2016 Mathematics. Advanced Higher. Finalised Marking Instructions

Transcription

1 National Qualifications Mathematics Advanced Higher Finalised ing Instructions Scottish Qualifications Authority 06 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from SQA s NQ Assessment team. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ Assessment team may be able to direct you to the secondary sources. These ing Instructions have been prepared by Examination Teams for use by SQA Appointed ers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 . (a) evidence of use of product rule,... tan x x... one resultant term of the product correct.tan x or x.. x complete differentiation x tan x x. Evidence for the award of should take the form f x... g x.... For a candidate who interprets. Accept x when awarding. tan x as tan x. is not available. (b) evidence of use of quotient or product rule and one term of numerator correct x x 5 complete differentiation correctly x 5 x....8x 6 simplify answer,5 0x 6 x 0x or x. Where a candidate uses the product rule, simplification to will be required in order to obtain 6. 0x x 0x x or 5. Incorrect working subsequent to a correct answer should be penalised in this instance eg an incorrect expansion of the denominator. (c) 7 correct derivatives 7 6 and sint 8 find dy dx 8 sin t 6 Notes Page 05

3 . (a) interpret geometric series ar 08 and ar evidence of strategy, ar ar value r 7 r. For accept r. 7. For a statement of the answer only, award and. To earn there must be evidence of a strategy eg 08 6 gives r. (b) know condition, < <. For may be replaced with a letter consistent with their answer to (a). However, in the case where a candidate obtains a value in (a) outside the open interval, will only be available where they also acknowledge that there is no sum to infinity.. Only award for a strict inequality, whether it is expressed algebraically or in words. (c) 5 calculate the first term 5 a 6 value 5, or equivalent leading to 5. For an incorrect value in (a) 6 will only be available provided the value satisfies the condition for convergence. 6. Where a candidate has used S a r r full credit is available. Page 06

4 . state general term r r C x x r 5 simplify powers of x r r r or x OR coefficients and signs,5 state simplified general term (completes simplification),5 C r r r r x determine value of r, r 9 r 5 evaluate term, x 9. Accept r r as bad form. and are r0 x available only to candidates who simplify a general term correctly. r. For accept the initial appearance of C x. and 5 are the only marks available to candidates who have not proceeded from a general term eg. an expansion using Pascal s Triangle. The required term must be explicitly identified in order for 5 to be awarded.. Starting with r x x 5. Accept r x when awarding or. Cr r leading to r can also gain full credit. Page 07

5 . Construct augmented matrix 5 Use row operations to establish first two zero elements Establish third zero element OR recognise linear relationship between two rows, OR State value of. Elementary row operations must be carried out correctly for and to be awarded.. is only available where a candidate s final matrix exhibits redundancy.. Disregard any working/statement subsequent to. Page 08

6 5. show true for n LHS: So true for n RHS: assume true for n k AND consider nk k and r r k k r k r r r k r... r r k k correct statement of sum to k terms using inductive hypothesis express explicitly in terms of k or achieve stated aim/goal, AND communicate k k k k k k k k k k k k, thus if true for n k then true for nk but since true for n, then by induction true for all n. RHS =, LHS = and/or True for n are insufficient for the award of. A candidate must demonstrate evidence of substitution into both expressions.. For acceptable phrases include: If true for... ; Suppose true for ; Assume true for. However, not acceptable: Consider n k, assume n k and True for n k. Allow if appears at conclusion.. Full marks are available to candidates who state an aim/goal earlier in the proof and who subsequently achieve the stated aim/goal.. Minimum acceptable form for : Then true for nk, but since true for n, then true for all n or equivalent. Page 09

7 6. Method for either function: first derivative and two evaluations OR all three derivatives OR all four evaluations f x x f f x x f f x x f f x x f sin 0 0 cos 0 9sin cos 0 7 f 0 f 0 f x f f x x x!! complete derivatives and evaluations AND substitute for second function: first derivative and two evaluations OR all three derivatives OR all four evaluations f x x 7 x! 9 x x x f x e f f x x e f f x x e f f x x e f complete derivatives and evaluations AND substitute 6x 6x f x x 6 x 8x x 5 multiply expressions 9 e sinx x x... x 8 x x... 9 x x x x... 5 x 6 multiply out and simplify Note x x 9 x... 6 Page 0

8 . If a candidate chooses to use the product rule to obtain the Maclaurin series for x e sinx x without first obtaining series for e and sinx separately then only 5 and 6 are potentially available. In this instance for the award of 5 apply the same principle as that used to award and. x sin x x sin cos 7 x x sin cos x x sin 7 cos f x e x f x e x e x f x e x e x f x e x e x. At 6 the appearance of terms in f 0 0 f 0 f 0 f 0 7 x or above should be disregarded. Method state the Maclaurin expansion for sin x substitute state the Maclaurin x expansion for e substitute 5 multiply expressions x sin x x... x sin x x... 9x sin x x... e x x x x... e x...!! x x e x 8 x... x x x x sin x e x x x x 9 6 multiply out and simplify 6 x 9x e sinx x x... x. For a candidate who writes down sin x x... without first writing down the series for sin x then may be awarded. A similar principle may be applied to the awarding of if required.. At 6 the appearance of terms in x or above should be disregarded. Page

9 7. (a) calculate determinant. If a candidate chooses to find identified. A then is only available where det A is clearly A... det A A... A... Award Do not award (b) Method find A use an appropriate method 0 A 0 0 A 0 A A I write in required form and explicitly state values of p and Note q Method find use an appropriate method write in required form and explicitly state values of p and q Note p and q 0 A 0 0 A p q 0 A A I p = and q = is acceptable for provided the values of p and q are explicitly stated. Page

10 (c) 5 square expression found in (b),, 5 A A I A AI I A I A I 6 substitute for A and complete process 6 5A 6I may be obtained by squaring to give and identifying the coefficient of 5 A as 5. This leads to 6 using the same method as in (b).. Accept equivalent expressions eg A A I.. Candidates may calculate A first so 5 can be awarded for A A I. Page

11 8. (a) correctly plot z on Argand diagram,,, Im O Re. Do not penalise the omission of the diagonal line.. Treat alternative axis labels as bad form (to include the case where there are no labels).. Accept a point labelled using coordinates:, and, in this instance,, i.. The minimum acceptable response for the award of is a point in quadrant together with and (or i ). (b) find modulus or argument,,,6 w a or arg w 6 complete and express in polar form,,5,6 w a cos isin 6 6. For the award of and accept any answer of the form π π, k k. 6. For the award of and accept any answer of the form k 0 60, k.. A candidate who chooses to work in degrees can only be awarded provided the degree symbol appears at some point within question 8.. Award for w acos isin At do not accept w a cos isin Working subsequent to the appearance of should be penalised where it leads to the 6 use of an incorrect argument. Page

12 (c) Method process modulus 5 process argument,,,,5 6 evaluate and express in form n ka x i y 56a cos isin w 8 8a 8 i. For the award of 5 accept any answer of the form π k π, k.. For the award of 5 accept any answer of the form k 0 60, k.. A candidate who chooses to work in degrees can only be awarded 5 provided the degree symbol appears at some point within question 8.. Do not penalise unsimplified fractions. 5. Award for... cos isin 6 6. Method find w correctly and attempt to find a higher power of w Note 5 obtain w 6 complete expansion and n express in form ka x i y eg w a i and w a i a i. 5 w a 8 8i 6 w 8 8a 8 i. Accept the omission of ' a ' at and 5 provided 8 a appears in the final answer. Page 5

13 Method write down full binomial, expansion i... i 8 i i 8 5 simplifies individual terms 6 complete expansion and n express in form ka x i y 5 8 6i i 60 68i 8 8i 6 w 8 8a 8 i. For the award of a full expansion must be written out.. Accept the omission of ' a ' at and 5 8 provided a appears in the final answer. Page 6

14 9. know to use integration by parts and start process,, ln... 8 x x 8 6 correct choice of functions to differentiate and integrate AND application thereof,, 8... ln d x x dx 8 dx differentiate ln x x x x x dx 8 8 ln 7 ln know to use second application and begin process,,, x... ln x 8 x x 8 dx 5 complete second application 8 8 x x x ln 6 simplify ln ln. For candidates who attempt to integrate ln x and differentiate x x x x x c 7 x then, and 6 may be awarded but not, and 5.. Evidence of use of integration by parts would be the appearance of an attempt to integrate one term and differentiate the other. 8t. For candidates who attempt to substitute for ln x eg t ln x leading to t e dt then becomes available upon evidence of using integration by parts ie.. 8 t t e is only available for a final answer expressed as a function of x.. For candidates who incorrectly differentiate ln x and do not require a second application of integration by parts, only, and 6 are available. 5. Do not penalise the omission of c. Page 7

15 0. give counterexample eg. choose p and since 5 5, hence not prime, statement is false set up n Notes, na, a 0 consider expansion of n Note complete proof with conclusion n 7a 7a 9a a a a 9 9 and statement such as so n has remainder when divided by statement is true.. Do not penalise the omission of a 0 in.. Treat a statement such as nn as bad form.. can only be awarded for the correct expansion of a.. Minimum statement of conclusion in is true. 5. Where a candidate invokes an incorrect use of proof by contradiction full credit may still be available provided all relevant steps are included. Page 8

16 . Method state differential, equation state relationship or apply chain rule dh 5 dt dv dv dh dt dh dt V h find the rate of change of volume with respect to height dv h dh evaluate Method dv h cm s dt express volume as a function of time V 5t find rate of change of volume with respect to time dv dt 75t find value of t t 5 evaluate dv dt 75 5 cm s 5. A candidate who assumes that only the height changes and that the length and breadth are constant can be awarded and only.. Where a candidate uses the wrong formula for the volume of a cube only and are available. dv dh. A candidate using Method who writes h can be awarded and. dt dt. To award there must be evidence of substituting and 5. Correct units must also be included. Page 9

17 . (a) correct shape, graph passes through c on the positive x- and y-axes. To award, the second arm must be sketched to within 5 of the reflected angle. Page 0

18 (b) graph of y f x passing through c on the positive y-axis, correct shape (symmetrical V) meeting positive x-axis at c. For a candidate who sketches the graph of y f x award for showing a straight line passing through 0, c.. To award, the second arm must be sketched to within 5 of the reflected angle. Page

19 . correct application of partial fractions x A B x 6 x x 6 x 9 starts process x A6 x Bx calculate one value A calculate second value B 5 5 re-state integral in partial fractions 5 x 6x 5 dx 6 one term correctly integrated 6 ln x... 7 Integrate second term correctly... 5ln 6 x 7 8 substitute limits 8 ln 5ln 6 ln 5ln 6 9 evaluate to expected form Note 86 9 ln 9. Do not penalise lack of modulus signs unless the candidate attempts to integrate rather than 6 x.. Award maximum [8/9] for appropriate working leading to 98 ln ( 9 lost) x 6 OR 08 ln 907 ( 7 lost).. Do not penalise unsimplified fractions in 9. Page

20 . (a) convert any two components of L to parametric form two from x, y 8, z 5 two linear equations involving two distinct parameters two from, 8, 7 find parameter values, verify third component in both equations or equivalent eg 7 z therefore the lines intersect z and 5 find point of intersection 5 7, 6, 7. A candidate who uses as the second parameter can only be awarded unless this is rectified later in the question.. Do not penalise the omission of the statement therefore the lines intersect. Page

21 (b) 6 identify first direction vector,, 6 d i j 7k 7 identify second direction,, vector 7 d i j k 8 calculate magnitudes and scalar product 9 calculate obtuse angle,5 8 d 7, d and d d 6 9 cos For L i j 7k and L i j k or equivalent, lose 6 but 7 is available (repeated error).. Do not penalise L i j7k and L i j k.. For L : i j7k and L : i j k or equivalent, 6 and 7 are both available.. For the award of 9 accept is not available to candidates who calculate an obtuse angle correctly but who subsequently calculate an acute angle. Page

22 5. state auxiliary equation solve auxiliary equation and state complementary, function construct particular integral m 5m 6 0 m, m y Ae Be x x y Cx Dx E 0 differentiate particular integral dy d y Cx D and C dx dx 5 calculate one coefficient of the particular integral 6 calculate remaining coefficients 5 C 6 D, E x x y Ae Be x x 7 differentiate general solution 8 construct equations using given conditions 9 Find one coefficient dy Ae Be x dx 7 x x 8 A B 7 and A B 6 or equivalent 9 A 8 or B 5 0 Find other coefficient and state particular solution y 8e 5e x x 0 x x. For do not penalise the omission of ' 0'.. can be awarded if the Complementary Function appears later as part of the general solution, as opposed to being explicitly stated immediately after solving the Auxiliary Equation.. A candidate who obtains m and mfrom a correct auxiliary equation, leading to y e e x x x x 0 7 cannot gain but all other marks are available.. Where a candidate substitutes the given conditions into the Complementary Function to obtain values of A and B and then finds the particular integral correctly 8 and 9 are unavailable. Page 5

23 6. Method working in minutes ( t 0 at noon) construct integral equation Note dt k dt T T F 9 integrate ln T T kt c F find constant, c substitute using given information 5 find constant, k 6 substitute given condition 7 know how to find time ln 9 8 k0 c c ln5 8 ln 65 5k ln5 8 ln 5 ln58 5 k ln t ln5 8 6 ln ln 58 7 t calculate time 8 t state the time to the nearest minute 9 The liquid was placed in the fridge at :7 (am) Page 6

24 Method working in minutes ( t 0 when T 5 ) construct integral equation Note dt k dt ( T T ) F integrate ln T T kt c F find constant, c. substitute using given information ln 5 k0 c ln c, kt ln 98 ln 5 know to use t 5 Note 5 5 appearance of t 5 6 use given condition 6 kt ln 65 5 ln 7 find constant, k Note k ln calculate time 8 t ln state the time to the nearest minute 9 The liquid was placed in the fridge at :7 (am). Page 7

25 Method working in hours ( t 0 at midnight) construct integral equation Note dt ( T T ) F k dt integrate ln T T kt c F use initial conditions ln58 k c interpret later time ln 5 5k c 5 find constant, k 5 ln58 ln 5 0 5k k find the constant, c 7 know to find time 6 ln c c 5... ln t calculate time 5 ln 8 t state the time to the nearest minute 9 The liquid was placed in the fridge at :7 (am). Page 8

26 Method working in minutes ( t 0 when T 5 ) construct integral equation Note dt k dt ( T T ) F integrate ln T T kt c F T T e kt c F kt T Ae T F T Ae kt use initial condition to calculate A 0 k 5 Ae A substitute using given information 98 e kt 5 know to use t 5 Note 7 5 appearance of t 5 6 substitute using given information 65 e 6 kt 5 7 find constant, k 58 5 ln ln 7 k calculate time 8 t ln state the time to the nearest minute 9 The liquid was placed in the fridge at :7 (am). Page 9

27 General note: Many candidates may use a combination of the given methods. For all methods the evidence for,, 8 and 9 is the same. To award up to 7 note that: two marks are awarded for using two different values of T one mark is awarded for finding the constant of integration one mark is awarded for finding or eliminating k (refer to Note 6) one mark is awarded for dealing with the elapsed time (noon until :5). Do not penalise the omission of integral symbols at. (All Methods). Do not penalise omission of c at. However, it is necessary to access some later marks. (All Methods). Where a candidate obtains an incorrect final answer because of earlier rounding, only 9 is unavailable. (All Methods). For Method, if the candidate works in hours: ln k ln k ln 5ln ln t ln ln ln 58 t t For Method, if the candidate works in hours: t appearance of 6 kt ln ln 7 5 k ln t ln In Method 7 can be awarded for eliminating k. 7. For Method, if the candidate works in hours: t appearance of 65 e 58 5 ln ln 7 k t ln kt 05 [END OF MARKING INSTRUCTIONS] Page 0