GCE AS Mathematics January 007 Mark Schemes Issued: April 007
NORTHERN IRELAND GENERAL CERTIFICATE OF SECONDARY EDUCATION (GCSE) AND NORTHERN IRELAND GENERAL CERTIFICATE OF EDUCATION (GCE) Introduction MARK SCHEMES (007) Foreword Mark Schemes are published to assist teachers and students in their preparation for examinations. Through the mark schemes teachers and students will be able to see what examiners are looking for in response to questions and exactly where the marks have been awarded. The publishing of the mark schemes may help to show that examiners are not concerned about finding out what a student does not know but rather with rewarding students for what they do know. The Purpose of Mark Schemes Examination papers are set and revised by teams of examiners and revisers appointed by the Council. The teams of examiners and revisers include experienced teachers who are familiar with the level and standards expected of 6- and 8-year-old students in schools and colleges. The job of the examiners is to set the questions and the mark schemes; and the job of the revisers is to review the questions and mark schemes commenting on a large range of issues about which they must be satisfied before the question papers and mark schemes are finalised. The questions and the mark schemes are developed in association with each other so that the issues of differentiation and positive achievement can be addressed right from the start. Mark schemes therefore are regarded as a part of an integral process which begins with the setting of questions and ends with the marking of the examination. The main purpose of the mark scheme is to provide a uniform basis for the marking process so that all the markers are following exactly the same instructions and making the same judgements in so far as this is possible. Before marking begins a standardising meeting is held where all the markers are briefed using the mark scheme and samples of the students work in the form of scripts. Consideration is also given at this stage to any comments on the operational papers received from teachers and their organisations. During this meeting, and up to and including the end of the marking, there is provision for amendments to be made to the mark scheme. What is published represents this final form of the mark scheme. It is important to recognise that in some cases there may well be other correct responses which are equally acceptable to those published: the mark scheme can only cover those responses which emerged in the examination. There may also be instances where certain judgements may have to be left to the experience of the examiner, for example, where there is no absolute correct response all teachers will be familiar with making such judgements. The Council hopes that the mark schemes will be viewed and used in a constructive way as a further support to the teaching and learning processes. iii
CONTENTS Page C: Module C C: Module C 7 F: Module FP : Module S: Module S 9
ADVANCED SUBSIDIARY (AS) General Certificate of Education January 007 Mathematics Assessment Unit C assessing Module C: AS Core Mathematics [AMC] WEDNESDAY 0 JANUARY, AFTERNOON Standardising Meeting Version MARK SCHEME AAMCW7P 96.0 AMCJ
Mathematics Assessment Unit C assessing Module C: AS Core Mathematics Mark Scheme (a) f() = 7 8 + 5 = 4 W (b) (x ) 9 + (x ) 7 5 (i) b 4ac = 44 44 = 0 (ii) root (a) 6( 7) 4( 7) 6 0 7 57 9 7 W (b) ( x )( x ) ( x )( x 5) x x W (x )(x 5) x (c) x x x x 4 W 4 AAMCW7P 96.0 [Turn over
4 (a) y 4x x x dy x x dx MW (b) dy x x dx m 4 48 4 4 y 4 0 8 ( y 8) 4( x ) y 4x 0 (c) dy dx 4x 4x > 0 x > 4 4 5 (i) x + x + 4 = 6 x = 5 y = 4 4 5 A = ( 5, 4 4 5 ) (ii) l m = m = (y 4 4 5 ) = (x 5 ) x + y 0 = 0 8 AAMCW7P 96.0
6 (i) y x MW (ii) y x MW (iii) x = x = x x x x = 0 x ( 4 4 ) x = + or x = y = or y = ( +, ) or (, ) 7 (i) h = 7 + 0 = 9 m (ii) dh dt (iii) 0t = 0 0t h t =. d dt 0max h = 7 +. 6.05 =.05 m 8 AAMCW7P 96.0 4 [Turn over
8 (i) Area of lawn = x Area of path = (x + ) x = 4x + 4 x < (4x + 4) MW4 (ii) x 8x 8 < 0 8 ± (64 + ) x = 8 ± 96 x = = 8 ± 4 6 y = 4 ± 6 W x 4 6 4 + 6 4 6 < x < 4 + 6 0 < x < 4 + 6 Total 75 AAMCW7P 96.0 5
ADVANCED SUBSIDIARY (AS) General Certificate of Education January 007 Mathematics Assessment Unit C assessing Module C: AS Core Mathematics [AMC] TUESDAY 6 JANUARY, MORNING Standardising Meeting Version MARK SCHEME Not to be circulated beyond the examining team AMCW7P AMCJ
Mathematics Assessment Unit C assessing Module C: AS Core Mathematics Mark Scheme (i) B 4 cm 5 cm A 4 y C Using sine rule sin 4 sin y 5 4 4sin 4 sin y 5 y. 6456. 4 (ii) Angle ABC 05. 654 Area of triangle ABC 4 5 sin 05. 654 9. 6cm 7 dy x x dx x y x x x 4 if y 0 when x 0 4 c 4 y x c x x 4 4 4 c 4 W 6 AMCW7P [Turn over
(i) L.H.S. tan sin sin cos cos R.H.S. cos (ii) 8 tan sin ( cos ) 8 cos 8cos cos cos 8cos 0 ( cos )(cos ) 0 cos, 70. 5, 89, no solution for cos MW 0 4 (i) r ( 5 ) ( ) r 0 (ii) Equation of circle (x + ) + (y ) = 0 (iii) If the line bisects the chord AB then the line is perpendicular to AB gradient of 4 AB 5 gradient of line = Equation of line through C y = (x + ) y = x 9 AMCW7P
5 Area under curve d x x x 0 x x 6 4 7 Area of rectangle 6 Required area 7 6 8 x 0 units squared units squared W 9 6 s r r r r. 4 r r. 4 [] r 0. 6 0. 7 r Substituting [] into [] 0. 7 r r. 4 r r. 4r 0. 7 0 00r 40r 7 0 5r 0r 9 0 ( 5r )( 5r ) 0 r 0. 6m [] 9 AMCW7P 0 [Turn over
7 (a) x 5 6 6 x 0 6 6 5 6 5 4 64 6 x 0 x 0 x 0... W 64 8. 4x 9. 6x. 8x... (b) (i) Arithmetic progression 0 + + 4... a = 0 and d = number of garments = 0 + (6 ) number of garments = 50 (ii) n 780 = ( 0 + ( n ) ) 780 = n( 0 + n ) n + 9n 780 = 0 ( n + 9)( n 0) = 0 n = 0 0 days in total AMCW7P
8 (a) log 0 x 4 5 x 4 log 5 ( x )log 4 log 5 log x log 0 0 0 0 0 5 4 log x log x. 6 0 0 5 4 (b) (i) 4 log ab log log ab 4 log log 8 log ab 4 log log log 8 ab 4log ab log ab a b a 4 a a a a a 4 (ii) log a log b a a log a log a log a log a a Total 75 AMCW7P [Turn over
ADVANCED SUBSIDIARY (AS) General Certificate of Education January 007 Mathematics Assessment Unit F assessing Module FP: Further Pure Mathematics [AMF] FRIDAY 6 JANUARY, AFtERNooN MARK SCHEME AMFPW7P 00.0 AMFJ
Mathematics Assessment Unit F assessing Module FP: Further Pure Mathematics Mark Scheme (i) We have M 4 4 7 5 0 and 5 5 0 5M + I 0 0 0 7 5 0 Hence M = 5M + I MW MW (ii) M = M(M ) M = M(5M + I) M = 5M + M M = 5(5M + I) + M M = 7M + 0I 7 AMFPW7P 00.0 4 [Turn over
a 4 (i) a 0 4 a a[( a)( + a) 4] +4[ + a + ] +[ 4 + a] = 0 a[ a ] +4[a + ] +[a 6] = 0 a + 7a 6 = 0 a 7a + 6 = 0 (a )(a + a 6) = 0 (a )(a + )(a ) = 0 a =,, W (ii) a = x 4y + z = 0 x + y z = 0 x 4y + z = 0 Equations and are the same Equations 5y + 4z = 0 z = 5 y 4 Using equation x 4y + 5 y = 0 4 x = 4 y Therefore the general solution is (t, 4t, 5t) a = x 4y + z = 0 x + 5y z = 0 x 4y z = 0 Equations 9y = 0 y = 0 Using equation x + z = 0 x = z Therefore the general solution is (t, 0, t) AMFPW7P 00.0 5 5
p q (i) r s 0 q = and s = p + q = r + s = q =, p = and s =, r = 4 Hence A 4 W (ii) 4 t x mt mx t mt = x 4t mt = mx 4 m m = m m m = 4 m m 4m + 4 = 0 (m ) = 0 m = Therefore the equation of the line is y = x (iii) A reflection in the x-axis (iv) A = MB AB = M 0 M 4 0 M 4 4 MW 5 AMFPW7P 00.0 6 [Turn over
4 (i) z i i i i i (ii) (iii) z 4 4 tan Hence arg z z = arg z = Im + z + z z z Re MW (iv) = 6 Therefore, angle = + Hence, = + Therefore, i.e. = 5 5 tan 5 tan 4 AMFPW7P 00.0 7 7
5 (i) Circle : Centre (, ) r = + + = 5 Radius = 5 MW Circle : Centre (5, 7) r = 5 + 7 + 6 = 00 Radius = 0 MW Distance between centres = + 4 = 5 Difference of radii = 5 Hence the circles touch internally. (ii) x + y 4x 6y = 0 x + y 0x 4y 6 = 0 Subtract to give 6x + 8y 86 = 0 Therefore, the common chord has equation x + 4y = 9 Since the circles touch internally then the equation of the common tangent is x + 4y = 9 0 AMFPW7P 00.0 8 [Turn over
6 (i) A B C D E F A A B C D E F B B C A F D E C C A B E F D D D F E B A C E E D F A C B F F E D C B A MW MW MW (ii) Identity element = A Since B = A then the period of B is Since F = A then the period of F is MW (iii) 7 4 5 6 4 5 6 4 6 5 6 5 4 4 4 5 6 5 5 6 4 6 6 5 4 (iv) The groups are isomorphic since the following isomorphism holds: A F 6 B C 4 E D 5 MW 7 Total 75 AMFPW7P 00.0 9 9
ADVANCED SUBSIDIARY (AS) General Certificate of Education January 007 Mathematics Assessment Unit assessing Module : Mechanics [AM] THURSDAY 8 JANUARY, AFTERNOON Standardising Meeting Version MARK SCHEME AMW7P 08.0 AMJ
Mathematics Assessment Unit assessing Module : Mechanics Mark Scheme 9 7 5 R magnitude : R angle N.6 tan 56. 8. y mg Mg moments clockwise = moments anticlockwise. mg = y Mg m. M y W 5 AMW7P 08.0 [Turn over
(i) R R T T 00 000 MW 750g 00g (ii) split up F ma 000 T 00a T 00 750a 700 950a a 0. 87ms T 954 N Alternative Solution whole F ma 000 00 950a a 0. 87 ms one part F = ma or F ma T 00 750a 000 T 00a T 954 N T 954 N 9 4 (i) 8 + g u 8 v 0 a 9. 8 t? v u at o 8 9. 8t t 0. 86s (ii) u 8 t 5 a 9. 8 s? s ut at s 40 4. 9 5 8. 5m cliff 8. 5m high 7 AMW7P 08.0
5 (i) momentum before = momentum after 6000 0 0 6960v ( mv) v 8. 6 ms (ii) I change in momentum 960 8. 6 0 875. 86 880 Ns 880 Ns in original direction of lorry W (iii) treat car/lorry as particles no air resistance something about elasticity 9 6 (i) a t 9 v t 9dt t 9t c p0 no c at t 0 v c t v 9t (ii) at rest v 0 t t 9t 0 6t 8 0 ( t )( t 4) 0 t or 4 s MW AMW7P 08.0 4 [Turn over
(iii) t s 9t dt t 9t t d [ ds cancel out for given times] MW 7 8 at t s 6 9 64 44 at t 4 s 48 8 5 5 at t 5 s 60 0 9 t = t = 4 8 z 0 t = 5 Total distance travelled + = m 6 AMW7P 08.0 5 5
7 (i) S B R R tan 4 5 4 MW (ii) R 0 R Ss Ss 0 (iii) A 0 6Ss 5 4 5 5 4 5 0 6 0 8 0 AMW7P 08.0 6 [Turn over
8 (i) 50 0 Q 0 40g (ii) at a constant speed a = 0 resolve along plane 50cos0 Q 40g sin0 Q 0. 5N F ma Q 40gsin0 40a 0. 5 0 9. 8 40a a 5. 4ms W MW u a 5. 4 v 0 s? v u as 0 5. 4 s s 0. 094 m Total 75 AMW7P 08.0 7 7
ADVANCED SUBSIDIARY (AS) General Certificate of Education January 007 Mathematics Assessment Unit S assessing Module S: Statistics [AMS] TUESDAY JANUARY, AFTERNOON Standardising Meeting Version MARK SCHEME AMSW7P 094.0 AMSJ 9
Mathematics Assessment Unit S assessing Module S: Statistics Mark Scheme (i) mid values.5, 5, 55, 75, 95 n 84 fx 9575 fx 5856. 5 fx x 9575 mean 5. 0 ( s.f.) n 84 = 45.70779 =. 5856 5 fx. x n 84 9575 84 (ii) Spending at both supermarkets is similar Spending less variable at supermarket B (iii) Amounts to be spent are estimated not exact Special offers not considered at either supermarket Monday lunchtime shopping. A representative of all shoppers etc. Any two valid reasons each M 9 AMSW7P 094.0 0 [Turn over
(i) Let X be the r.v. the diameter of golf balls in inches X N(, ), X N(.68, ) Z 0..68.7 Z P(X.7) = 0. P(X.7) = 0. = 0.87 P Z (. 7. 68 ) 0. 87 0. 05 ( 0. 87) From Normal Distribution table (0.87) =.7 0.05 =.7.7 = 0.05 = 0.05/.7 = 0.0444 AMSW7P 094.0
(ii) 0. d.68 X (z) = 0.8, z = 0.84 By symmetry d.68 = 0.84 0.0444 d =.64 7 AMSW7P 094.0 [Turn over
(i) x 4 5 f(x) k 4k 9k 6k 5k (ii) f(x) = k + 4k + 9k + 6k + 5k = 55k = k = 55 (iii) E(X) = xf(x) = 55 [ + 4 + 9 + 4 6 + 5 5] = 55 [5] = 4.09 (4 ) E(X ) = x f(x) = 55 [ + 4 + 9 + 4 6 + 5 5] = 55 [979] = 7.8 Var = E(X ) [E(X)] = 7.8 ( 45 ) =.06 (iv) E(4X ) = 4E(X) = 4(4 ) =.4 ( 4 ) Var(4X ) = 4 Var(X) = 6 Var(X) = 6.064 = 7.0 4 4 (i) X B(n, p) X B(, 6 ) P(X = ) = C ( 6 ) ( 5 6 ) 0 = 0.96 (ii) P(replacing at least vacuum cleaner) = P( or or or... replacements) = P(no replacements) = ( 5 6 ) = 0. = 0.888 (iii) Probability that at least 0 do not need replacement = = P(0 or or replacements) = C 0 ( 5 6 ) + C ( 6 ) ( 5 6 ) + C ( 6 ) ( 5 6 ) 0 MW = 0. + 0.69 + 0.96 = 0.678 (iv) For Binomial B(, 6 ), mean = np = 6 = Expected number of replacements = Total cost = 80 = 60 AMSW7P 094.0
5 (i) f(t) k t Area under graph = ( + ) k = 5 k = so k = 5 M W (ii) E(T) = tf(t) dt 5 t dt + 5 t( t) dt M 5 t dt + 5 (t t ) dt 5[ t ] 0 + 5[ t t ] 5 [ 0] + 5 [(() / /) (() / /)] 4 5 + 5 [ 7 9 (6 8 )] 4 5 + 5 ( 7 6 ) 4 5 minutes 0 AMSW7P 094.0 4 [Turn over
6 (i) A correct A 4 4 6 B correct B B 5 6 B P(A B ) = 5 6 = 5 8 P(at least one correct) = 5 8 = 8 (ii) P(A B) = P(A B)/P(B) P(A B) = ( 4 ) = P(B) = P(A B) + P(A B) = 4 + 6 = 5 9 P(A B) = ( )/( 5 9 ) = 9 0 9 AMSW7P 094.0 5 5
7 (i) Let r.v. X be number of lawnmowers demanded for hire on a particular day X Po(.4) P(X = 0) = e.4 = 0.0907 (ii) P(X ) = P(X = 0) P(X = ) P(X = ) = e.4.4e.4.4 e.4 /! = 0.0907 0.77 0.6 = 0.40 ( s.f.) (iii) Expected daily income =.4 = 8.80 (iv) Let r.v. Y be number of lawnmowers hired on a particular day. y 0 P(Y = y) 0.0907 0.77 0.6 0.40 MW Expected number hired = E(Y) Mean income =.0 = 4.7 = 0 0.0907 + 0.77 + 0.6 + 0.40 =.0 (v) If unlimited number available, mean income is 8.80, a difference of 4.4 Since the extra income exceeds, it is viable to avail of the offer 5 Total 75 AMSW7P 094.0 6 [Turn over