2026 Zhonghua AM Prelim P1 Solutions

1 Additional Mathematics Syllabus 4049 2026 Preliminary Exam Paper 4049/01 Level: Sec 4 Express/ 5 Normal Academic Qn Working 1(a) 1200lg 0.001M =  = 6.08 (to 3 s.f.) 1(b) 4.6 lg 0.001A=  4.6100.001A = 4.6 310 10A −= = 39.8 mm (to 3 s.f.) 1(c) 3log 9AM =  3 1200log 9M =  1200lg 9 lg 3M = 4.45M = (to 3 s.f.) 2 23 5 4y x x=− + − ………(1) 2yx=− ………(2) (1) = (2) 23 5 4 2x x x− + − = − 23 4 2 0xx− + − = Discriminant = 24 4( 3)( 2)− − − = – 8 Since discriminant < 0 and the coefficient of 2x is – 3 , which is negative ( or < 0), the curve lies below the line 2yx=− 3(a) Let 2e (3 1)xyx=− 22d 2e (3 1) 3ed xxy xx = − + 2 2 2d 6 e 2e 3ed x x xy xx = − + 22d 6 e ed xxy xx =+ 3(b) ( ) 2 2 26 e e d e (3 1)x x xx x x+ = −

2 ( ) 2 2 2 2 424 e d 6 e e d e d63x x x xx x x x x= + −   ( ) 2 22 2 2 e4 e d e 3 1 3 3 2 xxxx x x c = − − +   2 2 2 2 214 e d 2 e e e 33x x x xx x x c= − − + 2 2 24 e d 2 e ex x xx x x c= − + 4(a) LHS = 1 cos 2 sin 2 sin 2 1 cos 2 AA AA − + + = 2 2 1 (1 2sin ) 2sin cos 2sin cos 1 (2cos 1) A A A A A A −− + +− = 2sin sin 2sin cos 2sin cos 2cos cos A A A A A A A A+ = tan tanAA+ = 2 tan A 4(b) 1 cos 2 sin 2 2 cotsin 2 1 cos 2 7AA AAA− +=+ 212tan 7 tanA A=  2 1tan 7A= 1tan 7A= Basic angle, 1 1tan 7 −= 20.70 = (to 2 d. p.) Therefore A = – 20.7o (to 1 d.p.), 20.7o (to 1 d. p.) 5(a) xhyk += lg lg xhyk += lg ( )lgy x h k=+ lg lg lgy x k h k=+ Gradient of straight line = 23 3 10 0 − − = 2 lg 2k = 210k = 100k = lg 3hk = 3 2h=

3 5(b) 3lg 2 (2) 2yx=+ 4 2 3xx=+ 23x= 3 2x= 6(a) ( ) 2 2 22x yk+ − = ……..(1) 2yx= ………..(2) Substitute (2) into (1) ( ) 2 2 222x xk+ − = ( ) 2 2 22 4 4 4 0x x xk k+ − + − = 2 2 28 8 2 4 0x x xk k+ − + − = 229 8 2 4 0x xk k− + − = For the curve to meet the line at 1 point or intersect at two distinct points, Discriminant 0 ( ) 2 28 4(9)(2 4) 0kk − −  2264 72 144 0kk− +  28 144 0k− +  28 144k− − 2 18k  18 18k−   or 3 2 3 2k−   O.E. 6(b) 2 18k = 18k= 32k =

4 7(a) 5 2 4ax − = 0 1 2 3 5 4 3 25 5 5 52 2 2 2 ...0 1 2 3 4 4 4 4ax ax ax ax              − + − + − + − +                             = 2 2 3 3 532 20 5 ... 8ax a x a x− + − + B4 7(b) 2 2 3 3 5(5 2 )(32 20 5 ... 8x ax a x a x− − + − + = 64 100 ...x ax− − + No term in x, 64 100 0a− − = 100 64a−= 64 100a=− 16 25a=− M1 M1 A1 FT (7a) but must state “=0” 7 8(a) 2 2 2 (2 ) 7PQ x=+ 22 4 49PQ x=+ 24 49PQ x=+ B1 8(b) Let distance PQ be D ( ) 1 2 24 49Dx=+ ( ) ( ) 1 2 2d1 4 49 8d2D xxx −=+ = 2 4 4 49 x x + Rate of change, d d d d d d D D x t x t= 2 d4 5d 4 49Dxt x= + = 2 20 4 49 x x + At x = 12, 2 d 20(12) d 4(12) 49 D t = + = 9.6 m/s M1 M1 M1 A1 Differentiate composite ftn State correct chain rule for connected rate of change Substitute x = 12 8(c) Area of triangle PQR, A = ( ) 21 4 49 sin 602 x + = ( ) 23 4 494 x + O.E. d 23dA xx = Rate of change of area, M1 M1 Convert 3sin 60 2=

5 d d d d d d A A x t x t= = 2 3 5x = 10 3 x 10 3 250x 250 10 3 x x is increasing at a constant rate of 5 m/s, Therefore the shortest time = 250 5 10 3  = 2.89 s (to 3 s.f.) M1 M1 A1 10 9(a) 1 252y x x −= + − 3 2d5 1d2y xx −=− + For stationary point, d 0dyx = 3 5 10 2 x − + = 3 5 1 2 x = 325 x = 3 2 5 2x = 2 35 2x = 3 25 4x= M1 M1 A1

6 9(b) ( ) 3 2 d5 1d2y xx −=− + ( ) 2 5 2 2 d y 15 d4 xx − = 2 2 5 d y 15 d 4x x= For x > 0, 5 0x  , 2 2 dy 0dx  Therefore A is a minimum point. Alternative solution Using first derivative test x 3 25 4 −   3 25 4 3 25 4 +   d d y x < 0 or negative 0 > 0 or positive Sketch of tangent Outlne of tangent Therefore A is a minimum point. M1 M1 A1 Accept substituting 3 25 4x= to find 2 2 d 2.76d yx = and conclude 2 2 dy 0dx  M1 state the three x values M1 state the value of d d y x to be negative (< 0) , 0 , positive (> 0) for the corresponding x values A1 6 10(a) d 4cos3 ( sin 3 )(3) 4 cos 4dy x x b bxx = − − = 12cos3 sin 3 4 cos 4x x b bx−− When x = 0, d 4dy bx =− Gradient of normal = 1 4b 4 xyc b=+ Substitute (0, 4) 44xy b=+ M1 B1 A1 M1 M1 A1 Chain Rule: 4cos3 ( sin3 )(3)xx− 4 cos 4b bx Substitution “ x = 0” Product of 12 1mm =−

7 10(b) 86yx=+ 13 84yx=+ Gradient of the line = 1 8 11 48b = b = 2 M1 A1 8 11(a) 2 22 4 5 21 (2 3) 2 3 x x A B C x x x x x +− = + +−− 224 5 21 (2 3) (2 3)x x Ax x B x Cx+ − = − + − + 2 2 24 5 21 2 3 2 3x x Ax Ax Bx B Cx+ − = − + − + When x = 0, 21 3 B− =− B = 7 When x = 1, 12 2 3 2 3A A B B C− = − + − + 12 ABC− =− − + Sub B = 7, 5AC=+ …..(1) When x = 2, 25 2 (2) 3 (2) 4 3 4A A B B C= − + − + 5 8 6 4A A B C= − + + 2 2 4AC− = + ……(2) Substitute (1) into (2) 2 2( 5) 4CC− = + + 2 6 10C− = + 12 6 C−= 2C =− Substitute C = – 2 , A = 3 2 22 4 5 21 3 7 2 (2 3) 2 3 xx x x x x x +− = + −−− M1 M1 A3 Remove denominator Substitute any one factor of x 11(b) 244 2222 4 5 21 3 7 2 d d(2 3) 2 3xx xxx x x x x+− = + −−−  = 4 2 73ln ln(2 3)xx x − − − = 773ln 4 ln 5 3ln 2 ln142   − − − − −       = 33 7ln 4 ln 2 ln 5 4− − + M3 FT from 11(a)

8 = 87ln 54+ A2 10 12ai When y = 0, 2kx = or 3kx = 2x k= or 3x k= Therefore the coordinates of A are ( 2 k  , 0) and the coordinates of B are ( 3 k  , 0) . B1 B1 12aii Coordinates of C = 23 , 2kk a  +    = 5 , 2 ak tan 6 2 a k  = 1 23 a k=  12 3 ka = 3 2 3 3 a k =  3 6a k= (Shown) Award 3 marks regardless of the working

9 12b 3 6a k= 3 23 36k = 18k = Period = 2 k  = 2 18    = 36 M1 M1 A1 8

10 13a Gradient of AP = 2 Midpoint of AP = (0, 6) Equation of perpendicular bisector of AP: 16 ( 0)2yx− =− − 1 62yx=− + M1 A1 Use of 12 1mm =− , gradient of AP = 1 2− 13b 1 62yx=− + ……(1) Gradient of line AB = 3 4 Equation of AB : 32 ( 2)4yx− = + 332 42yx− = + 37 42yx=+ …… (2) (1) = (2) 1 3 762 4 2xx− + = + 2x= Substitute x = 2 into (1), y = 5 Centre of circle = (2, 5) Radius of circle = 22(2 ( 2)) (5 2)− − + − = 5 Equation of circle is 22( 2) ( 5) 25xy− + − = M1 M1 A1 M1 A1 Using gradient = 3 4 or the point A to find equation of line AB Equate both equations of line AP and AB Find centre of circle Apply distance formula 13c Coordinates of B = (6, 8) Substitute (6, 8) into 34y x k+= 3(8) 4(6)k =+ 48k = M1 A1 9