AMKSS Prelim 2024 AMath P2 MS

2024 AMKSS 4E5N Prelim AMP2 Marking Scheme Qn Solution Marks 1   1 2 100 10 24 0 10 10 10 24 0 x x x x       Let 10xy    2 10 24 0 12 2 0 y y y y       12y or 2y   10 12 rejected x  10 2x  lg10 lg 2x  lg 2x B1 (correct quadratic) M1 (factorise or quadratic formula) A1 (reject 10 12x  , do not accept if reject 12y ) A1 2 Let   3f 3 5 2x x x         3 f 1 3 1 5 1 2 0    By factor theorem, 1x is a factor of 33 5 2x x  .    3 23 5 2 1 3 2x x x x Ax      Comparing coefficient of x, 5 2 3 A A        21 3 3 2 0x x x    1x or        2 3 3 4 3 2 2 3x      3 33 6x   B1 M1 (or long division: must see 23x ) M1 (factors equate to 0) A1 (for 1x ) A1 (for exact values) 3(a)                   1 1 2 2 1 2 1 4 1 d 1 4 1 1 1 4 1 4d 2 4 1 4 1 2 1 4 1 2 2 4 1 6 1 4 1 y x x y x x xx x x x x x x x x                        M1 (for chain rule), M1 (product rule) M1 (factorise or combine to single fraction) A1

Qn Solution Marks 3(b)     66 2 2 66 6 2 2 2 6 1 d 1 4 1 4 1 6 1 d d 1 4 1 4 1 4 1 x x x x x x x x x x x x                                        6 1 266 2 2 2 6 6 2 2 6 2 4 16 d 1 4 1 14 1 42 1 4 1 d 1 4 16 24 1 4 6 16 1 4 6 1 21 d 64 1 4 2 12 1 4 2 1 2 53 6 xx x x x x x x x x x x x x x                                                         M1 (using (i)) M1 (integrate their 4 1 q x correctly) A1 (correct 6 2 d 4 1 x x x including 1 6 ) M1 (substitute correct values in correct order) A1 4(a) 2x or 3x   2 3 0x x    2 2 6 0 3 3 18 0 x x x x         3p 18q M1 (accept   2 3 0x x   ;   3 2 3y x x   ) A1 A1 4(b) (i)           2 2 2 2 2 1 3 3 1 1 0 1 4 3 1 0 2 1 12 12 0 10 11 0 11 1 0 qx x x q x q x q q q q q q q q q q                         11q or 1q M1 (eliminate x or y) M1 (correct D and = 0) M1 (factorise or quadratic formula) A1 4(b) (ii)     2 2 2 3 12 12 0 4 4 0 2 0 2 21 2, 21 x x x x x x y R            M1 (factorise or quadratic formula A1 (correct x) A1 (correct y)

Qn Solution Marks 5(a) (i) T 2.2 5.0 7.8 10 12.8 lg t 0.146 0.283 0.420 0.526 0.662 B2 (minus 1 for each incorrect point; minus 1 if axes not labelled) 5(a) (ii) (a) When 0T  lg 0.04 1.10 (accept 1.07 to 1.13) t t   M1 (equate vertical intercept to lg t) A1 (both marks only given if graph is extended to find lg t -intercept) 5(a) (ii) (b)     lg lg lg 1.064 lg lg lg 1.064 kT t a t a kT     Gradient = 0.0487       lg 1.064 0.0487 0.0487 lg 1.064 1.807611629 1.81 accept 1.7 to 1.9 k k     M1 (correct linear equation or equating gradient to  lg 1.064k ) M1 (find gradient using line drawn) A1 5(a) (iii) Physical attributes such body fat of the diver different, pre-existing health conditions, materials of diving suit. B1 (accept other logical reason based in context of question. Reject answers like values not accurate; different bodies take different time etc) y = 0.0487x + 0.0395 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 2 4 6 8 10 12 14 lg t

Qn Solution Marks 5(b) Gradient =     3 1 21 1       2 2 3 2 1 2 1 2 1 Y X Y X y x y        B1 (correct gradient) M1 (subs point into equation using their gradient) A1 6(a) sin 15 15sin DG DG     cos 4 4cos HB HB     Height of D from ground =  4cos 15sin cm  M1 (to find DG, only given with correct  and correct right-angle) M1 (to find HB, only given with correct  and correct right-angle) 6(b)  4cos 15sin cos cos cos sin sin R R R              cos 4 sin 15 R R     2 24 15 241 R   15tan 4 75.06858282       4cos 15sin 241cos 75.1      M1 (all correct, minus 1 if never write) M1 (find R) M1 (form trigo equation, accept 4 15cos ;sin 241 241    ) A1 (accept  15.5cos 75.1   )  A B C F D E 15 cm 4 cm G H 

Qn Solution Marks 6(c)     241cos 75.06 14 14cos 75.06 241 75.06 25.60175686 49.46682596 49.5 (1 d.p.)                 Since 40 49.5 50   , the chock can secure the aircraft wheel. M1 for   14cos their their R   A1 A1 (only given for correct ) 7(a)  0.24 0 12 8 12 8 4 ke k k      B1 (substitute 0t and equate to 12) 7(b) 0.24 0.24 0.24 21 4 8 13 4 13ln ln 4 130.24 ln 4 4.911062485 4.91 s (3sf) t t t e e e t t        Since 3t , he did not manage to pass point B before the traffic light turned red. B1 M1 (take ln on both sides) M1 A1 (only given for correct t and comparison with 3 s) 7(c) 0.24 0.24 4 8 d 50 83 t t s e t s e t c       Substitute s = 0 and t = 0     0.24 050 8 0 03 e c        0.24 0.24 4.911 50 3 50 50 83 3 50 50 8 4.9113 3 76.78849988 m t c s e t s e         Average speed 76.78849988 4.911062485 15.63582221 15.6 m/s (3sf)    M1 (without c) A1 (with correct c) M1 (sub their t from (ii)) M1 (divide their t by their s) A1

Qn Solution Marks 8(a) 2 r n n r r kC x x       Power of x 2 n r r n r      If n is an odd integer, since 2r is even, 2n r is an odd integer. Hence x only has odd powers in every term. B1 (correct general term, don’t need to expand) B1 (explain using power) 8(b) 11 2 7 2 4 2 r r r     Term in 7x   2 911 2 2 7 2 55 4 kC x x k x       Coefficient of 7x = 255 4 k M1 (using their general term to find 7x ) A1 (or B2) 8(c) 122 2 2 2 11 2 3 11 3 113 3 3 2 4 2 2 2 2 2 2 2 8 2 k k kx x x x k k k kx x x x x x x x k kx x x x k kx x x x                                                                        Alternative method: 122 2 2 112 2 2 112 2 3 3 3 113 3 3 2 4 2 M12 4 2 2 M1 (show expansion)2 2 4 4 8 2 8 2 k k kx x x x k k k kx x x x x x kx kx k k k kx x x x x x k kx x x x                                                      M1 (separate into 11 2 2 ky kyx x        ) M1 (using sum of cubes) No mark if they did not show 3 3 2 kyx           

Qn Solution Marks 8(d) In the expansion of 11 2 kx x     , Term in   5 611 5 5 231 2 16 kx C x k x x         3 2 5 5 5 55 231 1 5774 8 16 517 57732 18464 517 2.044405542 2.04 (3sf) kk k k k k           M1 (using their general term to find x ) M1 (correct products to form equation) A1 9(a) The points of intersection are  15, 15 and  5, 5 . Midpoint of these points of intersection =    15 5 15 5, 10, 102 2           Equation of line passing through the centres of C1 and C2:  10 1 10 20 y x y x      B1 (correct midpoint) M1 (find equation of line passing pass through centre) 9(b) Let the centre of C1 be  , 20a a              2 2 2 2 2 2 2 2 5 20 5 10 5 15 100 10 25 30 225 100 2 40 150 0 20 75 0 15 5 0 5 or 15 a a a a a a a a a a a a a a a                            Centres of C1 and C2 are  5, 15 and  15, 5 . M1 (sub their equation into length formula) M1 (factorise or quadratic formula) A1, A1

Qn Solution Marks 9(c) Centre of C3 is  10, 10 Distance from  10, 10 to  15, 5      2 2 10 15 10