FP Mark Schemes from old P, P5, P6 and FP, FP, FP papers (back to June 00) 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. FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
. [P January 00 Qn ]. [P January 00 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
. [P January 00 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
. [P January 00 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
5. ( > 0) 5 > or 5 = M ( + )( ), critical values ½ and A, A > A ft < 0 5 < M Using critical value 0: ½ < < 0 M, A ft Alt. 5 < 0 or ( 5) > M ( + )( ) > 0 or ( + )( ) > 0 M, A Critical values ½ and, > A, A ft Using critical value 0, ½ < < 0 M, A ft (7 marks) [P June 00 Qn ] y d sin 6. (a) + y = cos M d cos Int. factor Integrate: e tan d ln(cos ) = e = sec M, A y sec = cos d M, A y sec = sin + C A ( y = sin cos + C cos ) (6) (b) When y = 0, cos (sin + C) = 0, cos = 0 M solutions for this ( = π /, π / ) A () (c) y = 0 at = 0: C = 0 : y = sin cos M ( y = ½ sin ) Shape A A () Scales ( marks) [P June 00 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 5
7. (a) m + 7m + = 0 (m + )(m + ) = 0 m = ½, ½t C.F. is y = Ae t + Be M, A P.I. y = at + bt + c y = at + b, y = a B (a) + 7(at + b) + (at + bt + c) t + t M a =, a = + b =, b = A 7 + c = 0, c = M, A ½t t General solution: y = Ae + Be + ( t t + ) A ft (8) ½t t (b) y = ½Ae Be + (t ) M t = 0, y = : = ½A B t = 0, y = : = + A + B one of M, A these Solve: A + B = 0, A + 6B = A = / 5, B = / 5 M y 5 ½t t = ( t t + ) + (e e ) A (5) ½ (c) t = : y = (e e ) + (=.5 ) B () 5 ( marks) [P June 00 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 6
8. (a) y = r sinθ = a(sinθ + 5 sinθ cosθ ) dy = a(cosθ + dθ 5 cos θ ) M, A 5 cos θ + cosθ 5 = 0 ± 9 + 0 cosθ =, cos θ = 5 5 M, A θ = ±.07 A ft r = a A ft (6) (b) r sinθ = 0 M 8 a sinθ = 0, 0 a = =.795 8sinθ M, A () (c) ( + 5 cosθ ) = 9 + 6 5 cosθ + 5cos θ B sin θ θ Integrate: 9θ + 6 5 sinθ + 5 + Limits used: [...] 0 π = 8π + 5π (or upper limit: M, A 5π 9π + ) A ½ π 0 r dθ = a (π ) 8 m M, A (6) (5 marks) [P June 00 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 7
9. (a)(i) + (y )i = + (y + i) M + (y ) = ( + (y + ) ) (ii) so + y + y = 0 any correct from; terms; isw A () y Sketch circle B Centre (0, ) B r = or touches ais B () (b) w = (z 7 + i) B = z + i B () (7 marks) [P6 June 00 Qn ] d y dy d y dy d y dy 0. (a) y + ; + ; + = 0 marks can be awarded in(b) M A; B;B d d d d d d dy d y dy d y = d d d or sensible correct alternative d y B (5) (b) When = 0 d y d y =, and = 5 d d MA, A ft 5 y = + +... 6 M, A ft (5) (c) Could use for = 0. but not for = 50 as B approimation is best at values close to = 0 B () ( marks) [P6 June 00 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 8
. zw = π π π π cos cos sin sin = π cos + i sin π π π π π B + i sin cos + cos sin M A ( marks) [P January 00 Qn ]. (a) r + r + (b) n r + r + = + 5 B B () + 6 M + n n + + n + n + = + + n + n + A A 5 5n + 5n + 0 n 0 = 6 6( n + )( n + ) M n(5n + ) = 6( n + )( n + ) * A cso (5) (7 marks) [P January 00 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 9
. (a) y y = 5 y = shape points on aes B B () 5 (b) + = 5 - M = 7 > 7 A A ft () (5 marks). (a) (b) [P January 00 Qn ] d v v +,= ( + v)( + v) M, M d d v = v + 5v + v A d d v = (v + ) d dv = ( v + ) d * A () B, M (c) y = + v + v = v = = ln + c must have + c M A ln + c ln + c M A (5) B () ln + c (0 marks) [P January 00 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 0
5. (a) y = λ cos d y = λ cos λ sin M A d d d y = λ sin λ sin 9λ cos 6λ sin 9λ cos + 9λ cos = sin λ = A A cso () (b) λ 9 = 0 M λ = (±)i A y = A sin + B cos form M y = A sin + B cos + cos A ft on λ s () (c) y =, = 0 B = B d y M Aft on = A cos B sin + cos 6 sin d λ s = A + A = 0 y = cos + cos A () (d) y π π 5π 6 6 aes shape B B () ( marks) [P January 00 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
6. (a) a + cos θ + cosθ dθ M Acorrect with limits = cos θ + a + + cosθ dθ M A = a π sin θ θ θ + + + sinθ A 0 π πa = a = A (6) (b) = a cos θ + a cos θ r cos θ M d = a sin θ a cos θ sin θ dθ A d = 0 cos θ = dθ finding θ M θ = r = a π π or θ = or r = a a π A: r =, θ = finding r M a π B: r =, θ = both A and B A (5) (c) = a WX = a + a = a M A () (d) WXYZ = 7 a 8 B ft () (e) Area = 7 00 8 π 00 =. cm M A () (6 marks) [P January 00 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
7. (a) ( ) = ( ) ( ) r r r r r r r M A () (b) n r= r r ( r ) = n r= ( r ) r M M ( n ) n = + +... + = n (*) A cso () [P June 00 Qn ] 8. Identifying as critical values, B, B Establishing there are no further critical values Obtaining + or equivalent M = 6 < 0 A Using eactly two critical values to obtain inequalities M < < A Graphical alt. Identifying = and = as vertical asymptotes Two rectangular hyperbolae oriented correctly with respect to asymptotes in the correct half-planes. Two correctly drawn curves with no intersections As above y (6 marks) B, B M A M, A O [P June 00 QN n] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
9. (a) d t = d or equivalent M I = te t dt complete substitution M = te t dt + e t dt M A = te t e t ( + c) A = e e ( + c) A (6) (b) I.F. = e d = (or multiplying equation by ) B d ( y) = e d y = e or y = e e + C Alts (a) (i) mark t = similarly M (ii).(e d M ) d with evidence of attempt at integration by parts M Aft A () (0 marks) = ( e = e (iii) u = e ) +.e e, du d (+ c) d = e = ln u hence I = ln u du M A + A M A (6) M M = u ln u u. u du M A = u ln u u ( + c) A = e e ( + c) A (6) (The result ln u du = u ln u u may be quoted, gaining M A A but must be completely correct.) [P June 00 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
0. (a) A: (5a, 0) B: (a, 0) allow on a sketch (b) + cos θ = 5 cos θ cos θ = B, B () M M θ = π 5π, (allow π ) A π 5π Points are (a, ), (a, ) A () (c) ( ) r dθ = ( ) (5 cos θ ) dθ = ( ) (5 0 cos θ + cos θ ) dθ M = ( ) (5 0 cos θ + cos θ + ) dθ M = ( )[7θ 0 sin θ + sin θ ] A ( ) r dθ = ( ) ( + cosθ ) dθ = ( ) (9 + cos θ + cos θ ) dθ = ( ) ( + cos θ + cos θ ) dθ = ( ) [θ + sin θ + sin θ] nd integration A Area = (5 cosθ ) dθ + ( + cosθ ) dθ π = 0 π + π (addition; condone /½) M correctly identifying limits with s A = a π [7 0 + ] + a π [(π ) 6 ] dm = a [9 8 ] (*) A cso (8) ( marks) [P June 00 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 5
. (a) y = kt.e t + kt e t use of product rule M y = ke t + kt e t + 9t e t product rule, twice M substituting k + kt + 9kt kt 8kt + 9kt = M k = A () (b) Au. eqn. (if used) (m ) = 0 m =, repeated y C.F. = (A + Bt) e t M required form (allow just written down) M A G.S. y = (A + Bt) e t + t e t (ft on t e t ) A ft () (c) t = 0, y = A = B y = Be t + (A + Bt) e t + te t + 6t e t M y = 0, t = 0 = B + A B = 8 M A y = ( 8t + t )e t () (d) y shape crossing +ve -ais B, B y = ( + t)e t + ( t + t )e t = 0 6t 5t = 0 t = 6 5 M A y = 9 e.5 (.5) awrt.5 A (5) (6 marks) [P June 00 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 6
. (i)(a) Circle M A One half line correct B Second half line B () [SC Allow B for two full lines in correct position] (b) shading correct region A ft () (ii)(a) Rearrange z = w z w = to give z = f ( w) or z = f( w) M z w z =, or z = z w z w = A w z = w = w = w A () Completion ( ) (b) Correct line shown M Correct shading A () [0] [P6 June 00 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 7
. (a) (cos θ + i sin θ) 5 = cos 5θ + i sin 5θ M (cos θ + i sin θ) 5 = cos 5 θ + 5 cos θ (i sin θ) + 0 cos θ (i sin θ) + 0 cos θ (i sin θ) + 5 cos θ (i sin θ) + (i sin θ) 5 M A cos 5θ = cos 5 θ 0 cos θ sin θ + 5 cos θ sin θ M = cos 5 θ 0 cos θ ( cos θ) + 5 cos θ ( cos θ + cos θ) M = 6 cos 5 θ 0 cos θ + 5 cos θ (*) A cso (6) (b) cos 5θ = (or, or 0) 5θ = (n ± )80 θ = (n ± )6 = cos θ =, 0.09, 0.809 M A M A [0] () [P6 June 00 Qn 5] n. ( 6r + ) = 0 attempt to use an identity M r= n r= = : : (n ) (n ) n (n ) (n + ) (n ) differences (must see) M = (n + ) + n A 6 r = (n + ) + n n n or equiv. B n r= = n + n + n r = 6 n(n + )(n + ) ( ) Sub. Σ and 6 or equiv. c.s.o. M, A [P January 00 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 8
5. (a) IF = = d e + ln e + M A = e e (b) e y = y = ln must see = e A () e d M = e d = e e + c by parts M A c + e o.e. A () (c) I = ce c = e M y = - e.e + M 8 8 = 8 ( + e ) A () or = 0.7 0.7 or better (0 marks) [P January 00 Qn] 6. (a) y y = ( )( ) Line crosses aes B 8 Curve shape B 6 Aes contacts 6, 8, y = 6 B Cusps at and B () (b) 6 = ( )( ) and 6 + = ( )( ) M, M + = 0 8 + = 0 either M ± 6 8 8 ± 6 56 = = = = A, A (5) (c) < < M, A () ( marks) [P January 00 Qn5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 9
7. (a) m + m + 5 = 0 M ± m = = ± i A y = e (Acos ± Bsin ) PI = λ sin + µ cos M PI & attempt diff. M y = λ cos µ sin y = λ sin µ cos A λ 8µ + 5λ = 65 µ + 8λ + 5µ = 0 subst. in eqn. & equate M λ 8µ = 65 8λ + µ = 0 solving sim. eqn. M 6λ + 8µ = 0 65λ = 65 λ =, µ = 8 A y = e (Acos + Bsin ) + sin 8 cos ft on their λ and µ Aft (9) (b) As, e 0 y sin 8 cos y R sin( + α) Bft M R = 65 α = tan 8 =.6 or 8.9 A () (marks) [P January 00 Qn6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 0
8. (a) Shape + horiz. ais B B () (b) Area = r dθ = 9 cos θ dθ use of r M 9 cos θ + = dθ use of cosθ = cos θ M = 9 π sin θ θ + 8 dm, A π 6 9 π π = 8 6 subst. π and 6 π M 9 π = or 0.0 A (6) 6 (c) r sin θ = sin θ cos θ dy = cos θ cos θ 6 sin θ sin θ diff. r sin θ M, A dθ dy = 0 6 cos θ cos θ sin θ cos θ = 0 dθ use of dy = 0 dθ 6 cos θ cos θ ( cos θ )cos θ = 0 use double angle formula M 8 cos θ 5 cos θ = 0 solving M cos θ = 0 or cos θ = 6 5 or tan θ = 5 or sin θ = 6 A r = ( 65 ) = r sin θ = 6 use of d = r sin θ M M d = 6 A (8) (6 marks) [P January 00 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
9. [P June 00 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
0. [P June 00 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
. [P June 00 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
. [P June 00 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 5
. (a) f ( ) = sec f ( ) = sec (sec tan ) (or equiv.) M A f ( ) = sec (sec ) + tan (sec tan ) (or equiv.) A () (sec (sec + 6sec tan ) + sec tan ), (6sec sec ), ( + 8tan + 6tan ) π π (b) tan = or sec = (,,, 6) B π π π π π π π tan = f + f + f + f M 6 (c) π π 8 π = + + + (Allow equiv. fractions A(cso) () = π π π, so use π = M 0 0 0 π π π 8 π π π π tan + + = + + + 0 0 00 + 8000 (*) A(cso) () 0 00 000 8 [P6 June 00 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 6
FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 7 [P6 June 00 Qn ]. (a) n = : ( ) sin e cos e cos e d d = (Use of product rule) M ( ) sin cos sin sin cos cos cos = = + π π π M ( ) + = cos e cos e d d π True for n = (cso + comment) A Suppose true for n = k. ( ) + = + + cos e d d cos e d d k kπ k k M + + = sin e cos e π kπ k k A ( ) ( ) + + = + + = + cos e cos e π π π k k k k M A True for n = k +, so true (by induction) for all n. ( ) A(cso) (8) (b) ( ) cos cos 6 cos cos π π π π + + + + M () (0) ( ) ( ) 6 cos e + = (or equiv. fractions) A(,0) ().
π 5. (a) arg z = z = λ + λ i (or putting and y equal at some stage) B ( λ + ) + λ i w =, and attempt modulus of numerator or denominator. M λ + ( λ + )i (Could still be in terms of and y) + ) + λ i = λ + ( λ + )i = ( λ + ), = ( λ + λ (b) w (*) A, Acso () z + wi w = zw + wi = z + z = M z + i w z = wi = w M A For w = a + i b, ( + b) ai = ( a ) + ib M ( b + b ) + a = ( a ) + M (c) b = a Image is (line) y = A (6) B B () (d) z = i marked (P) on z-plane sketch. B + i i z = i w = = = i marked (Q) on w-plane sketch. B () i [P6 June 00 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 8
6. [FP/P January 005 Qn ] 7. [FP/P January 005 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 9
8. [FP/P January 005 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 0
9. [FP/P January 005 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
0. [FP/P January 005 Qn 7].(a) [FP/P June 005 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
. [FP/P June 005 Qn ].(a) [FP/P June 005 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
.(a) [FP/P June 005 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
5. [FP/P June 005 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 5
6. [FP/P6 June 005 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 6
7. [FP/P6 June 005 Qn 5] 8. [FP/P January 006 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 7
9. [FP/ P January 006 Qn ] 50. [FP/P January 006 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 8
5. [FP/P January 006 Qn 7] 5. [FP/P6 January 006 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 9
5. [FP/P6 January 006 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 0
5. [FP/P6 January 006 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
55. Use of r dθ B π π Limits are 8 and B ( θ) 6a cos θ 8a cos = + M sin θ ( + cos θ) dθ = θ + M A A a θ = + sin θ π π 8 π π = a + ( 0 ) 8 M π = a = a ( π ) cso A (7) [7] [FP June 006 Qn ] 56. (a) y = sin + 6cos M y = cos sin A Substituting cos sin + sin = k cos M k = A () (b) General solution is y = Acos + Bsin + sin B ( 0, ) A = B π π π π π, B B = + = M y = cos sin + sin Needs y = A () [8] [FP June 006 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
57. (a) (r+ ) = 8r + r + 6r+ (r ) = 8r r + 6r + = + ( A, B ) (r ) (r ) r Accept r = 0 B= and r = A+ B= 6 A= M for both = = M A () (b) (c) 5 ( n ) ( n ) = + = + Μ + = n + n+ = r + n ft their B M A Aft ( ) n r = n 8n + n + n r = M r = 0 0 r= r= = n( n + n+ ) = n( n+ )( n+ ) cso A (5) 6 6 (r ) = (9r 6r+ ) M = 9 0 8 6 0 + 0 6 M = 980 A () [0] [FP June 006 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
58. (a) 6 6 + = M Leading to + 6= 0 ( ) + = 7 = ± 7 surds required M A + 6= 6 M Leading to 0 0, = = A, A (6) (b) Accept if parts (a) and (b) done in reverse order y Curved shape B Line B At least intersections B () O (c) Using all CVs and getting all into inequalities > M 7, < 7 both Aft ft their greatest positive and their least negative CVs 0< < A () [] [FP June 006 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009
59. (a) dt = ln ( 0 t) B 0 t ln( 0 t) e = 0 t M A ( ) ds S + = ( 0 t) dt ( 0 t) ( 0 t) d S dt ( 0 t) ( 0 t) or integral equivalent S = 0 t 0 t + C ( ) ( ) ( ) ( 0, 6) 6 0 0 C C M M A = + = M 600 ( 0 t) 0 t S = accept C = awrt 0.007 A (8) 600 (b) Substituting ( t) ds 0 = + M dt 600 ds 0 t 5 dt = = M A S = 9 (kg) 8 A () [] [FP June 006 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 5
60. (a) f ( ) = cos, ( ) f 0 π = π ( ) =, ( ) f sin π ( ) =, ( ) f cos f = M f = 0 π ( ) =, ( ) f 8sin ( iv ) ( ) f = 6cos, ( v ) ( ) f = sin, f = 8 A ( iv ) ( ) f 0 π = ( v ) ( ) π f = A π π ( ) f ( ) ( ) ( ) f π π π π π cos = f ( ) + f ( )( ) + + +...! Three terms are sufficient to establish method M π π π ( ) ( ) ( ) 5 cos = + +... A (5) 5 π (b) Substitute = ( 0.60) B π π π ( ) ( ) ( ) 5 cos = + +... 5 0.67 cao M A () [8] [FP June 006 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 6
6. (a) In this solution cosθ = c and sinθ = s ( c s) 5 cos5θ + isin 5θ = + i M 5 ( c 5c is 0c ( is) 0c ( is) 5c( is) ( is) = + + + + + ) 5 I sin 5 5 0 5 θ = cs cs + s equate M A ( ) ( ) = 5cs 0c c s+ c s s = c M ( 6 ) = s c c + A (5) (b) ( ) sinθ 6 cos θ cos θ cos θ sinθ 0 + + = M sinθ = 0 θ = 0 B ( )( ) 6c 0c 8c c 0 + = = M c=±, c=± any two A θ.,.9; π π θ =, any two A all four A (6) accept awrt 0.79,.,.9,.6 [] Ignore any solutions out of range. [FP June 006 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 7
6. (a) Let z = + iy Leading to ( 6) + ( y+ ) = 9 ( + ) + ( y ) 8 8y 8 y 0 M + + = M A This is a circle; the coefficients of there is no y term. and y are the same and Allow equivalent arguments and ft their f (, y ) if appropriate. ( + 6+ y y = 0) Aft 5 y Leading to ( ) ( ) + + = M Centre: (, ) A Radius: 5 or equivalent A (7) (b) Circle B centre in correct quadrant B ft through origin B Line cuts ve and +ve y aes B 6 O intersects with circle on aes and all correct B (5) (c) Shading inside circle B and above line with all correct B () [] [FP June 006 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 8
6. [FP January 007 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 9
6. [FP Jan 007 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 50
65. FP January 007 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 5
66. [FP January 007 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 5
67. [FP January 007 Qn 8] 68. [FP June 007 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 5
69. [FP June 007 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 5
70. [FP June 007 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 55
7. [FP June 007 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 56
7. [FP June 007 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 57
7. (a) ( ) y y y y y + = 0 M d d d d d d d d d d At = 0, d y d dy = = d M A cso () (b) d y = d 0 ( ) ( ) f 0 f 0 y ( ) ( ) 6 =, + + 6 = f 0 + f 0 + + + Allow anywhere B M Aft, A (dep) () [FP June 007 Qn ] [7] n n 7. (a) z = ( cosθ + isinθ) = cos nθ + isin nθ n n ( ) ( ) ( ) both Adding cso z = cosθ + i sinθ = cos nθ + i sin nθ = cos nθ i sin nθ z n + = cos nθ n z M A () (b) two correct 6 z+ = z + 6z + 5z + 0 + 5z + 6z + z z 6 6 ( ) ( ) z z 6 z z 5 z z 0 M 6 6 = + + + + + + M 6 6cos θ = cos 6θ + cos θ + 0cos θ + 0 M 6 cos θ = cos 6θ + 6cos θ + 5cos θ + 0 A, A ( p=, q= 6, r = 5, s = 0) A any (5) 6 (c) = ( + + + ) cos θdθ cos 6θ 6cos θ 5cos θ 0 dθ sin 6θ 6sin θ 5sin θ = + + + 0θ 6 [ ] π 5 0π 5π... = 0 + + = + 8 equivalent or eact M Aft M A () [] [FP June 007 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 58
75. (a) Let z = λ+ λi ; w λ + λ + = λ ( i ) ( + i) ( λ ) ( ) ( λ ) λ+ + i i = λ + i i u+ i v= + + i λ A u, v λ λ M Eliminating λ gives a line with equation v= u or equivalent A (5) (b) Let z λ ( λ ) = + i; u = ( λ ) M M λ λi w = M λ + i ( ) ( ) ( λ ) ( ) λ λi λ+ + i = λ λ+ i λ+ λ+ i λ λ+ + λi u+ i v= ( λ+ ) λ + λ+ λ λ + λ+, λ v = λ + λ+ u = λ + v ( λ u ) v u ( ) ( ) λ + v = = = λ + λ+ λ + + v + Reducing to the circle with equation u + v u+ v= 0 cso M A (7) M A M M (c) v O ft their line Bft Circle through origin, centre in correct quadrant B u Intersections correctly placed B () [5] [FP June 007 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 59
76. [FP January 008 Qn ] 77.(a) [FP January 008 Qn ] 78.(a) [FP January 008 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 60
79.(a) [FP January 008 Qn 7] 80.(a) [FP January 008 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 6
8. [FP June 008 Qn ] 8. [FP June 008 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 6
8. [FP June 008 Qn 6] 8. [FP June 008 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 6
85. [FP June 008 Qn 8] 86. [FP June 008 QN ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 6
87. [FP June 008 Qn ] 88. [FP June 008 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version March 009 65