04a further maths - fp2 mark schemes v2

FP2 Mark Schemes from old P4, P5, P6 and FP1, FP2, FP3 papers (back to June 2002)

Please note that the following pages contain mark schemes for questions from past papers.

The standard of the mark schemes is variable, depending on what we still have – many are scanned, some are handwritten and some are typed.

The questions are available on a separate document, originally sent with this one.

FP2 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 1

[P4 January 2002 Qn 2]



1.

2.



3.



4.

5. (x > 0) 2x2 – 5x > 3 or 2x2 – 5x = 3 M1

(2x + 1)(x – 3) , critical values –½ and 3 A1, A1

x > 3 A1 ft

x < 0 2x2 – 5x < 3 M1

Using critical value 0: –½ < x < 0 M1, A1 ft

Alt. or (2x – 5)x2 > 3x M1

or x(2x + 1)(x – 3) > 0 M1, A1

Critical values –½ and 3, x > 3 A1, A1 ft

Using critical value 0, –½ < x < 0 M1, A1 ft

(7 marks)

[P4 June 2002 Qn 4]

6. (a) M1

Int. factor M1, A1

Integrate: M1 , A1

A1

(6)

(b) When y = 0, , M1

2 solutions for this (x = /2, 3/2) A1 (2)

(c) y = 0 at x = 0: C = 0 : M1

Shape

Scales

A1

A1 (3)

(11 marks)

[P4 June 2002 Qn 6]


7. (a) 2m2 + 7m + 3 = 0 (2m + 1)(m + 3) = 0

m = –½, –3

C.F. is M1, A1

P.I. y = at2 + bt + c B1

,

2(2a) + 7(2at + b) + 3(at2 + bt + c) 3t2 + 11t M1

3a = 3, a = 1 14 + 3b = 11 , b = –1 A1

4 – 7 + 3c = 0, c = 1 M1, A1

General solution: A1 ft (8)

(b) M1

t = 0, : 1 = –1 –½A – 3B

t = 0, y = 1: 1 = 1 + A + B one of

theseM1, A1

Solve: A + B = 0, A + 6B = –4

A = 4/5, B = – 4/5 M1

) A1 (5)

(c) t = 1: (= 1.445…) B1 (1)

(14 marks)

[P4 June 2002 Qn 7]


8. (a)

M1, A1

, M1, A1

= 1.107… A1 ft

r = 4a A1 ft (6)

(b) M1

, = 2.795… M1, A1 (3)

(c) B1

Integrate: M1, A1

Limits used: (or upper limit: ) A1

282 m2 M1, A1 (6)

(15 marks)

[P4 June 2002 Qn 8]


9. (a)(i) |x + (y – 2)i| = 2|x + (y + i)| M1

x2 + (y – 2)2 = 4(x2 + (y + 1)2)

(ii) so 3x2 + 3y2 + 12y = 0 any correct from; 3 terms; isw A1 (2)

Sketch circle B1

Centre (0,–2) B1

r = 2 or touches axis B1 (3)

(b) w = 3(z – 7 + 11i) B1

= 3z – 21 + 33i B1 (2)

(7 marks)

[P6 June 2002 Qn 3]

10. (a) marks can be awarded in(b) M1 A1; B1;B1

or sensible correct alternative B1 (5)

(b) When x = 0 , and M1A1, A1 ft

M1, A1 ft (5)

(c) Could use for x = 0.2 but not for x = 50 as B1

approximation is best at values close to x = 0 B1 (2)

(12 marks)

[P6 June 2002 Qn 4]


2

y x

11. zw =

12 + 12iB1

= 12 M1 A1

(3 marks)


12. (a) B1 B1 (2)

(b) =

+

+ M1

∶

+

+

= A1 A1

= M1

= * A1 cso (5)

(7 marks)



13. (a)

shape

points on axes

B1

B1 (2)

(b) 2x + 3 = 5x - 1 M1

x = A1

x > A1 ft (3)

(5 marks)


14.

(a)v + x ,= (4 + v)(1 + v) M1, M1

x = v2 + 5v + 4 – v A1

x = (v + 2)2 * A1 (4)

(b) = B1, M1

= ln x + c must have + c M1 A1

2 + v = M1

v = 2 A1 (5)

(c) y = 2x B1 (1)

(10 marks)



11

21

31

5x

y 5 1y x

2 3y x

15. (a) y = x cos 3x

= cos 3x – 3x sin 3x M1 A1

= 3 sin 3x – 3 sin 3x – 9x cos 3x A1

6 sin 3x – 9x cos 3x + 9x cos 3x = 12 sin 3x

= 2 csoA1

(4)

(b) 2 – 9 = 0 M1

= ()3i A1

y = A sin 3x + B cos 3x form M1

y = A sin 3x + B cos 3x + 2x cos 3xA1 ft on ’s

(4)

(c) y = 1, x = 0 B = 1 B1

= 3A cos 3x – 3B sin 3x + 2 cos 3x – 6x sin 3xM1 A1ft on

’s

2 = 3A + 2 A = 0

y = cos 3x + 2x cos 3xA1

(4)

(d)

B1B1 (2)

(14 marks)



1

y

x6

2 5

6

axesshape

16. (a) M1 A1correct

with limits

= M1 A1

= 2 a2 A1

= a2 = A1 (6)

(b) x = a cos + a cos2 r cos M1

= a sin 2a cos sin A1

= 0 cos = finding M1

= or =

r = or r = finding r M1

A: r = , =

B: r = , = both A and B A1 (5)

(c) x = WX = 2a + = 2 M1 A1 (2)

(d) WXYZ = B1 ft (1)

(e) Area = 100 = 113.3 cm2 M1 A1 (2)

(16 marks)



17. (a) M1 A1 (2)

(b) M1

= M1

= 1 – (*) A1 cso (3)

[P4 June 2003 Qn 1]

18. Identifying as critical values – , B1, B1

Establishing there are no further critical values

Obtaining 2x2 – 2x + 2 or equivalent M1

= 4 – 16 < 0 A1

Using exactly two critical values to obtain inequalities M1

– < x < A1

(6 marks)

Graphical alt.

Identifying x = – and x = as vertical asymptotes B1, B1

Two rectangular hyperbolae oriented correctly with respect to asymptotes in the correct half-planes.

M1

Two correctly drawn curves with no intersections A1

As above M1, A1

y

O

[P4 June 2003 QN n2]


19. (a) = 2x or equivalent M1

I = te–t dt complete substitution M1

= – te–t dt + e–t dt M1 A1

= – te–t – e–t ( + c) A1

= – x2e – e ( + c) A1 (6)

(b) I.F. = e = x3 (or multiplying equation by x2) B1

(x3y) = x3 e or x3y = x3 e dx M1

x3y = – x2 e – e + CA1ft A1

(4)

(10 marks)

Alts (a) (i) mark t = – x2 similarly M1

(ii) x2.(xe ) dx with evidence of attempt at integration by parts M1

= x2(– e ) + 2x.e dx M1 A1 + A1

= – x2e – e (+ c)M1 A1

(6)

(iii) u = e , = –2xe M1

x2 = ln u hence I = ln u du M1

= u ln u – u. du M1 A1

= u ln u – u ( + c) A1

= – x2 e – e ( + c) A1 (6)

(The result ln u du = u ln u – u may be quoted, gaining M1 A1 A1 but must be completely correct.)

[P4 June 2003 Qn 6]


20. (a) A: (5a, 0) B: (3a, 0) allow on a sketchB1, B1

(2)

(b) 3 + 2 cos = 5 – 2 cos M1

cos = M1

= (allow – ) A1

Points are (4a, ) , (4a, ) A1 (4)

(c) ( ) r2 d = ( ) (5 – 2 cos )2 d

= ( ) (25 – 20 cos + 4 cos2 ) d M1

= ( ) (25 – 20 cos + 2 cos 2 + 2) d M1

= ( )[27 – 20 sin + sin 2 ] A1

( ) r2 d = ( ) (3 + 2 cos )2 d

= ( ) (9 + 12 cos + 4 cos2 ) d

= ( ) (11 + 12 cos + 2 cos 2 ) d

= ( ) [11 + 12 sin + sin 2] 2nd integration A1

Area = 2 × (5 – 2 cos )2 d + 2 × (3 + 2cos )2 d

(addition; condone 2/½) M1

= … … + … … correctly identifying limits with s A1

= a2[27 × – 103 + ] + a2[11( – ) – 63 – ] dM1

= a2[49 – 483] (*) A1 cso

(8)

(14 marks)

[P4 June 2003 Qn 7]


1 1

21. (a) y = 2kt.e3t + 3kt2 e3t use of product rule M1

y = 2ke3t + 12kt e3t + 9t2 e3t product rule, twice M1

substituting 2k + 12kt + 9kt2 – 12kt – 18kt2 + 9kt2 = 4 M1

k = 2A1

(4)

(b) Aux. eqn. (if used) (m – 3)2 = 0 m = 3, repeated

yC.F. = (A + Bt) e3t M1 required form (allow just written down) M1 A1

G.S. y = (A + Bt) e3t + 2t2 e3t (ft on 2t2 e3t) A1 ft

(3)

(c) t = 0, y = 3 A = 3 B1

y = Be3t + 3(A + Bt) e3t + 4te3t + 6t2e3t M1

y = 0, t = 0 1 = B + 3A B = –8 M1

y = (3 – 8t + 2t2)e3tA1

(4)

(d) y

1 x

shape crossing +ve x-axis

, 1

B1

B1

y = (–3 + 4t)e3t + 3(1 – 3t + 2t2)e3t = 0

6t2 – 5t = 0 M1

t = A1

y = – e2.5 ( –1.35) awrt –1.35A1

(5)

(16 marks)

[P4 June 2003 Qn 8]


22. (i)(a) Circle M1 A1 One half line correct B1 Second half line B1 (4) [SC Allow B1 for two “full” lines in correct position]

(b) shading correct region A1 ft (1)

(ii)(a) Rearrange to give or M1

, or A1

Completion A1 (3)

(b)

Correct line shown M1 Correct shading A1 (2)

[10]

[P6 June 2003 Qn 4]


23. (a) (cos + i sin )5 = cos 5 + i sin 5 M1

(cos + i sin )5 = cos5 + 5 cos4 (i sin ) + 10 cos3 (i sin )2

+ 10 cos2 (i sin )3 + 5 cos (i sin )+ (i sin )5 M1 A1

cos 5 = cos5 – 10 cos3 sin + 5 cos sin4 M1

= cos5 – 10 cos3 (1 – cos2 ) + 5 cos (1 – 2cos2 + cos4) M1

= 16 cos5 – 20 cos3 + 5 cos (*) A1 cso (6)

(b) cos 5 = –1 (or 1, or 0) M1

5 = (2n 1)180 = (2n 1)36 A1

x = cos = –1, – 0.309, 0.809 M1 A1

(4)

[10]

[P6 June 2003 Qn 5]

24. = 23 – 03 attempt to use an identity M1

= 33 – 13

43 – 23

: :

(n – 1)3 – (n – 3)3

n3 – (n –2)3

(n + 1)3 – (n – 1)3 differences (must see) M1

= (n + 1)3 + n3 – 13 A1

= (n + 1)3 + n3 – 1 – 2n 2n or equiv. B1

= 2n3 + 3n2 + n

= n(2n + 1)(n + 1) () Sub. 2 and ÷ 6 or equiv. c.s.o. M1, A1



25. (a) IF = M1

= A1

= exe must see

= x3ex A1 (3)

(b) x3exy = M1

= xex dx

= xex – ex + c by parts M1 A1

y = o.e. A1 (4)

(c) I = ce–1 c = e1 M1

y = M1

= (1 + e–1) A1 (3)

or = 0.171 0.171 or better

(10 marks)

[P4 January 2004 Qn4]

26. (a) y y = (x – 2)(x – 4) 8 6 2 4 y = 6 – 2x

Line crosses axes B1

Curve shape B1

Axes contacts 6, 8,

3B1

Cusps at 2 and 4 B1 (4)

(b) 6 – 2x = (x – 2)(x – 4) and –6 + 2x = (x – 2)(x – 4) M1, M1

x2 – 4x + 2 = 0 x2 – 8x + 14 = 0 either M1

= 2 – 2 = 4 – 2 A1, A1 (5)

(c) 2 – 2 < x < 4 – 2 M1, A1 (2)

(11 marks)



27. (a) m2 + 4m + 5 = 0 M1

= –2 i A1

y = e–2x(Acos x Bsin x) M1

PI = sin 2x + cos 2x PI & attempt diff. M1

y´ = 2 cos 2x – 2 sin 2x

y´´ = –4 sin 2x – 4 cos 2x A1

–4 – 8 + 5 = 65

–4 + 8 + 5 = 0 subst. in eqn. & equate M1

– 8 = 65

8 + = 0 solving sim. eqn. M1

64 + 8 = 0

65 = 65

= 1, = –8 A1

y = e–2x(Acos x + Bsin x) + sin 2x – 8 cos 2x ft on their and A1ft (9)

(b) As x , e–2x 0 y sin 2x – 8 cos 2x B1ft

y R sin(2x + ) M1

R = 65

= tan–1–8 = –1.446 or –82.9° A1 (3)

(12marks)



28. (a)

Shape + horiz. axis B1

3 B1 (2)

(b) Area = r2 d

= 9 cos22 d use of r2 M1

= use of cos4 = 2cos22 –

1

M1

= dM1, A1

= subst. and M1

= or 0.103 A1 (6)

(c) r sin = 3 sin cos 2

= 3 cos cos 2 – 6 sin sin 2 diff. r sin M1, A1

= 0 6 cos2 – 3 cos – 12 sin2 cos = 0 use of = 0 M1

6 cos2 – 3 cos – 12(1 – cos2 )cos = 0 use double angle formula M1

18 cos3 – 15 cos = 0 solving M1

cos = 0 or cos2 = or tan2 = or sin2 = A1

r = 3(2 × ) – 1

= 2

r sin = 2 use of d = 2r sin M1

d = A1 (8)

(16 marks)



3

[P4 June 2004 Qn 4]


29.

[P4 June 2004 Qn 6]


30.

[P4 June 2004 Qn 7]


31.

[P4 June 2004 Qn 8]


32.

33. (a) (or equiv.) M1 A1

(or equiv.) A1 (3)

, ,

(b) (1, 2, 4, 16) B1

M1

(Allow equiv. fractions A1(cso) (3)

(c) M1

(*) A1(cso) (2)

8

[P6 June 2004 Qn 2]

34. (a) n = 1: (Use of product rule) M1

M1


True for n = 1 (cso + comment) A1

Suppose true for n = k.

M1

A1

M1 A1

True for n = k + 1, so true (by induction) for all n. ( 1) A1(cso) (8)

(b) M1

(1) (0) (–2) (–4)

(or equiv. fractions) A2(1,0) (3)

11

.

[P6 June 2004 Qn 4]


35. (a) arg z = (or putting x and y equal at some stage) B1

, and attempt modulus of numerator or denominator. M1

(Could still be in terms of x and y)

, (*) A1, A1cso (4) (b)

M1

M1 A1

For M1

M1

b = – a Image is (line) y = – x A1 (6)

(c)

B1 B1 (2)

(d) z = i marked (P) on z-plane sketch. B1

marked (Q) on w-plane sketch. B1 (2)

14

[P6 June 2004 Qn 7]


[FP1/P4 January 2005 Qn 1]



36.

37.



38.



39.


[FP1/P4 June 2005 Qn 1]


40.

41.(a)

[FP1/P4 June 2005 Qn 3]

[FP1/P4 June 2005 Qn 6]


42.

43.(a)

[FP1/P4 June 2005 Qn 7]


44.(a)

[FP1/P4 June 2005 Qn 7]


45.

[FP3/P6 June 2005 Qn 4]


46.

[FP3/P6 June 2005 Qn 5]



47.

48.

[FP1/ P4 January 2006 Qn 4]



49.

50.




51.

52.



53.



54.

55. Use of B1

Limits are and B1

M1

M1 A1

M1

cso A1 (7)

[7]

[FP1 June 2006 Qn 2]

56. (a) M1 A1Substituting M1 A1 (4)

(b) General solution is B1

B1

M1

Needs … A1 (4)

[8]



57. (a)

M1 A1 (2)

Accept and M1 for both

(b)

ft their B M1 A1 A1ft

M1

cso A1 (5)

(c) M1

M1

A1 (3)

[10]



58. (a) M1

Leading to

surds required M1 A1

M1

Leading to 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 (3)

x

(c) Using all 4 CVs and getting all into inequalities M1

, both A1ft

ft their greatest positive and their least negative CVs

A1 (3)

[12]



59. (a) B1

M1 A1

or integral equivalent M1

M1 A1

M1

accept C = awrt A1 (8)

(b) M1

M1 A1

Substituting (kg) A1 (4)

[12]



60. (a) ,

, M1

,

, A1

,

, A1

Three terms are sufficient to establish method M1

A1 (5)

(b) Substitute x = 1 B1

cao M1 A1 (3) [8]



61. (a) In this solution and

M1

( )

equate M1 A1

s2 = 1 – c2 M1

A1 (5)

(b) M1

B1

M1

any two A1

any two A1

all four A1 (6) accept awrt 0.79, 1.21,1.93,2.36 [11]Ignore any solutions out of range.



62. (a) Let

M1

Leading to M1 A1

This is a circle; the coefficients of and are the same and there is no xy term.

Allow equivalent arguments and ft their if appropriate. A1ft

Leading to M1

Centre: A1

Radius: or equivalent A1 (7)

(b) Circle B1 centre in correct quadrant B1 ft through origin B1 Line cuts ve x and +ve y axes B1 6 O intersects with circle on axes and all correct B1 (5)

(c) Shading inside circle B1 and above line with all correct B1 (2)

[14]



[FP1 January 2007 Qn 2]


63.

[FP1 Jan 2007 Qn 4]


64.

FP1 January 2007 Qn 5]


65.



66.




67.

68.



69.



70.



71.



72.

73. (a) M1

At M1 A1 cso (3)

(b) Allow anywhere B1

…

+ …M1 A1ft, A1 (dep) (4)

[7]


74. (a)

bothM1

Adding

csoA1 (2)

(b) M1

M1

M1 A1, A1

A1 any

two correct (5)

(c)

M1 A1ft

or exact

equivalent

M1 A1 (4)

[11]


u

v

O


75. (a) Let ; M1

M1

A1

M1

Eliminating gives a line with equation or equivalent A1 (5)

(b) Let ; M1

M1

A1

, M1

M1

Reducing to the circle with equation cso M1 A1 (7)

(c)

ft their line B1ft Circle through origin, centre in correct quadrant B1 Intersections correctly placed B1 (3) [15]







76.

78.(a)

77.(a)




79.(a)

80.(a)




81.

82.




83.

84.


[FP3 June 2008 QN 3]


85.

86.




87.

88.

04a further maths - fp2 mark schemes v2

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