YTSS 4NA Additional Mathematics Paper 2 Marking Scheme 2023

YISHUN TOWN SECONDARY SCHOOL Mathematics Department MARKING SCHEME Examination : 4NA Preliminary Examination Date: ___ Aug 2023 Subject : Additional Maths Paper No. : 2

Qn Key Steps / Solution Scheme 1a   22 42 5 3 1 d 2 1 d 2 53 xx x x x x x x c           For expansion 1b   32 2 3 2 32 d2 1 d 2 11 2 3 2 64 xx xx x x x x xc xx c           For simplifying fraction 2a 2 or 360 o 2b 1 cycle in 2 critical points ( 2  ,-2), ( 3 2  ,2), ( 2  ,2), ( 3 2  ,-2) asymptotes at  =  ,  =-  3a 2 2 22 2 2 f ( ) 3 9 7 3( 3 ) 7 3337 22 3 2737 24 313 24 x x x xx x x x                           factorizing out 3 completing the square multiplying 3 back 3b Since 2 33 0, 2x  2 3 1 13 ( 0). 2 4 4x    

4       3 3 2 1 2 d5 d 21 5 2 1 21( 5) 1 (2)2 5 21 q t t q t dt t c c t              5 21 55 2(0) 1 0 5 21 51 21 2 1 5 2 1 25 12 qc t c c q t t t t t             5a 2 2 2 22 2 2 2 2 6 25 d 12 (2 5 ) 6 ( 5) d (2 5 ) 24 60 30 (2 5 ) 24 30 (2 5 ) 6 (4 5 ) (2 5 ) xy x y x x x xx x x x x xx x xx x             a = 6 and b = 5 Quotient rule Expansion and simplification Factorisation

5b For decreasing function, 2 2 d 0d 6 (4 5 ) 0(2 5 ) (2 5 ) 0 6 (4 5 ) 0 40 5 y x xx x x xx x or x       Identifying d 0d y x  With indication of 2(2 5 ) 0x leading to 6 (4 5 ) 0xx  6a 2 2 d 32d dAt 1, 1, 4d Gradient of normal at 1 1 Equation normal at 1 4 1(1) 5 5 1Area (5)(5)2 12.5 units y xx yxy x x x c c yx               gradient function for value of y gradient of normal equation of normal 6b 2 2 3 2 1 33 1 When 1, 6 ( 1,6) x x x x y Q       7a 2 32 32 33 3 cos3 cos(2 ) cos 2 cos sin 2 sin (2cos 1)cos (2sin cos )sin 2cos cos 2sin cos 2cos cos 2(1 cos )cos 2cos cos 2cos 2cos 4cos 3cos                                           for addition formula for double angle for changing to cosine

7b 3 14cos 3cos 2 1cos3 2 Reference angle = 60 3 60 ,300 , 420 , 60 , 300 , 420 20 ,100 ,140 , 20 , 100 , 140 o o o o o o o o o o o o o               cos3 Ref angle 1st 3 angles next 3 angles 8a 2 2 108 108 rh h r    8b 2 22 2 2 2 (2 ) 2 10822 2163 723 A r r rh r r r r r r r r r                    surface area substituting h and simplifying to answer 8c 2 3 2 23 dAt minimum, 0d 723 2 0 2 72 3.30 d 144 32d 0 is a minimum at = 3.30 A r r r r r A rr Ar            differentiation and = 0 solving 2nd derivative conclusion 8d   23 108 36 9.9058 h  Height of container = 229.9058 (2(3.3019) (3.3019) 15.6   Since 15.6 < 35 cm, the container can be displayed in the shop.

9a 2 2 10 16 10 16 0 ( 2)( 8) 0 2,8 (2,16) xx xx xx x A         9b 2 2 0 2 23 0 23 2 2 10 15 3 15(2) (2)3 117 units3 Shaded area 11=17 (6)(16) (8)(16)32 149 units3 x x dx xx        correct limits integration substitution correct method to find area 2 triangles 10a         3 32 2 2 2 2 1 2 d 1 1 1 3d 2 2 2 1 1 3 2 2 2 1 22 dWhen 2, 0 d 1 (2) 2(2) 02 1 4 0 1, 4 (rej) y x x k y x k x x kx x k x k x x k x k yx x kk kk k                                               Product Rule Substitution and equating to 0 Solving and leading to shown + rejection

10b   2 d1 1 2 1d2 dWhen 0,d 12, 2 y xxx y x x      2 2 2 d d y x >0 0 >0 By 1st derivative test, x = 2 is a point of inflexion. 1 2  1 2 1 2  d d y x <0 0 >0 By 1st derivative test, the curve has a minimum point at 1 2x . x = ½ 1st derivative test 1st derivative test