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