2026 Zhonghua AM Prelim P2 solution

2026 4E5N Amath Paper 2 Mark Scheme (4049/02) Qn Mark Scheme 1 ( ) ( ) 5 103 108 75 5 108 5 75 108 5 75 75 108 5 3 108 5 108 5 108 5 5 3 108 5 103 90 5 15 x x xx x x x x x −= −= −= = − += −+ + = += 90, 5pq== 2a 32 3 3 3 3 2 2 2 0 f 2 17 421 30                     = − + − = By factor theorem, (2 3)x− is a factor of f ( )x . 2b 32 2 2 17 41 30 (2 3)( 10) x x x x x kx − + − = − + + Comparing coefficients of 2x , 17 3 2 2 14 7 k k k − =− + =− =− 32 2 f ( ) 2 17 41 30 (2 3)( 7 10) (2 3)( 2)( 5) x x x x x x x x x x = − + − = − − + = − − − 2c 3 or 2 or 5 f ( ) (2 3)( 2)( 5) 0 2 x x x x x xx = − − − = ===

2d g( ) ( 3)(3 1) 2x a x x= − − + Using 1a= (or any other non-zero value of a ) 2g( ) 3 10 5x x x= − + 2e 6 3 2 4 2 227 ( 3 )( 3 ) 9x y y yx x x y++ = + − 3 2 2 1 d 6sin 2 4cos 2 2 d d 4si6d s nco y xxx y x x xC− =− = − + At 20, ddx yx= = 1 1 5 2 4sin6cos02 0 C C−−= = + 2 d 4sin 5d 5 3co 3 sin 2 4cos2 s2y xx y x x x C x= − + = + + + − − Sub (0,5) 2 2 1 35 sin 2(0) 4cos(0) 5(0)2 C C =− + + = + Equation: 3 sin 2 4cos 5 12y x x x=− + + + 4ai 1sin 9090 x−  − 4aii 1tan 9090 x− −

4b 2sin 60 tan 45 cos 45cos30 3 12 3 2 2 1 3 2   +  =+ = 5a Total cost of reinforced fence 4 3 12x x= Total cost of long fence = L 12 120 120 12 xL Lx += =− 2 (120 12 ) 120 12 (shown) A Lx xx xx = =− =− 5b 2d 0d 1 24Ax x=− For stationary value d 0dAx = 120 24 0 24 120 5 x x x −= = = Showing that it is maximum, 2 2 0d 2d 4Ax =−  Hence, it is a maximum value. Maximum A 2 2 120(5) 12(5) 300 cm A=− = Overall units -1

6 2 3 29 3 22 9 20 0 v Ct tt st t = − + = − ++ At 0, 0ts== 0C = 3 2 2 2093s t t t−=+ Finding turning points, 2 0 9 20 0 ( 4)( 5) 0 4 or 5 v tt tt tt = − + = − − = == 6 3 (5 2 ( ( 0 ) )0 4 037) 88() 175 6 s s s s = = = = Total distance travelled 88 1 14 205 m or 34.2 m3 6 3 6+ + == Overall units -1 7a Let 3xu= 2 4 3 0 ( 1)( 3) 0 1 or 3 3 1 0 3 3 1 x x uu uu uu x x − + = − − = == =  = =  =

7b 2 2 2 2 2 2 2 2 2 22 log ( ) log ( 2) log ( ) 4 ( 2)log 4 ( 2) 16 2 ( 2) 16 2 ( 2) 16 ( 2) 8 2 8 0 ( 4)( 2) 0 4 (rej) or 2 2(2) 8 y yxx y x x yx x yx x xx x xx xx xx xx xx y =  = + + − = + = + = + = += += + − = + − = =− = == 8a   22 4 22 4 22 3 3 ( 2) (2 4) (d d ( 2) 2( 2) ( 2) ( 2) 2 ( 2) (2 4 3) 2( 2) 43 4 4 3 2 ) 4 xx x xx x x xy xx x x x x x x x x − − −= − − − − − + − +− = − + − =  −  + =  − − − 8b For increasing functions, d 0dyx 

3 2 0 0 ( 2 2) 2 x x x  −  − 8c 2 4 2 4 d 3( 2)d 2 6 ( 2)x y xx −−− =  −= − Since 4 4 06( 2) 0, ( 2)x x− − − . Hence, the gradient is a decreasing function for 2x . 9a (alt. segment theorem)XZY XYM = (alt. s / / )ZXY XYM ZX NM =  Hence, XZY ZXY = . Triangle XYZ is an isosceles triangle. 9b (common angle)XYV WYX = (alt. segment theorem) (alt s, / / ) XWY XYM VXY XYM ZX NM  =  =  XWY VXY = By AA similarity test, triangle YXW is similar to triangle YVX. 9c 223 4 5WY = += 5 4 5 12 3 XY VX WY V VX WX X = = = 242 5 4 ZX VX ZY XY =  = == 6 : 4: 245 5: ZY ZX = =

10a 2 2 2 2 2 3) 5 0 ( 2 1 1 3) 15 5 10 5( 1) 25( h t t t t t t t −− =− − = = − + =− +− − + − − 10b Maximum height = 20 Time = 1 10c 2 2 2 5 1 1 20 52 5 5 0 205 h k k k =− + − + =− + =− + 2 2 5 50 0 ( 10)( 5) 0 10 (rej) or 5 20 105 k kk k kk k k− − += + − = − + += == 11a 2 2 32 2 5 32 25 20 4 4 52 25 0 (2 1)(2 25) 0 1 25 or (rej)22 125 2 4 (0.5, 4) xx x x x xx xx xx y A =− + = − + − + = − − = == =− + =  Sub 1x= into 32y x= 32 4 2y= = (1,4 2)B

11b ( ) 2 1 0 3 . 1 2 0. 5 5 13 22 0.5 3 2 2 32 2 5 d 32 3 2 82 3 8 2 2 1 1153 2 4 8 2 2 1 9434 22 32 2 32 5 5 5 8 (0.5)3 8 3 37 12 12 1 2 37 xx x x x x x xx  =−   = + −   = − + − +− + +−  −    −+  = − = − =  11c Total area ( ) ( ) ) 4 0.5 132 2 3724 2.969 (3 d 2 .p = − = −− 12a 226 8 10R +== 1 08 0ta 73n 926 . −  = =  24 15 6sin 8cos12 12 24 15 10sin 0.9273012 9sin 0.9273012 10 tt t t       = + +        = + +   += Let reference angle be  1 9sin 10 1.1198   − = = 

0.92730 1.119812 0.735296 44.1min t t  += == It starts to flower at 00 44 (or 0045). 12b Maximum temp = 25 degree Celsius. sin 0.92730 112 0.9273012 2.45797 2 i 2 hr 27 m ns t t t    += += == Time = 02 27 or 02 28 12c 10 15 10sin 0.9273012 sin 0.92730 0.512 6 t t    = + +   + =− = 0.9273012t + lies in the third and fourth quadrant. 70.92730 ,12 10.5, 18.5 11 66 t t   += =