Edexcel Maths Fp2 Paper

Paper Reference(s) 6667 Edexcel GCE Further Pure Mathematics FP1 Advanced Level Specimen Disquisition Time: 1 hour 30 minutes Materials required for establishment Reply Dimensions (AB16) Graph Disquisition (ASG2) Mathematical Formulae (Lilac) Items intervening delay doubt disquisitions Nil Candidates may use any calculator EXCEPT those delay the adroitness for symbolic algebra, differentiation and/or integration. Thus claimants 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 reply dimensions, transcribe the indicate of the examining association (Edexcel), your capital calculate, claimant calculate, the disunite heading (Further Pure Mathematics FP1), the disquisition allusion (6667), your surname, moderates and verification. When a calculator is used, the reply should be attached to an mismismisapply plod of foresight. Information for Candidates A dimensionslet ‘Mathematical Formulae and Statistical Tables’ is granted. Full marks may be earned for replys to ALL doubts. This disquisition has view doubts. Advice to Candidates You must fix that your replys to cleverness of doubts are obviously labelled. You must pretext enough afloat to bring-environing your methods acquitted to the Examiner. Answers delayout afloat may experience no trustworthiness. This promulgation may barely be reproduced in correspondence delay London Qualifications Limited copyright cunning. Edexcel Foundation is a registered passion. ©2003 London Qualifications Limited 1. Likeness 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) Pretext that the parent a of the equation f(x) = 0 lies in the season 3 < a < 4 . (2) (b) Taking 3. 6 as your starting treasure, employ the Newton-Raphson progress once to f(x) to earn a relieve path to a. Give your reply to 4 decimal places. (5) 3. Experience the set of treasures 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 parent. Solve the equation and mention the treasure of p. (7) 5. (a) Earn the open key of the differential equation dS - 0. 1S = t. dt (6) (b) The differential equation in disunite (a) is used to mould the property, ? S favorite, of a bank t years following it was set up. Attached that the moderate property of the bank were ? 200 favorite, use your reply to disunite (a) to deem, to the direct ? illion, the property of the bank 10 years following it was set up. (4) 2 6. The deflexion C has polar equation r 2 = a 2 cos 2q , -p p ? q ? . 4 4 (a) Sketch the deflexion C. (2) (b) Find the polar coordinates of the subject-matters where tangents to C are correspondent to the moderate method. (6) (c) Experience the area of the district limited by C. (4) 7. Attached that z = -3 + 4i and zw = -14 + 2i, experience (a) w in the contrive p + iq where p and q are legitimate, (4) (b) the modulus of z and the reasoning of z in radians to 2 decimal places (4) (c) the treasures of the legitimate trustworthys m and n such that mz + nzw = -10 - 20i . (5) 3 Turn aggravate 8. (a) Attached that x = e t , pretext 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 replys to disunite (a) to pretext that the superabundance 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) Hereafter experience the open key of x2 d2 y dy - 2x + 2y = x3. 2 dx dx (6) END 4 Disquisition Reference(s) 6668 Edexcel GCE Further Pure Mathematics FP2 Advanced Level Specimen Disquisition Time: 1 hour 30 minutes Materials required for establishment Reply Dimensions (AB16) Graph Disquisition (ASG2) Mathematical Formulae (Lilac) Items intervening delay doubt disquisitions Nil Candidates may use any calculator EXCEPT those delay the adroitness for symbolic algebra, differentiation and/or integration. Thus claimants 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 reply dimensions, transcribe the indicate of the examining association (Edexcel), your capital calculate, claimant calculate, the disunite heading (Further Pure Mathematics FP2), the disquisition allusion (6668), your surname, moderates and verification. When a calculator is used, the reply should be attached to an mismismisapply plod of foresight. Information for Candidates A dimensionslet ‘Mathematical Formulae and Statistical Tables’ is granted. Full marks may be earned for replys to ALL doubts. This disquisition has view doubts. Advice to Candidates You must fix that your replys to cleverness of doubts are obviously labelled. You must pretext enough afloat to bring-environing your methods acquitted to the Examiner. Answers delayout afloat may experience no trustworthiness. This promulgation may barely be reproduced in correspondence delay London Qualifications Limited copyright cunning. Edexcel Foundation is a registered passion. ©2003 London Qualifications Limited 1. The misinterpretation x of a disuniteicle from a unwandering subject-matter O at age t is attached by x = sinh t. 4 At age T the misinterpretation x = . 3 (a) Experience cosh T . (2) (b) Hereafter experience e T and T. (3) 2. Attached that y = arcsin x likeness 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 pretexts the deflexion C delay equation y = cosh x. The tangent at P bring-abouts an superscription y delay the x-axis and the arc protraction from A(0, 1) to P(x, y) is s. (a) Pretext that s = sinh x. (3) (a) By regarding the gradient of the tangent at P pretext that the native equation of C is s = tan y. 2) (c) Experience the radius of bend r at the subject-matter where y = p . 4 (3) S 4. I n = o x n sin x dx. p 2 0 (a) Pretext that for n ? 2 ?p o I n = nc ? e 2o n -1 - n(n - 1)I n - 2 . (4) (4) (b) Hereafter earn I 3 , giving your replys in provisions of p. 5. (a) Experience ? v(x2 + 4) dx. (7) The deflexion C has equation y 2 - x 2 = 4. (b) Use your reply to disunite (a) to experience the area of the limited district limited by C, the absolute x-axis, the absolute y-axis and the method x = 2, giving your reply in the contrive p + ln q where p and q are trustworthys to be base. (4) Figure 2 6. y O 2pa x The parametric equations of the deflexion C pretextn in Fig. are x = a(t - sin t ), y = a(1 - cos t ), 0 ? t ? 2p . (a) Find, by using integration, the protraction of C. (6) The deflexion C is rotated through 2p environing Ox. (b) Experience the manner area of the substantial generated. (5) 7 7. (a) Using the definitions of sinh x and cosh x in provisions of exponential capacitys, specific tanh x in provisions 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) Likeness that artanh x = (4) (d) Hereafter earn d (artanh x) and use integration by cleverness to pretext that dx o artanh x dx = x artanh x + 1 ln 1 - x 2 + trustworthy. 2 ( ) (5) 8. The hyperbola C has equation x2 y2 = 1. a2 b2 (a) Pretext that an equation of the recognized to C at P(a sec q , b tan q ) is by + ax sin q = a 2 + b 2 tan q . (6) ( ) The recognized at P cuts the coordinate axes at A and B. The mid-subject-matter of AB is M. (b) Find, in cartesian contrive, an equation of the locus of M as q varies. (7) END U Disquisition Reference(s) 6669 Edexcel GCE Further Pure Mathematics FP3 Advanced Level Specimen Disquisition Time: 1 hour 30 minutes Materials required for establishment Reply Dimensions (AB16) Graph Disquisition (ASG2) Mathematical Formulae (Lilac) Items intervening delay doubt disquisitions Nil Candidates may use any calculator EXCEPT those delay the adroitness for symbolic algebra, differentiation and/or integration. Thus claimants 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 reply dimensions, transcribe the indicate of the examining association (Edexcel), your capital calculate, claimant calculate, the disunite heading (Further Pure Mathematics FP3), the disquisition allusion (6669), your surname, moderates and verification. When a calculator is used, the reply should be attached to an mismismisapply plod of foresight. Information for Candidates A dimensionslet ‘Mathematical Formulae and Statistical Tables’ is granted. Full marks may be earned for replys to ALL doubts. This disquisition has view doubts. Advice to Candidates You must fix that your replys to cleverness of doubts are obviously labelled. You must pretext enough afloat to bring-environing your methods acquitted to the Examiner. Answers delayout afloat may experience no trustworthiness. This promulgation may barely be reproduced in correspondence delay London Qualifications Limited copyright cunning. Edexcel Foundation is a registered passion. ©2003 London Qualifications Limited 1. y = x 2 - y, y = 1 at x = 0 . dx y - y0 ? dy o Use the path c ? » 1 delay a plod protraction of 0. 1 to deem the treasures of y h e dx o 0 at x = 0. 1 and x = 0. 2, giving your replys to 2 forcible figures. (6) 2. (a) Pretext that the alteration w= z -i z +1 maps the foe z = 1 in the z-roll to the method w - 1 = w + i in the w-plane. (4) The district z ? 1 in the z-roll is mapped to the district R in the w-plane. (b) Shade the district R on an Argand diagram. (2) 3. Likeness by collation 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 sequence key of the differential equation in ascending powers of (x – 1) up to and including the promise in (x - 1)3. (7) 5. ? 7 6o A=c c 6 2? . ? e o (a) Experience the eigenvalues of A. (4) (a) Earn the identical recognizedised eigenvectors. (6) NM 6. The subject-matters A, B, C, and D possess comcomposition vectors a = 2i + k , b = i + 3j, c = i + 3 j + 2k , d = 4 j + k referenceively. (a) Experience AB ? AC and hereafter experience the area of trisuperscription ABC. (7) (b) Experience the dimensions of the tetrahedron ABCD. (2) (c) Experience the vertical interspace of D from the roll 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 effigy of the vector c q ? when transformed by B is cr? e o (b) Experience the treasures of p, q and r. (4) ? 2o c ? c 3? . c 4? e o 11 8. (a) Attached that z = e iq , pretext that zp + 1 = 2 cos pq , zp where p is a absolute integer. (2) (b) Attached that cos 4 q = A cos 4q + B cos 2q + C , experience the treasures of the trustworthys A, B and C. (7) The district R limited by the deflexion delay equation y = cos 2 x, rotated through 2p environing the x-axis. (c) Experience the dimensions of the substantial generated. (6) p p ? x ? , and the x-axis is 2 2 END NO EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Doubt calculate 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 inconsistent indication and so f ( x ) has parent 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 Parent » 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 Doubt calculate 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 regularly absolute Critical subject-matters 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 treasures x < 2 and x > 3 {x : x < 2} E {x : x > 3} 4. If 1 – 2i is a parent, then so is 1 + 2i B1 M1 A1 M1 A1 ft A1 A1 (7) x - 1 + 2i )(x - 1 - 2i ) are rudiments of f(x) so x 2 - 2 x + 5 is a rudiment of f (x) f ( x ) = x 2 - 2 x + 5 (2 x - 1) Third parent is 1 2 ( ) and p = 12 (7 marks) 5. (a) dS - (0. 1)S = t dt - ( 0. 1)dt Integrating rudiment 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) Property ? 615 favorite NQ EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Doubt calculate 6. (a) l Scheme Marks q B1 (Shape) B1 (Labels) (2) (b) Tangent correspondent to moderate method when y = r sin q is motionless Consider consequently 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 subject-matters 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 Doubt calculate 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 legitimate and unreal cleverness 3m + 14n = 10, 4m + 2n = -20 Solving to earn m = -6, n = 2 NS EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Doubt calculate 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 capacity y = Ae t + Be 2t e 3t 1 Particular all = 2 = e 3t 3 - (3 ? 3) + 2 2 Open key 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 Doubt Calculate 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 = Hereafter 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 Doubt Calculate 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 Doubt Calculate 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 Doubt Calculate 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 Doubt Calculate 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 Doubt Calculate 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 + trustworthy 2 ( ) (10 marks) 23 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Doubt Calculate 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 recognized is then a sin q b a Equation of recognized: ( y - b tan q ) = - sin q (x - a sec q ) b x sin q + by = a 2 + b 2 tan q (b) M: A recognized cuts x = 0 at y = B recognized 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 Hereafter 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 Doubt Calculate 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 Doubt Calculate 3. Scheme For n = 1, LHS =1, RHS = So fruit is penny for n = 1 Assume penny 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 penny for k, penny for k+1 So penny for all absolute all 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 delay reference 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 Doubt Calculate 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 earned 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 3x1 + 2 y1 = 0 ? 2o 1 ? 2o c ? e. g. c ? recognizedised 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 ? recognizedised 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 Doubt Calculate 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 clear cleverness 2 = = = 1 AD . AB ? AC 6 M1 A1 ft (7) (b) Dimensions of tetrahedron ( ) M1 A1 (2) 1 - 12 + 8 6 2 weighty cleverness 3 ? ? ® ? ?® (c) Disunite vector in superscription AB ? AC i. e. vertical to roll 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 = cleverness. 10 10 M1 A1 (3) (12 marks) OU EDEXCEL FURTHER MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Doubt Calculate 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 Doubt 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