AMKSS Prelim 2019 4E5N AMath P1 Solution

1 AMKSS_2019_Prelim_4E5N_AM_P1 AMKSS 2019 Prelim 4E5N AM P1 Answer Scheme Answer Marks 1(i) [2]  f 2 8x px   2 2 8 4 3 p p   M1 (differentiate) A1 1(ii) [3]   32f4 x x x x c            32 32 3 4 3 3 0 6 f 4 6 c c x x x x          M1 (correct integration without c) M1 (substitute point to find c) A1 2[4] Gradient = 25 16 3 16 10 2   3 2Y X c  316 102 1 c c   3 12YX 2 3 12yx 2 321 2 2 4 x x x    M1 (find gradient) M1 (substitute a point to find c) M1 (form equation) A1

2 AMKSS_2019_Prelim_4E5N_AM_P1 3[7] 3 2 2 p     33 8 q  33 8  3 3 82 64 4 p p p        22 8 q             2 3 8 q          2 33 322 2 8 q            45q M1 M1 M1 (substitution) A1 M1 (factorise cubic) M1 (using their value of p) A1

3 AMKSS_2019_Prelim_4E5N_AM_P1 4(i) (ii)(b) [4] (i) B1 (correct shape passing through origin) (ii)(b) B1 (positive gradient straight line passing through origin) B1, B1 (indicate origin and (64, 32)) 4(ii) (a) [2] 3 1 1 3 2 2 3 2 3 2 2 4 12 2 1 2 xx xx xx xx yx      M1 (change to index form) A1 4(ii) (b) 1 1 3 2 1 1 3 2 11 36 2 20 20 xx xx xx      0x or 1 6 2x  0y 64 32 x y   Marks under 4(i) answer x O (64, 32) y = 0.5x 2 32yx

4 AMKSS_2019_Prelim_4E5N_AM_P1 5[5]   18 2 6 18 2 6 18 2 6 xx xx x    6 18 2 6 18 2 18 2 18 2 6 18 12 18 4 18 2 12 14 9 2 6 7 x        9, 6ab M1 (factorise) M1 (correct conjugate) M1 (expand) A1, A1 6(i) [1] Let   32f 2 11 12 9x x x x    32 1 1 1 1f 2 11 12 9 02 2 2 2                           Since remainder = 0,  21x is a factor. M1 (must give conclusion) 6(i) [3]    3 2 22 11 12 9 2 1 9x x x x x kx       2 1 11 6 k k         3 2 2 2 2 11 12 9 2 1 6 9 2 1 3 x x x x x x xx           M1 (using comparing coeff or long division) M1 A1 (no mark if no working) 6(iii) [4]      22 15 17 2 1 32 1 3 3 x A B C xxx x x              2 15 17 3 2 1 3 2 1x A x B x x C x        Let 3x , 28 7 4 C C    Let 1 2x , 1124 1224 2 A A   Comparing coeff. of 2x , 0 2 2 1 B B     232 15 17 2 1 4 2 11 12 9 2 1 3 3 x x x x x x x          M1 (or combine fractions) A1 A1 A1

5 AMKSS_2019_Prelim_4E5N_AM_P1 7(i) [3]            2 2 2 2 2 2 2 2 LHS cosec cot 1 cos sin sin 1 cos sin 1 cos sin 1 cos 1 cos 1 cos 1 cos 1 cos 1 cos 1 cos RHS                         Alternative method:               2 22 2 22 2 2 2 2 2 2 LHS cosec cot cosec 2cosec cot cot 1 2 cos cos M1sin sin sin sin 1 2cos cos sin 1 2cos cos M11 cos cos 1 1 cos cos 1 cos 1 1 cos 1 cos cos 1 M11 cos 1 cos 1 cos RHS                                           M1 (change to sin, cos) M1 (apply 22sin 1 cos ) M1 (factorise) 7(ii) [3]   2 1cosec cot 4 1 cos 1 1 cos 4 4 4cos 1 cos 3cos 5             Basic 0.9272952180 0.9272952180,5.355890089 0.927,5.36 rad(3sf)    M1 A1, A1

6 AMKSS_2019_Prelim_4E5N_AM_P1 8(i) [3] Let EAC  ACD  (alternate angles) ABC  (angles in alternate segments) DAC CAB  (common angle) Triangles ABC and ACD are similar (all corresponding angles are equal / AA) M1 M1 A1 (+ statement to get 3 marks) 8(ii) [3] AEC ABC    (base angles of isosceles triangle) 2BCA  (exterior angle of triangle) 2BCD ACD        CD bisects ACB . M1 M1 A1 9(i) [3] 2 2 2 2 2 75 2 75 2 75 75 2 r rh r rh rh r rh r         2 2 2 3 75 2 752 V r h rr r rr         M1 (form eqn) M1 (make h subject) M1 (substitute) 9(ii) [3]   275 32 dV rdr    2 2 75 3 02 25 5 5(rej) r r r or r           3 375 5 5 125 cm2V     B1 M1 (equate to 0) A1

7 AMKSS_2019_Prelim_4E5N_AM_P1 10(i) [1] 23 8 3v t t   B1 10(ii) [6]   2 62 32 v t dt t t c          2 2 2 3 0 2 0 2 3 2 2 c c v t t           2 32 1 3 2 2 2 s t t dt t t t c               32 1 1 0 0 0 2 0 0 c c      32 2s t t t   3 2 3 24 3 14 2t t t t t t         23 14 0 3 7 2 0 tt tt       12 3t or 2t  (rej) M1 (integrate without c) A1 M1 (integrate without 1c ) A1 M1 (equate) A1 10(iii) [2] When 12 3t , P: 2 113 2 8 2 3 32 033v               Q: 2 113 2 2 2 2 19 033v               P and Q are travelling in the same direction at the point of collision. M1 A1

8 AMKSS_2019_Prelim_4E5N_AM_P1 11(i) [3]       2 2 3 9 1 1 5 2 2 3 9 1 52 6 18 1 5 6 18 5 5 12 13 0 13 1 0 x xx xx x x x x x x x x xx xx                       13x or 1x (rej) 1y  1,1P  M1 (substitution) M1 (correct quadratic) M1 (solving) 11(ii) [7]   39 5 24 3 5 5 24 35 xy x x x x       39 55 3 9 25 5 8 16 2 x x xx x x                   2 2 1 2 1 24 35 24ln 5 3 24ln 5 2 3 2 24ln 5 1 3 7.635532 1 333 units dxx xx                       Area of rectangle    255 2 us1 it1 n    Area of shaded region 2 15 7.635532333 = 7.36446767 =7.36 units (3sf.)  M1 (or other valid method) B1 (correct x- coordinate of Q) M1, M1 (for ln and 3x ) M1 Evaluate from 1 to 2 B1 (for rectangle) A1

9 AMKSS_2019_Prelim_4E5N_AM_P1 12(i) [3] 4 163yx Gradient of normal = 3 4 3 4y x c   30 12 4 9 3 94 c c yx      B1 M1 (substitute point A) A1 12(ii) [5]          22 2 2 22 2 2 12 0 5 312 9 5 4 9 2724 144 81 2516 2 25 75 200 016 2 24 128 0 8 16 0 ab aa a a a a aa aa aa                         8a or 16a 3b 3b (rej) M1 (use distance formula or eqn of circle) M1 (substitute using equation of normal) M1 (form quadratic, RHS = 0) A1, A1 (must reject or else deduct 1 mark)