FFMS 2024 A Math Prelim P1 Answer Key
Uploaded by cazbrecker · 29 July 2025
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FMS(S) SEC 4 EXPRESS PRELIMINARY EXAMINATION 2024 ADDITIONAL MATHEMATICS PAPER 1 ANSWERS 1(i) ( ) 2 4 1 9x+− (ii) (–1, –9) 2 a = 4 and b = 2 3(i) (alt segment theorem/tangent chord thm)CAE ABC = (alternate s, // )ACB CAE BC AE = (base s of isos )ABC ACB = ( )AB AC shown= 3(ii) 180 2 ( sum of )BAC ABC = − ( s in same segment)BDC BAC = 180 (adj s on a straight line)CDE BDC = − 180 (180 2 ) (adj s on a straight line)CDE ABC = − − 2 (shown)CDE ABC = 4(a)(i) period = 8 8= b 1 8b= 4(a) (ii) 4 tan 38 xy =+ 4(b) 5(i) Length of rectangle = 20 3 2 x− Area = 220 3 1 sin 6022 x xx− − 223 1 310 2 2 2x x x= − − = 26310 4xx +− (shown) 5(ii) ( ) 20 =2.5866 = 2.6 (2 s.f.) 63 x= + A = 13 cm2 (2 s.f.) 6(a) ( )( ) ( ) ( ) ( ) 22 8 13 4 2 1 1 2 21 2 2 2 x xxx x x + = − ++++ + + 7(a) 19, 12ab== 6(b) 5 1 25 12ln or ln or 0.5304 12 16 12++ (3 s.f.) 7(b) 0.270x= or 4.94x=− (3 s.f.) (45º,7) (135º,1) (225º,7) (315º,1)
8(a) 3 4 2 1 32 5 16 5+− = x x x x 3 4 2 1 4 32 2 5 5 2 5 − = x x x x 4 3 3 216 2 2 5 55 x x x x= 16 255 xx= 16 105 x= 16lg lg105 x= 16lg10 lg 5x = 16lg 5x= (shown) 8(b) 222log log ( 4) 3xx− − = 2 22log log ( 4) 3xx− − = 2 2log 3 4 x x =− 2 324 x x =− 2 8 32 0xx− + = Either 22 4 ( 8) 4(1)(32) 64b ac− = − − =− < 0 no real solutions (shown) OR 2( 8) ( 8) 4(1)(32) 2(1)x − − − −= 8 64 2 −= 64− does exist, no real solutions (shown) 9(a) 8k = 9(b)(i) 24 g 9(b)(ii) 34.7 or 35 days 9(b)(iii) mass is decreasing at 0.475 g/day 10(i) LHS 22tan (1 sec ) tan (1 sec ) ++= + xx xx 22tan 1 2sec sec tan (1 sec ) + + += + x x x xx 22sec 2sec tan (1 sec ) += + xx xx 2sec (sec 1) tan (1 sec ) += + xx xx 2 sin cos cos= x xx 2 sin= x (shown) 10(ii) 41.8 ,138.2 (1 d.p.) 11(i) Sub ( , 3)−kk into (9 )( 3)= − −y x x 3 (9 )( 3)− = − −k k k 23 9 27 3− = − − +k k k k 2 11 24 0− + =kk ( 3)( 8) 0− − =kk 3( . .) or 8==k N A k (shown) 11(ii) 2212 units3 12(i) (2,6) 12(ii) Since OT = RT = 5 units, ORT is isosceles. 12(iii) ( 8,1)− 12(iv) 25 units2
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