Paper Reference(s) 6667 Edexcel GCE Further Pure Mathematics FP1 Advanced Level Specimen Paper Time: 1 hour 30 minutes Materials required for examination Answer Book (AB16) Graph Paper (ASG2) Mathematical Formulae (Lilac) Items included with question papers Nil Candidates may use any calculator EXCEPT those with the facility for symbolic algebra, differentiation and/or integration. Thus candidates may NOT use calculators such as the Texas Instruments TI-89, TI-92, Casio CFX-9970G, Hewlett Packard HP 48G.

Instructions to Candidates In the boxes on the answer book, write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Further Pure Mathematics FP1), the paper reference (6667), your surname, initials and signature. When a calculator is used, the answer should be given to an appropriate degree of accuracy. Information for Candidates A booklet ‘Mathematical Formulae and Statistical Tables’ is provided. Full marks may be obtained for answers to ALL questions. This paper has eight questions. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled.

You must show sufficient working to make your methods clear to the Examiner. Answers without working may gain no credit. This publication may only be reproduced in accordance with London Qualifications Limited copyright policy. Edexcel Foundation is a registered charity. ©2003 London Qualifications Limited 1. Prove that a (r r =1 n 2 – r -1 = ) 1 (n – 2)n(n + 2) . 3 (5) 2. 1 f ( x ) = ln x – 1 – . x (a) Show that the root a of the equation f(x) = 0 lies in the interval 3 < a < 4 . (2) (b) Taking 3. 6 as your starting value, apply the Newton-Raphson procedure once to f(x) to obtain a second approximation to a.

Give your answer to 4 decimal places. (5) 3. Find the set of values of x for which 1 x > . x -3 x -2 (7) 4. f ( x ) ? 2 x 3 – 5 x 2 + px – 5, p I ?. The equation f (x) = 0 has (1 – 2i) as a root. Solve the equation and determine the value of p. (7) 5. (a) Obtain the general solution of the differential equation dS – 0. 1S = t. dt (6) (b) The differential equation in part (a) is used to model the assets, ? S million, of a bank t years after it was set up. Given that the initial assets of the bank were ? 200 million, use your answer to part (a) to estimate, to the nearest ? illion, the assets of the bank 10 years after it was set up. (4) 2 6. The curve C has polar equation r 2 = a 2 cos 2q , -p p ? q ? . 4 4 (a) Sketch the curve C. (2) (b)

Find the polar coordinates of the points where tangents to C are parallel to the initial line. (6) (c) Find the area of the region bounded by C. (4) 7. Given that z = -3 + 4i and zw = -14 + 2i, find (a) w in the form p + iq where p and q are real, (4) (b) the modulus of z and the argument of z in radians to 2 decimal places (4) (c) the values of the real constants m and n such that mz + nzw = -10 – 20i . (5) 3 Turn over 8. (a) Given that x = e t , show 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 answers to part (a) to show 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 – 3 + 2 y = e 3t . 2 dt dt (3) (c) Hence find the general solution of x2 d2 y dy – 2x + 2y = x3. 2 dx dx (6) END 4 Paper Reference(s) 6668 Edexcel GCE Further Pure Mathematics FP2 Advanced Level Specimen Paper Time: 1 hour 30 minutes Materials required for examination Answer Book (AB16) Graph Paper (ASG2) Mathematical Formulae (Lilac) Items included with question papers Nil

Candidates may use any calculator EXCEPT those with the facility for symbolic algebra, differentiation and/or integration. Thus candidates may NOT use calculators such as the Texas Instruments TI 89, TI 92, Casio CFX-9970G, Hewlett Packard HP 48G. Instructions to Candidates In the boxes on the answer book, write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Further Pure Mathematics FP2), the paper reference (6668), your surname, initials and signature. When a calculator is used, the answer should be given to an appropriate degree of accuracy.

Information for Candidates A booklet ‘Mathematical Formulae and Statistical Tables’ is provided. Full marks may be obtained for answers to ALL questions. This paper has eight questions. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You must show sufficient working to make your methods clear to the Examiner. Answers without working may gain no credit. This publication may only be reproduced in accordance with London Qualifications Limited copyright policy. Edexcel Foundation is a registered charity. ©2003 London Qualifications Limited 1.

The displacement x of a particle from a fixed point O at time t is given by x = sinh t. 4 At time T the displacement x = . 3 (a) Find cosh T . (2) (b) Hence find e T and T. (3) 2. Given that y = arcsin x prove that (a) dy = dx (1 – x ) 2 1 , (3) (b) (1 – x 2 ) d2 y dy -x = 0. 2 dx dx (4) Figure 1 3. y P(x, y) s A y O x Figure 1 shows the curve C with equation y = cosh x. The tangent at P makes an angle y with the x-axis and the arc length from A(0, 1) to P(x, y) is s. (a) Show that s = sinh x. (3) (a) By considering the gradient of the tangent at P show that the intrinsic equation of C is s = tan y. 2) (c) Find the radius of curvature r at the point where y = p . 4 (3) S 4. I n = o x n sin x dx. p 2 0 (a) Show that for n ? 2 ?p o I n = nc ? e 2o n -1 – n(n – 1)I n – 2 . (4) (4) (b) Hence obtain I 3 , giving your answers in terms of p. 5. (a) Find ? v(x2 + 4) dx. (7) The curve C has equation y 2 – x 2 = 4. (b) Use your answer to part (a) to find the area of the finite region bounded by C, the positive x-axis, the positive y-axis and the line x = 2, giving your answer in the form p + ln q where p and q are constants to be found. (4) Figure 2 6. y O 2pa x The parametric equations of the curve C shown in Fig. are x = a(t – sin t ), y = a(1 – cos t ), 0 ? t ? 2p . (a) Find, by using integration, the length of C. (6) The curve C is rotated through 2p about Ox. (b) Find the surface area of the solid generated. (5) 7 7. (a) Using the definitions of sinh x and cosh x in terms of exponential functions, express tanh x in terms 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) Prove that artanh x = (4) (d) Hence obtain d (artanh x) and use integration by parts to show that dx o artanh x dx = x artanh x + 1 ln 1 – x 2 + constant. 2 ( ) (5) 8.

The hyperbola C has equation x2 y2 = 1. a2 b2 (a) Show that an equation of the normal to C at P(a sec q , b tan q ) is by + ax sin q = a 2 + b 2 tan q . (6) ( ) The normal at P cuts the coordinate axes at A and B. The mid-point of AB is M. (b) Find, in cartesian form, an equation of the locus of M as q varies. (7) END U Paper Reference(s) 6669 Edexcel GCE Further Pure Mathematics FP3 Advanced Level Specimen Paper Time: 1 hour 30 minutes Materials required for examination Answer Book (AB16) Graph Paper (ASG2) Mathematical Formulae (Lilac) Items included with question papers Nil

Candidates may use any calculator EXCEPT those with the facility for symbolic algebra, differentiation and/or integration. Thus candidates may NOT use calculators such as the Texas Instruments TI 89, TI 92, Casio CFX 9970G, Hewlett Packard HP 48G. Instructions to Candidates In the boxes on the answer book, write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Further Pure Mathematics FP3), the paper reference (6669), your surname, initials and signature. When a calculator is used, the answer should be given to an appropriate degree of accuracy.

Information for Candidates A booklet ‘Mathematical Formulae and Statistical Tables’ is provided. Full marks may be obtained for answers to ALL questions. This paper has eight questions. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You must show sufficient working to make your methods clear to the Examiner. Answers without working may gain no credit. This publication may only be reproduced in accordance with London Qualifications Limited copyright policy. Edexcel Foundation is a registered charity. ©2003 London Qualifications Limited 1. y = x 2 – y, y = 1 at x = 0 . dx y – y0 ? dy o Use the approximation c ? » 1 with a step length of 0. 1 to estimate the values of y h e dx o 0 at x = 0. 1 and x = 0. 2, giving your answers to 2 significant figures. (6) 2. (a) Show that the transformation w= z -i z +1 maps the circle z = 1 in the z-plane to the line w – 1 = w + i in the w-plane. (4) The region z ? 1 in the z-plane is mapped to the region R in the w-plane. (b) Shade the region R on an Argand diagram. (2) 3. Prove by induction that, all integers n, n ? 1 , ar > 2 n r =1 n 1 2 . (7) 4. dy d2 y dy +y = x, y = 0, = 2 at x = 1. 2 dx dx dx

Find a series solution of the differential equation in ascending powers of (x – 1) up to and including the term in (x – 1)3. (7) 5. ? 7 6o A=c c 6 2? . ? e o (a) Find the eigenvalues of A. (4) (a) Obtain the corresponding normalised eigenvectors. (6) NM 6. The points A, B, C, and D have position vectors a = 2i + k , b = i + 3j, c = i + 3 j + 2k , d = 4 j + k respectively. (a) Find AB ? AC and hence find the area of triangle ABC. (7) (b) Find the volume of the tetrahedron ABCD. (2) (c) Find the perpendicular distance of D from the plane containing A, B and C. (3) 7. ? 1 x – 1o c ? 5 A( x) = c 3 0 2 ? , x ? 2 c1 1 0 ? e o (a) Calculate the inverse of A(x). (8) ? 1 3 – 1o c ? B = c3 0 2 ? . c1 1 0 ? e o ? po c ? The image of the vector c q ? when transformed by B is cr? e o (b) Find the values of p, q and r. (4) ? 2o c ? c 3? . c 4? e o 11 8. (a) Given that z = e iq , show that zp + 1 = 2 cos pq , zp where p is a positive integer. (2) (b) Given that cos 4 q = A cos 4q + B cos 2q + C , find the values of the constants A, B and C. (7) The region R bounded by the curve with equation y = cos 2 x, rotated through 2p about the x-axis. (c) Find the volume of the solid 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 number 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 – 8 6 [ ] M1 A1 A1 (5) (5 marks) 1 n(n – 2 )(n + 2 ) 3 2. (a) f ( x) = ln x – 1 – 1 x f (3) = ln 3 – 1 – 1 = -0. 2347 3 f (4) = ln 4 – 1 – 1 = 0. 1363 4 f (3) and f (4) are of opposite sign and so f ( x ) has root in (3, 4) (b) x 0 = 3. 6 f ? (x ) = 1 1 + x x2 M1 A1 (2) M1 A1 f ? (3. 6 ) = 0. 354 381 f (3. 6) = 0. 003 156 04 Root » 3. – f (3. 6) f ? (3. 6) M1 A1 ft A1 (5) (7 marks) » 3. 5911 13 EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question number 3. Scheme x x x 2 – 3x + 3 1 1 > ? >0 ? >0 x-3 x-2 x-3 x-2 (x – 3)(x – 2 ) Marks M1 A1 B1 B1 Numerator always positive Critical points of denominator x = 2, x = 3 x < 2 : den = (- ve)(- ve) = + ve 2 < x < 3 : den = (- ve)(+ ve) = – ve 3 < x : den = (+ ve)(+ ve) = + ve M1 A1 A1 (7) (7 marks) Set of values x < 2 and x > 3 {x : x < 2} E {x : x > 3} 4. 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 factors of f(x) so x 2 – 2 x + 5 is a factor 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 – (0. 1)S = t dt – ( 0. 1)dt Integrating factor e o = e -(0. 1)t M1 d Se – (0. 1)t = te – (0. 1)t dt Se – (0. 1)t = o te – (0. 1)t dt = -10te – (0. 1)t – 100e – (0. 1)t + C [ ] A1 A1 M1 A1 A1 (6) S = Ce (0. 1)t – 10t – 100 (b) S = 200 at t = 0 ? 200 = C – 100 i. e. C = 300 S = 300e (0. 1)t – 10t – 100 M1 A1 At t = 10, S = 300e – 100 – 100 = 615. 484 55 M1 A1 ft (4) (10 marks) Assets ? 615 million NQ EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667)

SPECIMEN PAPER MARK SCHEME Question number 6. (a) l Scheme Marks q B1 (Shape) B1 (Labels) (2) (b) Tangent parallel to initial line when y = r sin q is stationary Consider therefore d 2 a cos 2q sin 2 q dq ( ) M1 A1 = -2 sin 2q sin 2 q + cos 2q (2 sin q cos q ) =0 2 sin q [cos 2q cos q – sin 2q sin q ] = 0 sin q ? 0 ? cos 3q = 0 ? q = p -p or 6 6 M1 A1 o ? ? o ? 1 p o? 1 -p Coordinates of the points c c a, ? c a, ? c 6 6 oe 2 e 2 A1 A1 (6) 1 o4 2 1 2o4 (c) Area = o r dq = a o cos 2q dq 2 o -p 2 o -p 4 4 p p M1 A1 a2 a2 1 2 e sin 2q u = a e = [1 – (- 1)] = 2 e 2 u -4p 4 2 u p 4 M1 A1 (4) (12 marks) 15

EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question number 7. (a) z = -3 + 4i, zw = -14 + 2i Scheme Marks w= = = – 14 + 2i (- 14 + 2i )(- 3 – 4i ) = (- 3 + 4i )(- 3 – 4i ) – 3 + 4i M1 A1 A1 A1 M1 A1 M1 A1 M1 A1 A1 M1 A1 (5) (13 marks) (4) (42 + 8) + i(- 6 + 56) 9 + 16 50 + 50i = 2 + 2i 25 (4) (b) z = (3 2 + 42 = 5 4 = 2. 21 3 ) arg z = p – arctan (c) Equating real and imaginary parts 3m + 14n = 10, 4m + 2n = -20 Solving to obtain m = -6, n = 2 NS EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question number 8. (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 (3) d2 y dy – 3 + 2 y = e 3t 2 dt dt (c) Auxiliary equation m 2 – 3m + 2 = 0 (m – 1)(m – 2) = 0 Complementary function y = Ae t + Be 2t e 3t 1 Particular integral = 2 = e 3t 3 – (3 ? 3) + 2 2 General solution y = Ae t + Be 2t + 1 e 3t 2 = Ax + Bx 2 + 1 x 3 2 M1 A1 M1 A1 M1 A1 ft 6) (14 marks) 17 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 1. cosh 2 T = 1 + sinh 2 T = 1 + 16 25 = 9 9 Scheme Marks M1 A1 (2) M1 A1 A1 ft (3) cosh T = ± 5 5 = since cosh T > 1 3 3 4 5 + =3 3 3 e T = cosh T + sinh T = Hence T = ln 3 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 (3) (b) d2 y dx 2 = – 1 1- x2 2 ( ) -3 2 (- 2 x ) M1 A1 = x 1- x2 ( ) -3 2 (1 – x ) 2 d2 y dy -x = 1 – x2 x 1 – x2 2 dx dx ( )( ) -3 2 – x 1- ( 1 2 -2 x ) =0 M1 A1 (4) (7 marks) NU EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668)

SPECIMEN PAPER MARK SCHEME Question Number 3. Scheme x 0 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 0 [ ] 1 = o cosh x dx = sinh x 0 x M1 A1 (3) (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 4 4 (3) (8 marks) 19 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 4. Scheme I n = o x n sin x dx = x n (- cos x ) p 2 0 Marks (a) [ ] p 2 0 – o 2 nx n -1 (- cos x )dx 0 p M1 A1 i i = 0 + 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 (4) (b) ?p o I 3 = 3c ? – 3. 2 I 1 e2o I 1 = o x sin x dx = [x(- cos x )] + o cos x dx 0 p 2 0 p 2 p 2 0 M1 = [sin x ] = 1 0 p 2 A1 3p ? p o I 3 = (3)c ? – 6 = -6 4 e 2o 2 2 M1 A1 (4) (8 marks) OM EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 5. Scheme x = 2 sinh t Marks B1 (a) (x 2 + 4 = 4 sinh 2 t + 4 ) ( 2 ) 1 2 = 2 cosh t dx = 2 cosh t dt I =o (x + 4 dx = 4 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 + 4 + 2arsinh c ? + c e 2o 2 0 ) (b) Area = o y dx = o 0 (x ) 2 + 4 dx 2 ) M1 e1 =e x e2 = 2 ( xu u e x + 4 u + e 2arsinh u 2u0 u0 e 2 2 1 2 2 8 + 2arsinh (1) 2] = 2 2 + ln 3 + 2 A1 2 + 2 ln[1 + ( 2 ) M1 A1 (4) (11 marks) 21 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 6. Scheme 2p 0 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 0 M1 A1; A1 2p 0 a (1 – cos t ) + sin 2 t 2 dt = a o 2 p ? 2 sin c 0 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 0 1 o2 (b) s = 2p o = 2p o 2p 0 ? yc x + y ? dt c ? e o 1 22 2p · 2 · 2 2p 0 a 2 (1 – cos t ) 2 dt M1 A1 M1 3 = 8pa 2 o 0 2p 0 ?to sin 3 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 3 t u = 8pa e – 2 cos + cos u = 2 3 2u0 3 e A1 A1 ft (5) (11 marks) OO EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 7. Scheme tanh x = sinh x e x – e – x = cosh x e x + e – x B1 Marks (1) (a) (b) 1 y 0 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 (4) (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 + constant 2 ( ) (10 marks) 23 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 8. Scheme x2 y2 =1 a2 b2 2 x 2 y dy =0 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 normal is then a sin q b a Equation of normal: ( y – b tan q ) = – sin q (x – a sec q ) b x sin q + by = a 2 + b 2 tan q (b) M: A normal cuts x = 0 at y = B normal cuts y = 0 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 Hence 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 Number 1. Scheme Marks ? dy o x 0 = 0, y 0 = 1, c ? = 0 – 1 = -1 e dx o 0 ? dy o y1 – y 0 = hc ? ? y1 = 1 + (0. 1)(- 1) = 0. e dx o 0 ? dy o x1 = 0. 1, y1 = 0. 9, c ? e dx o 1 ? dy o y 2 = y1 + hc ? e dx o 1 = (0. 1) – 0. 9 2 B1 M1 A1 ft A1 = -0. 89 = 0. 9 + (0. 1)(- 0. 89) = 0. 811 » 0. 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 (4) B1 (line) B1 (shading) (2) (6 marks) OR qiea=liEe EDEXCEL FURTHER PURE MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number 3. Scheme For n = 1, LHS =1, RHS = So result is true for n = 1 Assume true for n = k. Then k +1 r =1 Marks 1 2 M1 A1 r > 2 k2 + k +1 = = 1 2 1 k + 2k + 1 + 2 2 1 (k + 1)2 + 1 2 2 1 M1 A1 ( ) M1 A1 A1 (7) (7 marks) If true for k, true for k+1 So true for all positive integral n d2 y dy dy +y = x, y = 0, = 2 at x = 1 2 dx dx dx d2 y = 0 +1=1 dx 2 Differentiating with respect to x d 3 y ? dy o d2 y + c ? + y 2 =1 dx 3 e dx o dx 2 4. B1 M1 A1 d3 y dx 3 = -(2) + 0 + 1 = -3 2 A1 x =1 By Taylor’s Theorem y = 0 + 2(x – 1) + = 2(x – 1) + 1 1 2 3 1( x – 1) + (- 3)(x – 1) 3! 2! M1 A1 A1 (7) (7 marks) 1 (x – 1)2 – 1 (x – 1)3 2 2 OS EDEXCEL FURTHER MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number 5.

Scheme A – lI = 0 Marks (a) (7 – l ) 6 6 =0 (2 – l ) M1 A1 (7 – l )(2 – l ) – 36 = 0 l2 – 9l + 14 – 36 = 0 l2 – 9l – 22 = 0 (l – 11)(l + 2) = 0 ? l1 = -2, l2 = 11 (b) l = -2 Eigenvector obtained from M1 A1 (4) 6 o ? x1 o ? 0 o ? 7 – (- 2) c ? c ? =c ? c 6 2 – (- 2)? c y 1 ? c 0 ? e oe o e o 3×1 + 2 y1 = 0 ? 2o 1 ? 2o c ? e. g. c ? normalised c – 3? c ? 13 e – 3o e o M1 A1 M1 A1 ft ? – 4 6 o ? x2 o ? 0o c ? c ? =c ? l = 11 c ? c ? c ? e 6 – 9o e y2 o e 0o – 2 x2 + 3 y 2 = 0 ? 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 Number 6. (a) AB = (- 1, 3, – 1) ; AC = (- 1, 3, 1) . i j k Scheme Marks M1 A1 AB ? AC = – 1 3 – 1 -1 3 1 = i (3 + 3) + j (1 + 1) + k (- 3 + 3) = 6i + 2 j M1 A1 A1 Area of D ABC = = 1 AB ? AC 2 1 36 + 4 = 10 square units 2 = = = 1 AD . AB ? AC 6 M1 A1 ft (7) (b) Volume of tetrahedron ( ) M1 A1 (2) 1 – 12 + 8 6 2 cubic units 3 ? ? ® ? ?® (c) Unit vector in direction AB ? AC i. e. perpendicular to plane 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 + 4 j) = 1 2 -6+4 = units. 10 10 M1 A1 (3) (12 marks) OU

EDEXCEL FURTHER MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number Scheme ? 1 x – 1o c ? A( x ) = c 3 0 2 ? c1 1 0 ? e o 3 o ? – 2 2 c ? Cofactors c – 1 1 x – 1? c 2 x – 5 – 3x ? e o Determinant = 2 x – 3 – 2 = 2 x – 5 ? – 2 1 c A (x ) = c 2 2x – 5 c e 3 -1 Marks 7. (a) M1 A1 A1 A1 M1 A1 M1 A1 (8) -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 3? c q ? = B c 3? = c 2 c 4? 1 c 3 cr? 2 – 9? c 4? e o e o e oe o M1 A1 ft M1 A1 = (17, – 13, – 24 ) (4) (12 marks) 29 EDEXCEL FURTHER PURE MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number

Scheme zp + Marks 8. (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 4 1o ? 4 p = 1 : (2 cos q ) = c z + ? zo e M1 A1 M1 A1 1 1 1 1 = z 4 + 4 z 3 . + 6 z 2 2 + 4 z. 3 + 4 z z z z 1 o ? 1 o ? = c z 4 + 4 ? + 4c z 2 + 2 ? + 6 z o e z o e = 2 cos 4q + 8 cos 2q + 6 M1 A1 3 8 cos 4 q = 1 cos 4q + 1 cos 2q + 8 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 4 x dx =p o 3o 1 ? 1 c cos 4q + cos 2q + ? dq 8o 2 e8 p M1 A1 ft 1 3 u 2 e1 = p e sin 4q + sin 2q + q u 4 8 u-p e 32 2 M1 A1 ft 3 = p2 8 M1 A1 (6) (15 marks) PM

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