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 equivalent 9ai 6  9aii 3 4 

Qn Key Steps / Solution Scheme 9b 111 2 22 x x       9c sin(45 2 ) 0.75 Ref angle = 48.590 0 180 315 45 2 45 45 2 48.590 (rej), (180 48.590 ), (360 48.590 ) 273.590 , 311.410 136.8 ,178.2 o o oo o o o o o o o o o oo oo x x x x x                Ref angle 10 22 (sin 45 cos30 )(sin 60 cos 45 ) 2 3 3 2 2 2 2 2 32 22 32 44 1 4 o o o o                           Special angles 11a ( 2) 2 1 dP 2 21d 21 2 2 1 21 31 21 y x x x xx x xx x x x              Product rule Same base 11b   44 00 4 0 2 0 6 2 3 1d 2 d 2 1 2 1 2 ( 2) 2 1 2 2(3) ( 2)(1) 16 xx xx xx xx           

Qn Key Steps / Solution Scheme 12 3 2 2 2 2 3 2 30 2 (2.5) 30 2 (2.5) 12 5 Vh dV hdh dV dV dh dt dh dt dh dt dh dt            Differentiation 13a Let  3 cos sin cos R        2 231R 2R 1 1tan 3   6   2sin 2cos 2cos 66                 Finding R 13b Minimum value 2 Occurs when 7 66    13c 2cos 1 6    1cos 62    (2nd / 3rd quads) Domain: 06  0.52360 5.47646     Reference angle of 63   ,6 3 3 53 ,62            Finding reference angle

Qn Key Steps / Solution Scheme 14a 31gradient of = 93 1gradient of the perpendicular bisector = 1 3 = 3 13 1midpoint of , 22 Equation of perpendicular bisector is AB AB     1 13 3 22 3 20 yx yx     Gradient Midpoint Form equation 14b       22 4For centre of circle, 10 = 3 203 5 103 6 18 20 2 centre 6, 2 radius, = 6 2 2 1 or xx x x y r                22 11 6 = 5 = 5 Equation of circle is 6 2 25 r xy      Equating Centre Radius 14c  point 6, 7