Edexcel GCE Final
Further Pure Maths Unit no. 6674/01 June 2006
Further Pure Maths 6674/01
Edexcel GCE
Mark Scheme (Results)
General Instructions 1.
The total number of marks for the paper is 75.
2.
Method (M) marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated.
3.
Accuracy (A) marks can only be awarded if the relevant method (M) marks have been earned.
4.
(B) marks are independent of method marks.
5.
Method marks should not be subdivided.
6.
For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. Indicate this action by ‘MR’ in the body of the script (but see also note 10).
7.
If a candidate makes more than one attempt at any question: (a) If all but one attempt is crossed out, mark the attempt which is NOT crossed out. (b) If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt.
8.
Marks for each question, or part of a question, must appear in the right-hand margin and, in addition, total marks for each question, even where zero, must be ringed and appear in the right-hand margin and on the grid on the front of the answer book. It is important that a check is made to ensure that the totals in the right-hand margin of the ringed marks and of the unringed marks are equal. The total mark for the paper must be put on the top right-hand corner of the front cover of the answer book.
9.
For methods of solution not in the mark scheme, allocate the available M and A marks in as closely equivalent a way as possible, and indicate this by the letters ‘OS’ (outside scheme) put alongside in the body of the script.
10.
All A marks are ‘correct answer only’ (c.a.o.) unless shown, for example, as A1 f.t. to indicate that previous wrong working is to be followed through. In the body of the script the symbol should be used for correct f.t. and for incorrect f.t. After a misread, however, the subsequent A marks affected are treated as A f.t., but manifestly absurd answers should never be awarded A marks.
11.
Ignore wrong working or incorrect statements following a correct answer.
June 2006 6674 Further Pure Mathematics FP1 Mark Scheme Question Number 1.
Scheme (a) Adding
(b)
2.
Marks
2 z + i w = −1 i z − i w = 3i− 3 2 z + i z = −4 + 3 i Eliminating either variable −4 + 3i z= 2+i −4 + 3i 2 − i z= × 2+i 2−i −8 + 3 + 4i + 6i = 5 = −1 + 2 i arg z = π − arctan 2 ≈ 2.03
Use of π
1 2
∫r
2
cao
dθ
A1 M1
A1 M1 A1
(4)
(2) [6]
B1
π
Limits are 8 and 4
B1
16a 2 cos 2 2θ = 8a 2 (1 + cos 4θ )
∫ (1 + cos 4θ ) dθ = θ +
M1
M1
sin 4θ 4
M1 A1
π
sin 4θ ⎤ 4 ⎡ A = 4a ⎢θ + 4 ⎥⎦π ⎣ 8 2
⎡ ⎛π π ⎞ ⎤ = a 2 ⎢ 4 ⎜ − ⎟ + ( 0 − 1) ⎥ ⎣ ⎝4 8⎠ ⎦ ⎛π ⎞ 1 = a 2 ⎜ − 1⎟ = a 2 (π − 2 ) ¿ ⎝2 ⎠ 2
M1 cso
A1
(7) [7]
June 2006 6674 Further Pure Mathematics FP1 Mark Scheme
Question Number 3.
Scheme y ′ = 3sin 2 x + 6 x cos 2 x y ′′ = 12 cos 2 x − 12 x sin 2 x 12cos 2 x − 12 x sin 2 x + 12 x sin 2 x = k cos 2 x k = 12
M1 A1 M1 A1
General solution is y = A cos 2 x + B sin 2 x + 3 x sin 2 x
B1
(a) Substituting
(b)
Marks
( 0, 2 )
⇒
A=2
(4)
B1
3π π ⎛π π ⎞ ⎜ 4, 2 ⎟ ⇒ 2 = B+ 4 ⎝ ⎠
⇒
B=−
π
y = 2 cos 2 x − sin 2 x + 3 x sin 2 x 4
π
M1
4
Needs y = … A1
(4) [8]
4.
(a)
3 + 2i is a solution ( x − 3 − 2i )( x − 3 + 2i ) = x 2 − 6 x + 13
f ( x ) = ( x 2 − 6 x + 13)( x 2 + ax + b )
b=6 a − 6 = −6 Coefficients of x or equivalent a=0 2 x + 6 = 0 ⇒ x = √ 6i, − √ 6i 3
B1 M1 B1 M1 A1 M1A1 (7)
ℑ
(b)
O
ℜ
Conjugate complex pair on imaginary axis B1 Conjugate complex B1 pair in correct quadrants
(2) [9]
June 2006 6674 Further Pure Mathematics FP1 Mark Scheme Question Number 5.
Scheme
Marks
(2r + 1)3 = 8r 3 + 12r 2 + 6r + 1 (2r − 1)3 = 8r 3 − 12r 2 + 6r − 1
(a)
( A = 24, B = 2 )
(2r + 1)3 − (2r − 1)3 = 24r 2 + 2 Accept r = 0 ⇒
B = 2 and r = 1 ⇒
M1 A1 (2)
A + B = 26 ⇒ A = 24 M1 for both
33 − 13 = 24 ×12 + 2
(b)
53 − 33 = 24 × 22 + 2 M
( 2n + 1)
3
− ( 2n − 1) = 24 × n 2 + 2 3
( 2n + 1)
3
n
− 13 = 24∑ r 2 + 2n
ft their B
M1 A1 A1ft
r =1
8n3 + 12n 2 + 4n 24 1 1 = n ( 2n 2 + 3n + 1) = n ( n + 1)( 2n + 1) ¿ 6 6
n
∑ r2 = r =1
40
(c)
40
∑ (3r − 1) = ∑ (9r 2
r =1
2
− 6r + 1)
M1 cso A1
(5)
M1
r =1
1 1 = 9 × × 40 × 41× 81 − 6 × × 40 × 41 + 40 6 2 = 194 380
M1 A1
(3) [10]
June 2006 6674 Further Pure Mathematics FP1 Mark Scheme Question Number 6.
Scheme (a)
Marks
f ( 0.24 ) ≈ −0.058, f ( 0.28 ) = 0.089
accept 1sf M1
Change of sign (and continuity) ⇒ α ∈ ( 0.24, 0.28 )
(b)
( ⇒ α ∈ ( 0.24, 0.26 ) ) f ( 0.25) ≈ −0.020 ( ⇒ α ∈ ( 0.25, 0.26 ) )
f ( 0.26 ) ≈ 0.017
A1
accept 1sf
M1
f ( 0.255 ) ≈ −0.001 ⇒ α ∈ ( 0.255, 0.26 ) (c)
f(11) ≈ 0.0534
2 cos √ x 1 + 4 √x f ′ (11) ≈ −0.3438
M1 A1 (3) at least 3sf B1
f ′( x) =
β ≈ 11 +
M1 A1 at least 2sf A1
0.0534 ≈ 11.16 0.3438
If f ′ (11) ≈ −0.3438 is produced without working, this is to be accepted for three marks M1 A1 A1.
(2)
cao
M1 A1 (6) [11]
June 2006 6674 Further Pure Mathematics FP1 Mark Scheme Question Number 7.
Scheme (a)
Marks
2 x 2 + x − 6 = 6 − 3x x2 + 2 x − 6 = 0
Leading to 2 ( x + 1) = 7 ⇒
M1
x = −1 ± √ 7
surds required
−2 x 2 − x + 6 = 6 − 3 x Leading to 2 x 2 − 2 x = 0 ⇒ x = 0, 1
M1 A1 M1 A1, A1 (6)
(b) Accept if parts (a) and (b) done in reverse order y
Curved shape B1 Line B1 At least 3 intersections B1
O
(3)
x
(c) Using all 4 CVs and getting all into inequalities x > √ 7 −1 , x < − √ 7 −1 both ft their greatest positive and their least negative CVs 0 < x <1
M1 A1ft A1
(3) [12]
June 2006 6674 Further Pure Mathematics FP1 Mark Scheme Question Number 8.
Scheme
∫
(a)
1
(120 − t )
2
Marks
2 dt = −2 ln (120 − t ) 120 − t
B1
e −2ln (120−t ) = (120 − t )
M1 A1
dS 2S 1 + = 3 2 d t (120 − t ) 4 (120 − t )
⎞ d⎛ S 1 ⎜ ⎟= 2 dt ⎜⎝ (120 − t ) ⎟⎠ 4 (120 − t )2 S
(120 − t ) ( 0, 6 )
−2
2
=
or integral equivalent
1 ( +C ) 4 (120 − t )
⇒ 6 = 30 + 1202 C ⇒ C = −
120 − t (120 − t ) S= − 4 600
Substituting
S =9
3 8
(kg)
M1 A1 1 600
M1
2
accept C = awrt −0.0017
dS 1 2 (120 − t ) =− + dt 4 600 dS = 0 ⇒ t = 45 dt
(b)
M1
A1
(8)
M1 M1 A1 A1
(4) [12]
June 2006 6674 Further Pure Mathematics FP1 Mark Scheme
Question Number 8.Contd.
Scheme
Marks
Alternative forms for S are ( t + 30 )(120 − t ) 3t t2 − = S = 6+ 20 600 600 3600 + 90t − t 2 5625 − ( t − 45 ) = = 600 600
2
Alternative for part (b) S can be found without finding t dS = 0 in the original differential equation dt 2S 1 = 120 − t 4 Substituting for t into the answer to part (a) 64 S 2 S = 2S − 600
Using
Solving to
S = 9 83 (kg)
M1
M1 A1 A1
(4)