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 ang