MARK SCHEMES FOR FP1 PRACTICE PAPERS A TO G Practice paper ... · PDF fileMARK SCHEMES FOR FP1 PRACTICE PAPERS A TO G Practice paper A mark scheme ... FP2, FP3 papers – Version 2

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]


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.


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]


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


(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]


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]


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]


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]


[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]


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]


16.

18.

[P4 June 2004 Qn 3]

[P4 June 2004 Qn 5]


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


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


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]


24.

25.

[FP3/P6 June 2005 Qn 1]

[FP1/P4 January 2006 Qn 1]


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]


28.

[*FP2/P5 Jan 2006 Qn 9]

[*FP3/P6 Jan 2006 Qn 3]


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]


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]


35.

[FP1 Jan 2007 Qn 3]


36.

[FP1 June 2007 Qn 4]


37.



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]


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]


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 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]


42.

[FP1 January 2008 Qn 6]


44.



45.



46.



47.

48.


49.


50.


MARK SCHEMES FOR FP1 PRACTICE PAPERS A TO G Practice paper ... · PDF fileMARK SCHEMES FOR FP1 PRACTICE PAPERS A TO G Practice paper A mark scheme ... FP2, FP3 papers – Version 2

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