Paper Reference(s)



FP1 PAST PAPERSwith mark schemesJune 2014 back to January 2010Included:June 2014June 2014 (R)June 2013June 2013 (R)Jan 2013June 2012Jan 2012June 2011Feb 2010Omitted:June 2013 (withdrawn paper)January 2011June 2010Paper Reference(s)6667/01Edexcel GCEFurther Pure Mathematics FP1Advanced/Advanced SubsidiaryTuesday 10 June 2014 MorningTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) NilCandidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation or symbolic differentiation/integration, or haveretrievable mathematical formulae stored in them.-825520066000Instructions to CandidatesIn the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions.You must write your answer for each question in the space following the question.When a calculator is used, the answer should be given to an appropriate degree of accuracy.-762019558000Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 9 questions in this question paper. The total mark for this paper is 75.There are 28 pages in this question paper. Any blank pages are indicated.-889019939000Advice to CandidatesYou 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 not gain full credit.1.The complex numbers z1 and z2 are given byz1 = p + 2i and z2 = 1 – 2iwhere p is an integer.(a)Find in the form a + bi where a and b are real. Give your answer in its simplest form in terms of p.(4)Given that ,(b)find the possible values of p.(4)3175106680002.,x > 0(a)Show that the equation f(x) = 0 has a root α in the interval [1.1, 1.5].(2)(b)Find f ?(x).(2)(c)Using x0 = 1.1 as a first approximation to α, apply the Newton-Raphson procedure once to f(x) to find a second approximation to α, giving your answer to 3 decimal places.(3)23495108585003. Given that 2 and 1 – 5i are roots of the equationx3 + px2 + 30x + q = 0,(a)write down the third root of the equation.(1)(b)Find the value of p and the value of q.(5)(c)Show the three roots of this equation on a single Argand diagram.(2)-17145102870004.(i) Given that and ,(a) find AB.(b) Explain why AB ≠ BA.(4)(ii) Given that, where k is a real numberfind C–1, giving your answer in terms of k.(3)-22225121285005.(a) Use the standard results for and to show that(6)(b) Hence show thatwhere a and b are constants to be found.(3)-11430155575006.The rectangular hyperbola H has cartesian equation xy = c2.The point P, t > 0, is a general point on H.(a) Show that an equation of the tangent to H at the point P ist2y + x = 2ct(4)An equation of the normal to H at the point P is t3x – ty = ct4 – c.Given that the normal to H at P meets the x-axis at the point A and the tangent to H at P meets the x-axis at the point B,(b)find, in terms of c and t, the coordinates of A and the coordinates of B.(2)Given that c = 4,(c)find, in terms of t, the area of the triangle APB. Give your answer in its simplest form.(3)3175100330007.(i) In each of the following cases, find a 2 × 2 matrix that represents(a)a reflection in the line y = –x,(b)a rotation of 135° anticlockwise about (0, 0),(c)a reflection in the line y = –x followed by a rotation of 135° anticlockwise about (0, 0).(4)(ii) The triangle T has vertices at the points (1, k), (3, 0) and (11, 0), where k is a constant. Triangle T is transformed onto the triangle T ? by the matrixGiven that the area of triangle T ? is 364 square units, find the value of k.(6)3175100330008.The points P(4k2, 8k) and Q(k2, 4k), where k is a constant, lie on the parabola C with equation y2 = 16x.The straight line l1 passes through the points P and Q.(a)Show that an equation of the line l1 is given by3ky – 4x = 8k2(4)The line l2 is perpendicular to the line l1 and passes through the focus of the parabola C.The line l2 meets the directrix of C at the point R.(b)Find, in terms of k, the y coordinate of the point R.(7)3175100330009.Prove by induction that, for ,f(n) = 8n – 2nis divisible by 6.(6)2159012954000TOTAL FOR PAPER: 75 MARKSENDPaper Reference(s)6667/01REdexcel GCEFurther Pure Mathematics FP1 (R)Advanced/Advanced SubsidiaryTuesday 10 June 2014 MorningTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) NilCandidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation or symbolic differentiation/integration, or haveretrievable mathematical formulae stored in them.-825520066000Instructions to CandidatesIn the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions.You must write your answer for each question in the space following the question.When a calculator is used, the answer should be given to an appropriate degree of accuracy.-762019558000Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 9 questions in this question paper. The total mark for this paper is 75.There are 28 pages in this question paper. Any blank pages are indicated.-889019939000Advice to CandidatesYou 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 not gain full credit.1.The roots of the equation2z3 – 3z2 + 8z + 5 = 0are z1, z2 and z3.Given that z1 = 1 + 2i, find z2 and z3.(5)3175106680002.f(x) = 3cos 2x + x – 2,–π ≤ x < π(a)Show that the equation f(x) = 0 has a root α in the interval [2, 3].(2)(b)Use linear interpolation once on the interval [2, 3] to find an approximation to α.Give your answer to 3 decimal places.(3)(c)The equation f(x) = 0 has another root β in the interval [–1, 0]. Starting with this interval, use interval bisection to find an interval of width 0.25 which contains β.(4)23495108585003.(i)(a)Describe fully the single transformation represented by the matrix A.(2)The matrix B represents an enlargement, scale factor –2, with centre the origin.(b)Write down the matrix B.(1)(ii),where k is a positive constant.Triangle T has an area of 16 square units.Triangle T is transformed onto the triangle T? by the transformation represented by the matrix M.Given that the area of the triangle T? is 224 square units, find the value of k.(3)-17145102870004.The complex number z is given bywhere p is an integer.(a)Express z in the form a + bi where a and b are real. Give your answer in its simplest form in terms of p.(4)(b)Given that arg(z) = θ, where tan θ = 1 find the possible values of p.(5)-22225121285005.(a)Use the standard results for and to show that(5)(b) Calculate the value of .(3)-11430155575006. and Given that M = (A + B)(2A – B),(a) calculate the matrix M,(6)(b) find the matrix C such that MC = A.(4)3175100330007.The parabola C has cartesian equation y2 = 4ax, a > 0.The points P(ap2, 2ap) and P?(ap2, –2ap) lie on C.(a) Show that an equation of the normal to C at the point P isy + px = 2ap + ap3(5)(b) Write down an equation of the normal to C at the point P?.(1)The normal to C at P meets the normal to C at P? at the point Q.(c)Find, in terms of a and p, the coordinates of Q.(2)Given that S is the focus of the parabola,(d)find the area of the quadrilateral SPQP?.(3)3175100330008.The rectangular hyperbola H has equation xy = c2, where c is a positive constant.The point , t ≠ 0 is a general point on H.An equation for the tangent to H at P is given byThe points A and B lie on H.The tangent to H at A and the tangent to H at B meet at the point .Find, in terms of c, the coordinates of A and the coordinates of B.(5)3175100330009.(a)Prove by induction that, for ,(5)(b)A sequence of numbers is defined byu1 = 0,u2 = 32,un+2 = 6un+1 – 8un n ≥ 1Prove by induction that, for ,un = 4n+1 – 2n+3(7)2159012954000TOTAL FOR PAPER: 75 MARKSENDPaper Reference(s)6667/01Edexcel GCEFurther Pure Mathematics FP1Advanced/Advanced SubsidiaryMonday 10 June 2013 MorningTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) NilCandidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation or symbolic differentiation/integration, or haveretrievable mathematical formulae stored in them.-825520066000Instructions to CandidatesIn the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions.You must write your answer for each question in the space following the question.When a calculator is used, the answer should be given to an appropriate degree of accuracy.-762019558000Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 9 questions in this question paper. The total mark for this paper is 75.There are 32 pages in this question paper. Any blank pages are indicated.-889019939000Advice to CandidatesYou 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 not gain full credit.1.M = Given that the matrix M is singular, find the possible values of x.(4)3175106680002.f(x) = cos(x2) – x + 3,0 < x < π(a)Show that the equation f(x) = 0 has a root α in the interval [2.5, 3].(2)(b)Use linear interpolation once on the interval [2.5, 3] to find an approximation for α, giving your answer to 2 decimal places.(3)23495108585003. Given that x = is a root of the equation2x3 – 9x2 + kx – 13 = 0,find(a)the value of k,(3)(b)the other 2 roots of the equation.(4)-17145102870004.The rectangular hyperbola H has Cartesian equation xy = 4.The point lies on H, where t ≠ 0.(a) Show that an equation of the normal to H at the point P isty – t3x = 2 – 2t4(5)The normal to H at the point where t = meets H again at the point Q.(b) Find the coordinates of the point Q.(4)-22225121285005.(a) Use the standard results for and to show thatfor all positive integers n.(6)(b) Hence show thatwhere a, b and c are integers to be found.(4)-11430155575006.A parabola C has equation y2 = 4ax,a > 0The points P(ap2, 2ap) and Q(aq2, 2aq) lie on C, where p ≠ 0, q ≠ 0, p ≠ q.(a) Show that an equation of the tangent to the parabola at P ispy – x = ap2(4)(b)Write down the equation of the tangent at Q.(1)The tangent at P meets the tangent at Q at the point R.(c)Find, in terms of p and q, the coordinates of R, giving your answers in their simplest form.(4)Given that R lies on the directrix of C,(d)find the value of pq.(2)3175100330007.z1 = 2 + 3i, z2 = 3 + 2i, z3 = a + bi, a, b (a) Find the exact value of |z1 + z2|.(2)Given that w = ,(b)find w in terms of a and b, giving your answer in the form x + iy,x, y .(4)Given also that w = ,(c)find the value of a and the value of b,(3)(d)find arg w, giving your answer in radians to 3 decimal places.(2)3175100330008.A = and I is the 2 × 2 identity matrix.(a)Prove thatA2 = 7A + 2I(2)(b)Hence show thatA–1 = (A – 7I)(2)The transformation represented by A maps the point P onto the point Q.Given that Q has coordinates (2k + 8, –2k – 5), where k is a constant,(c)find, in terms of k, the coordinates of P.(4)3175100330009.(a) A sequence of numbers is defined byu1 = 8un + 1 = 4un – 9n,n ≥ 1Prove by induction that, for n ,un = 4n + 3n +1(5)(b) Prove by induction that, for m ,(5)2159012954000TOTAL FOR PAPER: 75 MARKSEND Paper Reference(s)6667/01REdexcel GCEFurther Pure Mathematics FP1 (R)Advanced/Advanced SubsidiaryMonday 10 June 2013 MorningTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) NilCandidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation or symbolic differentiation/integration, or haveretrievable mathematical formulae stored in them.This paper is strictly for students outside the UK.-825520066000Instructions to CandidatesIn the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions.You must write your answer for each question in the space following the question.When a calculator is used, the answer should be given to an appropriate degree of accuracy.-762019558000Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 10 questions in this question paper. The total mark for this paper is 75.There are 36 pages in this question paper. Any blank pages are indicated.-889019939000Advice to CandidatesYou 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 not gain full credit.1.The complex numbers z and w are given byz = 8 + 3i, w = –2iExpress in the form a + bi, where a and b are real constants,(a)z – w,(1)(b)zw.(2)3175106680002.(i), where k is a constantGiven thatB = A + 3Iwhere I is the 2 × 2 identity matrix, find(a)B in terms of k,(2)(b)the value of k for which B is singular.(2)(ii)Given that, D = (2 –1 5)andE = CDfind E.(2)23495108585003. f(x) = (a)Show that the equation f(x) = 0 has a root α between x = 2 and x = 2.5.(2)(b)Starting with the interval [2, 2.5] use interval bisection twice to find an interval of width 0.125 which contains α.(3)The equation f(x) = 0 has a root β in the interval [–2, –1].(c)Taking –1.5 as a first approximation to β, apply the Newton-Raphson process once to f(x) to obtain a second approximation to β.Give your answer to 2 decimal places.(5)-17145102870004.f(x) = (4x2 +9)(x2 – 2x + 5)(a) Find the four roots of f(x) = 0.(4)(b) Show the four roots of f(x) = 0 on a single Argand diagram.(2)-22225121285005.Figure 1Figure 1 shows a rectangular hyperbola H with parametric equationsx = 3t, y = , t ≠ 0The line L with equation 6y = 4x – 15 intersects H at the point P and at the point Q as shown in Figure 1.(a) Show that L intersects H where 4t2 – 5t – 6 = 0.(3)(b) Hence, or otherwise, find the coordinates of points P and Q.(5)-11430155575006., The transformation represented by B followed by the transformation represented by A is equivalent to the transformation represented by P.(a) Find the matrix P.(2)Triangle T is transformed to the triangle T? by the transformation represented by P.Given that the area of triangle T? is 24 square units,(b)find the area of triangle T.(3)Triangle T? is transformed to the original triangle T by the matrix represented by Q.(c)Find the matrix Q.(2)3175100330007.The parabola C has equation y2 = 4ax, where a is a positive constant.The point P(at2, 2at) is a general point on C.(a) Show that the equation of the tangent to C at P(at2, 2at) isty = x + at2(4)The tangent to C at P meets the y-axis at a point Q.(b)Find the coordinates of Q.(1)Given that the point S is the focus of C,(c)show that PQ is perpendicular to SQ.(3)3175100330008.(a) Prove by induction, that for ,(6)(b)Hence, show thatwhere a, b and c are integers to be found.(4)3175100330009.The complex number w is given byw = 10 – 5i(a) Find .(1)(b) Find arg w, giving your answer in radians to 2 decimal places(2)The complex numbers z and w satisfy the equation(2 + i)(z + 3i) = w(c)Use algebra to find z, giving your answer in the form a + bi,where a and b are real numbers.(4)Given thatarg(λ + 9i + w) = where λ is a real constant,(d)find the value of λ.(2)31751003300010.(i)Use the standard results for and to evaluate(2)(ii)Use the standard results for and to show that for all integers n ≥ 0, where a, b and c are constant integers to be found.(6)2159012954000TOTAL FOR PAPER: 75 MARKSENDPaper Reference(s)6667/01Edexcel GCEFurther Pure Mathematics FP1Advanced Subsidiary/Advanced LevelMonday 28 January 2013 MorningTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) NilCandidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation or integration, or have retrievable mathematical formulae stored in them. -71183520066000Instructions to CandidatesWrite 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.-71183519558000Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 9 questions in this question paper. The total mark for this paper is 75.-69405519939000Advice to CandidatesYou 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 not gain full credit.1.Show, using the formulae for and , that = n(2n + 1)(2n – 1), for all positive integers n.(5)3175106680002.z = .Find, in the form a + ib where a, b ?,(a) z,(2)(b) z2.(2)Find(c) z,(2)(d) arg z2, giving your answer in degrees to 1 decimal place.(2)23495108585003. f(x) = + ? 5, x > 0.(a) Find f ′(x).(2)The equation f(x) = 0 has a root in the interval [4.5, 5.5].(b) Using x0 = 5 as a first approximation to , apply the Newton-Raphson procedure once to f(x) to find a second approximation to , giving your answer to 3 significant figures.(4)-17145102870004.The transformation U, represented by the 2 2 matrix P, is a rotation through 90° anticlockwise about the origin.(a) Write down the matrix P.(1)The transformation V, represented by the 2 × 2 matrix Q, is a reflection in the line y = ?x.(b) Write down the matrix Q.(1)Given that U followed by V is transformation T, which is represented by the matrix R,(c) express R in terms of P and Q,(1)(d) find the matrix R,(2)(e) give a full geometrical description of T as a single transformation.(2)3175100330005.f(x) = (4x2 + 9)(x2 ? 6x + 34).(a) Find the four roots of f (x) = 0.Give your answers in the form x = p + iq , where p and q are real.(5)(b) Show these four roots on a single Argand diagram.(2)-11430155575006.X = , where a is a constant.(a) Find the value of a for which the matrix X is singular.(2)Y = .(b) Find Y?1.(2)The transformation represented by Y maps the point A onto the point B.Given that B has coordinates (1 – , 7 – 2), where is a constant,(c) find, in terms of , the coordinates of point A.(4)-1143056515007.The rectangular hyperbola, H, has cartesian equation xy = 25.The point P and the point Q , where p, q 0, p q, are points on the rectangular hyperbola H.(a) Show that the equation of the tangent at point P isp2y + x = 10p.(4)(b) Write down the equation of the tangent at point Q.(1)The tangents at P and Q meet at the point N.Given p + q 0,(c) show that point N has coordinates .(4)The line joining N to the origin is perpendicular to the line PQ.(d) Find the value of p2q2. (5)4445-5080008.(a) Prove by induction that, for n ?+, = n(n + 1)(n + 5). (6)(b) A sequence of positive integers is defined byu1 = 1,un + 1 = un + n(3n + 1), n ?+.Prove by induction thatun = n2(n – 1) + 1, n ?+. (5)825572390009.Figure 1Figure 1 shows a sketch of part of the parabola with equation y2 = 36x.The point P (4, 12) lies on the parabola.(a) Find an equation for the normal to the parabola at P.(5)This normal meets the x-axis at the point N and S is the focus of the parabola, as shown in Figure?1.(b) Find the area of triangle PSN. (4)2159012954000TOTAL FOR PAPER: 75 MARKS ENDPaper Reference(s)6667/01Edexcel GCEFurther Pure Mathematics FP1Advanced SubsidiaryFriday 1 June 2012 MorningTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) NilCandidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation, differentiation and integration, or have retrievablemathematical formulae stored in them.-71183520066000Instructions to CandidatesWrite the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Core Mathematics FP1), the paper reference (6667), your surname, initials and signature.-71183519558000Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.There are 10 questions in this question paper. The total mark for this paper is 75.-69405519939000Advice to CandidatesYou 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 not gain full credit.1. f(x) = 2x3 – 6x2 – 7x ? 4.(a) Show that f(4) = 0.(1)(b) Use algebra to solve f(x) = 0 completely.(4)3175106680002.Given thatA = and B = ,(a)find AB.(2)Given thatC = and D = , where k is a constantandE = C + D,(b)find the value of k for which E has no inverse.(4)23495130175003. f(x) = x2 + – 3x – 7, x > 0.A root of the equation f(x) = 0 lies in the interval [3, 5].Taking 4 as a first approximation to , apply the Newton-Raphson process once to f(x) to obtain a second approximation to . Give your answer to 2 decimal places.(6)-762017145004.(a) Use the standard results for and to show that = n2(n2 + 2n + 13)for all positive integers n.(5)(b) Hence find the exact value of. (2)4000566675005.Figure 1Figure 1 shows a sketch of the parabola P with equation y2 = 8x. The point P lies on C, where y?>?0, and the point Q lies on C, where y < 0. The line segment PQ is parallel to the y-axis.Given that the distance PQ is 12,(a) write down the y-coordinate of P,(1)(b) find the x-coordinate of P.(2)Figure 1 shows the point S which is the focus of C.The line l passes through the point P and the point S.(c)Find an equation for l in the form ax + by + c = 0, where a, b and c are integers.(4)31115132080006. f(x) = tan + 3x – 6, – < x < .(a) Show that the equation f(x) = 0 has a root in the interval [1, 2].(2)(b) Use linear interpolation once on the interval [1, 2] to find an approximation to . Give your answer to 2 decimal places.(3)7620-635007. z = 2 ? i√3.(a) Calculate arg z, giving your answer in radians to 2 decimal places.(2)Use algebra to express(b) z + z2 in the form a + bi√3, where a and b are integers,(3)(c) in the form c + di√3, where c and d are integers.(4)Given thatw = – 3i,where is a real constant, and arg (4 – 5i + 3w) = –,(d) find the value of .(2)-10795127000008.The rectangular hyperbola H has equation xy = c2, where c is a positive constant.The point P, t ≠ 0, is a general point on H.(a) Show that an equation for the tangent to H at P isx + t 2 y = 2ct.(4)The tangent to H at the point P meets the x-axis at the point A and the y-axis at the point B.Given that the area of the triangle OAB, where O is the origin, is 36,(b) find the exact value of c, expressing your answer in the form k√2, where k is an integer.(4)15875118110009. M = .(a) Find det M.(1)The transformation represented by M maps the point S(2a – 7, a – 1), where a is a constant, onto the point S (25, –14).(b) Find the value of a.(3)The point R has coordinates (6, 0).Given that O is the origin,(c) find the area of triangle ORS.(2)Triangle ORS is mapped onto triangle OR 'S ' by the transformation represented by M.(d) Find the area of triangle OR 'S '.(2)Given thatA =(e) describe fully the single geometrical transformation represented by A.(2)The transformation represented by A followed by the transformation represented by B is equivalent to the transformation represented by M.(f) Find B.(4)38735990600010. Prove by induction that, for n ?+,f(n) = 22n – 1 + 32n – 1is divisible by 5.(6)2159012954000TOTAL FOR PAPER: 75 MARKSENDPaper Reference(s)6667/01Edexcel GCEFurther Pure Mathematics FP1Advanced SubsidiaryMonday 30 January 2012 AfternoonTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) NilCandidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation or integration, or have retrievable mathematical formulae stored in them. -71183520066000Instructions to CandidatesIn 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.-71183519558000Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for individual questions and the parts of questions are shown in round brackets: e.g. (2).There are 9 questions on this paper. The total mark for this paper is 75.-69405519939000Advice to CandidatesYou 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 not gain full credit.1.Given that =1 ? i,(a)find arg ().(2)Given also that = 3 + 4i, find, in the form a + ib, a, b ?,(b),(2)(c) .(3)In part (b) and part (c) you must show all your working clearly.46355111760002.(a) Show that f(x) = x4 + x ? 1 has a real root in the interval [0.5, 1.0].(2)(b) Starting with the interval [0.5, 1.0], use interval bisection twice to find an interval of width 0.125 which contains .(3)(c) Taking 0.75 as a first approximation, apply the Newton Raphson process twice to f(x) to obtain an approximate value of . Give your answer to 3 decimal places.(5)46355144780003. A parabola C has cartesian equation y2 = 16x. The point P(4t2, 8t) is a general point on C.(a) Write down the coordinates of the focus F and the equation of the directrix of C.(3)(b) Show that the equation of the normal to C at P is y + tx = 8t + 4t3.(5)27305121920004. A right angled triangle T has vertices A(1, 1), B(2, 1) and C(2, 4). When T is transformed by the matrix P = , the image is T ′.(a) Find the coordinates of the vertices of T ′.(2)(b) Describe fully the transformation represented by P.(2)The matrices Q = and R = represent two transformations. When T is transformed by the matrix QR, the image is T .(c) Find QR.(2)(d) Find the determinant of QR.(2)(e) Using your answer to part (d), find the area of T .(3)27305121920005. The roots of the equationz3 ? 8z2 + 22z ? 20 = 0are , and .(a) Given that = 3 + i, find and .(4)(b) Show, on a single Argand diagram, the points representing , and .(2)1968593345006.(a) Prove by induction = n2(n + 1)2.(5)(b) Using the result in part (a), show that = n(n3 + 2n2 + n – 8). (3)(c) Calculate the exact value of . (3)7.A sequence can be described by the recurrence formula = 2 + 1, n 1, = 1.(a)Find and .(2)(b) Prove by induction that = 2n ? 1.(5)19685142240008.A = .(a) Show that A is non-singular.(2)(b) Find B such that BA2 = A.(4)1206592710009.The rectangular hyperbola H has cartesian equation xy = 9.The points Pand Q lie on H, where p ≠ ± q.(a)Show that the equation of the tangent at P is x + p2y = 6p.(4)(b) Write down the equation of the tangent at Q.(1)The tangent at the point P and the tangent at the point Q intersect at R.(c)Find, as single fractions in their simplest form, the coordinates of R in terms of p and q.(4)-222259334500TOTAL FOR PAPER: 75 MARKSENDPaper Reference(s)6667/01Edexcel GCEFurther Pure Mathematics FP1Advanced/ Advanced SubsidiaryWednesday 22 June 2011 MorningTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) NilCandidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation, differentiation and integration, or have retrievablemathematical formulae stored in them.-71183520066000Instructions to CandidatesWrite the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Core Mathematics FP1), the paper reference (6667), your surname, initials and signature.-71183519558000Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.There are 9 questions in this question paper. The total mark for this paper is 75.-69405519939000Advice to CandidatesYou 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 not gain full credit.1. f (x) = 3x + 3x ? 7(a) Show that the equation f (x) = 0 has a root between x =1 and x = 2.(2)(b) Starting with the interval [1, 2], use interval bisection twice to find an interval of width 0.25 which contains .(3)3175106680002. = ? 2 + i(a) Find the modulus of .(1)(b) Find, in radians, the argument of , giving your answer to 2 decimal places.(2)The solutions to the quadratic equationz2 ? 10z + 28 = 0are and .(c) Find and , giving your answers in the form p iq, where p and q are integers.(3)(d) Show, on an Argand diagram, the points representing your complex numbers , and .(2))23495130175003. (a) Given thatA = , (i) find A2,(ii) describe fully the geometrical transformation represented by A2.(4)(b) Given thatB = , describe fully the geometrical transformation represented by B.(2)(c) Given thatC = , where k is a constant, find the value of k for which the matrix C is singular.(3)-762017145004. f(x) = x2 + – 3x – 1, x 0.(a) Use differentiation to find f ′(x).(2)The root of the equation f(x) = 0 lies in the interval [0.7, 0.9].(b) Taking 0.8 as a first approximation to , apply the Newton-Raphson process once to f(x) to obtain a second approximation to . Give your answer to 3 decimal places.(4)4000566675005.A = , where a and b are constants.Given that the matrix A maps the point with coordinates (4, 6) onto the point with coordinates (2, ?8),(a) find the value of a and the value of b.(4)A quadrilateral R has area 30 square units.It is transformed into another quadrilateral S by the matrix A.Using your values of a and b,(b) find the area of quadrilateral S.(4)31115132080006.Given that z = x + iy, find the value of x and the value of y such thatz + 3iz* = ?1 + 13iwhere z* is the complex conjugate of z.(7)7620-635007.(a) Use the results for and to show that = n(2n + 1)(2n – 1)for all positive integers n.(6)(b) Hence show that = n(an2 + b)where a and b are integers to be found.(4)-10795127000008.The parabola C has equation y2 = 48x.The point P(12t 2, 24t) is a general point on C.(a) Find the equation of the directrix of C.(2)(b) Show that the equation of the tangent to C at P(12t 2, 24t) isx ? ty + 12t 2 = 0.(4)The tangent to C at the point (3, 12) meets the directrix of C at the point X.(c) Find the coordinates of X.(4)15875118110009.Prove by induction, that for n ?+,(a) = (6)(b) f(n) = 72n ? 1 + 5 is divisible by 12.(6)2159012954000TOTAL FOR PAPER: 75 MARKSEND Paper Reference(s)6667/01Edexcel GCEFurther Pure Mathematics FP1Advanced/Advanced SubsidiaryMonday 1 February 2010 AfternoonTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Orange) NilCandidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation or integration, or have retrievable mathematical formulae stored in them. -71183520066000Instructions to CandidatesIn 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.-71183519558000Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for individual questions and the parts of questions are shown in round brackets: e.g. (2).There are 9 questions on this paper. The total mark for this paper is 75.-69405519939000Advice to CandidatesYou 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 not gain full credit.1.The complex numbers and are given by = 2 + 8i and = 1 – iFind, showing your working,(a) in the form a + bi, where a and b are real,(3)(b) the value of ,(2)(c) the value of arg , giving your answer in radians to 2 decimal places.(2)46355111760002.f(x) = 3x2 – .(a) Write down, to 3 decimal places, the value of f(1.3) and the value of f(1.4).(1)The equation f(x) = 0 has a root α between 1.3 and 1.4(b) Starting with the interval [1.3,?1.4], use interval bisection to find an interval of width?0.025 which contains α.(3)(c) Taking 1.4 as a first approximation to α, apply the Newton-Raphson procedure once to?f(x) to obtain a second approximation to α, giving your answer to 3 decimal places.(5)46355144780003. A sequence of numbers is defined by = 2, = 5 – 4, n 1.Prove by induction that, for n ?, = 5n – 1 + 1.(4)27305121920004. Figure 1Figure 1 shows a sketch of part of the parabola with equation y2 = 12x .The point P on the parabola has x-coordinate .The point S is the focus of the parabola.(a) Write down the coordinates of S.(1)The points A and B lie on the directrix of the parabola.The point A is on the x-axis and the y-coordinate of B is positive.Given that ABPS is a trapezium,(b) calculate the perimeter of ABPS.(5)27305121920005. A = , where a is real.(a) Find det A in terms of a.(2)(b) Show that the matrix A is non-singular for all values of a.(3)Given that a = 0,(c) find A–1.(3)1968581280006.Given that 2 and 5 + 2i are roots of the equationx3 ? 12x2 + cx + d = 0, c, d ∈?,(a) write down the other complex root of the equation.(1)(b) Find the value of c and the value of d.(5)(c) Show the three roots of this equation on a single Argand diagram.(2)2349546355007.The rectangular hyperbola H has equation xy = c2, where c is a constant.The point P is a general point on H.(a) Show that the tangent to H at P has equationt2y + x = 2ct.(4)The tangents to H at the points A and B meet at the point (15c, –c).(b) Find, in terms of c, the coordinates of A and B.(5)19685142240008.(a) Prove by induction that, for any positive integer n, = n2(n + 1)2.(5)(b) Using the formulae for and , show that= n(n + 2)(n2 + 7).(5)(c) Hence evaluate .(2)1206592710009.M = .(a) Describe fully the geometrical transformation represented by the matrix M.(2)The transformation represented by M maps the point A with coordinates (p, q) onto the point?B with coordinates (3√2, 4√2).(b) Find the value of p and the value of q.(4)(c) Find, in its simplest surd form, the length OA, where O is the origin.(2)(d) Find M2.(2)The point B is mapped onto the point C by the transformation represented by M2.(e) Find the coordinates of C.(2)-222259334500TOTAL FOR PAPER: 75 MARKSENDJUNE 2014 MARK SCHEMEQuestionNumberSchemeMarks1.(a)M1M1A1, A1(4)(b)`M1dM1dM1A1ORM1 oedM1oedM1A1(4)Total 8QuestionNumberSchemeMarks2.(a)M1Sign change (and is continuous) therefore a root / α is between and A1(2)(b)M1A1(2)(c)M1M1A1(3)Total 7Question NumberSchemeMarks3.(a)B1 (1)(b)M1A1 M1A1, A1ORf(1+5i)=0 or f(1-5i)=0M1 and A1M1A1, A1(5)(c)9563106851650010083803347720001020445351409000103187543815000144399022313902002889635222504010011020444206692500-1333532639000072390622300500515284452037080002482852070099002216152022475O00O33020152082500-450852194560002002790202247500156845234950003162303402329003162307499340044830942100500B1B1(2)Total 8Question NumberSchemeMarks4.A = , B = (i)(a)M1A2(b)B1(4)(ii)M1M1A1(3)Total 75.(a)B1Proof by induction will usually score no more marks without use of standard results M1A1B1M1A1(6)(b)M1A1A1(3)Total 9Question NumberSchemeMarks6.(a)M1 or equivalent expressionsA1dM1A1*(4)(b)B1B1(2)(c) or PA= and PB= M1Area APB = M1= A1(3)Total 9Question NumberSchemeMarks7.(i)(a)B1(b)B1(c)M1A1(4)(ii)Area triangle T = M1A1M1A1Area triangle T = M1A1(6)Total 108.(a)M1 or orM1A1orA1*(4)(b)(Focus) (4, 0)B1(Directrix) x = -4B1Gradient of l2 is M1M1, A1M1A1(7)Total 11Question NumberSchemeMarks9. is divisible by 6.B1Assume that for is divisible by 6.M1M1A1A1If the result is true for then it is now true for As the result has been shown to be true for then the result is true for all A1cso(6)Total 6JUNE 2014(R) MARK SCHEMEQuestion NumberSchemeMarks1.B1 M1A1M1 A1(5)Total 52.(a)f(2) = -1.9609......f(3) = 3.8805......M1Sign change (and is continuous) therefore a rootis between and A1(2)(b)M1If any “negative lengths” are used, score M0A1ftA1(3)(c)f(0) = +(1) or f(-1) = -(4.248)B1f(-0.5) (= -0.879.....)M1f(-0.25) (= 0.382....)M1A1(4)Total 9Question NumberSchemeMarks3.(i)(a)Rotation of 45 degrees anticlockwise, about the originB1B1(2) (b)B1(1) (ii)B1M1A1(3)Total 64.(a)M1M1A1, A1(4)(b)M1M1A1M1A1(5)Total 9Question NumberSchemeMarks5.(a)B1M1A1M1A1(5)(b)M1A1= 1621800 - 1890= 1619910A1(3)Total 86.(a)M1A1M1A1M1A1(6)(b)B1M1dM1A1(4)Total 10Question NumberSchemeMarks7.(a)M1A1At P, gradient of normal = -pA1M1A1*(5)(b)B1(1)(c)M1A1(2)(d)S is (a, 0)B1Area SPQP’ = M1A1(3)Total 118.M1A1M1M1A1(5)Total 5Question NumberSchemeMarks9.(a)When n = 1, rhs = lhs = 2B1M1A1A1If the result is true for n = k then it has been shown true for n = k + 1. As it is true for n = 1 then it is true for all n (positive integers.)A1(5)(b)When n = 1 B1When n = 2 B1True for n = 1 and n = 2Assume and M1A1M1So A1If the result is true for n = k and n = k + 1 then it has been shown true for n = k + 2. As it is true for n = 1 and n = 2 then it is true for all n (positive integers.)A1(7)Total 12JUNE 2013 MARK SCHEMEQuestion NumberSchemeMarks1.detM = x(4x – 11) – (3x – 6)(x – 2)M1x2 + x – 12 (=0)A1(x + 4)(x – 3) (= 0 ) x = ...M1 A1[4]2(a)f(2.5) = 1.499.....f(3) = -0.9111.....M1Sign change (positive, negative) (and is continuous) therefore root or equivalent.A1Use of degrees gives f(2.5) = 1.494 and f(3) = 0.988 which is awarded M1A0(2)(b) M1 A1ft(2d.p.)A1 cao(3)[5]Question NumberSchemeMarks3(a)Ignore part labels and mark part (a) and part (b) together.M1dM1k = 30A1 caoAlternative using long division:M1dM1A1Alternative by inspection: M1dM1k = 30A1(3)(b)M1or A1or equivalentM1A1 oe(4)[7]Question NumberSchemeMarks4(a)M1 or equivalent expressionsA1M1M1A1* cso(5)(b)M1or or.M1 ororM1A1(4)[9]Question NumberSchemeMarks5(a)B1M1,B1ftM1 A1A1*cso(6)5(b)M1A13f(n) – f(n or n+1) is M0dM1A1 (4)[10]Question NumberSchemeMarks6(a)M1or A1 M1py – x = ap2 *A1 cso(4)(b)qy – x = aq2 B1(1)(c)qy – aq2= py – ap2M1M1 A1,A1 (4)(d) M1 A1(2)[11]Question NumberSchemeMarks7(a)M1 A1 cao(2)(b)M1B1dM1A1(4)(c)M1 dM1a = 1, b = -1A1(3)(d)M1 A1(2)[11]Question NumberSchemeMarks8(a)M1A1OR (2)(b)M1*A1* csoNumerical approach award 0/2.(2)(c)B1M1A1,A1Or:B1M1A1,A1(4)[8]Question NumberSchemeMarks9(a) B1Assume true for n = k so that M1 A1A1If true for n = k then true for n = k + 1 and as true for n = 1 true for all nA1 cso(5)(b)Condone use of n here.B1M1A1A1If true for m = k then true for m = k + 1 and as true for m = 1 true for all mA1 cso(5)[10]JUNE 2013 (R) MARK SCHEMEQuestion NumberSchemeMarks1.(a)B1(1)(b)M1A1(2)[3]2. (i)(a)M1A1(2) (b)B is singular M1 A1cao(2) (ii) M1A1(2) [6]Question NumberSchemeMarks3. (a)M1Sign change (and is continuous) therefore a root exists between and A1(2)(b)B1M1A1(3)(c)M1A1 B1M1 A1 cao(5)[10]Question NumberSchemeMarks4.(a)M1A1M1A1(4)(b)21463012065Oxy00OxyB1ftB1ft(2)[6]Ignore part labels and mark part (a) and part (b) together5. (a)M1 A1 *A1 cso(3)(b)M1A1When WhenM1A1A1 (5) [8]Question NumberSchemeMarks6. (a)M1 A1(2)(b)M1dM1A1ft(3)(c)M1A1ft(2)[7]7. (a)or (implicitly) or (chain rule) M1When or A1T: M1T: T: A1 cso *(4)(b)At Q, B1(1)(c)M1A1A1 cso (3) [8]Question NumberSchemeMarks8. (a)B1As the summation formula is true for Assume that the summation formula is true for With terms the summation formula becomes:M1dM1A1dM1If the summation formula is true for then it is shown to be true for As the result is true for , it is now also true for all and by mathematical induction.A1 cso(6)8. (b)M1A1dM1A1(4)[10]Question NumberSchemeMarks9.(a)11.1803...B1(b)M1A1 oe(2)(c)B1M1M1 (Note: A1 (4)(d)M1So, A1(2)[9]10. (i)M1A1 cao(2)(ii)M1A1B1B1M1A1(6)[8]JANUARY 2013 MARK SCHEMEQuestion NumberSchemeMarks 1. M1 = A1, B1 M1 A1 cso [5]2.(a) M1 A1cao(2)(b) = = M1 A1 (2)(c) =10M1 A1ft(2)(d) M1 so or A1 cao (2)[8]3.(a) M1 A1(2)(b) f(5) = - 0.0807 B1M1 M1 =5.2(0)A1(4)[6]Question NumberSchemeMarks 4.(a) B1 (1)(b) B1 (1)(c) B1 (1)(d) M1 A1 cao(2)(e) Reflection in the y axisB1 B1 (2)[7]5.(a) , or equivalentM1, A1 Solving 3-term quadratic by formula or completion of the square or M1 A1 A1ft (5)5O(b) Two roots on imaginary axisB1ft Two roots – one the conjugate of the other B1ft Accept points or vectors(2)[7]Question NumberSchemeMarks 6.(a) Determinant: 2 – 3a = 0 and solve for a = M1 So or equivalent A1(2)(b) Determinant: M1A1 (2) (c) M1depM1A1A1(4)[8]Question NumberSchemeMarks 7.(a) M1 A1 (*) M1 A1 (4)(b) onlyB1 (1)(c) so M1 A1cso = M1 A1 cso(4)(d) Line PQ has gradient M1 A1ON has gradient or could be as unsimplified equivalents seen anywhereB1As these lines are perpendicular so OR for ON with gradient (equivalent to) pq and sub in points O AND N to give OR for PQ with gradient (equivalent to) –pq and sub in points P AND Q to give M1 A1(5)[14]Question NumberSchemeMarks 8.(a) If n =1, and ,B1(so true for n = 1. Assume true for n = k) SoM1 = = A1 = which implies is true for dA1 As result is true for this implies true for all positive integers and so result is true by induction dM1A1cso(6)(b) B1(so true for . Assume true for n = k)M1, which implies is true for n = k + 1A1As result is true for n = 1 this implies true for all positive integers and so result is true by induction M1A1cso(5)[11]Question NumberSchemeMarks 9.(a) so M1Gradient when x = 4 is and gradient of normal is M1 A1So equation of normal is (or ) M1 A1 (5)(b) S is at point (9,0)B1N is at (22,0), found by substituting y = 0 into their part (a)B1ftBoth B marks can be implied or on diagram.So area is M1 A1 cao(4)[9]JUNE 2012 MARK SCHEMEQuestion NumberSchemeMarks1.B1(a)[1](b)M1 A1So, M1A1[4]2. (a) M1 A1[2] where k is a constant,(b)M1E does not have an inverse M1M1A1 oe[4]6 marksQuestion NumberSchemeMarks3.M1A1 B1M1M1 A1 cao[6]6 marks Question NumberSchemeMarks4. (a)M1A1B1dM1 (AG)A1 *[5](b)M1A1 cao[2]7 marksQuestion NumberSchemeMarks5.(a)B1[1](b)M1(So P has coordinates )A1 oe[2](c)Focus B1Gradient M1Either;M1or and ;l: A1 [4]7 marksQuestion NumberSchemeMarks6.(a)M1Sign change (and is continuous) therefore a rootis between and A1[2]or M1(b)A1A1[3]5 marksQuestion SchemeMarks7. (a)M1A1[2](b) M1 (Note: M1A1[3]M1(c)dM1M1 (Note: A1[4](d), and So real part of = 0 or M1So, A1[2]11 marksQuestion NumberSchemeMarks8.(a)M1or equivalent expressionsA1M1A1 *[4](b)B1B1Area M1A1[4]8 marksQuestion NumberSchemeMarks9. (a) B1[1](b)Therefore, M1Either, or orA1giving A1[3](c)M1A1[2](d)M1A1[2](e)Rotation; anti-clockwise (or clockwise) about B1;B1[2](f) M1A1M1A1[4]14 marksQuestion NumberSchemeMarks is divisible by 5.10. B1Assume that for is divisible by 5 for M1A1 M1 A1If the result is true for then it is now true for As the result has shown to be true for then the result is true for all n.A1 cso[6]6 marksJANUARY 2012 MARK SCHEMEQuestion NumberSchemeNotesMarks1(a) or or M1 or -45 or awrt -0.785 (oe e.g )A1Correct answer only 2/2(2)(b)At least 3 correct terms (Unsimplified)M1 caoA1(2)(c)Multiply top and bottom by (1 + i)M1 A1 or A1 (3)Correct answers only in (b) and (c) scores no marks(7 marks)Question NumberSchemeNotesMarks2 (a)f(0.5) = -0.4375 (-) f(1) = 1Either any one of f(0.5) = awrt -0.4 or f(1) = 1M1Sign change (positive, negative) (and is continuous) therefore (a root) is between and f(0.5) = awrt -0.4 and f(1) = 1, sign change and conclusionA1(2)(b)f(0.75) = 0.06640625()Attempt f(0.75)M1f(0.625) = -0.222412109375()f(0.75) = awrt 0.07 and f(0.625) = awrt -0.2A1 or or or equivalent in words.A1In (b) there is no credit for linear interpolation and acorrect answer with no working scores no marks.(3)(c)Correct derivative (May be implied later by e.g. 4(0.75)3 + 1)B1Attempt Newton-RaphsonM1Correct first application – a correct numerical expression e.g. or awrt 0.725 (may be implied)A1Awrt 0.724A1caoA1A final answer of 0.724 with evidence of NR applied twice with no incorrect work should score 5/5(5)(10 marks)Question NumberSchemeNotesMarks3(a)Focus B1Directrix x + “4” = 0 or x = - “4”M1x + 4 = 0 or x = - 4A1(3)(b) or their M1Correct differentiationA1At P, gradient of normal = -tCorrect normal gradient with no errors seen.A1Applies or using in an attempt to find c. Their mN must be different from their mT and must be a function of t. M1cso **given answer**A1Special case – if the correct gradient is quoted could score M0A0A0M1A1(5)(8 marks)Question NumberSchemeNotesMarks4(a)Attempt to multiply the right way round with at least 4 correct elementsM1has coordinates (1,1), (1,2) and (4,2) or NOT just Correct coordinates or vectorsA1(2)(b)Reflection in the line y = xReflectionB1y = xB1Allow ‘in the axis’ ‘about the line’ y = x etc. Provided both features are mentioned ignore any reference to the origin unless there is a clear contradiction.(2)(c)2 correct elementsM1Correct matrixA1Note that scores M0A0 in (c) but allow all the marks in (d) and (e)(2)(d)“-2”x”2” – “0”x”0”M1-4A1Answer only scores 2/2 scores M0(2)(e)Area of T = Correct area for TB1Area of Attempt at M16 or follow through their det(QR) x Their triangle area provided area > 0A1ft(3)(11 marks)Question NumberSchemeNotesMarks5(a)B1Attempt to expand or any valid method to establish the quadratic factor e.g.Sum of roots 6, product of roots 10M1Attempt at linear factor with their cd in Or Or attempts f(2)M1A1Showing that f(2) = 0 is equivalent to scoring both M’s so it is possible to gain all 4 marks quite easily e.g. B1, shows f(2) = 0 M2, A1.Answers only can score 4/4(4)5(b)Argand Diagram3, 13, -12, 0-1.5-1-0.500.511.500.511.522.533.5ReIm First B1 for plotting (3, 1) and (3, -1) correctly with an indication of scale or labelled with coordinates (allow points/lines/crosses/vectors etc.) Allow i/-i for 1/-1 marked on imaginary axis.Second B1 for plotting (2, 0) correctly relative to the conjugate pair with an indication of scale or labelled with coordinates or just 2B1B1(2)(6 marks)Question NumberSchemeNotesMarks6(a)Shows both LHS = 1 and RHS = 1B1Assume true for n = kWhen n = k + 1Adds (k + 1)3 to the given resultM1 Attempt to factorise out dM1Correct expression with factorised out.A1 Must see 4 things: true for n = 1, assumption true for n = k, said true for n = k + 1 and therefore true for all nFully complete proof with no errors and comment. All the previous marks must have been scored.A1csoSee extra notes for alternative approaches(5)(b)Attempt two sumsM1 is M0Correct expressionA1 Completion to printed answer with no errors seen. A1(3)(c)Attempt S50 – S20 or S50 – S19 and substitutes into a correct expression at least once.M1Correct numerical expression (unsimplified)A1 = 1 589 463caoA1(3)(11 marks)Question NumberSchemeNotesMarks7(a)B1, B1(2)(b)At n =1, and so result true for n = 1B1Assume true for n = k; and so Substitutes uk into uk+1 (must see this line)M1Correct expressionA1 Correct completion to A1Must see 4 things: true for n = 1, assumption true for n = k, said true for n = k + 1 and therefore true for all nFully complete proof with no errors and comment. All the previous marks in (b) must have been scored.A1csoIgnore any subsequent attempts e.g. etc. (5)Total 7Question NumberSchemeNotesMarks8(a)Correct attempt at the determinantM1 (so A is non singular)det(A) = -2 and some reference to zeroA1 scores M0(2)(b)Recognising that A-1 is requiredM1At least 3 correct terms in M1B1ftFully correct answerA1Correct answer only score 4/4(4)Ignore poor matrix algebra notation if the intention is clear(6 marks)Question NumberSchemeNotesMarks9 (a)M1Correct use of product rule. The sum of two terms, one of which is correct.or their Correct differentiation. A1Applies or using in an attempt to find c. Their m must be a function of p and come from their dy/dx.M1 *Cso **given answer**A1Special case – if the correct gradient is quoted could score M0A0M1A1(4)(b)Allow this to score here or in (c)B1(1)(c)Attempt to obtain an equation in one variable x or yM1Attempt to isolate x or y – must reach x or y = f(p, q) or f(p) or f(q)M1One correct simplified coordinateA1Both coordinates correct and simplifiedA1(4)(9 marks)JUNE 2011 MARK SCHEMEQuestion NumberSchemeNotesMarks1. (a)Either any one of f(1) = -1 or f(2) = 8.M1Sign change (positive, negative) (and is continuous) therefore (a root) is between and Both values correct, sign change and conclusionA1(2)(b) (or truncated to 2.6)B1Attempt to find.M1 with or or or equivalent in words.A1 (3)5Question NumberSchemeNotesMarks2. (a) or awrt 2.24B1(1)(b) or or oror or M1awrt 2.68A1 oe (2)(c) An attempt to use the quadratic formula (usual rules)M1Attempt to simplify their in terms of i,. e.g. i or i M1So, . A1 oe (3)(d)yxNote that the points are andThe point plotted correctly on the Argand diagram with/without label.B1The distinct points and plotted correctly and symmetrically about the x-axis on the Argand diagram with/without label.B1 (2)8Question NumberSchemeNotesMarks3. (a)A = (i)A2 = = A correct method to multiply out two matrices. Can be implied by two out of four correct elements.M1 Correct answerA1(2)(ii)Enlargement; scale factor 3, centre Enlargement;B1;scale factor 3, centre (0, 0)B1(2)(b)Reflection; in the line Reflection;B1;y = –xB1 (2)(c) k is a constant.C is singular (Can be implied) B1 Applies M1 A1 (3)9Question NumberSchemeNotesMarks4. (a)At least two of the four terms differentiated correctly.M1Correct differentiation. (Allow any correct unsimplified form)A1(2)(b) A correct numerical expression for f(0.8)B1Attempt to insert x = 0.8 into their f’(x). Does not require an evaluation.(If is incorrect for their derivative and there is no working score M0)M1Correct application of Newton-Raphson using their values. Does not require an evaluation.M10.869A1 cao (4)6Question NumberSchemeNotesMarks5. where a and b are constants.(a)Therefore, Using the information in the question to form the matrix equation. Can be implied by both correct equations below.M1So, and Allow Any one correct equation.Any correct horizontal lineM1giving and Any one of or A1Both and A1(4)(b)Finds determinant by applying M1A1 or M1150 or ft answerA1 (4)8Question NumberSchemeNotesMarks6.B1Substituting and their z* into M1Correct equation in x and y with i2 = -1. Can be implied.A1An attempt to equate real and imaginary parts.M1Correct equations.A1Attempt to solve simultaneous equations to find one of x or y. At least one of the equations must contain both x and y terms.M1Both and A1(7)7Question NumberSchemeNotesMarks7. (a)Multiplying out brackets and an attempt to use at least one of the two standard formulae correctly.M1First two terms correct.A1B1Attempt to factorise out M1Correct expression with factorised out with no errors seen.A1Correct proof. No errors seen.A1 *(6)(b)Use of or with the result from (a) used at least once.M1Correct unsimplified expression. E.g. Allow 2(3n) for 6n.A1Factorising out ( or )dM1A1(4)10Question NumberSchemeNotesMarks8. with general point (a)Using to find a.M1So, directrix has the equation A1 oe (2)(b)or (implicitly) or (chain rule) their M1When or A1T: Applies or using in an attempt to find c. Their mT must be a function of t . M1T: T: Correct solution. A1 cso* (4)(c)Compare with gives B1NB with x = 3 and y = 12 gives into T gives Substitutes their t into T.M1At X, Substitutes their x from (a) into T.M1So, So the coordinates of X are A1 (4)10Question NumberSchemeNotesMarks9. (a) Check to see that the result is true for B1As the matrix result is true for Assume that the matrix equation is true for With the matrix equation becomes by M1Correct unsimplified matrix with no errors seen.A1 Manipulates so that on at least one term.dM1Correct result with no errors seen with some working between this and the previous A1A1If the result is true for then it is now true for (2) As the result has shown to be true for then the result is true for all n. (4) Correct conclusion with all previous marks earnedA1 cso(6)Question NumberSchemeNotesMarks (b) Shows that B1{which is divisible by 12}.{ is divisible by 12 when }Assume that for is divisible by 12 for So, Correct unsimplified expression for B1giving, Applies No simplification is necessary and condone missing brackets. M1Attempting to isolate M1A1cso, which is divisible by 12 as both and are both divisible by 12.(1) If the result is true for (2) then it is now true for (3) As the result has shown to be true for then the result is true for all n. (5). Correct conclusion with no incorrect work. Don’t condone missing brackets.A1 cso (6)12ask LEVEL "WHAT LEVEL?" ask SUBJECT "WHAT SUBJECT?" Comparison of key skills specifications 2000/2002 with 2004 standardsask CODE "WHAT CODE?" X015461ask DATE "WHAT DATE?" July 2004ask ISSUE "WHAT ISSUE?" Issue 1JANUARY 2010Mark SchemeQuestion NumberSchemeMarksQ1(a) M1A1 A1(3)(b) (or awrt 5.83)M1 A1ft(2)(c) M1 A1(2)[7] Notes(a) and attempt to multiply out for M1-3 for first A1, +5i for second A1(b) Square root required without i for M1 award M1 for attempt at Pythagoras for both numerator and denominator(c) tan or , or seen with their 3 and 5 award M12.11 correct answer only award A1Question NumberSchemeMarksQ2(a) and (allow awrt)B1(1)(b) M1 A1 A1(3)(c) M1 A1M1 A1, A1(5)[9] NotesBoth answers required for B1. Accept anything that rounds to 3dp values above.(b) f(1.35) or awrt -0.6 M1(f(1.35) and awrt -0.6) AND (f(1.375) and awrt -0.1) for first A1 or expression using brackets or equivalent in words for second A1(c) One term correct for M1, both correct for A1Correct formula seen or implied and attempt to substitute for M1awrt 16.4 for second A1 which can be implied by correct final answerawrt 1.384 correct answer only A1Q3For n = 1: Assume true for n = k:True for n = k + 1 if true for n = k. True for n = 1, true for all n.B1M1 A1 A1 cso[4] NotesAccept or above B1 seen award M1 or award first A1All three elements stated somewhere in the solution award final A1Q4(a) (3, 0) caoB1(1)(b) P: A and B lie on or a correct method to find both PB and PSPerimeter = B1B1M1M1 A1(5)[6] Notes(b) Both B marks can be implied by correct diagram with lengths labelled or coordinates of vertices stated.Second M1 for their four values added together. or awrt 14.7 for final A1Q5(a) det A = M1 A1(2)(b) Positive for all values of a, so A is non-singularM1 A1ft A1cso(3)(c) B1 for B1 M1 A1(3)[8] Notes(a) Correct use of for M1(b) Attempt to complete square for M1Alt 1Attempt to establish turning point (e.g. calculus, graph) M1Minimum value 6 for A1ftPositive for all values of a, so A is non-singular for A1 csoAlt 2Attempt at for M1. Can be part of quadratic formulaTheir correct -24 for first A1No real roots or equivalent, so A is non-singular for final A1cso(c) Swap leading diagonal, and change sign of other diagonal, with numbers or a for M1Correct matrix independent of ‘their award’ final A1Q6(a) is a root B1 (1)(b) M1 M1M1A1, A1(5)(c) Conjugate pair in 1st and 4th quadrants(symmetrical about real axis) Fully correct, labelledB1B1(2)[8] (b) 1st M: Form brackets using and expand. 2nd M: Achieve a 3-term quadratic with no i's.(b) Alternative: Substitute a complex root (usually 5+2i) and expand brackets M1 M1 (2nd M for achieving an expression with no powers of i) Equate real and imaginary partsM1 A1, A1Q7(a) without x or y (*)B1M1M1 A1cso(4)(b) Substitute : Points are and bothM1A1M1 A1A1(5)[9] Notes(a) Use of where m is their gradient expression in terms of c and / or t only for second M1. Accept and attempt to find k for second M1.(b) Correct absolute factors for their constant for second M1.Accept correct use of quadratic formula for second M1.Alternatives:(a) and B1 M1, then as in main scheme.(a) B1 M1, then as in main scheme.Q8(a) and Assume true for n = k : True for n = k + 1 if true for n = k. True for n = 1, true for all n.B1B1M1 A1A1cso(5)(b) (*)B1, B1M1 A1 A1cso(5)(c) with attempt to sub in answer to part (b) M1A1(2)[12] Notes(a) Correct method to identify as a factor award M1award first A1All three elements stated somewhere in the solution award final A1(b) Attempt to factorise by n for M1 and for first A1(c) no working 0/2Q9(a) 45 or rotation (anticlockwise), about the originB1, B1(2)(b) and or equivalent p = 7 and q = 1 both correctM1M1 A1A1(4)(c) Length of OA (= length of OB) = M1, A1(2)(d) M1 A1(2)(e) so coordinates are M1 A1(2)[12] NotesOrder of matrix multiplication needs to be correct to award Ms(a) More than one transformation 0/2(b) Second M1 for correct matrix multiplication to give two equations Alternative:(b) First M1 A1 Second M1 A1(c) Correct use of their p and their q award M1(e) Accept column vector for final A1. ................
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