SMSS 2024 PRELIM AMATHV4 P1 Solutions

SMSS/PRELIM/2024 Secondary 4E5N Additional Mathematics 1 St Margaret School (Sec) Preliminary Examinations 2024 Sec 4E5N Additional Mathematics Paper 1 Question Answer 1 (a) ( ) ( ) 2 2 2 2 126 2 12[ 3 3 ] 2 2 3 19 xx x x − + − =− + − − =− + + Max y value is 19 Corresponding x value is -3. (b) 22 12 1 (1) 2 1 (2) y x x yx =− − + −−− =− + −−−−−−− Sub (1) into (2) ( ) 2 2 2 1 2 12 1 2 10 0 2 5 0 x x x xx xx − + =− − + += += 05x or x= =− 1 11y or y== ( ) ( ) 22 0 5 1 11 125 AB= + + − = 125k= 2 (a) Refer to last page (b) lg lg lgT x B A=+ lg 1.78 ( 0.01)A= ( )60.3 2A= 0.63 1.55 25 5 0.046 Gradient −= − =− lg 0.046 0.899 ( 0.01) B B =− = (c) 13min, lg 1.18 15.1 16 ( ) At x T T C C FreezingPt == =    Hence the chocolate is frozen. Alternatively At T = 16°C, lg 1.2041T = . x = 12.5 mins < 13mins Hence the chocolate is frozen.

SMSS/PRELIM/2024 Secondary 4E5N Additional Mathematics 2 3 (a) ( ) 2 22 31 2 1 2 1 x A B C x x x x x − = + +−− By Comparison ( ) 223 1 2 1 (2 1)x Ax x B x Cx− = − + − + Let x = 0, 1 ( 1)B− = − B = 1 Let x = 1 2 , 11 44 C−= C = −1 Let x = 1, 2 1 1A= + − A = 2 ( ) 2 22 3 1 2 1 1 2 1 2 1 x x x x x x − = + −−− 2 (c) 112ln ln(2 1) 2x x Cx− − − + 4 (a) 10 1 k k −  ( )( ) ( ) ( )( ) 2 2 2 2 2 40 4 4 1 2 0 16 4( 2) 0 4 4 24 0 4 6 0 3 2 0 32 b ac kk kk kk kk kk k or k − − − +  − + −  +− + −  + −  −  Ans 3k − (b) 2( 1) 4 2 (1) 2 3 (2) y k x x k yx = − + + + −−− =− + −−−−−−−−−− Sub (1) into (2) ( ) ( ) 2 2 2 2 2 ( 1) 4 2 2 3 ( 1) 6 1 0 40 6 4 1 ( 1) 0 19 13 42 k x x k x k x x k b ac kk k k k or − + + + =− + − + + − = −= − − − = −= − = =

SMSS/PRELIM/2024 Secondary 4E5N Additional Mathematics 3 5 (a) ( ) ( ) 8 1 8 8 2 8 23 8 32 r r r rrr Tx r x xr − + −−  =−  =− Since power of ( )8 2 2 4x r r= − = − . Power is a multiple of 2 for all integer values of r. Hence it is always even, no odd powers of x. 5 (b) In 8 23x x  − , For 4x power, 8 2 4 2 r r −= = For 2x power, 8 2 2 3 r r −= = ( ) ( ) 232 6 4 5 2 88( 1)(... 3 2 3 2 ...)23ax x x    − + − + − +       ( ) ( ) ( ) ( ) 6 2 5 388 3 2 3 2 023 a   − − + − =       81648 108864 0 3 4 a a − − = =− 6 2 2 sin 1 cos 2 sin 1 cos 2 1cos sin 2 2 dy xxdx dy x x dxdx x x x C = + + = + + =− + + +  At 2x = , 2 dy dx = 1 sin 2 cos2 2 2 2 2 C       = − + +       22 0 C C =+ = 2 1 sin 2 cos2 11cos 2 sin42 y x x x dx x x x D = − + =− − + +  At 2x = , 3 4y=−

SMSS/PRELIM/2024 Secondary 4E5N Additional Mathematics 4 6 2 2 3 1 1 cos 2 sin4 4 2 2 2 2 8 D D          − =− − + +            =− 2 211cos 2 sin4 2 8Eqn y x x x  =− − + − 7 (a) R is the midpoint of AB, S is the midpoint of AC By Midpoint Theorem, RS//BC and 2BC RS= Hence, SRC RCB = (Alternate angles, RS//BC) (b) RTB CTR = (Common angle) TRB TCR = (Alternate Segment Theorem)  Triangle TBR and Triangle TRC are similar (AA Similarity) (c) From (b) Triangle TRB and Triangle TRC are similar. TR TB TC TR= (Corresponding sides of similar  s) ( ) 2 2 22 TR TB TC TR TB TB BC TR TB TB BC = =  + = +  From (a), 2BC RS= ( ) 22 22 2 2 TR TB TB RS TR TB TB RS =+ − =  8 (a) ( ) ( ) ( ) ( ) ( ) 2 2 ' 2 22 2 2 2 2 2 2 2 1f ( ) 33 (2 1)(3 3) 3( 1)f ( ) 33 6 6 3 3 3 3 3 33 36 33 3( 2 ) 91 2 31 xxx x x x x xx x x x x x x x xx x xx x xx x −+= − − − − − += − − − + − + −= − −= − −= − −= −

SMSS/PRELIM/2024 Secondary 4E5N Additional Mathematics 5 8 (b) ( ) 2 2 22 4 4 22 2 22 2'( ) 31 3(2 2)( 1) 6( 1)( 2 )''( ) 0 9( 1) Since 9( 1) 0, 3(2 2)( 1) 6( 1)( 2 ) 0 ( 1)[(6 6)( 1) 6 12 ] 0 ( 1)[(6 12 6 6 12 ] 0 6( 1) 0 1 xxfx x x x x x xfx x x x x x x x x x x x x x x x x x x x −= − − − − − −= − − − − − − −  − − − − +  − − + − +  −  9 (a)(i) 2 (a)(ii) 180or  (b) (c) 4 solution 10 (a)(i) 2ACm =− 26Eqn AC is y x=− + (a)(ii) 2 6 (1) 5 6 (2) yx xy =− + −−−−−−− + =− −−−−−−− Sub (1) into (2) 5( 2 6) 6 4 xx x + − + =− = 2y=− (4, 2)C − 2sinyx=− 2cos 2 1yx=−

SMSS/PRELIM/2024 Secondary 4E5N Additional Mathematics 6 10 (b) ( )6,0D − Area of ACD = 0 6 4 01 6 0 2 62 − − = ( )1 12 24 ( 36)2 + − − = 36 units² Area of 36 1.5ABC= =24 units Let ( ),6Bk 0 4 01 246 2 6 62 k =− ( )1 24 6 24 2 242 6 kk k + − + = = ( )6,6B Alternative Method Area of ACD = 0 6 4 01 6 0 2 62 − − = ( )1 12 24 ( 36)2 + − − = 36 units² Area of 36 1.5ACD= =24 units ( )1 8 242 6 AB AB = = ( )6,6B 11 (a) 12dva t kdt= = + At t = 1, a = 6− 6 12(1) k− = + 18k =− (Shown) (b)(i) For min velocity, 0a= 12 18 0 1.5 t t −= = At t = 1.5, 26(1.5) 18(1.5) 12v= − + = 1.5− m/s

SMSS/PRELIM/2024 Secondary 4E5N Additional Mathematics 7 (b)(ii) For instantaneous rest, v = 0 ( ) ( )( ) 2 2 6 18 12 0 6 3 2 0 2 1 0 12 tt tt tt t or t − + = − + = − − = == 2 32 6 18 12 2 9 12 s t t dt t t t C = − + = − + +  At t = 0, s = 0 0 = 0 + C C = 0 322 9 12s t t t = − + At t = 1, 322(1) 9(1) 12(1) 5s= − + = At t = 2, 322(2) 9(2) 12(2) 4s= − + = At t = 4, 322(4) 9(4) 12(4) 32s= − + = Total distance = 5 1 (32 4)+ + − =34 m 12 (a) ( ) ( ) ( ) 2 2 3 2 (3 1) 2 5 2 xxdy dx x x + − += + = + For x > -2, 50 ( ) 2 20x+ ( ) 2 5 0 2x  + 00dy dx  Hence, the curve does not have a stationary point. 12 (b) Gradient of normal = -5 ( ) ( ) ( ) 2 2 1 5 51 52 25 2 25 3 7( , 0) 3 3 1 232 dy dx x x x x or reject x y = = + =+ + = = −  +== + Coordinate of Q is (3, 2) Since 5y x c=− + passes through (3,2) 2 5(3) 17 c c =− + = Eqn AB is 5 17yx=− +

SMSS/PRELIM/2024 Secondary 4E5N Additional Mathematics 8 12 (b) At x = 0, y = 17 At y = 0, x = 3.4 Area of triangle AOB = ( )( )1 17 3.42 =28.9 units² (c) 3 1 3 1 36 5 xx x ++ + − 3 1 5 322 x xx + =−++ Given that 2x− 5 02 533 2 x x + − + 3c such that Line yc= does not intersect the curve.

SMSS/PRELIM/2024 Secondary 4E5N Additional Mathematics 9 lg lg lgT x B A=+ (5, 1.55) (25, 0.63) 1.78 1.18 B1 - Correct Plot & Axes B1 – Correct straight line