AQA w TRB PRACT PAPERS

abc General Certificate of Education Mathematics PRACTICE PAPERS ADVANCED SUBSIDIARY MATHEMATICS (5361) ADVANCED SUBSIDIARY PURE MATHEMATICS (5366) ADVANCED SUBSIDIARY FURTHER MATHEMATICS (5371) ADVANCED MATHEMATICS (6361) ADVANCED PURE MATHEMATICS (6366) ADVANCED FURTHER MATHEMATICS (6371) © Assessment and Qualifications Alliance 2004 COPYRIGHT AQA retains the copyright on all its publications, including the specimen units and mark schemes/teachers’ guides However, registered centres of AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to centres to photocopy any material that is acknowledged to a third party even for internal use within the centre The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales 3644723 and a registered charity number 1073334 Registered address AQA Devas Street, Manchester M15 6EX Dr Michael Cresswell Director General Contents Introduction Abbreviations use in the Mark Schemes AS Module Pure Core MPC1 AS Module Pure Core MPC2 13 AS Module Further Pure MFP1 .23 A2 Module Statistics MS03 .29 A2 Module Statistics MS04 .40 A2 Module Mechanics MM03 48 A2 Module Mechanics MM04 55 A2 Module Mechanics MM05 65 Introduction These practice papers are additional to the Specimen Units and Mark Schemes for the AQA GCE Mathematics specification (6360) The Specimen Units booklet, which contains an example of each question paper for the specification, is available from the AQA Publications Department and can also be downloaded from the AQA website (www.aqa.org.uk) This booklet of practice papers contains a further question paper and mark scheme for selected units Practice papers are included for: MPC1 because a non-calculator paper has not recently formed part of AS and A Level Mathematics; MPC2 because the equivalent paper in the previous specifications was problematical; MFP1 because the combination of subject content is different to any module of the previous specifications; MS03, MS04, MM03, MM04 and MM05 because there will be fewer , if any, past papers for the first cohorts of students taking these units Live papers are subject to many quality control checks before examinations to ensure that they are technically correct, within the specification and at the right level of demand These practice papers have not been subject to the same degree of scrutiny They are provided mainly to demonstrate the range of questions that could appear in a particular unit, rather than to illustrate the level of demand [3] Abbreviations used in the Mark Schemes M Method mark for any acceptable method, even though numerical errors may occur A method mark is not awarded until the stage referred to in the scheme is reached Once awarded, a method mark cannot be lost Method marks are not divisible when more than one is allocated, i.e M2 can only result in the award of or marks m Dependent method mark A method mark which is only awarded if a previous M or m mark has been awarded Where necessary the circumstances are specified in the scheme A Accuracy mark A mark which is awarded for accurate use of a correct method An accuracy mark is dependent on all relevant M or m marks being gained B Accuracy mark which is independent of any M or m mark E Explanation mark A mark for a response requiring explanation or comment by the candidate This mark can be independent of, or dependent on, previous marks being gained The circumstances are specified in the scheme ft or Follow through from candidate’s previous answer Follow-through marks may be given where at least one previous result which would have gained an A, B or E mark has been incorrect The candidate’s work is marked as though that previous result were correct These marks are dependent on all relevant correct methods being used Exact circumstances are specified if necessary in the scheme cao Correct answer only The accuracy mark depends on completely correct working to that stage An exception is that answers given in the question paper can be used without loss of cao marks even if the candidate has not succeeded in obtaining the given answer An accuracy mark is usually cao unless specified otherwise cso Mark(s) can only be awarded if the specified method is used awfw Anything which falls within the acceptable stated range of the answer awrt Anything which rounds or truncates to the stated answer ag Answer given (i.e printed in the question paper) Beware faked solutions However, a printed answer may be used in a later section of a question without penalty oe Or equivalent There are obvious alternative acceptable answers which can be given equivalent credit Details are specified in the scheme if necessary sc Special case Where a particular solution given by candidates needs a different mark scheme to enable appropriate credit to be given [4] abc General Certificate of Education Practice paper Advanced Subsidiary Examination MATHEMATICS Unit Pure Core MPC1 Dateline In addition to this paper you will require: • an 8-page answer book; • the blue AQA booklet of formulae and statistical tables You must not use a calculator Time allowed: hour 30 minutes Instructions • • Use blue or black ink or ball-point pen Pencil should only be used for drawing Write the information required on the front of your answer book The Examining Body for this paper is AQA The Paper Reference is MPC1 • Answer all questions • All necessary working should be shown; otherwise marks for method may be lost • The use of calculators (scientific and graphics) is not permitted Information • • The maximum mark for this paper is 75 Mark allocations are shown in brackets Advice • Unless stated otherwise, formulae may be quoted, without proof, from the booklet MPC1 [5] Answer all questions Express each of the following in the form p + q , where p and q are integers (a) (b) (4 − )(3 + ) ; 22 4− (3 marks) (3 marks) (a) Express x + x + in the form (x + p ) + q , where p and q are integers (2 marks) (b) Hence describe geometrically the transformation which maps the graph of y = x onto the (3 marks) graph of y = x + x + The points A and B have coordinates (1, ) and (7, − ) respectively (a) Find the length of AB (2 marks) (b) Show that the line AB has equation x + y = k , stating the value of the constant k (3 marks) (c) The line AC is perpendicular to the line AB Find the equation of AC (3 marks) Two numbers x and y are such that x + y = 12 The product P is formed by multiplying the first number by the square of the second number, so that P = xy (a) Show that P = x − 48 x + 144 x (b) Find the two values of x for which (2 marks) dP =0 dx (5 marks) (c) The values of x and y must both be positive (i) Show that there is only one value of x for which P is stationary (1 mark) d2P at this stationary value and hence show that it gives a (ii) Find the value of dx maximum value (3 marks) (iii) Find the maximum value of P (1 mark) [6] The polynomial p(x) is given by ( p ( x) = (x + 1) x − x + ) (a) Find the remainder when p ( x) is divided by x − (2 marks) (b) Express p (x ) in the form x + mx + nx + , stating the value of each of the integers m and n (2 marks) (c) Show that the equation x − x + = has no real roots (2 marks) ( ) (d) Find the coordinates of the points where the curve with equation y = ( x + 1) x − x + intersects the coordinate axes (2 marks) A curve has equation y = x − x − 24 x − 18 (a) Find (b) dy dx (3 marks) (i) Show that y is increasing when x − x − > (2 marks) (ii) Hence find the possible values of x for which y is increasing (3 marks) (c) Find an equation for the tangent to the curve at the point (-1, 2) (3 marks) A circle with centre C has equation x + y + x − y = (a) (i) Find the coordinates of C (2 marks) (ii) Find the radius of the circle, leaving your answer in the form k , where k is an integer (3 marks) (b) The line with equation y = mx − intersects the circle (i) Show that the x–coordinates of any points of intersection satisfy the equation (1 + m )x 2 + 4(1 − 3m )x + 20 = (3 marks) (ii) Show that the quadratic equation + m x + (1 − m ) x + 20 = has equal roots when 2m − 3m − = (3 marks) ( ) (iii) Hence find the values of m for which the line is a tangent to the circle (2 marks) Turn over ► [7] The diagram shows a curve C and a line L The curve C has equation y = 3x − x + and the line L has equation x + y = and they intersect at the points P and Q (a) The point P has coordinates (–1, 11) Find the coordinates of Q (b) (i) Find ∫ (3x ) − x + dx (4 marks) (3 marks) (ii) Find the area of the shaded region enclosed by the curve C and the line L (5 marks) END OF QUESTIONS [8] Mathematics MPC1 Practice Paper Question 1(a) Solution ( ) 12 + − − ( ) Marks M1 2 22 A1 4− B1 A1 Answer = + Total (x + 2) B1 (b) Translation M1 A1 A1 ⎡ − 2⎤ ⎢3⎥ ⎣ ⎦ Total AB = (1 − ) + (6 + 2) (b) Gradient = y−6 = − A1 Use of m1m2 = −1 M1 A1 ⇒ x + y = 22 , (k = 22) y−6 = M1 (x − 1) Gradient AC = M1, A1, A0 for ‘shift’ A1 ∆y − = ∆x M1 AB = 100 ⇒ AB = 10 (c) B1 +3 3(a) Multiply top and bottom by conjugate 4+ (4 − )( + ) = 11 2(a) M1 4+ × Comments At least terms B1 = 10 Answer = + 5 (b) Total Must be in this form with correct gradient A1 (x − 1) A1 Total from their gradient [9] oe e.g y − x = 21 MPC1 (cont) Question 4(a) Solution Marks M1 Total = x − 48 x + 144 x A1 dP = 12 x − 96 x + 144 dx M1 A1 A1 Decrease power by One term correct All correct M1 Attempt to factorise / solve y = 12 − x ⇒ P = x(12 − x ) Comments P = x(144 − 48 x + x ) (b) dP = ⇒ 12 (x − )( x − ) dx x = 2, x = (c)(i) A1 E1 ag x = ⇒ y = rejected ⇒x=2 is only value y =8 (ii) M1 d2P = 24 x − 96 dx When x = 2, d2P < ⇒ Maximum dx (iii) Max P = × 64 = 128 5(a) (b) p(2 ) = (2 + 1) (2 − + 5) E1 B1 12 cso Must have statement Total M1 Remainder = A1 p( x ) = x − x + x + x − x + M1 = x − 3x + x + A1 (c) Discriminant = 16 − 20 = − Multiplying out M1 < ⇒ no real roots (d) ft their stationary value A1 d2P = − 48 dx A1 (− 1, 0) and (0, 5) Or ( x − 2) = −1 B1 B1 Total [10] m = –3, n = A uniform circular disc of radius a can rotate freely in a vertical plane about a fixed horizontal axis through its centre and perpendicular to its plane The moment of inertia of the disc about this axis is 5ma2 A light inextensible string passes over the rough rim of the disc and two particles A and B, of masses m and 3m respectively, are attached to its ends Assume that in the subsequent motion the string does not slip around the disc Initially the system is at rest with the particles hanging freely in equilibrium The system is then released and after time t the wheel has turned through an angle θ In the subsequent motion, the particle at A remains below the disc and no slipping occurs between the string and the disc (a) Explain why the speed of the particles is aθ& (b) By conservation of energy, or otherwise, show that (1 mark) aθ& = 94 gθ (8 marks) A body is composed of a uniform wire and three particles The wire of length 6l and mass 3m is bent to form an equilateral triangle ABC The three particles of masses 2m, 4m and 4m are fixed at the vertices A, B and C respectively The body can rotate in a vertical plane about a horizontal axis through A perpendicular to the triangle (a) Show that the moment of inertia of the section BC of the wire about the axis is (b) Hence show that the moment of inertia of the body about the axis is 38ml2 10 ml2 (4 marks) (4 marks) (c) The body is released from rest with BC horizontal and above A Find the maximum angular velocity of the body in the subsequent motion (5 marks) END OF QUESTIONS [59] Mathematics MM04 Practice paper Question Solution Marks Comments AB = b – a = ⎛ ⎞ ⎛5⎞ ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ − ⎜ 4⎟ ⎜ − 3⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ −3 ⎞ ⎜ ⎟ = ⎜ ⎟ ⎜ − 12 ⎟ ⎝ ⎠ B1 Moment is r × F ⎛ −3 ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ × ⎜ − 6⎟ ⎜ − 12 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ i j k = − − 12 −6 M1 A1 = – 64i – 12j + 15k A1 Total Total M1 A1 M1 A1 M1 Resolve vertically at A TAB cos45 + TAC cos30 = 300 Resolve horizontally at A TAB sin45 = TAC sin30 T AC (1 + ) = 300 TAC = 1600 = 220 N + T AB = TAB = A1 Accept 219.6 T AC 300 1+ A1 = 155 N Total [60] MM04 (cont) Question 3(a) Solution ∫ (b) 64 x = (c) ∫ A1 A1 y = 18.3 − 3.2 tan θ = 18.286 A1 M1 Finding y A1 A1 A1 Total [61] Correct answer (awrt 18.3) Using tan with a fraction Numerator M1 A1 θ = 2.51° Correct answer from correct working Integrating M1 16384 14 cao x x7 Obtaining 14 ⎡ x7 ⎤ x dx = ⎢ ⎥ 64 y = 02 ⎣ 14 ⎦ Integrating x x5 Obtaining Finding x A1 x = 3.2 ∫ M1 M1 Comments Integrating x ⎡ x5 ⎤ x dx = ⎢ ⎥ ⎣ ⎦0 41 Total M1 1024 64 x = 64 y = (d) Marks ⎡x ⎤ x 3dx = ⎢ ⎥ = 64 ⎣ ⎦0 4 14 ft Denominator ft Angle MM04 (cont) Question Solution Marks Total Comments (a) X = – + cos θ = 1+5 × =5 Y = + + sin θ = + 5× = 12 Correct at this stage M1A1 M1 (both X and Y correct) A1 ∴ Resultant = 52 + 12 (b)(i) = 13 Xd = – × – × + 19 5d = – 15 d = –3 ∴ line cuts axis at (0, – 3) A1 M1A1 A1 A1 A1 M1 (ii) Gradient line of action + Y = 12 A1F X 12 ∴y = x –3 A1F (or any acceptable equivalent e.g 5y = 12x – 15 etc) Total [62] cao (for Xd) 1st terms RHS (+ 19) cao ft from (a) 13 MM04 (cont) Question Solution 5(a) Resolve vertically R = Mg – P sinθ Resolve horizontally F = P cosθ On point of sliding F = µ R µ ( Mg – P sinθ) = P cosθ µMg P= cosθ + µ sin θ (b) Taking moments about A 5l × P cos θ = l × W Mg cosθ (c) If slides before it topples, Mg µMg < cosθ + µ sin θ cosθ 5µ cosθ < cosθ + µ sin θ 5µ < + µ tan θ tanθ = 17 , 5µ < + 17 µ µ < 347 P= Marks Total M1 A1 B1 B1 M1 A1 M1 A1 A1 B1 M1 M1 A1 M1 A1 Total 15 6(a) Since string does not slip, speed of particle is the same as the speed of the rim of the disc ∴ speed of particle is aθ& (b) T2 T1 A m mg T1 B 3m 3mg By conservation of energy M1 A1A1 A1 + 12 mv + 12 5ma 2θ& = 3mga θ − mga θ Using v = aθ& , x = aθ ⇒ 2ma 2θ& + 52 ma 2θ& = 2mgaθ 3mv 2 M1A1 aθ& = gθ ∴ aθ& = gθ θ&& a T2 B1 A1 Total [63] Comments MM04 (cont) Question Solution 7(a) Marks D B Total Comments C 2l A Each of AB, BC, AC has mass m and length 2l M of I of BC about axis through D is ml AD = 3l By parallel axis theorem, M of I of BC about axis through A is ( ) ml + m 3l = B1 M1 10 ml A1 (b) M of I of AB about axis through A is ml M of I of AC is also ml M of I of system is + 4m(2l ) + 4m(2l ) B1 4 10 ml + ml + ml 3 = 38ml B1 M1 m1 20 g A1 19l rods or particles all parts M1 A1 A1 19ml 2ω = 20 gl ω= ⎫ ⎪ ⎪⎪ ⎬ for either ⎪ ⎪ ⎪⎭ B1 A1 (c) Using conservation of energy 38ml 2ω = 4m.2 gl + 4m gl + m gl + m gl + m gl Total TOTAL A1 left, A1 right dependent on first M1 13 75 [64] abc General Certificate of Education Practice paper Advanced Level Examination MATHEMATICS Unit Mechanics MM05 In addition to this paper you will require: • an 8-page answer book; • the AQA booklet of formulae and statistical tables You may use a graphics calculator Time allowed: hour 30 minutes Instructions • • • • • • • Use blue or black ink or ball-point pen Pencil should only be used for drawing Write the information required on the front of your answer book The Examining Body for this paper is AQA The Paper Reference is MM05 Answer all questions All necessary working should be shown; otherwise marks for method may be lost The final answer to questions requiring the use of tables or calculators should be given to three significant figures, unless stated otherwise Take g = 9.8 m s –2, unless stated otherwise Information • • • The maximum mark for this paper is 75 Mark allocations are shown in brackets • Unless stated otherwise, formulae may be quoted, without proof, from the booklet Advice MM05 [65] A particle moves with simple harmonic motion on a straight line between two points A and B which are 0.4 metres apart The maximum speed of the particle is 10 m s–1 π (a) Show that the period of the motion is seconds (4 marks) 25 (b) Find the speed of the particle when it is 0.04 metres from A (c) The distance, s , of the particle from A at time t is given by (3 marks) s = p − q cos(ωt ) where ω, p and q are constants (i) State the values of ω and q (2 marks) (ii) When t = 0, the particle is at A Find the value of p (2 marks) The polar coordinates of a particle at time t are r = 2t + and ⎛ tπ ⎞ ⎟ ⎝4⎠ θ = 24 sin ⎜ Find the radial and transverse components of: (a) the velocity of the particle when t = 2; (5 marks) (b) the acceleration of the particle when t = (5 marks) A hailstone falls vertically under gravity through still air As it falls, water vapour from the surrounding still air condenses on the hailstone causing its mass to increase The hailstone is modelled as a uniform sphere, and at time t it has mass m and radius r (a) (b) Given that dr = λr , where λ is a positive constant, show that dt dm = 3λm dt (4 marks) Assume that the only external force acting on the hailstone is gravity If the speed of the hailstone at time t is v, show that dv = g − 3λv dt [66] (4 marks) (c) If the initial speed of the hailstone is u, show that g ( g − 3λu )e −3λt − 3λ 3λ g Hence show that the limiting value of v is 3λ v= (d) (7 marks) (2 marks) A particle, P, of mass 3m can move freely around a smooth circular ring of radius l and centre Q The circular ring is in a vertical plane The particle is attached by a light elastic string, of natural length 2l and modulus of elasticity 4mg, to a fixed point A, where A is a vertical distance 3l above Q The radius PQ makes an angle θ with the downward vertical (a) Show that the potential energy , V, of the system may be given by V = mgl ( 10 + cosθ − 2) − 3mgl cosθ + constant (b) (c) Show that ( dV 3mgl sin θ = − 10 + cos θ dθ 10 + cos θ ) Hence find the values of θ for which the system is in equilibrium (5 marks) (6 marks) (3 marks) Turn over ► [67] A spring of natural length 4a and modulus λ has one end attached to a fixed support A, and a particle P of mass m is attached to its other end Another spring of natural length 2a and modulus 4mg has one end attached to P and the other end attached to a fixed support B which is situated at a distance of 10a vertically below A The system is in equilibrium in a vertical line with the upper spring stretched to a length of 7a and the lower spring stretched to a length 3a as shown in the diagram (a) (b) Show that λ = 4mg (4 marks) a below its equilibrium position and released from rest The subsequent motion of P is subject to a resistance of magnitude mk v, where At time t = 0, the particle is lowered to a distance k2 = 6g and v is the speed of the particle at time t a (i) Given that x is the downward displacement of P from its equilibrium position at time t, show that dx d2x + 5k x = (6 marks) 10 + 2k dt dt (ii) Hence find x in terms of a, k and t (10 marks) (iii) Is the damping of the motion of the particle light, critical or heavy? Give a reason for your answer (3 marks) END OF QUESTIONS [68] Mathematics MM05 Practice Paper Question 1(a) a = 0.2 0.2ω = 10 ω = 50 π 2π P= = 50 25 (b) Solution = m s –1 (ii) Total A1 Correct period from correct working Using x = 0.16 in SHM formula Correct substitution of all values Correct speed M1 A1 A1 v = 50 0.2 − 0.16 (c)(i) Marks B1 M1 A1 ω = 50 , q = 0.2 B1 B1 M1 A1 = p − 0.2 cos p = 0.2 Total 2(a) (b) r& = 6t tπ θ& = π cos & r& = rrˆ + rθ θˆ when t = , r = 20 r& = 24, θ& = ∴ Components of velocity are 24 radially, transversely &r& = 12t tπ 3π θ&& = − sin M1 A1 A1 B1 B1 M1 Components of acceleration are 24 radially, − 30π transversely A1 ) ( 11 B1 B1 when t = 2, &r& = 24 , θ&& = − π 2 & & & & &r& = &r& − rθ rˆ + 2r&θ + rθ θˆ ( A1 ) Total [69] 10 Comments Stating amplitude Using v = aω Correct value of ω Correct ω Correct q Using s = Correct p MM05 (cont) Question (a) Solution Marks dm dm dr = ⋅ dt dr dt Comments M1 =4 πr ρ × λr =4 πρλ × Total Using m= πr ρ B1 A1F 3m 4πρ =3m λ (b) Change in momentum=impulse of external force (m+ δm)(v + δv) − mv = (m + δm) gδ t As δt → mv+m δv + vδm − mv = mgδt dv dm +v = mg dt dt dv m + 3mvλ = mg dt dv = g − 3λv dt Alternative to part (b) A1 cao cao B1 M1 A1 m A1 (B1) Change in momentum = Impulse d ( mv) = mg dt dv dm = mg m +v dt dt dv = g − 3λv dt (c) dv = dt g − 3λv (M1) (A1) (A1) ∫ ∫ M1 − ln( g − 3λv) = t + c 3λ A1 (4) For ln form t=0, v=u m1 A1F ln( g − 3λu ) 3λ g − 3λu t= ln 3λ g − 3λv g g − 3λu − 3λt e v= − 3λ 3λ c= − m1 A1 A1 (d) As t → ∞, e −3λt → Therefore v → g 3λ Attempting ln form cao Printed result M1 A1 17 Total [70] MM05 (cont) Question Solution Marks Total Comments (a) M1 AP = (3l ) + l + 2.3l l cosθ (use of cosine rule) = 10l + 6l cosθ ∴ Extension is l 10 + cos θ − 2l EPE is A1 λ x2 2l 4mgl ⎧ ⎫ = ⎨(10 + cosθ ) − 2⎬ ⎭ ⎩ P.E of particle below Q is − 3mgl cosθ ∴V = mgl (b) ( 10 + cosθ − 2) − 3mgl cos θ dV − sin θ = mgl ⋅ dθ 10 + cos θ M1 dep A1 A1 ( 10 + cosθ − 2) + 3mgl sin θ M1 A1 B1 = −mgl sin θ + mgl 12 sin θ 10 + cos θ sin θ ( (c) + 3mgl sin θ − 3mgl sin θ 10 + cos θ 3mgl sin θ = − 10 + cos θ 10 + cos θ dV When in equilibrium, =0 dθ ∴ sin θ = or − 10 + cos θ = = 12mgl ∴θ = 0, π B1 (3mgl sin θ ) M1 M1 ) A1 B1 16 = 10 + cos θ cos θ = M1 (or 2π) A1 Total [71] 14 MM05 (cont) Question Solution (a) TAP = λ 3a = λ 4a a TPB = 4mg = 2mg 2a Using F =ma vertically mg + TPB = TAP Marks B1 A1 λ = 4mg A1 (b) (i) When particle is moved a distance x below the equilibrium position, forces acting on it are λ.(3a + x) mg (3a + x) , mg, TAP = = 4a a (a − x) 2mg = (a–x) TPB = 4mg 2a a and resistance 15 mkx& [forces and are upwards] Using F = ma vertically downwards m&x& = mg + TPB − T AP – 15 mkx& Either M1 M1 All four forces m1 Dependent on both M1 above m&x& = 2mg mg (3a + x) (a–x) – – mkx& a a 2g g (3a + x) &x& − g – (a–x)+ + kx& = a a gx &x& + 15 kx& + = a 10 d 2x + 2k dx + 5k x = dt dt mg + Comments M1 ∴mg + 2mg = λ Total A1 A1 A1 [72] MM05 (cont) Question Solution (b) (ii) Substituting x = Aent, 10n2 + 2kn + 5k2 = Marks − 2k ± 4k − 200k 20 = (− k ± ki) M1 n= A1 10 x = e −kt 10 ( A cos kt 10 Total + B sin 10 M1 A1 kt ) When t = 0, x = a , A = a 2 Differentiating B1 k dx = − k e −10 t ( A cos kt + B sin kt ) 10 10 dt 10 + e −kt 10 (− 10 kA sin 10 When t = 0, dx = 0, dt a B= 14 kt ) M1 A1 A1 = − k A + kB 10 10 M1 kt + kB cos 10 10 k − t x = a e 10 (7 cos kt + sin kt ) 10 10 14 (iii) The damping is light damping since the motion is oscillating with the amplitude reducing to zero A1 B1 B1 B1 Total TOTAL [73] 10 23 75 Comments [...]... All necessary working should be shown; otherwise marks for method may be lost Information • The maximum mark for this paper is 75 • Mark allocations are shown in brackets Advice • Unless stated otherwise, formulae may be quoted, without proof, from the booklet MPC2 [13] 2 Answer all questions 1 The diagram shows a circle with centre O and radius 8 cm The angle between the radii OP and OQ is θ radians... you will require: • an 8-page answer book; • the blue AQA booklet of formulae and statistical tables You may use a graphics calculator Time allowed: 1 hour 30 minutes Instructions • • Use blue or black ink or ball-point pen Pencil should only be used for drawing Write the information required on the front of your answer book The Examining Body for this paper is AQA The Paper Reference is MS03 • Answer... fill bags with compost The weight, X kilograms, in a bag filled by this machine can be modelled by a normal distribution with mean µ and standard deviation 0.125 An inspector wishes to calculate a 95% confidence interval for µ with a width of approximately 0.05 kilograms Calculate, to the nearest 10, the sample size necessary 2 (5 marks) A fruit grower, who suspects that apples of Variety D weigh, on... (c) Hence, or otherwise, find the quadratic equation with roots in the form x2 + px + q = 0 1 α and 1 β , writing your answer (2 marks) [24] 3 4 Given that f(x) = x4 − 1: (a) write down the value of f(−1) ; 5 (1 mark) (b) show that f(−1 + h) = −4h + 6h2 − 4h3 + h4 ; (3 marks) (c) hence find the value of f'(−1) (2 marks) (a) Use the identity n ∑ r 3 = 14 n 2 (n + 1) 2 r =1 to show that n ∑ (r 3 − 1)... (b) (−1 + h) = 1 − 4h + 6h − 4h + h 4 2 4 6 1 B1F 1 ft wrong answer to (a)(i) M1A1F 2 ft wrong answers in (a) M1A1F 2 7 1 ft wrong answers in (b) M1A1 A1 f (−1 + h) − f (−1) = −4 + terms in h (−1 + h) − (−1) M1 So f' (−1) = −4 A1 Total 5(a) (n + 1)2 = n2 + 2n + 1 oe; i2 = −1 must be used B1 Hence result M1 for two correct terms 3 ag convincingly shown 2 6 B1 Σ(r3 − 1) = (Σr3) − n = 14 n(n 3 + 2n 2 +... • the blue AQA booklet of formulae and statistical tables You may use a graphics calculator Time allowed: 1 hour 30 minutes Instructions • Use blue or black ink or ball-point pen Pencil should only be used for drawing • Write the information required on the front of your answer book The Examining Body for this paper is AQA The Paper Reference is MPC2 • Answer all questions • All necessary working should... necessary working should be shown; otherwise marks for method may be lost • The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures Information • • The maximum mark for this paper is 75 Mark allocations are shown in brackets Advice • Unless stated otherwise, formulae may be quoted, without proof, from the booklet MS03 [29] 2 Answer all... • the blue AQA booklet of formulae and statistical tables You may use a graphics calculator Time allowed: 1 hour 30 minutes Instructions • Use blue or black ink or ball-point pen Pencil should only be used for drawing • Write the information required on the front of your answer book The Examining Body for this paper is AQA The Paper Reference is MFP1 • Answer all questions • All necessary working should... curve 9 (6 marks) (3 marks) (3 marks) (a) Sketch the ellipse C which has equation x2 y2 + =1 9 4 showing the coordinates of the points where the ellipse intersects the axes (4 marks) (b) Describe a sequence of geometrical transformations which would transform the unit circle x2 + y2 = 1 into the ellipse C (4 marks) (c) Show that, if the line L which has equation 8 x + 9 y = 30 intersects the ellipse C,... no difference between the proportions of males in the two populations State, giving a reason, whether or not you agree with this claim 5 (2 marks) A Passenger Transport Executive (PTE) carries out a survey of the commuting habits of city centre workers The PTE discovers that 40% of city centre workers travel by bus, 25% travel by train and the remainder use private vehicles Of those who travel by bus,