(b) Explain why AB ≠ BA. (4) (ii) Given that , where k is a real number find C–1, giving your answer in terms of k. (3) 5. (a) Use the standard results for and to show that (6) (b) Hence show that where a and b are constants to be found. (3) 6. 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 is t2y + 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)
7. (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 ʹ by the matrix Given that the area of triangle T ʹ is 364 square units, find the value of k. (6) 8. 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 by 3ky – 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) 9. Prove by induction that, for , f(n) = 8n – 2n is divisible by 6. (6) TOTAL FOR PAPER: 75 MARKS
where 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) 5. (a) Use the standard results for and to show that (5) (b) Calculate the value of . (3) 6. and Given that M = (A + B)(2A – B), (a) calculate the matrix M, (6) (b) find the matrix C such that MC = A. (4) 7. The parabola C has cartesian equation y2 = 4ax, a > 0. The points P(ap2, 2ap) and Pʹ(ap2, –2ap) lie on C. (a) Show that an equation of the normal to C at the point P is y + px = 2ap + ap3 (5) (b) Write down an equation of the normal to C at the point Pʹ. (1) The normal to C at P meets the normal to C at Pʹ 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 SPQPʹ. (3) 8. 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 by The 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) 9. (a) Prove by induction that, for , (5) (b) A sequence of numbers is defined by u1 = 0, u2 = 32, un+2 = 6un+1 – 8un n ≥ 1 Prove by induction that, for , un = 4n+1 – 2n+3 (7) TOTAL FOR PAPER: 75 MARKS END Paper Reference(s) 6667/01 Edexcel GCE
(6) (b) Hence show that where a, b and c are integers to be found. (4) 6. A parabola C has equation y2 = 4ax, a > 0 The 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 is py – 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) 7. 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)
(c) Find the matrix Q. (2) 7. 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) is ty = 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) 8. (a) Prove by induction, that for , (6) (b) Hence, show that where a, b and c are integers to be found. (4) 9. The complex number w is given by w = 10 – 5i (a) Find . (1) (b) Find arg w, giving your answer in radians to 2 decimal places (2)
Advanced Subsidiary/Advanced Level Monday 28 January 2013 Morning Time: 1 hour 30 minutes Materials required for examination Items included with question papers Mathematical Formulae (Pink) Nil Candidates 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. Instructions to Candidates 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. Information for Candidates A 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. 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 not gain full credit.
Figure 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) 6. 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) 7. 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 that w = – 3i, where is a real constant, and arg (4 – 5i + 3w) = –, (d) find the value of .
1. The complex numbers and are given by = 2 + 8i and = 1 – i Find, 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) 2. 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) 3. A sequence of numbers is defined by = 2, = 5 – 4, n 1. Prove by induction that, for n ℤ, = 5n – 1 + 1. (4) 4. Figure 1 Figure 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)
8. (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) 9. 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) TOTAL FOR PAPER: 75 MARKS END JUNE 2014 MARK SCHEME Question Number Scheme Marks 1.(a) M1 M1 A1, A1 (4) (b) ` M1 dM1 dM1A1 OR M1
or equivalent expressions A1 M1 M1 A1* cso (5) (b) M1 or or . M1 or or M1 A1 (4) [9] Question Number Scheme Marks 5(a) B1 M1,B1ft M1 A1 A1*cso (6) 5(b) M1A1 3f(n) – f(n or n+1) is M0
dM1 A1 (4) [10] Question Number Scheme Marks 6(a) M1 or A1 M1 py – x = ap2 * A1 cso (4) (b) qy – x = aq2 B1 (1) (c) qy – aq2= py – ap2 M1 M1 A1,A1 (4) (d) M1 A1 (2) [11]
B1 M1 A1,A1 (4) [8] Question Number Scheme Marks 9(a) B1 Assume true for n = k so that M1 A1 A1 If true for n = k then true for n = k + 1 and as true for n = 1 true for all n A1 cso (5) (b) Condone use of n here. B1 M1 A1 A1 If true for m = k then true for m = k + 1 and as true for m = 1 true for all m A1 cso (5) [10] JUNE 2013 (R) MARK SCHEME Question Number Scheme Marks 1. (a) B1
M1 = A1, B1 M1 A1 cso [5] 2. (a) M1 A1cao (2) (b) = = M1 A1 (2) (c) =10 M1 A1ft (2) (d) M1 so or A1 cao (2) [8] 3. (a) M1 A1(2) (b) f(5) = - 0.0807 B1 M1 M1 =5.2(0) A1 (4) [6] Question Number Scheme Marks 4. (a) B1 (1)
[14] Question Number Scheme Marks 8. (a) If n =1, and , B1 (so true for n = 1. Assume true for n = k) So M1 = = 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 + 1 A1 As result is true for n = 1 this implies true for all positive integers and so result is true by induction M1A1cso (5) [11] Question Number Scheme Marks 9. (a) so M1 Gradient when x = 4 is and gradient of normal is M1 A1 So equation of normal is (or ) M1 A1 (5) (b) S is at point (9,0) B1 N is at (22,0), found by substituting y = 0 into their part (a) B1ft
(AG) A1 * [5] (b) M1 A1 cao [2] 7 marks Question Number Scheme Marks 5. (a) B1 [1] (b) M1 (So P has coordinates ) A1 oe [2] (c) Focus B1 Gradient M1 Either ; M1 or and ; l: A1 [4] 7 marks Question Number Scheme Marks 6.
Scheme Notes Marks 2 (a) f(0.5) = -0.4375 (-) f(1) = 1 Either any one of f(0.5) = awrt -0.4 or f(1) = 1 M1 Sign 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 conclusion A1 (2) (b) f(0.75) = 0.06640625() Attempt f(0.75) M1 f(0.625) = -0.222412109375() f(0.75) = awrt 0.07 and f(0.625) = awrt -0.2 A1 or or or equivalent in words. A1 In (b) there is no credit for linear interpolation and a correct answer with no working scores no marks. (3) (c) Correct derivative (May be implied later by e.g. 4(0.75)3 + 1) B1 Attempt Newton-Raphson M1 Correct first application – a correct numerical expression e.g. or awrt 0.725 (may be implied) A1 Awrt 0.724 A1 cao A1 A final answer of 0.724 with evidence of NR applied twice with no incorrect work should score 5/5 (5) (10 marks) Scheme Notes Marks 3(a) Focus B1 Directrix x + “4” = 0 or x = - “4” M1 x + 4 = 0 or x = - 4 A1
(e) Area of T = Correct area for T B1 Area of Attempt at M1 6 or follow through their det(QR) x Their triangle area provided area > 0 A1ft (3) (11 marks) Scheme Notes Marks 5(a) B1 Attempt to expand or any valid method to establish the quadratic factor e.g. Sum of roots 6, product of roots 10 M1 Attempt at linear factor with their cd in Or Or attempts f(2) M1 A1 Showing 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) B1 B1
and come from their dy/dx. * Cso **given answer** A1 Special 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 y M1 Attempt to isolate x or y – must reach x or y = f(p, q) or f(p) or f(q) M1 One correct simplified coordinate A1 Both coordinates correct and simplified A1 (4) (9 marks) JUNE 2011 MARK SCHEME Question Number Scheme Notes Marks 1. (a) Either any one of f(1) = -1 or f(2) = 8. M1 Sign change (positive, negative) (and is continuous) therefore (a root) is between and Both values correct, sign change and conclusion A1 (2) (b) (or truncated to 2.6) B1 Attempt to find. M1 with or or or equivalent in words. A1 (3) 5 Question Number Scheme Notes Marks 2. (a) or awrt 2.24 B1 (1) (b) or or or or or M1
awrt 2.68 A1 oe (2) (c) An attempt to use the quadratic formula (usual rules) M1 Attempt to simplify their in terms of i,. e.g. i or i M1 So, . A1 oe (3) (d) Note that the points are and The point plotted correctly on the Argand diagram with/without label. B1 The distinct points and plotted correctly and symmetrically about the x-axis on the Argand diagram with/without label. B1 (2) 8 Question Number Scheme Notes Marks 3. (a) A = (i) A2 = = A correct method to multiply out two matrices. Can be implied by two out of four correct elements. M1 Correct answer A1 (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 = –x B1 (2) (c) k is a constant. C is singular (Can be implied) B1 Applies M1 A1 (3) 9 Question Number Scheme Notes Marks 4. (a) At least two of the four terms differentiated correctly. M1 Correct differentiation. (Allow any correct unsimplified form) A1 (2) (b) A correct numerical expression for f(0.8) B1 Attempt 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) M1 Correct application of using their values. Does not require an evaluation. M1 0.869 A1 cao (4) 6
Question Number Scheme Notes Marks 5. 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. M1 So, and Allow Any one correct equation. Any correct horizontal line M1 giving and Any one of or A1 Both and A1 (4) (b) Finds determinant by applying M1 A1 or M1 150 or ft answer A1 (4) 8 Question Number Scheme Notes Marks 6. B1 Substituting and their z* into M1 Correct equation in x and y with i2 = -1. Can be implied. A1 An attempt to equate real and imaginary parts. M1 Correct equations. A1 Attempt to solve simultaneous M1
equations to find one of x or y. At least one of the equations must contain both x and y terms. Both and A1 (7) 7 Question Number Scheme Notes Marks 7. (a) Multiplying out brackets and an attempt to use at least one of the two standard formulae correctly. M1 First two terms correct. A1 B1 Attempt to factorise out M1 Correct expression with factorised out with no errors seen. A1 Correct proof. No errors seen. A1 * (6) (b) Use of or with the result from (a) used at least once. M1 Correct unsimplified expression. E.g. Allow 2(3n) for 6n. A1 Factorising out ( or ) dM1 A1 (4)
10 Question Number Scheme Notes Marks 8. with general point (a) Using to find a. M1 So, directrix has the equation A1 oe (2) (b) or (implicitly) or (chain rule) their M1 When or A1 T: Applies or using in an attempt to find c. Their mT must be a function of t . M1 T: T: Correct solution. A1 cso* (4) (c) Compare with gives B1 NB with x = 3 and y = 12 gives into T gives Substitutes their t into T. M1 At X, Substitutes their x from (a) into T. M1 So, So the coordinates of X are A1 (4) 10 Question Number Scheme Notes Marks 9. (a) Check to see that the result is true B1 As the matrix result is true for
for Assume that the matrix equation is true for With the matrix equation becomes by M1 Correct unsimplified matrix with no errors seen. A1 Manipulates so that on at least one term. dM1 Correct result with no errors seen with some working between this and the previous A1 A1 If 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 earned A1 cso (6) Question Number Scheme Notes Marks (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 B1 giving, Applies No simplification is necessary and condone missing M1
True for n = k + 1 if true for n = k. True for n = 1, true for all n. A1 cso [4] Notes Accept or above B1 seen award M1 or award first A1 All three elements stated somewhere in the solution award final A1 Q4 (a) (3, 0) B1 (1) (b) P: A and B lie on or a correct method to find both PB and PS Perimeter = B1 B1 M1 M1 A1 (5) [6]
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. B1 B1 M1 A1 A1cso (5) (b) (*) B1, B1 M1 A1 A1cso (5)