2022 ACSBR AMath Prelims P1 Ans

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 1 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) 1 (a)  180C A B     tan tan 180C A B    M1     tan180 tan 1 tan180 tan A B A B       M1 Use of addition formula  tan 1 A B   tan A B  AG (b)   tan tantan 1 tan tan A BA B A B    4 5 3 12 4 51 3 12        M2 Subst. using addition formula With correct 4tan 3A and 5tan 12B 63 16 A1 63tan 16C  A1 [6] 2 (a) 24 6 4 6 12x x x    M1 Line is above curve 26 12 4 6 4x x x    24 18 10 0x x   22 9 5 0x x     5 2 1 0x x   M1 15 2x   A1 B1 correct number line (b)  45 3 45 3 h     3 5 3 3 5 3 h    3 5 3 3 5 3 h  

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 1 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) 3 5 3 3 5 3 3 5 3 3 5 3      M1 45 3 45 3 45 3 45 3 h      45 6 15 3 45 3    M1 48 6 15 42  8 1 15 cm7 7      A1 [7] 3 (a) 40a  B1 (b) 0 40sin (0.25)b M1 sin (0.25) 0b  0.25 0, b  M1 0.25b  4b  (shown) AG (c) 20 40sin 4 t  M1 1sin 4 2t  1 1sin 2 6         M1 4 6t    7 24t s A1 [6] 4 (a) angle BDC = angle BAD (tangent chord) B1 = angle ADE (alternate angles, CA and DE are parallel) B1 (b) From (a) angle BDC = angle ADE B1 angle ADE 180 angle DBC  B1 either one angle DAE = angle DBC Hence triangles BDC and EDA are similar (c) BD = BC BDC and EDA are similar isosceles triangles angle DAE = angle ADE B1

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 1 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) = angle BAD (alternate angles ,CA and DE are parallel) B1 [6] 5 (a) 1 sin cos 2 tancos sin 2 A A AA A      21 sin 1 2sin cos 2sin cos A A A A A      M1 M1 21 2sin A 2sin cosA A 2sin 2sin cos 2sin cos A A A A A       sin 1 2sin cos 1 2sin A A A A   M1 tan A =RHS (Shown) (b) cos 2 sin 4 21 sin 2 cos 4 B B B B    1 2tan 2B  M1 1tan 2 2B 2 26.565 ,206.565 ,386.565 ,566.565B    M1 M1 answers for 1st and 3rd quadrant answers for extended quadrants 13.3 ,103.283 ,193.3 ,283.3B      A1 All answers [7] 6 (a)     2 2 2 25 ln 5d ln d 5 5 x xx x x x x          M2     2 2 25 ln 5 5 x xx x      2 2 10 5ln 25 x x  M1   2 2 5 2 ln 25 x x   2 2 2 ln 5 x x  (shown) AG

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 1 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) (b) 2 2 21 1 2 ln ln d5 5 e e x x xx x      2 2 2 21 1 1 2 ln lnd d5 5 5 e e e x xx xx x x        M1 split   2 2 2 2 21 1 1 ln ln ln1 2 d d5 5 5 1 5 e e x e x xx e x           2 2 21 1 1 ln 2 2 d 0 d5 5 5 e e x x x xx e          M1 Evaluate 2 1 ln 5 e x x      2 1 21 1 1 ln 2 2 d5 5 5 e e x x xx e         M1 Integrate 2 1 2 5 e x  correctly 2 21 1 1 ln 2 2 d5 5 5 e e x xx e x            2 21 1 ln 2 2 2 d5 5 5 5 1 e x xx e e          2 21 1 ln 4 2 d5 5 5 e x xx e   M1 Simplify 2 21 ln 4 2 d 5 5 5 e x xx e      4 2e  A1 [8] 7 (a)    4 2 21 1 1x x x        2 1 1 1x x x    B1 (b) 4 4 4 111 1 x x x    M1     2 11 1 1 1x x x          22 1 1 1 11 1 1 A B Cx D x x xx x x        M1          2 2 21 1 1 1 1 1A x x B x x Cx D x         Subst. 1x  , 1 4A A1 Subst. 1x  , 1 4B A1

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 1 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) Comparing coefficient of x2: 0 A B C   1 10 4 4 C   0C  A1 Comparing constant: 1 A B D   1 11 4 4 D   1 2D A1       4 4 2 1 1 111 4 1 4 1 2 1 x x x x x        A1 [8] 8 (a)   34 33V  36 cm3 B1 34 363 8117 r     M1   34 36 8 1173 r      34 8 117 363 r    3 972 4 3 r    3 729r  M1 9r  AG (b) 2d 4d V rr  dA 8d rr  B2 seen d d d d d d A A r r r t  M1 Chain rule d d d d A dV V r dt r       M1 d d d d V dV t r dt r    28 8 4r r    

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 1 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) 16 r  when 9r  d 16 5.59d 9 A r   cm2/s A1 [8] 9 (a) Let D be (a, b) and B be (c, 0) Midpoint AC = 0 5 1 4,2 2        5 3,2 2     Midpoint BD = 0,2 2 a c b      ,2 2 a c b    Midpoint AC = Midpoint BD 5 2 2 a c 3 2 2 b M1 Midpoint concept 5 a c  ---(1) 3b Gradient BD = 0b a c   0 1b a c   M1 Gradient 3 1a c  3 a c  ----(2) (1) +(2) 8 2 c 4c 1a  B(4, 0) A1 D(1, 3) A1 (b) Length AB =      22 4 0 0 1    = 17 Length BC =     2 2 5 4 4 0   = 17 AB = BC M1 Since ABCD is a parallelogram, AB = BC = DC = AD R1

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 1 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) Hence ABCD is a rhombus. (c) Area ABCD = 0 4 5 1 01 1 0 4 3 12   =    1 0 16 15 1 4 0 4 02           M1 15 units2 A1 [8] 10 (a) 1a b y x  1 1 1b y a x a      M1 Gradient = 0.625 0.4 3 0.5 0.2 4   M1 3 4 b a  Incorrect conclude 4, 3a b  at this stage 1 3 1 1 4y x a       3 10.4 0.24 a  M1 1 1 4a  3 4 4 b  4, 3a b  A2 (b) 1 1 1b y a x a      x b x y a a  (vertical axis)x y B1 (c) 1gradient a 1 gradienta B1

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 1 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) vertical-intercept b a = vertical-intercept b a  vertical-intercept= gradient B1 [8] 11 (a) 2d 3sec (3 1)d y xx   B1   2 d 3 d cos 3 1 y x x  Since   2cos 3 1 0x  , B1   2cos 3 1 0x  d 0d y x  , y has no stationary points. R1 d 0d y x  (b) 3 = cos sin 2 d 4 y k x x x     3sin cos 28 ky k x x c   M1 At  0,4 3sin cos 28 ky k x x c