Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 93/ Paper MARK SCHEME Maximum Mark: 00 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components. IGCSE is a registered trademark. This document consists of 5 printed pages. UCLES 07 [Turn over
Mark Scheme Notes Marks are of the following three types: M A B Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). Mark for a correct result or statement independent of method marks. When a part of a question has two or more method steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. The symbol FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously correct answers or results obtained from incorrect working. Note: B or A means that the candidate can earn or 0. B//0 means that the candidate can earn anything from 0 to. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded ( d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.8 instead of 0. UCLES 07 Page of 5
The following abbreviations may be used in a mark scheme or used on the scripts: AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG CAO CWO ISW SOI SR Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) Correct Answer Only (emphasising that no follow through from a previous error is allowed) Correct Working Only often written by a fortuitous answer Ignore Subsequent Working Seen or implied Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) Penalties MR A penalty of MR is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become follow through marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the coordination meeting. PA This is deducted from A or B marks in the case of premature approximation. The PA penalty is usually discussed at the meeting. UCLES 07 Page 3 of 5
n+ r = 4 + 3 + 3 Method mark for using S n S n (i) n u ( n ) ( n ) n ( n ) 3 = 4n + 9n (ii) u r ( r 3) ( r ) ( r ) r Total: = + + Method mark for using S r S r OE = 6r SR: CAO without wrong working Let P n be the proposition that 5 n +3 is divisible by 4 5 0 +3 =4 P 0 is true (allow P ) Some explanation of what P k being true means Assume that P k is true for some non-negative integer k. or e.g. 5 k +3 = 4α for nd ( α ) k 5 + + 3= 5 4 3 + 3 Alt method: Use f(k+) f(k) ( α ) = 0α = 4 5 3 k k k (or shows that ( ) 5 + + 3 = 5.5 + 5.3 4.3 = 5 5 + 3 4.3) P 0 is true and P k P k+, hence P n is true for all non-negative integers n. Total: 5 UCLES 07 Page 4 of 5
3(i) ( x ) y + = or dx d dy sec d y = dx ( x ) y + = or dx dy cos dx = y Using implicit differentiation dy x ( x ) d y + + = 0 dx dx d d y dy = x x dx (AG) for good attempt at product rule Alt method: ( ) dy d y dy dy = cos y = cos y sin y = x dx dx dx dx for good attempt at implicit differentiation 3(ii) π y' ( ) cos ( ) y' () 4 = = ( ) y" = FT Total: 4(i) 3 k 5 = 0 Using the determinant Alt method: Uses row operations k = 8 UCLES 07 Page 5 of 5
4(ii) x+ 3y+ 8z = 4 () 4x y 0z = 5 () From () and () obtain y= 3x or other correct expression x + y+ z = (3) Substitute x = t and y = 3t in (3) to obtain z. FT Alternative: Find REF for augmented matrix and form equations () Correctly () x 0 y = 0 + t 3 or z 3 t 3 + 0 (OE) 5(i) Eigenvectors are i + j + k, i + k and j + k (OE) for λ =, and respectively. (Award for any one correct and for each of the other two.) Total: 4 5(ii) Eigenvalues for B are, 3 and 4. Eigenvectors are the same as for A respectively. FT Total: UCLES 07 Page 6 of 5
6(i) π π n n In = x cos x + nx cos xdx 0 0 π π n n = 0+ nx sinx n( n ) x sinx dx 0 0 n ( π) ( ) n n ( ) n ( π) n = n n n I I + n n I = n (AG) Total: 4 Uses integration by parts with u n = x 6(ii) π π π = sin d = [ cos ] + cos d I x x x x x x x 0 0 0 π [ x] = sin = 0 ( ) 3 π n= 3 I = 3 3 = π 6 3 4 FT UCLES 07 Page 7 of 5
7 αβ = 9 αβ = 4 αβ 4 = = αβγ = 4 αβγ αβγ FT 3 x + x 4x 4 = 0 ( x )( x 4) + Attempt to factorise cubic Solution is, in ± 4 any order. Accept ±3.74 (awrt) SR for correct roots without working 8(i) ( ) ( ) Total: 8 z z = cosθ + isinθ cos θ isin θ = isinθ ( z ) 5 ( 5 5) ( 3 = z 5 z 3) + 0( z ) z z z z 5 3sin θi = i sin5θ 0isin3θ + 0isinθ Grouping not required at this stage. 5 6sin θ = sin 5θ 5sin3θ + 0sinθ for grouping and applying initial result Total: 6 UCLES 07 Page 8 of 5
8(ii) π 5 π 3 cos5θ 5cos3θ 3 6sin θ dθ = + 0cosθ 5 3 0 0 FT 5 5 53 5 0 0 3 5 3 30 = + = 9(i) x = y = x 9(ii) ( ) y' = + 4 x = 0 x =,3 Turning points are (,5) and ( 3, 3) 9(iii) (0, 6) y= 0 x 3x+ 6 ; = 9 4 No intersection with x-axis Total: UCLES 07 Page 9 of 5
93/ Cambridge International A Level Mark Scheme Question Answer Marks Guidance 9(iv) FT Asymptotes correct. Each branch. Total: 3 0(i) dy dx d y d t d y = = t dt dx d t (AG) d y d dy dt dy = = t + t dx dt dx dx dt d y dy d y t 4t = + dt dt dt (AG) Total: 3 0(ii) Substitute in (*): d y d y d y dy + 4t 6t + ty = 4 d t d t d t dt te t d y dy 4 + 3y = dt dt t e (AG) Total: UCLES 07 Page 0 of 5
93/ Cambridge International A Level Mark Scheme Question Answer Marks Guidance 0(iii) CF: ( m )( m 3) = 0 y = Ae + Be t 3t PI: t t y = ke y' = ke y" = ke t t t ke + 4ke + 3 ke k 8 t = e t GS: t 3t y= Ae + Be + e t 8 FT x y= Ae + Be 3x + 8 e x FT Total: 7 (i) Sketch of a cardioid Correct orientation and labelled Total: UCLES 07 Page of 5
(ii) a π A = ( sinθ sin θ) dθ + + π a π = + sinθ + cosθ dθ π a 3θ π 3π a = cosθ sinθ = 4 π Total: 4 (iii) π π Show that when r=0, θ =, when r=a θ = and that d r acosθ dθ = π s = π a ( + sinθ + a sin θ) + a cos θ dθ = a sin θ dθ π π + Uses correct formula for arc length; AG (iv) du dθ ( ) u = + sinθ = cosθ = u = u u so s= a du AG 0 u ( ) = a u = 4a 0 Total: 4 Including limits UCLES 07 Page of 5
E(i) x x x x 4 ( ) ( ) 4 4 4 0 0 ydx= e + e dx= e e = e e 4 x x xy dx= 0 x( e + e ) dx x 4 4 4 ( x ) ( x+ ) = ( + ) e e 3e 5e x 0 4 x x y dx= 0( e + + e ) dx 8 ( e ) 4 x x 8 8 e x e 6 e 8 + + = + 0 6 Uses correct formulae for and 3e + 5e 3e + 5e x = = e e e + e 4 4 4 4 8 e + 6 e 8 = 4 4 8 e e ; y ( OE) Total: 0 E(ii) ds = + ' = e + + e = e + e = e + e (AG) dx x x x x ( ) ( ) ( ) x x y 4 4 ( ) 4 x x π 4 x x S = π ( e + e ) dx = ( e + + e ) dx 04 0 x x 8 8 π e e 4 π e e = + x = + 8 0 Total: 4 UCLES 07 Page 3 of 5
O(i) uuur uuur uuur AC = AB = CD = 4 α i j k 5 α α 5 uuur uuur AB CD = = 3 α ~ α 3 4 α May find, instead of AC α 5 α 3 ( α ) ( α ) 5 + 3 + 4 = α 6α + 38 = 8 Substitutes their vectors into correct formula α 6α + 38= 8 α 8α + 5= 0 ( )( ) α 3 α 5 = 0 α = 3or5 (AG) Total: 7 UCLES 07 Page 4 of 5
O(ii) 4 uuur uuur AD = 6 or CD = 4 0 Alt method: Let P be point on AC with parameter i j k 56 Distance of D from AC = 4 6 0 = = 4.3 3 + + (or with CD) Use DP.AC = 0 to find (= -5/3) Find length O(iii) i j k i j k 3 ABC: n = = ABD: n = = 0 3 0 cosθ = θ = 9. or 0.333 rads 3 + 0 4 Total: 4 UCLES 07 Page 5 of 5