NBSS Prelims 2021 - 4E Add Maths Paper 1 (Solution)

Secondary 4E Prelims 2021 Add Math Paper 1 (Solution) Q. Solution Remarks 1(i) (ii)     22 228 (12 11) 8 0 (11 2)p       M1   2 8 1 145p     2 8 144p M1 8 12p or –12 20p or –4 (reject) A1 (must reject –4) OR 2 16 64 1 145 0pp     2 16 80 0pp   M1 ( 20)( 4) 0pp   20p or –4 (reject) A1 Gradient of EF = 12 2 1 20 0 2   (allow ecf) Gradient of perp. bisector = –2 M1 2y x c  Subst (8, 11) into equation, 11 2(8) c  27c Equation of bisector is 2 27yx  A1 OR Midpoint =  0 20 2 12, 10,722   Gradient of perp. bisector = 11 7 28 10   M1 (ecf) 2y x c  Subst (10, 7) [or (8, 11)] into equation, 7 2(10) c  27c Equation of bisector is 2 27yx  A1

11 sin 2yx 2(i) (ii) For each graph, B2 (–1 mark for incorrect interval (x-axis) –1 mark for incorrect min/max (y-axis) –1 mark for incorrect graph shape ) 13cos2 sin 1 2xx for 2  x  2 13cos 2 1 sin 2xx Number of solution is 2×4 = 8 B1 3(i)       4 3 2 42 441 (1) 1 ( ) 1 ( ) ...12ax ax ax                 M1 221 4 6 ...ax a x    Comparing coeff. of x, 4 12a  M1 3a (shown) A1 OR     44 1 44 1 ( ) 1 ( ) rr r r r rT ax a xrr                Let rx = x 1r M1   41 14 1 ( ) 121 a    M1 4 12a  3a (shown) A1 3cos2yx

(ii)   4 221 3 1 4(3) 6(3) ...x x x     21 12 54 ...xx    M1 4(2 1)(1 )x ax 2(2 1)(1 12 54 ...)x x x     Coefficient of 2x = (2)(–12) + (1)(54) M1 = 30 A1 4(i) (ii) 2 2 3f ( ) 5 xx x   22 22 ( 5)(2 ) ( 3)(2 )f '( ) ( 5) x x x xx x     M1 22 22 (2 )( 5 3) ( 5) x x x x     22 16 ( 5) x x  M1 Given that 0x 16 0x M1 22 16 0( 5) x x  Since, f '( ) 0x  , f is an increasing function. A1 22 d 16 d ( 5) yx xx  At x = 1, 22 d 16(1) 4 d (1 5) 9 y x   d d d d d d x x y t y t d9 0.2 0.45d4 x t    units/s M1, A1 5(i) P = 3.75ekt 5.79 = 3.75 ek(20) e20k = 5.79 3.75 M1 k = 5.79ln 3.75   ÷ 20 = 0.0217188 M1

(ii) When t = 30 P = 3.75 e0.0217188 (30) M1 = 7.19 population = 7.19 million A1 10 = 3.75 e0.0217188 t e0.0217188 t = 10 3.75 t = 10ln 3.75   ÷ 0.0217188 M1 = 45.16 years the year = 2045 (Note: no need to round up) A1 6(i) (ii)   32 2 2 18 d d   xx y   3d 18 2 dd y xxx     2 18 2 2(1) x c      2 92xc     M1 Given d 2d y x  and 5x ,   2 2 9 5 2