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 c      3c M1   2d 9 2 3d y xx       2 [ 9 2 +3] dy x x       1 92 31(1) x xc       1 9 2 3x x c      M1 Given 5x and 20y ,   1 20 9 5 2 3(5) c       2c Equation of the curve is   1 9 2 3 2y x x      A1   2d 9 5 2 3 2d y x      Gradient of normal = 1 2 M1

1 2y x c  120 (5) 2 c  122 2c Equation of the normal is 11 2222yx  A1 7(i) (ii) 23 6 5xx = 2 532 3xx  or   23 2 5xx =   2 2 53 1 (1) 3x    or   2 23 1 (1) 5x    M1 =   2 3 1 2x A1 2 43xx   =   2 43xx   or   2 43xx   =   2 22 (2) 3x    or   2 22 (2) 3x    M1 =   2 21x   A1 23 6 5xx =   2 3 1 2x Graph is curving upwards, with the turning point at (1, 2) M1 2 43xx   =   2 21x   Graph is curving downwards, with the turning point at (–2, 1) M1 Hence, as seen from the diagram, they will not intersect. A1 OR The y-coordinate of the minimum value of 23 6 5xx is larger than the maximum value of 2 43xx   .

8(i) (ii) sin 75 sin(30 45 )   sin 30 cos 45 cos30 sin 45     M1 1 2 3 2 2 2 2 2    26 4  (shown) A1 1cosec75 sin 75  4 26   M1 (for 1 sin 75 )     264 26 26     M1 4 2 4 6 26   26  A1   22cosec 75 2 6   2 12 12 6    M1 8 2 4 3   8 4 3 A1 OR 1cosec75 sin 75  4 26   M1 2 2 4cosec 75 26    16 2 12 12 6   M1 16 8 2 4 3       8 4 316 8 4 3 8 4 3     M1, M1  16 8 4 3 64 16(3)    8 4 3 A1

9(i) (ii) (iii) 2(3 ) 3 36xy M1 3 36 6yx 12 2yx A1 1 ( )( )sin 602V x x y  M1   213 12 222xx    233 62 xx A1 23 333 2V x x 2d 3 3 63d2 V xxx  M1 Let d 0d V x  33 6 0 2xx  M1 0x (reject) or 4x     23 33 3 4 4 16 32V    or 27.7 cm3 A1 2 2 d 6 3 3 3d V xx  When 4x 2 2 d 6 3 3 3(4) 6 3d V x    (or –10.4) Since 2 2 d 0d V x  , V = 27.7 cm3 is a maximum value. A1 10(i) 32f ( ) 2 10x x x ax b    Let 1x Remainder =       32 2 1 10 1 1 ab      3 12 ab   M1 15ab   --- (1)

(ii) Let 2x Remainder =       32 2 2 10 2 2 ab   3 24 2 ab    M1 2 21ab --- (2) Solving the simultaneous equations, 2a and 17b A1, A1 f ( ) 3 0x  322 10 (2) (17) 3 0x x x     322 10 2 14 0x x x    Let 1x Remainder =       32 2 1 10 1 2 1 14 0       Hence, ( 1)x is a factor M1 Using long division or comparing coefficient, 3 2 22 10 2 14 ( 1)(2 12 14)x x x x x x       M1 322 10 2 14 0x x x    2( 1)(2 12 14) 0x x x    1x or 22 12 14 0xx   A1   2 12 12 4(2)(14) 2(2)x    12 32 4x  12 4 2 324x    A1 11(i) (ii) Angle A is common ADB ACD  (Alternate segment theorem) M1 Hence triangle ADB is similar to triangle ACD (AA) A1 Since triangle ADB is similar to triangle ACD , AC AD AD AB 2AC AB AD M1

(iii) Given that AD = 2AB   2 2AC AB AB M1 24 4ABAC AB AB (shown) A1 DF = FC means F the midpoint of DC. DE is the diameter means O is the midpoint of DE. Hence, OF is parallel to EC (midpoint theorem) M1 DEC DOF  (corresponding angles) DEC CDG  (alternate segment theorem) M1 Hence, angle CDG = angle DOF (shown) A1 12(i) (ii) cossec 1 sin xLHS x x  1 cos cos 1 sin x xx          1 1 sin cos cos cos 1 sin 1 sin cos x x x x x x x   M1   2sin (1 cos ) cos 1 sin xx xx     2sin sin cos 1 sin xx xx   M1     sin 1 sin cos 1 sin xx xx   M1 tan x RHS (proved) A1 cos cotsec 1 sin 3 xxx x 1tan 3tanx x 2 1tan 3x 1tan 3x (Q1, Q2, Q3, Q4) M1 Acute angle of x = 1 1tan 63   A1

(iii) cossec 1 sin xyx x  --- (1) cot 3 xy --- (2) Subst (1) into (2), cos cotsec 1 sin 3 xxx x  1tan 3tanx x  2 1tan 3x  M1 1tan 3x  (no solution) Hence, there is no intersection between the 2 curves. A1 13(i) (ii) (iii) Let v = 0 22 5 2 0tt    M1 0)2)(12(  tt 2 1t or 2t A1 54  tdt dva M1 Let a < 0 (decelerating) 4 5 0t M1 4 5t seconds A1 Hence, 50 4t 2 (2 5 2)s v dt t t dt    = 3225 232 tt tc   M1 When t =0, s =0 , 322(0) 5(0)0 2(0)32 c     0c Hence,