YTSS 4NA Additional Mathematics Paper 1 Marking Scheme 2023

YISHUN TOWN SECONDARY SCHOOL Mathematics Department MARKING SCHEME Examination : 4NA Preliminary Examination Date: 14 Aug 2023 Subject : Additional Maths Paper No. : 1 Qn Key Steps / Solution Scheme 1   2 22 2 ( 2) ( 4) 2 4 2 x A B C x x x x x                222 2 4 4 2x A x x B x C x        Sub 2x , 1B Sub 4x , 1 2C  Sub 0x , 1 2A       2 22 2 1 1 1 ( 2) ( 4) 2 2 2 4 2 x x x x x x         For decomposition of partial fractions A1 for each fraction 2               1 8 48 12 12 12 11 32 1 8 2 12 12 12 12 11 32 20 3 12 24 22 3 20 6 3 24 22 3 10 3 312 11 3 10 3 3 10 3 3 120 36 3 110 3 99 100 27 219 146 3 73 3 2 3 height height height height height                         Rationalise 48 to simplify expression For conjugate surd Numerator Denominator 3 22 70x xy y    ----------------(1) 3 7 0yx   37yx ---------------------------(2) Sub (2) to (1)     22 3 7 3 7 7 0x x x x      2 2 23 7 9 42 49 7 0x x x x x       Substitution from (2) Expansion and

Qn Key Steps / Solution Scheme 27 35 42 0xx   2 5 6 0xx   3x or 2x 2y  or 1y simplification Solving quadratic equation Both sets of answers 4     32 32 Let f 3 + 4 (3 1) is a factor of f 1f0 3 1 1 13 4 03 3 3 1 409 9 3 3 35 ------------(1) x x hx kx xx hk hk hk                                  Divisor = 1, reminder = 4 f 1 4 3 4 4 3 -----------(2) x hk hk          (1) (2), 4 32 8 3 = 3 + 8 = 11 k k hk    Use of Factor Theorem Use of Remainder Theorem Solving For values of k and h 5   2 2 2 2 2 912 4 41 4 25(12 ) 41 0 4 2512 4(1)( 41) 0 4 144 24 25 164 0 20 0 ( 4)( 5) 0 54 x x k kx k x k x k kk k k k kk kk k                              For equating the 2 expressions and equating to 0 For discriminant<0 For factorisation 6a 327 125x  = 33(3 ) 5x  2(3 5)(9 15 25)x x x    6b Consider 29 15 25xx 2 4b ac 2( 15) 4(9)(25)   675 Since 2 40b ac , therefore 29 15 25xx cannot be factorised. Finding value of discriminant

Qn Key Steps / Solution Scheme Thus, 5 3x Explanation 7a sin( ) sin( ) cos( ) cos( ) sin cos cos sin sin cos cos sin cos cos sin sin cos cos sin sin 2sin cos 2cos cos tan A B A B A B A B A B A B A B A B A B A B A B A B AB AB A             7b sin105 sin15 cos105 cos15 105 15 2 120 60 sin105 sin15 tan 60 3cos105 cos15 oo oo o o o o oo o oo AB AB A A         8a 2 cos 2 2sin 1cot tan 4 2 pecA A p A A p     For sqrt 8b 2 2 2 2 2 2 2 cos 2 2cos 1 421 28 1 8 AA p p p p p p        Or equi