
1 ⎡( 2π a 2 + 0 + 9π + 0 ) − ( 0 ) ⎤ ⎦ 2⎣
= π a2 +
Hence,
9π 2
π a2 +
π a2 + 9π 107 = π 2 2
a2 +
Integrated expression equal to
9π 2
107 2
A1
π . dM1*
9 107 = 2 2 a 2 = 49 a=7
As a > 0, a = 7
A1 cso [8]
Some candidates may achieve a = 7 from incorrect working. Such candidates will not get full marks
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Question Number
Scheme
Marks
y = sec 2 x = (sec x) 2
Q5 (a)
Either 2(sec x)1 (sec x tan x)
dy = 2(sec x)1 (sec x tan x) = 2sec 2 x tan x dx Apply product rule: ⎧ u = 2sec 2 x ⎪ ⎨ du 2 ⎪ = 4sec x tan x ⎩ dx
or 2sec 2 x tan x
B1 aef
v = tan x ⎫ ⎪ dv 2 ⎬ = sec x ⎪ dx ⎭
Two terms added with one of either A sec 2 x tan 2 x or B sec 4 x M1 in the correct form.
2
d y = 4sec 2 x tan 2 x + 2sec4 x dx 2
Correct differentiation A1 = 4sec 2 x(sec 2 x − 1) + 2sec 4 x
Applies tan 2 x = sec 2 x − 1 leading to the correct result. A1 AG
d2 y = 6sec 4 x − 4sec 2 x dx 2
Hence,
(4) (b)
yπ = 4
( 2)
2
⎛ dy ⎞ = 2, ⎜ ⎟ = 2 ⎝ dx ⎠ π4
⎛d y⎞ ⎜ 2⎟ = 6 ⎝ dx ⎠ π 2
( 2)
4
−4
( 2)
2
( 2 ) (1) = 4
⎛ dy ⎞ Both y π = 2 and ⎜ ⎟ = 4 4 ⎝ dx ⎠ π4
2
B1
Attempts to substitute x = π4 into both terms in the expression for M1 d2 y . dx 2
= 24 − 8 = 16
4
Two terms differentiated with either 24sec 4 x tan x or M1 − 8sec2 x tan x being correct
d3 y = 24sec3 x(sec x tan x) − 8sec x(sec x tan x) 3 dx
= 24sec 4 x tan x − 8sec 2 x tan x ⎛ d2 y ⎞ ⎜ 2 ⎟ = 24 ⎝ dx ⎠ π4
( 2 ) (1) − 8 ( 2 ) (1) = 96 − 16 = 80
sec x ≈ 2 + 4 ( x −
4
2
⎛ d3 y ⎞ ⎜ 3 ⎟ = 80 ⎝ dx ⎠ π
B1
4
π
{sec x ≈ 2 + 4 ( x −
4
) + 162 ( x − π4 ) π
4
2
) + 8( x − 4 )
π 2
+
80 6
(x − )
+
40 3
(x − 4 )
π 3 4
π 3
Applies a Taylor expansion with at least 3 out of 4 terms ft M1 correctly. Correct Taylor series expansion. A1
+ ...
}
(6)
+ ...
[10]
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Question Number Q6
Scheme
w= (a)
z , z = −i z+i Complete method of rearranging M1 to make z the subject. iw z= A1 aef (1 − w)
w( z + i) = z ⇒ wz + i w = z ⇒ i w = z − wz iw ⇒ i w = z (1 − w) ⇒ z = (1 − w) iw =3 1− w
z =3 ⇒
Putting z in terms of their w = 3
⎧⎪ i w = 3 1 − w ⇒ w = 3 w − 1 ⇒ w 2 = 9 w − 1 2 ⎨ 2 2 ⎪⎩ ⇒ u + i v = 9 u + i v − 1
2 2 2 2 ⎪⎧ ⇒ u + v = 9u − 18u + 9 + 9v ⎨ 2 2 ⎪⎩ ⇒ 0 = 8u − 18u + 8v + 9
⇒ 0 = u 2 − 94 u + v 2 + ⇒ (u −
9 2 8
)
−
⇒ (u −
9 2 8
)
+ v2 =
81 64
{Circle} centre
Simplifies down to dddM1 u 2 + v 2 ± α u ± β v ± δ = 0. 9 8
=0
9 64
( 98 , 0 ) , radius
One of centre or radius correct. A1 Both centre and radius correct. A1
3 8
(8)
Circle indicated on the Argand diagram in the correct position B1ft in follow through quadrants. Ignore plotted coordinates.
v
O
⎫⎪ ⎬ ⎪⎭
⎫⎪ ⎬ ⎪⎭
9 8
+ v2 +
dM1
Applies w = u + i v , and uses Pythagoras correctly to get an ddM1 equation in terms of u and v without any i 's. Correct equation. A1
⇒ u 2 + v 2 = 9 ⎡⎣ (u − 1) 2 + v 2 ⎤⎦
(b)
Marks
Region outside a circle indicated B1 only.
u
(2) [10]
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Question Number
Scheme
Marks
y = x2 − a2 , a > 1
Q7 (a)
y a2
−a
O
Correct Shape. Ignore cusps. B1 Correct coordinates. B1 a
x
(2) (b)
x − a = a − x , a >1 2
2
2
{ x > a},
x2 − a2 = a2 − x
x2 − a2 = a2 − x
M1 aef
⇒ x 2 + x − 2a 2 = 0 −1 ±
⇒ x=
−1 ±
⇒ x=
{ x < a},
Applies the quadratic formula or completes the square in order to M1 find the roots.
1 − 4(1)(−2a 2 ) 2
Both correct A1 “simplified down” solutions.
1 + 8a 2 2
− x 2 + a 2 = a 2 − x or
− x2 + a2 = a2 − x
{⇒ x
2
x2 − a2 = x − a2
M1 aef
− x = 0 ⇒ x( x − 1) = 0} x=0 x =1
⇒ x = 0 ,1
B1 A1 (6)
(c)
x2 − a2 > a2 − x , a > 1
x<
−1 −
{or}
1 + 8a 2 2
{or} x >
−1 +
x is less than their least value B1 ft x is greater than their maximum B1 ft value
1 + 8a 2 2
For { x < a} , Lowest < x < Highest
0 < x <1
0 < x <1
M1 A1 (4) [12]
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Question Number Q8
Scheme
Marks
d2 x dx dx + 5 + 6 x = 2e − t , x = 0, = 2 at t = 0. 2 dt dt dt (a)
AE , m 2 + 5m + 6 = 0 ⇒ (m + 3)(m + 2) = 0 ⇒ m = − 3, − 2. So, xCF = Ae
−3 t
+ Be
Ae m1 t + Be m2 t , where m1 ≠ m2 .
−2 t
Ae
−3 t
+ Be
−2 t
M1 A1
⎧ ⎫ dx d2 x −t = − k e−t ⇒ 2 = k e− t ⎬ ⎨ x = ke ⇒ dt dt ⎩ ⎭
⇒ k e − t + 5(− k e − t ) + 6 k e − t = 2e − t ⇒ 2 k e − t = 2e− t ⇒ k =1
{ So,
Substitutes k e − t into the differential equation given in the M1 question. Finds k = 1. A1
xPI = e − t }
their xCF + their xPI
So, x = Ae−3t + Be −2t + e − t Finds
dx = = − 3 Ae −3t − 2 Be −2t − e − t dt
t = 0, x = 0 ⇒ dx t = 0, =2 ⇒ dt
M1*
dx by differentiating dM1* dt their xCF and their xPI
Applies t = 0, x = 0 to x dx dx = 2 to to ddM1* and t = 0, dt dt form simultaneous equations.
0 = A+ B +1 2 = − 3 A − 2B − 1
⎧ 2 A + 2 B = − 2⎫ ⎨ ⎬ ⎩−3 A − 2 B = 3 ⎭ ⇒ A = − 1, B = 0
So, x = − e−3t + e− t
x = − e −3t + e− t
A1 cao (8)
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Question Number
Scheme
Marks
x = − e −3t + e− t (b)
Differentiates their x to give
dx = 3e − 3t − e − t = 0 dt
dx and puts equal to 0. dt
So, x = − e x = −3 =−
− 32
+e
− 12 ln 3
1 2
= −e
ln 3
−3 2
+e
−1 2
ln 3
Substitutes their t back into x and an attempt to eliminate out the ln’s. ddM1
− 12
+3
1 3 3
1 2 2 3 = = 9 3 3 3
+
uses exact values to give
d2 x = − 9e − 3t + e− t dt 2
At t = 12 ln 3, = − 9(3)
As
− 32
− 12
=−
2 3 9
d2 x dt 2 d2 x and substitutes their t into 2 dt
A1 AG
Finds
d2 x − 3 ln 3 − 1 ln 3 = − 9e 2 + e 2 2 dt
+3
M1
A credible attempt to solve. dM1* t = ln 3 or t = ln 3 or awrt 0.55 A1
3 − e 2t = 0 ⇒ t = 12 ln 3 − 32 ln 3
dx dt
9 3 3
+
dM1*
1 3 1 = − + 3 3 3
d2 x 9 1 ⎧ 2 ⎫ = − + = ⎨− ⎬ <0 2 dt 3 3 3 ⎩ 3⎭ then x is maximum.
−
9 3 3
+
1 < 0 and maximum 3 conclusion.
A1 (7) [15]
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