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          M