Paper Reference(s) 6667 Edexcel GCE Additional Pure Arithmetic FP1 Superior Stage Specimen Paper Time: 1 hour 30 minutes Supplies required for examination Reply E book (AB16) Graph Paper (ASG2) Mathematical Formulae (Lilac) Gadgets included with Question Assignment papers Nil Candidates could use any calculator EXCEPT these with the ability for symbolic algebra, differentiation and/or integration. Thus candidates could NOT use calculators such because the Texas Devices TI-89, TI-92, Casio CFX-9970G, Hewlett Packard HP 48G.
Directions to Candidates Within the packing containers on the reply ebook, write the title of the inspecting physique (Edexcel), your centre quantity, candidate quantity, the unit title (Additional Pure Arithmetic FP1), the paper reference (6667), your surname, initials and signature. When a calculator is used, the reply must be given to an acceptable diploma of accuracy. Info for Candidates A booklet ‘Mathematical Formulae and Statistical Tables’ is supplied. Full marks could also be obtained for solutions to ALL questions. This paper has eight questions. Recommendation to Candidates You will need to make sure that your solutions to components of questions are clearly labelled.
You will need to present enough working to make your strategies clear to the Examiner. Solutions with out working could acquire no credit score. This publication could solely be reproduced in accordance with London Qualifications Restricted copyright coverage. Edexcel Basis is a registered charity. ©2003 London Qualifications Restricted 1. Show (r r =1 n 2 – r -1 = ) 1 (n – 2)n(n + 2) . Three (5) 2. 1 f ( x ) = ln x – 1 – . x (a) Present that the basis a of the equation f(x) = Zero lies within the interval Three < a < four . (2) (b) Taking Three. 6 as your beginning worth, apply the Newton-Raphson process as soon as to f(x) to acquire a second approximation to a.
Give your reply to four decimal locations. (5) Three. Discover the set of values of x for which 1 x > . x -Three x -2 (7) four. f ( x ) ? 2 x Three – 5 x 2 + px – 5, p I ?. The equation f (x) = Zero has (1 – 2i) as a root. Remedy the equation and decide the worth of p. (7) 5. (a) Get hold of the final answer of the differential equation dS – Zero. 1S = t. dt (6) (b) The differential equation partially (a) is used to mannequin the property, ? S million, of a financial institution t years after it was arrange. On condition that the preliminary property of the financial institution had been ? 200 million, use your reply to half (a) to estimate, to the closest ? illion, the property of the financial institution 10 years after it was arrange. (four) 2 6. The curve C has polar equation r 2 = a 2 cos 2q , -p p ? q ? . four four (a) Sketch the curve C. (2) (b)
Discover the polar coordinates of the factors the place tangents to C are parallel to the preliminary line. (6) (c) Discover the realm of the area bounded by C. (four) 7. On condition that z = -Three + 4i and zw = -14 + 2i, discover (a) w within the kind p + iq the place p and q are actual, (four) (b) the modulus of z and the argument of z in radians to 2 decimal locations (four) (c) the values of the actual constants m and n such that mz + nzw = -10 – 20i . (5) Three Flip over eight. (a) On condition that x = e t , present that (i) y dy = e -t , dx dt 2 dy o d2 y – 2t ? d y c 2 – ?. =e c 2 dt ? dx o e dt (ii) (5) (b) Use you solutions to half (a) to point out that the substitution x = e t transforms the differential equation d2 y dy x 2 2 – 2x + 2y = x3 dx dx into d2 y dy – Three + 2 y = e 3t . 2 dt dt (Three) (c) Therefore discover the final answer of x2 d2 y dy – 2x + 2y = x3. 2 dx dx (6) END four Paper Reference(s) 6668 Edexcel GCE Additional Pure Arithmetic FP2 Superior Stage Specimen Paper Time: 1 hour 30 minutes Supplies required for examination Reply E book (AB16) Graph Paper (ASG2) Mathematical Formulae (Lilac) Gadgets included with Question Assignment papers Nil
Candidates could use any calculator EXCEPT these with the ability for symbolic algebra, differentiation and/or integration. Thus candidates could NOT use calculators such because the Texas Devices TI 89, TI 92, Casio CFX-9970G, Hewlett Packard HP 48G. Directions to Candidates Within the packing containers on the reply ebook, write the title of the inspecting physique (Edexcel), your centre quantity, candidate quantity, the unit title (Additional Pure Arithmetic FP2), the paper reference (6668), your surname, initials and signature. When a calculator is used, the reply must be given to an acceptable diploma of accuracy.
Info for Candidates A booklet ‘Mathematical Formulae and Statistical Tables’ is supplied. Full marks could also be obtained for solutions to ALL questions. This paper has eight questions. Recommendation to Candidates You will need to make sure that your solutions to components of questions are clearly labelled. You will need to present enough working to make your strategies clear to the Examiner. Solutions with out working could acquire no credit score. This publication could solely be reproduced in accordance with London Qualifications Restricted copyright coverage. Edexcel Basis is a registered charity. ©2003 London Qualifications Restricted 1.
The displacement x of a particle from a set level O at time t is given by x = sinh t. four At time T the displacement x = . Three (a) Discover cosh T . (2) (b) Therefore discover e T and T. (Three) 2. On condition that y = arcsin x show that (a) dy = dx (1 – x ) 2 1 , (Three) (b) (1 – x 2 ) d2 y dy -x = Zero. 2 dx dx (four) Determine 1 Three. y P(x, y) s A y O x Determine 1 reveals the curve C with equation y = cosh x. The tangent at P makes an angle y with the x-axis and the arc size from A(Zero, 1) to P(x, y) is s. (a) Present that s = sinh x. (Three) (a) By contemplating the gradient of the tangent at P present that the intrinsic equation of C is s = tan y. 2) (c) Discover the radius of curvature r on the level the place y = p . four (Three) S four. I n = o x n sin x dx. p 2 Zero (a) Present that for n ? 2 ?p o I n = nc ? e 2o n -1 – n(n – 1)I n – 2 . (four) (four) (b) Therefore acquire I Three , giving your solutions by way of p. 5. (a) Discover ? v(x2 + four) dx. (7) The curve C has equation y 2 – x 2 = four. (b) Use your reply to half (a) to search out the realm of the finite area bounded by C, the constructive x-axis, the constructive y-axis and the road x = 2, giving your reply within the kind p + ln q the place p and q are constants to be discovered. (four) Determine 2 6. y O 2pa x The parametric equations of the curve C proven in Fig. are x = a(t – sin t ), y = a(1 – cos t ), Zero ? t ? 2p . (a) Discover, through the use of integration, the size of C. (6) The curve C is rotated by means of 2p about Ox. (b) Discover the floor space of the strong generated. (5) 7 7. (a) Utilizing the definitions of sinh x and cosh x by way of exponential features, categorical tanh x by way of e x and e – x . (1) (b) Sketch the graph of y = tanh x. (2) 1 ? 1 + x o lnc ?. 2 e1 – x o (c) Show that artanh x = (four) (d) Therefore acquire d (artanh x) and use integration by components to point out that dx o artanh x dx = x artanh x + 1 ln 1 – x 2 + fixed. 2 ( ) (5) eight.
The hyperbola C has equation x2 y2 = 1. a2 b2 (a) Present that an equation of the traditional to C at P(a sec q , b tan q ) is by + ax sin q = a 2 + b 2 tan q . (6) ( ) The conventional at P cuts the coordinate axes at A and B. The mid-point of AB is M. (b) Discover, in cartesian kind, an equation of the locus of M as q varies. (7) END U Paper Reference(s) 6669 Edexcel GCE Additional Pure Arithmetic FP3 Superior Stage Specimen Paper Time: 1 hour 30 minutes Supplies required for examination Reply E book (AB16) Graph Paper (ASG2) Mathematical Formulae (Lilac) Gadgets included with Question Assignment papers Nil
Candidates could use any calculator EXCEPT these with the ability for symbolic algebra, differentiation and/or integration. Thus candidates could NOT use calculators such because the Texas Devices TI 89, TI 92, Casio CFX 9970G, Hewlett Packard HP 48G. Directions to Candidates Within the packing containers on the reply ebook, write the title of the inspecting physique (Edexcel), your centre quantity, candidate quantity, the unit title (Additional Pure Arithmetic FP3), the paper reference (6669), your surname, initials and signature. When a calculator is used, the reply must be given to an acceptable diploma of accuracy.
Info for Candidates A booklet ‘Mathematical Formulae and Statistical Tables’ is supplied. Full marks could also be obtained for solutions to ALL questions. This paper has eight questions. Recommendation to Candidates You will need to make sure that your solutions to components of questions are clearly labelled. You will need to present enough working to make your strategies clear to the Examiner. Solutions with out working could acquire no credit score. This publication could solely be reproduced in accordance with London Qualifications Restricted copyright coverage. Edexcel Basis is a registered charity. ©2003 London Qualifications Restricted 1. y = x 2 – y, y = 1 at x = Zero . dx y – y0 ? dy o Use the approximation c ? » 1 with a step size of Zero. 1 to estimate the values of y h e dx o Zero at x = Zero. 1 and x = Zero. 2, giving your solutions to 2 vital figures. (6) 2. (a) Present that the transformation w= z -i z +1 maps the circle z = 1 within the z-plane to the road w – 1 = w + i within the w-plane. (four) The area z ? 1 within the z-plane is mapped to the area R within the w-plane. (b) Shade the area R on an Argand diagram. (2) Three. Show by induction that, all integers n, n ? 1 , ar > 2 n r =1 n 1 2 . (7) four. dy d2 y dy +y = x, y = Zero, = 2 at x = 1. 2 dx dx dx
Discover a sequence answer of the differential equation in ascending powers of (x – 1) as much as and together with the time period in (x – 1)Three. (7) 5. ? 7 6o A=c c 6 2? . ? e o (a) Discover the eigenvalues of A. (four) (a) Get hold of the corresponding normalised eigenvectors. (6) NM 6. The factors A, B, C, and D have place vectors a = 2i + okay , b = i + 3j, c = i + Three j + 2k , d = four j + okay respectively. (a) Discover AB ? AC and therefore discover the realm of triangle ABC. (7) (b) Discover the quantity of the tetrahedron ABCD. (2) (c) Discover the perpendicular distance of D from the airplane containing A, B and C. (Three) 7. ? 1 x – 1o c ? 5 A( x) = c Three Zero 2 ? , x ? 2 c1 1 Zero ? e o (a) Calculate the inverse of A(x). (eight) ? 1 Three – 1o c ? B = c3 Zero 2 ? . c1 1 Zero ? e o ? po c ? The picture of the vector c q ? when reworked by B is cr? e o (b) Discover the values of p, q and r. (four) ? 2o c ? c Three? . c four? e o 11 eight. (a) On condition that z = e iq , present that zp + 1 = 2 cos pq , zp the place p is a constructive integer. (2) (b) On condition that cos four q = A cos 4q + B cos 2q + C , discover the values of the constants A, B and C. (7) The area R bounded by the curve with equation y = cos 2 x, rotated by means of 2p in regards to the x-axis. (c) Discover the quantity of the strong generated. (6) p p ? x ? , and the x-axis is 2 2
END NO EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question Assignment #1. Scheme Marks M1 B1 a (r r =1 n 2 – r -1 = a r2 – a r – a1 r =1 r =1 r =1 ) n n n ? n o c a1 = n ? e r =1 o = = = n (n + 1)(2n + 1) – ? 1 on(n + 1) – n c ? 6 e 2o n 2n 2 – eight 6 [ ] M1 A1 A1 (5) (5 marks) 1 n(n – 2 )(n + 2 ) Three 2. (a) f ( x) = ln x – 1 – 1 x f (Three) = ln Three – 1 – 1 = -Zero. 2347 Three f (four) = ln four – 1 – 1 = Zero. 1363 four f (Three) and f (four) are of reverse signal and so f ( x ) has root in (Three, four) (b) x Zero = Three. 6 f ? (x ) = 1 1 + x x2 M1 A1 (2) M1 A1 f ? (Three. 6 ) = Zero. 354 381 f (Three. 6) = Zero. 003 156 04 Root » Three. – f (Three. 6) f ? (Three. 6) M1 A1 ft A1 (5) (7 marks) » Three. 5911 13 EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question Assignment quantity Three. Scheme x x x 2 – 3x + Three 1 1 > ? >Zero ? >Zero x-Three x-2 x-Three x-2 (x – Three)(x – 2 ) Marks M1 A1 B1 B1 Numerator all the time constructive Vital factors of denominator x = 2, x = Three x < 2 : den = (- ve)(- ve) = + ve 2 < x < Three : den = (- ve)(+ ve) = – ve Three < x : den = (+ ve)(+ ve) = + ve M1 A1 A1 (7) (7 marks) Set of values x < 2 and x > Three x : x < 2 E four. If 1 – 2i is a root, then so is 1 + 2i B1 M1 A1 M1 A1 ft A1 A1 (7) x – 1 + 2i )(x – 1 – 2i ) are elements of f(x) so x 2 – 2 x + 5 is an element of f (x) f ( x ) = x 2 – 2 x + 5 (2 x – 1) Third root is 1 2 ( ) and p = 12 (7 marks) 5. (a) dS – (Zero. 1)S = t dt – ( Zero. 1)dt Integrating issue e o = e -(Zero. 1)t M1 d Se – (Zero. 1)t = te – (Zero. 1)t dt Se – (Zero. 1)t = o te – (Zero. 1)t dt = -10te – (Zero. 1)t – 100e – (Zero. 1)t + C [ ] A1 A1 M1 A1 A1 (6) S = Ce (Zero. 1)t – 10t – 100 (b) S = 200 at t = Zero ? 200 = C – 100 i. e. C = 300 S = 300e (Zero. 1)t – 10t – 100 M1 A1 At t = 10, S = 300e – 100 – 100 = 615. 484 55 M1 A1 ft (four) (10 marks) Property ? 615 million NQ EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667)
SPECIMEN PAPER MARK SCHEME Question Assignment quantity 6. (a) l Scheme Marks q B1 (Form) B1 (Labels) (2) (b) Tangent parallel to preliminary line when y = r sin q is stationary Think about subsequently d 2 a cos 2q sin 2 q dq ( ) M1 A1 = -2 sin 2q sin 2 q + cos 2q (2 sin q cos q ) =Zero 2 sin q [cos 2q cos q – sin 2q sin q ] = Zero sin q ? Zero ? cos 3q = Zero ? q = p -p or 6 6 M1 A1 o ? ? o ? 1 p o? 1 -p Coordinates of the factors c c a, ? c a, ? c 6 6 oe 2 e 2 A1 A1 (6) 1 o4 2 1 2o4 (c) Space = o r dq = a o cos 2q dq 2 o -p 2 o -p four four p p M1 A1 a2 a2 1 2 e sin 2q u = a e = [1 – (- 1)] = 2 e 2 u -4p four 2 u p four M1 A1 (four) (12 marks) 15
EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question Assignment quantity 7. (a) z = -Three + 4i, zw = -14 + 2i Scheme Marks w= = = – 14 + 2i (- 14 + 2i )(- Three – 4i ) = (- Three + 4i )(- Three – 4i ) – Three + 4i M1 A1 A1 A1 M1 A1 M1 A1 M1 A1 A1 M1 A1 (5) (13 marks) (four) (42 + eight) + i(- 6 + 56) 9 + 16 50 + 50i = 2 + 2i 25 (four) (b) z = (Three 2 + 42 = 5 four = 2. 21 Three ) arg z = p – arctan (c) Equating actual and imaginary components 3m + 14n = 10, 4m + 2n = -20 Fixing to acquire m = -6, n = 2 NS EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question Assignment quantity eight. (a)(i) x = et , dy dy dy dt = = e -t dt dx dt dx
Scheme Marks M1 A1 ? dx t o c =e ? e dt o (ii) d 2 y dt d e – t dy u e = dt u dx 2 dx dt e u e M1 e dy d2 yu = e – t e – e -t + e -t 2 u dt dt u e e d 2 y dy u = e – 2t e 2 – u dt u e dt (b) x2 2t A1 A1 (5) d2 y dy – 2x + 2y = x3 2 dx dx – 2t e e e d 2 y dy u t – t dy + 2 y = e 3t e 2 – u, – 2e e dt u dt e dt M1 A1, A1 (Three) d2 y dy – Three + 2 y = e 3t 2 dt dt (c) Auxiliary equation m 2 – 3m + 2 = Zero (m – 1)(m – 2) = Zero Complementary perform y = Ae t + Be 2t e 3t 1 Specific integral = 2 = e 3t Three – (Three ? Three) + 2 2 Common answer y = Ae t + Be 2t + 1 e 3t 2 = Ax + Bx 2 + 1 x Three 2 M1 A1 M1 A1 M1 A1 ft 6) (14 marks) 17 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Assignment Number one. cosh 2 T = 1 + sinh 2 T = 1 + 16 25 = 9 9 Scheme Marks M1 A1 (2) M1 A1 A1 ft (Three) cosh T = ± 5 5 = since cosh T > 1 Three Three four 5 + =Three Three Three e T = cosh T + sinh T = Therefore T = ln Three 2. (5 marks) (a) y = arcsin x ? sin y = x M1 cos y dy =1 dx dy 1 1 = = dx cos y 1- x2 M1 A1 (Three) (b) d2 y dx 2 = – 1 1- x2 2 ( ) -Three 2 (- 2 x ) M1 A1 = x 1- x2 ( ) -Three 2 (1 – x ) 2 d2 y dy -x = 1 – x2 x 1 – x2 2 dx dx ( )( ) -Three 2 – x 1- ( 1 2 -2 x ) =Zero M1 A1 (four) (7 marks) NU EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668)
SPECIMEN PAPER MARK SCHEME Question Assignment Quantity Three. Scheme x Zero Marks (a) s=o e ? dy o 2 u 2 e1 + c ? u dx e e dx o u u e dy = sinh x dx 1 y = cosh x, x B1 s = o 1 + sinh 2 x 2 dx Zero [ ] 1 = o cosh x dx = sinh x Zero x M1 A1 (Three) (b) Gradient of tangent dy = tan y = sinh x = s dx s = tan y M1 A1 M1 A1 A1 (2) (c) r= ds = sec2 y dy At y = p , r = sec2 p = 2 four four (Three) (eight marks) 19 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Assignment Quantity four. Scheme I n = o x n sin x dx = x n (- cos x ) p 2 Zero Marks (a) [ ] p 2 Zero – o 2 nx n -1 (- cos x )dx Zero p M1 A1 i i = Zero + ni x n -1 sin x i i [ -o 0 p 2 p 2 0 = n (p ) 2 [ n -1 – (n – 1)I n -2 n -1 ] u i (n – 1)x n- 2 sin x dxy i ? A1 So I n = n(p ) 2 2 – n(n – 1)I n -2 A1 (four) (b) ?p o I Three = 3c ? – Three. 2 I 1 e2o I 1 = o x sin x dx = [x(- cos x )] + o cos x dx Zero p 2 Zero p 2 p 2 Zero M1 = [sin x ] = 1 Zero p 2 A1 3p ? p o I Three = (Three)c ? – 6 = -6 four e 2o 2 2 M1 A1 (four) (eight marks) OM EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Assignment Quantity 5. Scheme x = 2 sinh t Marks B1 (a) (x 2 + four = four sinh 2 t + four ) ( 2 ) 1 2 = 2 cosh t dx = 2 cosh t dt I =o (x + four dx = four o cosh 2 t dt ) M1 A1 = 2 o (cosh 2t + 1) dt = sinh 2t + 2t + c
M1 A1 M1 A1 ft (7) = 1 x 2 (x 2 2 ? xo + four + 2arsinh c ? + c e 2o 2 Zero ) (b) Space = o y dx = o Zero (x ) 2 + four dx 2 ) M1 e1 =e x e2 = 2 ( xu u e x + four u + e 2arsinh u 2u0 u0 e 2 2 1 2 2 eight + 2arsinh (1) 2] = 2 2 + ln Three + 2 A1 2 + 2 ln[1 + ( 2 ) M1 A1 (four) (11 marks) 21 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Assignment Quantity 6. Scheme 2p Zero Marks (a) s=o e e x + y u dt e u e u · 2 1 · u2 2 dy · dx · = x = a (1 – cos t ); = y = a sin t dt dt s=o 2p Zero M1 A1; A1 2p Zero a (1 – cos t ) + sin 2 t 2 dt = a o 2 p ? 2 sin c Zero 2p [ ] 1 [2 – 2 cos t ]2 dt M1 A1, A1 ft (6) 1 = 2a o e ? t ou to ? t , = -4a ecosc ? u = 8a e 2o e e 2 ou Zero 1 o2 (b) s = 2p o = 2p o 2p Zero ? yc x + y ? dt c ? e o 1 22 2p · 2 · 2 2p Zero a 2 (1 – cos t ) 2 dt M1 A1 M1 Three = 8pa 2 o Zero 2p Zero ?to sin Three c ? dt e 2o = 8pa 2 o 2 e t 2 ? t ou e1 – cos c 2 ? u sin 2 dt e ou e 2p 64pa 2 t 2 e Three t u = 8pa e – 2 cos + cos u = 2 Three 2u0 Three e A1 A1 ft (5) (11 marks) OO EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Assignment Quantity 7. Scheme tanh x = sinh x e x – e – x = cosh x e x + e – x B1 Marks (1) (a) (b) 1 y Zero x -1 B1 B1 (2) (c) artanhx = z ? tanh z = x e z – e-z e z + e -z =x M1 A1 e z – e-z = x e z + e-z ( ) 1 – x )e z = (1 + x )e – z e2z = z= 1+ x 1- x 1 ? 1 + x o lnc ? = artanh x 2 e1- x o M1 A1 M1 A1 1 x dx (four) (d) dz 1 ? 1 1 o 1 = c + ? = dx 2 e 1 + x 1 – x o 1 – x 2 o artanh x dx = (x artanh x ) – o 1 – x = (x artanh x ) + 2 M1 A1 A1 (5) 1 ln 1 – x 2 + fixed 2 ( ) (10 marks) 23 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Assignment Quantity eight. Scheme x2 y2 =1 a2 b2 2 x 2 y dy =Zero a 2 b 2 dx Marks (a) M1 A1 M1 A1 dy 2 x b 2 b 2 a sec q b = 2 = 2 = dx a 2 y a b tan q a sin q Gradient of regular is then a sin q b a Equation of regular: ( y – b tan q ) = – sin q (x – a sec q ) b x sin q + by = a 2 + b 2 tan q (b) M: A traditional cuts x = Zero at y = B regular cuts y = Zero at x = ( ) M1 A1 (6) (a 2 + b2 tan q b ) M1 A1 (a = ( ) a2 + b2 tan q a sin q + b2 a cos q 2 ) A1 e a2 + b2 u a2 + b2 sec q , tan q u Therefore M is e 2b e 2a u Eliminating q sec 2 q = 1 + tan 2 q 2 2 ( ) M1 M1 e 2aX u e 2bY u =1+ e 2 e u u ea2 + b2 u ea + b2 u A1 2 4a 2 X 2 – 4b 2Y 2 = a 2 + b 2 [ ] A1 (7) (15 marks) OQ EDEXCEL FURTHER MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Assignment Number one. Scheme Marks ? dy o x Zero = Zero, y Zero = 1, c ? = Zero – 1 = -1 e dx o Zero ? dy o y1 – y Zero = hc ? ? y1 = 1 + (Zero. 1)(- 1) = Zero. e dx o Zero ? dy o x1 = Zero. 1, y1 = Zero. 9, c ? e dx o 1 ? dy o y 2 = y1 + hc ? e dx o 1 = (Zero. 1) – Zero. 9 2 B1 M1 A1 ft A1 = -Zero. 89 = Zero. 9 + (Zero. 1)(- Zero. 89) = Zero. 811 » Zero. 81 z -i ? w( z + 1) = ( z – i ) z +1 M1 A1 (6) (6 marks) 2. (a) w= z (w – 1) = -i – w z= -i-w w -1 -i-w =1 w -1 M1 A1 z =1? i. e. w – 1 = w + i (b) z ? 1? w + i ? w -1 M1 A1 (four) B1 (line) B1 (shading) (2) (6 marks) OR qiea=liEe EDEXCEL FURTHER PURE MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Assignment Quantity Three. Scheme For n = 1, LHS =1, RHS = So result’s true for n = 1 Assume true for n = okay. Then okay +1 r =1 Marks 1 2 M1 A1 r > 2 k2 + okay +1 = = 1 2 1 okay + 2k + 1 + 2 2 1 (okay + 1)2 + 1 2 2 1 M1 A1 ( ) M1 A1 A1 (7) (7 marks) If true for okay, true for okay+1 So true for all constructive integral n d2 y dy dy +y = x, y = Zero, = 2 at x = 1 2 dx dx dx d2 y = Zero +1=1 dx 2 Differentiating with respect to x d Three y ? dy o d2 y + c ? + y 2 =1 dx Three e dx o dx 2 four. B1 M1 A1 d3 y dx Three = -(2) + Zero + 1 = -Three 2 A1 x =1 By Taylor’s Theorem y = Zero + 2(x – 1) + = 2(x – 1) + 1 1 2 Three 1( x – 1) + (- Three)(x – 1) Three! 2! M1 A1 A1 (7) (7 marks) 1 (x – 1)2 – 1 (x – 1)Three 2 2 OS EDEXCEL FURTHER MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Assignment Quantity 5.
Scheme A – lI = Zero Marks (a) (7 – l ) 6 6 =Zero (2 – l ) M1 A1 (7 – l )(2 – l ) – 36 = Zero l2 – 9l + 14 – 36 = Zero l2 – 9l – 22 = Zero (l – 11)(l + 2) = Zero ? l1 = -2, l2 = 11 (b) l = -2 Eigenvector obtained from M1 A1 (four) 6 o ? x1 o ? Zero o ? 7 – (- 2) c ? c ? =c ? c 6 2 – (- 2)? c y 1 ? c Zero ? e oe o e o 3×1 + 2 y1 = Zero ? 2o 1 ? 2o c ? e. g. c ? normalised c – Three? c ? 13 e – 3o e o M1 A1 M1 A1 ft ? – four 6 o ? x2 o ? 0o c ? c ? =c ? l = 11 c ? c ? c ? e 6 – 9o e y2 o e 0o – 2 x2 + Three y 2 = Zero ? 3o 1 ? 3o c ? e. g. c ? normalised c 2? c ? 13 e 2 o e o A1 A1 ft (6) (10 marks) 27 EDEXCEL FURTHER PURE MATHEMATICS FP3 (6669)
SPECIMEN PAPER MARK SCHEME Question Assignment Quantity 6. (a) AB = (- 1, Three, – 1) ; AC = (- 1, Three, 1) . i j okay Scheme Marks M1 A1 AB ? AC = – 1 Three – 1 -1 Three 1 = i (Three + Three) + j (1 + 1) + okay (- Three + Three) = 6i + 2 j M1 A1 A1 Space of D ABC = = 1 AB ? AC 2 1 36 + four = 10 sq. items 2 = = = 1 AD . AB ? AC 6 M1 A1 ft (7) (b) Quantity of tetrahedron ( ) M1 A1 (2) 1 – 12 + eight 6 2 cubic items Three ? ? ® ? ?® (c) Unit vector in path AB ? AC i. e. perpendicular to airplane containing A, B, and C is 1 n= (6i + 2 j) = 1 (3i + j) 10 40 M1 p = n ? AD = 1 10 (3i + j) ? (- 2i + four j) = 1 2 -6+four = items. 10 10 M1 A1 (Three) (12 marks) OU
EDEXCEL FURTHER MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Assignment Quantity Scheme ? 1 x – 1o c ? A( x ) = c Three Zero 2 ? c1 1 Zero ? e o Three o ? – 2 2 c ? Cofactors c – 1 1 x – 1? c 2 x – 5 – 3x ? e o Determinant = 2 x – Three – 2 = 2 x – 5 ? – 2 1 c A (x ) = c 2 2x – 5 c e Three -1 Marks 7. (a) M1 A1 A1 A1 M1 A1 M1 A1 (eight) -1 1 (x – 1) 2x o ? -5 ? – 3x ? o (b) ? 2o ? po ? – 2 – 1 6 o ? 2o c ? 1c c ? ?c ? -1 1 – 5? c Three? c q ? = B c Three? = c 2 c four? 1 c Three cr? 2 – 9? c four? e o e o e oe o M1 A1 ft M1 A1 = (17, – 13, – 24 ) (four) (12 marks) 29 EDEXCEL FURTHER PURE MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Assignment Quantity
Scheme zp + Marks eight. (a) 1 1 = e ipq + ipq p z e = e ipq + e -ipq = 2 cos pq ( ) M1 A1 (2) (b) By De Moivre if z = e iq zp + 1 = 2 cos pq zp four 1o ? four p = 1 : (2 cos q ) = c z + ? zo e M1 A1 M1 A1 1 1 1 1 = z four + four z Three . + 6 z 2 2 + four z. Three + four z z z z 1 o ? 1 o ? = c z four + four ? + 4c z 2 + 2 ? + 6 z o e z o e = 2 cos 4q + eight cos 2q + 6 M1 A1 Three eight cos four q = 1 cos 4q + 1 cos 2q + eight 2 A1 ft (7) (c) V =p o p 2 p 2 p 2 p 2 y dx = p o 2 p 2 p 2 cos four x dx =p o 3o 1 ? 1 c cos 4q + cos 2q + ? dq 8o 2 e8 p M1 A1 ft 1 Three u 2 e1 = p e sin 4q + sin 2q + q u four eight u-p e 32 2 M1 A1 ft Three = p2 eight M1 A1 (6) (15 marks) PM
