Mark Scheme (Results) January 2010
GCE
Further Pure Mathematics FP1 (6674)
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January 2010 6674 Further Pure Mathematics FP1 Mark Scheme Question Number Q1
Scheme
(a)
a + ib 3 + i (3a − b) + i (a + 3b) × = 3−i 3+i 10
Marks
(A1 for numerator, A1 for 10)
(b) z1 = a 2 + (−2a ) 2 = 5a 2 = a √ 5 (c) arg
(*) 7π ⎛ ⎞ , or − 45°, or 315° ⎟ ⎜ or 4 ⎝ ⎠
z1 π a + 3b = arctan (-1), = − = arctan 3a − b 4 z2
M1 A1 A1
(3)
M1 A1
(2)
M1 A1ft, A1 (3) [8]
(c) The final A1 requires a single answer, so for example: 3π π arctan (-1) = − or is A0 4 4
GCE Further Pure Mathematics FP1 (6674) January 2010
3
Question Number Q2
Scheme
(= 0.1677...)
(a) f (2) = 2 cos 2 − 4 + 5
(= −0.2601...)
f (2.1) = 2.1 cos 2.1 − 4.2 + 5 Values correct (to 1 s.f.).
Change of sign ⇒ Root
(b) f ′( x) = cos x − x sin x − 2 f ( x0 ) 0.1677 = 2− , x1 = x0 − f ′( x0 ) − 4.2347
(c) f (2.035) = ...... and
Marks
= 2.04
M1 A1, A1
(5)
M1
Change of sign ⇒ Correct to 2 d.p.
(c) The M1 is also given for evaluating f at the ends of a 'tighter' interval.
GCE Further Pure Mathematics FP1 (6674) January 2010
(2)
M1 A1
f (2.045) = ......
0.0189... and − 0.0238...
M1 A1
4
A1
(2) [9]
Question Number Q3
Scheme
Marks
(a) 5 − 2 i is a root (b)
B1
(x − (5 + 2 i) )(x − (5 − 2 i) ) = x 2 − 10 x + 29 x 3 − 12 x 2 + cx + d = (x 2 − 10 x + 29)( x − 2)
M1
c = 49,
A1, A1
d = −58 Conjugate pair in 1st and 4th quadrants, (symmetrical about real axis).
4 y
(c)
3
(1)
M1 M1
(5)
B1
2
1
x −1
1
2
3
4
5
Fully correct, labelled.
B1
(2)
6
−1
−2
−3
−4
[8]
(b) 1st M: Form brackets using (x − α )( x − β ) and expand. 2nd M: Achieve a 3-term quadratic with no i's. (b) Alternative: Substitute a root (usually 5 + 2 i ) and expand brackets
M1
(5 + 2i) 3 − 12(5 + 2i) 2 + c(5 + 2i) + d = 0 (125 + 150i − 60 − 8i) − 12(25 + 20i − 4) + (5c + 2ci) + d = 0 (2nd M for achieving an expression with no powers of i) Equate real and imaginary parts c = 49, d = −58
GCE Further Pure Mathematics FP1 (6674) January 2010
5
M1 M1 A1, A1
Question Number Q4
Scheme
Marks B1
m 2 + 6m + 9 = 0 m = −3 C.F. (x = ) ( At + B )e −3t P.I. x = p cos t + q sin t
M1 A1 B1
dx d2x = − p cos t − q sin t = − p sin t + q cos t dt dt 2 − p cos t − q sin t − 6 p sin t + 6q cos t + 9 p cos t + 9q sin t = 5 cos t − 6 p + 8q = 0 and 8 p + 6q = 5 Solve simultaneously to find either p or q: 2 3 p = and q = 5 10 2 3 General solution: ( x =) ( At + B )e −3t + cos t + sin t 5 10 The final A1ft is dependent on the 3 preceding M marks (for the P.I.)
GCE Further Pure Mathematics FP1 (6674) January 2010
6
M1 M1 A1 M1 A1 A1ft [10]
Question Number Q5
Scheme
Marks
(a) 1 = A(2r + 5) + B(2r + 1) , and find values of A and B 1 1 − 4(2r + 1) 4(2r + 5) 1⎛1 1⎞ ⎜ − ⎟ 4⎝3 7⎠ 1⎛1 1⎞ r = 2: ⎜ − ⎟ 4⎝5 9⎠ 1⎛ 1 1 ⎞ r = n −1: − ⎜ ⎟ 4 ⎝ 2n − 1 2n + 3 ⎠ 1⎛ 1 1 ⎞ r = n: − ⎜ ⎟ 4 ⎝ 2n + 1 2n + 5 ⎠ 1⎛1 1 1 1 ⎞ Sum: − ⎜ + − ⎟ 4 ⎝ 3 5 2n + 3 2n + 5 ⎠ 1 ⎛ 8(2n + 3)(2n + 5) − 15(2n + 5) − 15(2n + 3) ⎞ ⎟⎟ = ⎜⎜ 4⎝ 15(2n + 3)(2n + 5) ⎠
(b) r = 1 :
1 ⎛ 32n 2 + 128n + 120 − 30n − 75 − 30n − 45) ⎞ ⎟⎟ = ⎜⎜ 4⎝ 15(2n + 3)(2n + 5) ⎠
=
⎞ 1⎛ 32n 2 + 68n ⎟ ⎜⎜ 4 ⎝ 15(2n + 3)(2n + 5) ⎟⎠
=
n(8n + 17) 15(2n + 3)(2n + 5)
(c = 17)
M1 A1 (2) B1ft
M1 A1 M1 A1
A1
(6) [8]
(b) B1ft for one correct difference (ft A and B). 1 M1 A1 M1 A1: The is not needed for these marks, only for the final A1. 4
GCE Further Pure Mathematics FP1 (6674) January 2010
7
Question Number Q6
Scheme
(a) r =
a 2
(b) sin 2 2θ =
sin 2θ =
1 2
2θ =
π 6
or
Marks
5π 6
θ=
π
5π 12 12 ,
1 (1 − cos 4θ ) 2
a2 sin 2 2θ dθ = ......... 2 sin 4θ ⎤ ⎡ ± ⎢θ − 4 ⎥⎦ ⎣
M1 A1, A1
(3)
B1
∫
M1
(Correct integration of ± (1 − cos 4θ ) )
A1
5π
⎛ ⎡ ⎤ 12 a 2 ⎛⎜ 5π ⎛ 3⎞ π 3 ⎞⎟ 3⎞ 2 π ⎟ ⎟− + ⎜ ,+ a = = = − ⎜⎜ − .......... . ......., ⎢ ⎥ ⎟ ⎜ 12 4 ⎜⎝ 12 ⎝ 8 ⎟⎠ 12 8 ⎟⎠ 16 ⎣ ⎦ π 12 ⎠ ⎝
M1, A1, A1
(6) [9]
1 2 r dθ with some integration attempt. 2 2nd M: Correct use of their limits. π ⎛ ⎡ ⎤ 4 ⎞⎟ ⎜ N.B. Other methods are possible, e.g. ⎜ e.g. 2 ⎢ ...........⎥ ⎟ ⎣ ⎦ "π 12" ⎠ ⎝ Slips such as omitting the a or not squaring the a: just the final A mark is lost.
(b) 1st M: Use of
∫
GCE Further Pure Mathematics FP1 (6674) January 2010
8
Question Number Q7
Scheme
(a) 10 + 3x − x 2 = 3 x − 1
Marks
11 ( β )
x 2 = ... or x = ... ,
M1, A1 M1
10 + 3 x − x 2 = 1 − 3 x
x 2 − 6x − 9 = 0
x=
6 ± 72 or equiv. 2 3 − 3 2 or exact equiv. ( α )
A1
(b) 3 − 3 2 < x < 11 (c) Forming inequalities using all their four x values − 11 < x < 3 − 3 2 ,
(±
11 and 3 ± 3 2
11 < x < 3 + 3 2
Answers with decimals (3 s.f. accuracy) are acceptable in (b) and (c). (b) M: Answer including x < β (positive β ) or x > α (negative α ). A1ft requires negative α and positive β .
GCE Further Pure Mathematics FP1 (6674) January 2010
9
)
A1
(5)
M1 A1ft
(2)
M1 B1, B1
(3) [10]
Question Number Q8
Scheme
(a) z =
1 y2
y3 2 y2 − 2 −
Marks
dz 2 dy =− 3 dx y dx
dz + y = 4 xy 3 dx dz + 1 = 4 xy 2 dx
B1 M1
−
1 dz 4x , +1 = z 2 z dx
dz − 2 z = −8 x dx
(*)
− 2 dx (b) Integrating factor e ∫ = e −2 x d ze −2 x = −8 xe −2 x dx or ze − 2 x = −8 xe −2 x dx ⎧ xe −2 x 1 − 2 x ⎫ xe − 2 x dx = ⎨ e dx ⎬ + 2 ⎭ ⎩ −2
(
∫
∫
B1
)
M1
∫
M1 A1
ze −2 x = 4 xe −2 x + 2e −2 x + C , z = 4 x + 2 + Ce 2 x (The second of these M marks is dependent on the first, and both are dependent on the use of an integrating factor). 1 y= (or equiv.) 2x 4 x + 2 + Ce
(c)
dy = 0: dx
y=
y = 4xy 3
M1, A1 (4)
1
(*)
2 x
M1, dM1
A1 (7)
M1 A1 (2) [13]
(b) Alternative for first 6 marks: C.F. z = Ce 2 x
B1
P.I. z = px + q ,
p − 2 px − 2q = −8 x p=4 q=2 z = 4 x + 2 + Ce 2 x
GCE Further Pure Mathematics FP1 (6674) January 2010
10
dz =p dx
M1 M1 M1 A1 M1
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