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1 National Quali cations AHEXEMPLAR PAPER ONLY EP/AH/0 Mathematics Date Not applicable Duration hours Total marks 00 Attempt ALL questions. You may use a calculator. Full credit will be given only to solutions which contain appropriate working. State the units for your answer where appropriate. Write your answers clearly in the answer booklet provided. In the answer booklet, you must clearly identify the question number you are attempting. Use blue or black ink. Before leaving the eamination room you must give your answer booklet to the Invigilator; if you do not, you may lose all the marks for this paper. *EPAH0*

2 FORMULAE LIST Standard derivatives Standard integrals f ( ) f ( ) f ( ) f ( ) d sin cos sec ( a) a a tan( a) + c sin a + c tan + a + tan c a a + tan sec ln + c cot cosec a e a e a + c sec sectan cosec ln coseccot e e Summations (Arithmetic series) S = n n a + ( n ) d (Geometric series) S n n ( ) a r = r n n n nn ( + ) nn ( + )( n + ) n ( n+ ) r =, r =, r = 6 r= r= r= Binomial theorem n ( a+ b) = n r= 0 n r a n r b r where n n n Cr r = =! r!( n r)! Maclaurin epansion iv f ( ) f ( ) f ( ) f( ) = f( ) + f ( ) !!! Page 0

3 FORMULAE LIST (continued) De Moivre s theorem Vector product [ (cos sin )] n n r θ + i θ = r ( cosnθ + isinnθ) i j k a a a a a a a b= absinθnˆ = a a a = i j + k b b b b b b b b b Matri transformation Anti-clockwise rotation through an angle, θ, about the origin, cosθ sinθ sin θ cosθ Page 0

4 Total marks 00 Attempt ALL questions MARKS. Use the binomial theorem to epand and simplify.. Given cos y= e sin (a) Find dy d. Given f( ) = + (b) Obtain f ( ) and simplify your answer.. Use Gaussian elimination on the system of equations below to give an epression for z in terms of λ. + y+ z = + y λz = + 6y+ 8z = Determine the value(s) of λ for which this system does not have a solution.. The velocity, v, of a particle, P, at time, t, is given by t t v e e = +. (a) Find the acceleration of P at time t. (b) Find the distance covered by P between t = 0 and t = ln.. Given that z = i, write down the conjugate z and epress z in polar form. 6. The equation + y + 9 6y = defines a curve passing through the point,. A( ) Find the equation of the tangent to the curve at A. Page 0

5 MARKS 7. Matrices A and B are defined by (a) Find A. p A = and 6 B =. (b) Find the value of p for which A is singular. (c) Find the values of p and if B = A. 8 (a) Give the first three non-zero terms of the Maclaurin series for cos. (b) Write down the first four terms of the Maclaurin series for e. (c) Hence, or otherwise, determine the Maclaurin series for including, the term in. e cos up to, and 9. Prove by contradiction, that if is irrational then is irrational. 0. Find the coordinates of the point of infleion on the graph of y= sin + tan, where π π < <.. (a) Write down the matri, M, associated with a reflection in the y-ais. (b) Write down a second matri, M, associated with an anti-clockwise rotation through an angle of π radians about the origin. (c) Find the matri, M, associated with an anti-clockwise rotation through π radians about the origin followed by a reflection in the y-ais. (d) State the single transformation associated with M.. Prove by induction that, for all positive integers n, n =. r rr ( + ) n+ = Page 0

6 . A semi-circle with centre ( 0, ) and radius, lies on the -ais as shown. Find the volume of the solid of revolution formed when the shaded region is rotated completely about the -ais. MARKS y O. Part of the straight line graph of a function f( ) is shown. y (0, c) O (, 0) (a) Sketch the graph of f ( ) showing points of intersection with the aes. (b) State the value of k for which f( ) + k is an odd function. (c) Find the value of h for which f ( + h) is an even function.. Use the substitution = tanθ to determine the eact value of ( + ) 0 d. Page 06

7 6. A line, L, passes through the point P(,, ) and is parallel to u = i+ j k MARKS and a second line, L, passes through Q(,, ) and is parallel to u = i+ j+ k. (a) Determine the vector equations for L and L. (b) Show that the lines L and L intersect and find the point of intersection. (c) Determine the equation of the plane containing L and L. 7. (a) Find the general solution of the differential equation d y dy y = e +. d d 7 (b) Find the particular solution for which y = and dy d = when = Vegetation can be irrigated by putting a small hole in the bottom of a cylindrical tank, so that the water leaks out slowly. Torricelli s law states that the rate of change of volume, V, of water in the tank is proportional to the square root of the height, h, of the water above the hole. This is given by the differential equation: dv dt = k h, k > 0. (a) For a cylindrical tank with constant cross-sectional area, A, show that the rate of change of the height of the water in the tank is given by dh k = dt A (b) Initially, when the height of the water is cm, the rate at which the height is changing is 0 cm/hr. By solving the differential equation in part (a), show that h. h= t. 80 (c) How many days will it take for the tank to empty? (d) Given that the tank has radius of 0 cm, find the rate, in cm /hr, at which the vegetation was receiving water from the tank at the end of the fourth day. [END OF EXEMPLAR QUESTION PAPER] Page 07

8 National Quali cations AHEXEMPLAR PAPER ONLY EP/AH/0 Mathematics Marking Instructions These Marking Instructions have been provided to show how SQA would mark this Eemplar Question Paper. The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purpose, written permission must be obtained from SQA s Marketing team on permissions@sqa.org.uk. Where the publication includes materials from sources other than SQA (ie secondary copyright), this material should only be reproduced for the purposes of eamination or assessment. If it needs to be reproduced for any other purpose it is the user s responsibility to obtain the necessary copyright clearance.

9 General Marking Principles for Advanced Higher Mathematics This information is provided to help you understand the general principles you must apply when marking candidate responses to questions in this Paper. These principles must be read in conjunction with the Detailed Marking Instructions, which identify the key features required in candidate responses. (a) Marks for each candidate response must always be assigned in line with these General Marking Principles and the Detailed Marking Instructions for this assessment. (b) Marking should always be positive. This means that, for each candidate response, marks are accumulated for the demonstration of relevant skills, knowledge and understanding: they are not deducted from a maimum on the basis of errors or omissions. (c) Candidates may use any mathematically correct method to answer questions ecept in cases where a particular method is specified or ecluded. (d) Working subsequent to an error must be followed through, with possible credit for the subsequent working, provided that the level of difficulty involved is approimately similar. Where, subsequent to an error, the working is easier, candidates lose the opportunity to gain credit. (e) Where transcription errors occur, candidates would normally lose the opportunity to gain a processing mark. (f) Scored-out or erased working which has not been replaced should be marked where still legible. However, if the scored-out or erased working has been replaced, only the work which has not been scored out should be judged. (g) Unless specifically mentioned in the Detailed Marking Instructions, do not penalise: working subsequent to a correct answer correct working in the wrong part of a question legitimate variations in solutions repeated errors within a question Definitions of Mathematics-specific command words used in this Eemplar Question Paper Determine: determine an answer from given facts, figures, or information. Epand: multiply out an algebraic epression by making use of the distributive law or a compound trigonometric epression by making use of one of the addition formulae for sin( A ± B) or cos( A ± B). Epress: use given information to rewrite an epression in a specified form. Find: obtain an answer showing relevant stages of working. Hence: use the previous answer to proceed. Hence, or otherwise: use the previous answer to proceed; however, another method may alternatively be used. Prove: use a sequence of logical steps to obtain a given result in a formal way. Show that: use mathematics to show that a statement or result is correct (without the formality of proof) all steps, including the required conclusion, must be shown. Page 0

10 Sketch: give a general idea of the required shape or relationship and annotate with all relevant points and features. Solve: obtain the answer(s) using algebraic and/or numerical and/or graphical methods. Page 0

11 Detailed Marking Instructions for each question Question Epected response (Give one mark for each ) Ans: Ma mark Additional guidance (Illustration of evidence for awarding a mark at each ) correct unsimplified epansion = C0 + C + C C C C fully simplified powers of = ,, 0 7 powers of and binomial coefficients OR powers of correct complete and simplifies correctly. Accept negative powers of.. Coefficients must be fully processed to simplified fractions and whole numbers. a Ans: cos cos ( ) e. sin cos + sin e. sin evidence of product rule see. first term cos e. sin cos +... second term cos... + sin. e ( sin ) b Ans: f ( ) = ( + ) know to use quotient ( + ). ( ). (or product) rule f ( ) = ( + ) correct derivative, using either + + rule, unsimplified = ( + ) Page 0

12 6 simplify answer use polynomial division (or inspection) correctly simplify f ( ) = 6 ( ) OR f ( ) + = + correctly complete first step f ( ) = ( )( + )... ( ) 6 apply chain rule and simplify answer 6 f ( ) = ( + ) = ( ). Evidence of method: statement of the rule and evidence of progress in applying it. OR Application showing the sum of two terms, both involving differentiation. cos cos. Sign switched available for e sin e sin cos or equivalent.. Evidence of method: statement of the rule and evidence of progress in applying it. OR Application showing the difference of two terms, both involving differentiation and a denominator.. Accept use of product rule with equivalent criteria for. Ans: λ = + set up augmented matri λ 6 8 obtain zeroes in first elements of second and third rows complete elimination to upper triangular form 0 + λ λ λ obtain simplified epression for z ( + λ )z =, z = + λ statement based on epression λ =, accept ( λ ) at. Row operations commentary not required for full credit. Methods not using augmented matri may be acceptable. Page 0

13 . Not necessary to have unitary values for second elements for.. Accept lower triangular form.. If lower triangular form used, will need to have simplified epression for z.. Accept z =. λ.6 Also accept: when λ = there are no solutions λ < and λ > ; λ < or λ >. a Ans: t t e + evidence of knowing to differentiate e, dy t e... d = complete differentiation t t e + e b Ans: 8 or or equivalent correctly set up integral ln ln t t = = ( + ) s vdt e e dt 0 0 Note: integrate and evaluate correctly. Accept rounded answers to sf or better. = + + ln ln e e 8 or or equivalent Ans: π π cos + i sin correct statement of conjugate z = + i one of r, θ correct complete substitution process answer OR, π π z = cos + i sin π π π π z = cos + i sin = cos + i sin obtain z in Cartesian form ( ) one of r,θ correct z = + i = + i = + i ( cosθ sinθ) z = + i= r + i Page 06

14 second substitution r =, π θ=, π π z = cos + i sin. Accept cis π or the eponential form of a comple number z = e i at,.. Accept angles epressed in degrees, ie 60, 0.. Where a candidate has applied De Moivre s theorem to k( cosθ isinθ) do not penalise.. Correct polar form only. Answer in form k( cosθ isinθ) loses unless correct form appears also.. Accept answers from π to π as being in polar form. For answers outside this range do not award. π 6 Ans: y= + or y= + or y + = 0 terms and constant = 0 y terms dy dy + y = 0 d d gradient equation of tangent at A (, ) dy d y = ( ) + 9 = = = Rearrangement and eplicit statement of dy not required for full marks. d d 6. Where candidates assert that ( ) =, not given, but, and all possible, leading to d 6y= + (or equivalent) for. 7 a Ans: 6 p p 0 p correct answer 7 b Ans: p = property stated or implied (see note 7.) correct value for p (see note 7.) OR + p = 0 p = 6 p p+ p 6 p p = 8 p+ 0 p A is singular when det A = 0 p = OR A is singular when A is singular, ie when det A = 0 Eplicitly states property (not essential but preferred) correct value of p (see note 7.) Page 07

15 7 c Ans: = and p = T A transpose ( A ) correct. Does not have to be eplicitly stated. values of p and correct p 6 = p 6 6 = p = and p = 7. For (a) and (c), statement of answers only: award full marks. For (b), p = only, award only ( out of ). 7. Misinterpretation of T A as inverse leading to p = 0 and = OR to 8 p = and 9 = OR to p = and =. OR any other set of inconsistent equations: do not award or, ie 0 out of. 7. Accept unsimplified answers. 7. Usually implied by net line. 7. For any equation based on answer to (a), correctly obtaining all possible solutions, including comple, both available. No solutions, not possible etc not available, even if true. 8 a Ans: correct statement of series for cos cos = +...!! substitute and evaluate coefficients ( ) ( ) cos = +...!! 9 8 = = OR correct differentiation and evaluation if doing from first principles f( ) = cos f ( ) = sin f ( ) = 9cos f ( ) = 7sin f ( ) = 8cos f ( 0) = f ( 0) = 0 f ( 0) = 9 f ( 0) = 0 f ( 0) = 8 8 b Ans: e = Page 08

16 8 c state series with correct substitution e = Ans: = + know to multiply the two previously obtained series together correctly multiply out brackets 6 simplify to lowest terms 9 7 cos... e = = = +... OR either all three derivatives correct OR first derivative and first two evaluations (above ) OR all evaluations [not eased if at least one each of e and sin/ cos ] OR last two derivatives and last two evaluations correct f( ) = e cos f ( ) = e cos e sin f ( 0) = f ( 0) = f ( ) = e ( cos sin ) f ( 0) = f ( ) = e ( 6cos 9sin ) f ( 0) = 6 remainder correct 6 correct substitution of coefficients obtained at into formula and simplify to lowest terms ( ) ( 6) cos e = !! 6 e cos = = Award for substitution of into series for cos. 8. Must have at least three terms for if no further working. 8. Candidates may differentiate from first principles for any or all of the three required series for full credit. 8. For and 6 ignore additional terms in or higher. 9 Ans: proof start proof (Given that is irrational) assume that is rational Page 09

17 set up number and process a = ( a,b natural numbers with no b common factor) a = b complete proof is rational, which is a contradiction, therefore the original statement is true 0 Ans: (0, 0) with justification find st derivative and begin to find nd derivative dy cos sec d = + d y sin... d = + complete nd derivative set nd derivative to zero and process d y = sin + (cos ) sin or equivalent d sin sin cos = 0 solve to find potential POI sin = 0leads, to = 0, y= 0 Note: check for nd derivative change of sign d y d 0. Candidates may refer to the curve s change of concavity Ans: reflection in the line with equation y = a correct reflection matri M b correct rotation matri M c evaluation of M d correct transformation consistent with candidate s M = reflection in the line with equation y = Page 0

18 Ans: proof test for n = state assumption consider n= k + LHS = = RHS= = so true when n = assume true for n= k, = rr ( + ) k+ k+ k k r = = + rr ( + ) rr ( + ) ( k+ )( k+ ) r= r= = + k + ( k + )( k + ) process k + k + = = ( k + )( k + ) ( k + ) (( k + ) + ) complete proof Thus, if true for n= k, statement true for n= k +, but since true for n =, by mathematical induction, true for all n.. Statement of conclusion can only gain if clear attempt at processing is shown at.. Candidates may approach the question by stating the target result with n= k + before starting processing for. This is a valid method. Ans: 9 π units correctly identify circle equation correct form of integral and limits substitute ( ) + y = or equivalent V = π y d. = π ( ) 0 ( ) 0 d integrate = π ( ) 0 evaluate = 9π units. Accept any version of circle equation.. Circle may be translated unit left with appropriate equation and limits used.. For a numerical answer of 8 or better to gain the result in terms of must be shown.. units not required. Page

19 a Ans: diagram straight line with negative gradient crossing the positive sections of the and y aes both intersections correctly annotated, b Ans: k = c correct value of k y = f( ) c is odd therefore, k = c c Ans: h = sketch of y= f( ) with point of reflection marked eplicit statement of answer y = f( + ) is even therefore, h = Ans: process solution to obtain both limits for θ = and = 0 become π θ = and θ = 0 correctly replace all terms π sec θ dθ 0 ( + tan θ ) replace + tan θ with sec θ π sec θ d 0 ( sec θ ) θ simplify to integrable form integrate and evaluate correctly Note: 0 π cos θ dθ ] π 0 [ sinθ =. limits can be kept during working provided integral epression is epressed back in terms of. Page

20 6 a Ans: r = + λ r = + λ L equation correct (can be written using column vectors or i, j,k ) r = p+ λu where p = L equation correct (can be written using column vectors or i, j,k ) r = q+ μu where q = 6 b Ans: lines intersect at (,, ) two equations for two parameters + λ = μ μ + λ = 7 two parameter solutions λ = + μ λ = + λ = + μ μ λ = μ = check third component in both z = ( ) z = + ( ) equations = = 6 point of intersection Since z 6 at (,, ) = z, the lines intersect 6 c Ans: + y+ z = or equivalent 7 correct strategy to find normal 7 i j k i j k i j or 8 normal vector = i( + ) j( ) + k( + 8) 8 = 6i+ j+ k Page

21 9 find value for constant and equation 6 Point of 6+ y+ z =. 9 intersection = 6 6. In (a), lines written in parametric or symmetric forms would gain only mark out of available. 6. In (c), the plane equation can be given in vector form, eg r = a+ su + tu where a is position vector of a point on the plane and st, R. 7 a Ans: y = Ae + Be e 6 7 set up auilary equation m m solutions to AE ( )( ) m m + = 0 m= or m= state complementary function y Ae = + Be state particular integral y= Ce + D process dy = d y Ce = Ce d d 6 calculate C and D 6 d y dy y e d d = + Ce Ce Ce D + = e + Ce D = e + Hence, C = and D = 6 7 state general solution 7 y = Ae + Be e 6 7 b Ans: y = e + e e 6 8 set up equations 8 = 0 and y = A+ B 6 = = 0 and dy d = A+ B = Page

22 9 process to find A and B 9 B 7= B= A= 0 state particular solution 0 y = e + e e 6 8 a Ans: proof epress related rates of change complete show that Method V=Ah dh dh dv =. dt dv dt dv = A* dh dh = * dv A =. k h A k = h A or Method V=Ah dv d ( Ah = ) dt dt dh k h = A dt Adh = k h dt dh k = h dt A 8 b Ans: proof substitute values dh = 0 cm/hr whenh= dt k 0= A k = A= 0k A 0 separate variables k dh = dt h A dh = dt h 0 or integrate correctly k h = t+ c A Page

23 6 evaluate constant and complete rearrangement 8 c Ans: 0 days 6 = c c= k h = t+ A k h = t+ A k h= t+ A h= t know to set to zero 7 0= t calculate the number of days 8 d Ans: rate to vegetation is 08π( cm /hr) 9 find k t = 960 hours number days = = 0 days k 9 A = 00 π, so = A 0 k = 0π 0 calculate h or h 0 h = = 80 process to and interpret answer for dv dt dv = 08 π[ cm /hr] dt 8. In (a), one or both of the * lines needed for method. 8. In (c), accept any numerical answer rounding to 9. Do not penalise the omission of units. 8. In (c), accept the omission of a negative sign for dv provided interpretation, eg as in the answer dt above demonstrates understanding of the contet of the problem. dv 8. A = = 9 πcm /hr which comes from taking t =. Do not award 0. dt 8. Using h = in part (d) leading to 77, do not award 0 or. 8.6 Do not penalise the omission of integration symbols. 8.7 Where candidates use instead of 0 initially, 7 lost, but 8 available if resulting quadratic solved correctly to obtain both t = 0 and t =90, discarding t = 0 answer and converting to 80 days. [END OF EXEMPLAR MARKING INSTRUCTIONS] Page 6

2014 Mathematics. Advanced Higher. Finalised Marking Instructions

2014 Mathematics. Advanced Higher. Finalised Marking Instructions 0 Mathematics Advanced Higher Finalised ing Instructions Scottish Qualifications Authority 0 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial