mark schemes for fp1 practice papers a to g practice paper ... · pdf filemark schemes for fp1...
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MARK SCHEMES FOR FP1 PRACTICE PAPERS A TO G
Practice paper A mark scheme
Question
numberScheme Marks
1. (a) AB = 2 1 3 1 10 0
4 3 4 2 0 10
M1
= 10I, c = 10 A1 (2)
(b) A–1 = 3 114 210
M1 A1
(2)(4
marks)2. f(2) = 1
91 f(3) = 2627 B1
f(2.5) = 0.06415… 2.5 < < 3 M1 A1f(2.75) = –0.45125… 2.5 < < 2.75 A1
(4
marks)3. (a) f(k + 1) = (2k + 3)7k+1 – 1 B1
f(k + 1) – f(k) = (2k + 3)7k+1 – 1 – [(2k + 1)7 k –
1] = (12k + 20)7k a = 12, b =
20
M1 A1
(3)(b) f(1) = 3 7 – 1 = 20 ; divisible by 4 B1
f(k + 1) – f(k) = 4 (3k + 5)7k M1 true for n = k + 1 if true for n = k A1Conclusion, with no wrong working seen. A1 (4)
(7
marks)4. (a) f (x) = 3x2 + 1 M1 A1
(2)(b) f(1.2) = –0.072 f (x) = 5.32 B1
= 1.2 – 0.0725.32
= 1.21353… = 1.21 (3 sf) M1 A1
= 1.21 (3 sig figs) A1 cso
(4) (6
marks)
GCE Further Pure Mathematics FP1 practice paper A mark scheme 1
Question
numberScheme Marks
5. (a) Second root = 3 – i B1Product of roots = (3 + i)(3 – i) = 10
or quadratic factor is x2 – 6x + 10 M1 A1Complete method for third root or linear factor M1Third root = 1
2 A1 (5)
(b)Use candidate’s 3 roots to find cubic with real
coefficientsM1
(x2 – 6x + 10)(2x – 1) = 2x3 – 13x + 26x – 10Equating coefficients M1 a = –13, b = 26 A1 (3)
(8
marks)
6. (a)0
0
k
k
B1 (1)
(b)0 1
1 0
M1 A1
(2)
(c)3 0 0 1 0 3
0 3 1 0 3 0
M1 A1
(2)
(d)0 3 2 3
3 0 3 6
a b
b a
M1
3b = 5a + 2, –3a – 6 = 2 – b M1Eliminate a or b M1a = –5.5, b = –8.5 A1 A1 (5)
(10
marks)7. (a) w = (2 + 2i)(1 + 3i) M1
(2 2 3) (2 3 2) iA1, A1
(3)
(b)arg w =
2 3 2arctan
2 2 3
or
adds two
M1
args e.g. O O60 45
2 GCE Further Pure Mathematics FP1 practice paper A mark scheme
Question
numberScheme Marks
= 712
or 105° or 1.83 radians A1 (2)
(c) 32w ( = 4 2)M1 A1
(2)
(d)
ft in quadrant other than first
B1 B1 (2)
(e)2
4 32 16 2cos45AB (=20), then square root M1
AB = 2 5 A1 (2)
(e) Alternative: w – z = 1 2 3 (2 3)i 2 2(1 2 3) (2 3)AB w z M1
= 20 2 5 A1 (2)(11
marks)8. (a) a = 3 B1 (1)
(b)1 1 1 12 2 2 2d
2 dy
y a x a xx
M1
y = and attempt ddyx
d 1dyx t
sub x = 23t M1
Tangent is 216 3y t x t
t M1
23ty x t () A1 cso
(4)(c) Equation of tangent at Q is 23qy x q B1
At R, y = 0 0 = x + 3q2 M1 x = –3q2 A1 (3)
(d) Equation of directrix is x = –3 B1
GCE Further Pure Mathematics FP1 practice paper A mark scheme 3
Im z
Re z
A
B
O
Question
numberScheme Marks
RD = 3q2 – 3 M1 3q2 – 3 = 12 q2 = 5 M1 q = 5 A1 (4)
(12
marks)
9. (a)If n = 1, 2
1
n
r
r
= 1, 1 1
( 1)(2 1) 1 2 36 6
n n n =
1
B1
true for n = 11
2 2
1
1( 1)(2 1) ( 1)
6
k
r
r k k k k
M1 A1
= 1
( 1)[ (2 1) 6( 1)]6
k k k k M1
= 21( 1)(2 7 6)
6k k k
=1
( 1)( 2)(2 3)6
k k k
= 1
( 1)([ 1] 1)(2[ 1] 1)6
k k k A1
true for n = k + 1 if true for n = k, true for n ℤ+ by induction A1 cso
(6)
(b)Expand brackets and attempt to use
appropriate formulae.M1
2
1
1 1( 6 5) ( 1)(2 1) 6 ( 1) 5
6 2
n
r
r r n n n n n n
M1 A1
= 2[2 3 1 18 18 30]6n
n n n A1
= 2[2 21 49]6n
n n A1
= ( 7)(2 7)6n
n n () A1 cso
(5)
(c)Use S(40) – S(9) =
40 947 87 16 25
6 6 , = 26
660
M1, A1
(2)(13
marks)
4 GCE Further Pure Mathematics FP1 practice paper A mark scheme
Further Pure Mathematics FP1 (6667)
Practice paper B mark scheme
Question
numberScheme Marks
1. (a) det A = q(q – 1) + 6M1 A1
(2)(b) 2 2 31
2 46 ( ) 5q q q M1 A1
> 0 for all real q A is non-
singular ()
A1 cso
(3)(b) Alternative: If A is singular, det A = 0
q2 – q + 6 = 0 ‘b2 – 4ac’ = 1 – 24 < 0
no real roots M1 A1
A is non-singular
A1 (3)
(5
marks)
2. (a) 2 26 ( 8)zw
M1
= 10 A1 (2)
(b)22 4i 6 8i6 8i 6 8i
w
M1
= 100 200i
100
= 1 + 2i A1 A1 (3)
(c)4
arg arctan22
z M1
= 0.18
A1 (2)
(7
marks)
3. (a) 2
1 1 1 1
( 1)( 2) 2n n n n
r r r r
r r r r
M1
1 11 2 1 1 2
6 2n n n n n n
[A1 for –2n, A1 for rest]
A1 A1
212 6 8
6n n n
Use factor n and use common denominator.
(e.g.3, 6, 12) M1
21 13 4 1 4
3 3n n n n n n ()
Attempt complete factorisation
A1 cso
(5)
GCE Further Pure Mathematics FP1 practice paper B mark scheme 1
Question
numberScheme Marks
(b) Use S(20) – S(4) = 1 1
19 20 24 3 4 83 3
M1
= 3008 A1 (2)
(7
marks)4. (a) f(0.5) = –0.75, f(0.6) = 0.36 B1 B1
0.5 0.36 0.6 0.750.36 0.75
M1
= 0.568 (3sf) A1 (4)
(b) f (x) = 2
32x
x M1 A1
f(0.55) = –0.1520… f (0.55) = 11.0173… B1
= 0.55 – 0.1520...
11.0173...
M1
= 0.564 (3sf) A1 (5)
(9
marks)5. (a) (i) 2 – i B1
(ii) (2 – i)2 + b(2 + i) + c = 0 o.e. M1Imaginary parts b = – 4 B1Real parts c + 3 + 2b = 0 M1 c = 5 A1 (5)
(b) (2 + i)3 = 2 + 11i B1 = + m(3 + 4i) + n(2 + i) – 5 = 0 M1Real parts 3m + 2n = 3,
Imaginary parts 8m + 2n = –22M1
m = –5, n = 9 A1 A1 (5)
(10
marks)6. (a) A–1AB = A–1C M1
B = A–1C ()A1 cso
(2)(b) det C = 2 B1 (1)(c) Area of T2 = det C× 1
2 ×5×3 = 2× 12 ×5×3 M1 M1
= 15 A1 (3)
(c) Alternative 1 1 0 5 0 0 5 3
1 1 0 0 3 0 5 3
M1
Area = 12 50 18= 15
M1 A1
(3)
2 GCE Further Pure Mathematics FP1 practice paper B mark scheme
Question
numberScheme Marks
(d) det A = 1 A–1 =
1 1
2 21 1
2 2
M1 A1
B =
1 11 1 2 02 2
1 1 1 1 0 22 2
M1 A1
(4)(e) Enlargement centre (0, 0), scale factor 2 B1 B1
(2)
(12
marks)
7. (a)
4y
x 2
d 4dyx x
y = and
attempt ddyx
M1
2
d 1dyx t
substitute x = 2t M1
Gradient of normal = t2 M1
Equation of normal is y – 2t
= t2(x – 2t) M1
y = t2x + 2t
2t3
()
A1 cso
(5)
(b) At A 2t = –4, t = –2 B1Equation of normal is y = 4x + 15 M1
At B 4x + 15 = 4x
M1 A1
4x2 + 15x – 4 = 0 (4x – 1)(x + 4) = 0
At B x = 14
M1 A1
y = 16 A1 (7)
(12
marks)
8. (a) f(k + 1) – f(k) M1
GCE Further Pure Mathematics FP1 practice paper B mark scheme 3
Question
numberScheme Marks
(i) = k3 + 3k2 + 3k + 1 – 10k – 10 + 15 – (k3 – 10k
+ 15)
= 3k2 + 3k – 9A2, 1, 0
(3)(b) f(1) = 6 = 3×2 true for n = 1 B1
f(k + 1) – f(k) = 3k2 + 3k – 9 = 3(k2 + k – 3) M1 A1true for n = k + 1 if true for n = k, true for n ℤ+ by induction
A1 cso(4)
(ii) When n = 1, LHS = 1(2)1 = 2; RHS = 2{1 +
0} = 2
true for n = 1
B1
11
1
2 2{1 ( 1)2 } ( 1)2k
r k kr k k
M1 A1
= 2 + k 2k + 1 – 2k + 1 + k 2k + 1 + 2k + 1
= 2(1 + k 2k + 1) = 12{1 [( 1) 1]2 }kk M1 A1true for n = k + 1 if true for n = k, true for n ℤ+ by induction A1 cso
(6)
(12marks)
4 GCE Further Pure Mathematics FP1 practice paper B mark scheme
[P4 January 2002 Qn 1]
[P4 January 2002 Qn 3]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 2
2.
MARK SCHEME FOR PRACTICE PAPERS C TO G.
[P4 January 2002 Qn 4]
5. 6r2 – 6 = n(n + 1)(2n + 1) , – 6n M1, A1 = n(2n2 + 3n – 5) M1 = n(n – 1)(2n – 5) (*) A1
(4
marks)[P4 June 2002 Qn 1]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 3
4.
[*P4 January 2002 Qn 5]
3.
6.
(a)|w| = 50 (or equivalent) B1 (1)
(b)
B ×
× AB1 (1)
(c)5|| OA B1
3
4BA 5|| BA , isosceles
M1, A1
52 + 52 = (50)2, right-angled (or gradient method) M1, A1 (5)
(d)wz
wz
argargarg
M1
= (–)AOB = 4
M1, A1 (3)
(10 marks)[P4 June 2002 Qn 5]
7. 2
4dd
xxy
; at x = 2p 2
1dd
pxy
M1, A1
Equation of tangent at P, )2(12
2px
ppy M1
pxypp
xp
y 4 , 41
( 22
etc)
At Q q2y + x + 4q Two correct
equations in any formA1
(p2 – q2)y = 4(p – q) M1
qpy
4 () A1
qp
pqqp
ppx
444
2
() M1, A1 (8)
[*P5 June 2002 Qn 7]
8. For n = 1 25 + 52 = 57, which is divisible by 3 M1, A1
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 4
(a)
Assume true for n = k (k + 1)th term is 23k +5 + 5k+2 B1
(k + 1)th term kth term = 23k +5 + 5k+2 23k +2 + 5k+1 M1
= 23k +2(23 1) + 5k+1(5 1) M1, A1
= 6(23k +2 + 5k+1) + 3. 23k +2 or = 4(23k +2 + 5k+1) + 3. 23k +2 M1
which is divisible by 3 (k + 1)th term is divisible by 3 A1
Thus by induction true for all n csoB1
(9) (
b)For n = 1 RHS =
49
12B1
Assume true for n = k
412999
13232
139
31
49
12
49
121
kkk
kkk
kk
kkk
M1 A3/2/1/0
(-1 each
error)
1)1(3)1(9
)1()1(31
kk
kk
B1
If true for k then true for k + 1 by induction true
for all n
B1
(7)(16 marks)
[P6 June 2002 Qn 6]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 5
3.
9. f(2) = 1.514
B1f() = 1.142 B1
514.1142.1
2
M1
1.514 + 2 1.142 = (1.142 + 1.514) = 2.65 A1 (4)
[*P4 January 2003 Qn 4]
10. (a)
z2 = (3 – 3i)(3 – 3i) = 18i M1 A1
(2)
(b) z1
= )i33)(i33()i33(
=
18i33
= 6
i1 M1 A1
(2)(c) z = (9 + 9) = 18 = 32
z = 18
two correctM1
z1
= 181
= 23
1 =
62
all three correct
A1
(2)
(d) C
D O
two correct A
four correct
B
B1
B1 (2)
(e) 6/2
23,18
OCOA
ODOB
= 18 M1 A1
AOB = COD = 45 similarB1
(3)(11
marks)[P4 January 2003 Qn 6]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 6
2 1.142
1.514
11. (a) x3 – 27 = (x – 3)(x2 + 3x + 9) M1
(x = 3 is one root). Others satisfy (x2 + 3x + 9) = 0 (*)A1
(2) (b) Roots are x = 3 B1
and x = 2
3693 M1
= i233
23
, i233
23
A1
(3) (c)
3
3 and one other
root in correct
quad
Root in complex
conjugate posn.
B1
B1 ft
(2)
(7
marks)[#P4 June 2003 Qn 3]
12. (a) f(1) = –4 f(2) = 1 M1
Change of sign (and continuity) implies (1, 2)A1
(2) (b) f(1.5) = –2.3… B1
f(1.75) = –0.9…
f(1.875) = –0.03…
B1
(2)NB Exact answer is 1.8789…
Alt to (a) y y = 3x
98 y = x + 67
3
O 1 2 x
Two graphs with single point of intersection (x > 0)
Two calculations at both x = 1 and x = 2
M1
A1
[*P4 June 2003 Qn 4]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 7
13.
(a) i42i34
=
20)i42)(i34(
= 20
i1020 ( = 1 – 2
1 i ) M1
| z | = 2212 )(1 , =
25 awrt 1.12, accept
exact equivalents
M1, A1
(3)
(b) )i2)(i2()i2)(i3(
aaaa
= 2
2
4
i)6(5
a
aa
accept in (a) if clearly
applied to (b)
M1
)(tan4 1 = aa
56 2
obtaining
quadratic or equivalent
M1 A1
a2 + 5a – 6 = (a + 6)(a – 1) a = –6, 1 M1 A1
Reject a = –6, wrong quadrant/– 43 , one value
A1
(6)(9
marks)Alt.
(a) | 4 + 3i | = 5, | 2 + 4i | = 20 M1
| z | = 20
5 ( =
25 )
M1 A1
(3)(b) arg z = arg (a + 3i) – arg (2 + ai)
2arctan
3arctan
4a
a
M1
1 =
23
1
23
a
a M1 A1
(3)
leading to a2 + 5a – 6 = 0, then as before[P4 June 2003 Qn 5]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 8
[P6 June 2003 Qn 2]
15.
(a)| z | = 22 | w | = 2 M1, A1
| wz2 | = (22)2 × 2 = 16 M1, A1
arg z = – 4 arg w = 6
5 ; arg wz2 = – 4 – 4
+ 65 = 3
, 60°
M1, A1
(6)
ALT z2 = –8i; z2w = 8 + 83i M1, A1| z2w | = 388 22 M1 = 16 A1arg z2w = tan–13 M1
= 3 A1
(b) C
B
A
Points A and B B1Point C B1ft
angle BOC = 65 – 3
M1
= 2 , 90° A1 (4)
(10
marks)[P4 January 2004 Qn 3]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 9
14.
[P4 June 2004 Qn 1]
17. f 1 1 and f 2 2 B1
13
2 21
1 1
B1
(2)
[*P4 June 2004 Qn 2]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 10
16.
18.
[P4 June 2004 Qn 3]
[P4 June 2004 Qn 5]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 11
19.
20.
2
2
d 1d
dy cy dt t
dxx c tdt
The normal to the curve has gradient 2t .
The equation of the normal is 2( )c
y t x ctt
The equation may be written 2 3cy t x ct
t
M1 A1
B1
M1A1 (5)
[*P5 June 2004 Qn 8]
21. (a) 2 222 2 2 2 16z , 22 21 3 4w M1
4
22
zzw w
M1 A1 (3)
(b) 2 2 2 2i 2 2 2 2i 1 i 3
1 i 3 1 i 3 1 i 3zw
M1
21 3 i 3 1
4
3 1
arctan 153 1
M1
In correct quadrant arg 165zw
1112
A1 (3)
(c) A
B1
B
B1
C ft
B1ft (3)
(d) 60DOB B1 Correct method for AOC M1
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 12
A
CB
A
B
C
45 15 60AOC A1 (3)
(e) 12 4 2 sin60 2 3AOC !
awrt 3.46M1 A1 (2)
[#FP1/P4 January 2005 Qn 8]
22. (a) 3 26 20 2 2 10z z z z z
Long division or any complete method
M1
2 4 40
1 3i2
z
M1 A1 (3)
(b)
In correct quadrants and on negative
B1ft (1)
x- axis only required
(c) 3
1, 13AB ACm m
Full methodM1
1AB ACm m triangle is right angledA1 (2) (6)
Alternative to (c) 2 2 218 18 36AB AC BC Result follows by (converse of) Pythagoras, or any complete method.
M1 A1
[#FP1/P4 June 2005 Qn 2]
23. f 1.2 0.2937 … f 1.2 to
1sf or betterB1
f 1.1 0.42 ... , f 1.15 2.05 … Attempt at f 1.1 , M1
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 13
f 1.15
1.2 A1 (3)
[*FP1/P4 June 2005 Qn 4]
[FP1/P4 June 2005 Qn 5]
[*FP2/P5 June 2005 Qn 5]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 14
24.
25.
[FP3/P6 June 2005 Qn 1]
[FP1/P4 January 2006 Qn 1]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 15
26.
27.
[FP1/P4 January 2006 Qn 3]
29. (a) f 1.8 19.6686 ... 20 0.3313 … awrt 0.33
B1
f 2 20.6424 ... 0.6424 … awrt 0.64
B1
1.8 2
"0.33" "0.64"
or equivalentM1
1.87 cao
A1 (4)
(b) 112 (min) (1 hr 52 min)B1 (1) (5)
[#*FP1/P4 January 2006 Qn 5]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 16
28.
[*FP2/P5 Jan 2006 Qn 9]
[*FP3/P6 Jan 2006 Qn 3]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 17
30.
31.
33 . (a) 2 i 1z w i i 3i 3z w Adding 2 i 4 3iz z Eliminating
either variableM1
4 3i2 i
z
A1
4 3i 2 i2 i 2 i
z
M1
8 3 4i 6i
5
= 1 2i A1
(4)(b) argz arctan 2 M1 2.03
cao
A1
(2) (6
marks)[FP1 June 2006 Qn 1]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 18
32.
[*FP3/P6 January 2006 Qn 5]
34.(a) f 0.24 0.058, f 0.28 0.089 accept
1sf M1
Change of sign (and continuity) 0.24,0.28 A1 (2)
(b) f 0.26 0.017 0.24,0.26 accept
1sfM1
f 0.25 0.020 0.25,0.26
f 0.255 0.001 0.255,0.26 M1 A1 (3)
(5)[*FP1 June 2006 Qn 6]
[FP1 Jan 2007 Qn 1]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 19
35.
[FP1 Jan 2007 Qn 3]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 20
36.
[FP1 June 2007 Qn 4]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 21
37.
[FP1 June 2007 Qn 6]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 22
38.
39. (a) 2 2
2 2 2Gradient of
ap aqPQ
ap aq p q Can be
impliedB1
Use of any correct method or formula to obtain an equation of PQ in any form.
M1
Leading to 2p q y x apq A1 (3)
(b) Gradient of normal at P is p . Given or implied at any stage
B1
Obtaining any correct form for normal at either point. Allow if just written down.
M1 A1
32y px ap ap 32y qx aq aq Using both normal equations and eliminating x or y. Allow in any unsimplified form.
M1
)()(2)( 33 qpaqpaxqp Any correct formfor x or y
A1
Leading to )2( 22 pqqpax cso
A1
)( qpapqy cso
A1 (7)
[*FP2 June 2007 Qn8]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 23
40. 1n : 2 11 1 1 3
3 B1
(Hence result is true for 1n .)
1
22 2
1 1
2 1 2 1 2 1k k
r r
r r k
212 1 2 1 2 1
3k k k k , by
induction hypothesis
M1
212 1 2 6 3
3k k k k
212 1 2 5 3
3k k k
12 1 2 3 1
3k k k M1 A1
]1)1(2][1)1(2)[1(31
kkk
(Hence, if result is true for n k , then it is true
for 1.n k )By Mathematical Induction, above implies the result is true
for all .n ¢
cso
A1
(5)(5
marks) [FP3 June 2007 Qn5]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 24
41. (a) 4 4 4 6 4 4 2f 1 f 3 2 3 2k k k kk k
4 4 4 2 43 3 1 2 2 1k k M1
4 4 23 80 2 15k k can be
impliedA1
4 1 4 2 4 1 4 23 240 2 15 15 16 3 2k k k k M1
Hence 15|f 1 fk k
cso
A1
(4)
Note: f 1 fk k is divisible by 240 and other appropriate multiples of 15 lead to the required result.
(b) 1n : 4 6f 1 3 2 145 5 29 5|f 1 B1 (Hence result is true for 1n .)From (a) f 1 f 15 ,k k say. By induction hypothesis
f 5 ,k say.
f 1 f 15 5 3 5|f 1k k k M1 (Hence, if result is true for n k , then it is true
for 1.n k )By Mathematical Induction, above implies the result is true for
all .n ¢ Accept equivalent arguments
cso
A1
(3)(7
marks)[*FP3 June 2007 Qn6]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 25
[FP1 January 2008 Qn 2]
43. f 0.7 0.195028497, f 0.8 0.297206781x 3 dp or
betterB1, B1
Using
f 0.80.8
0.7 f 0.7
to obtain
0.8f 0.7 0.7f 0.8
f 0.8 f 0.7
M1
0.739620991 3dp or better
A1 (4)
[*FP1 January 2008 Qn 4]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 26
42.
[FP1 January 2008 Qn 6]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 27
44.
[FP1 June 2008 Qn 1]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 28
45.
[FP1 June 2008 Qn 3]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 29
46.
[FP3 June 2008 Qn 5]
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 30
47.
48.
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 31
49.
FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 32