Dunman 2026 Amaths P1 MS

DUNMAN SECONDARY SCHOOL CANDIDATE NAME CLASS INDEX NUMBER PRELIMINARY EXAMINATION 2026 SECONDARY 4 EXPRESS/ 5 NORMAL ACADEMIC ADDITIONAL MATHEMATICS 4049/01 Paper 1 2 hours 15 minutes Marking Scheme

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Question Answer Marks Partial Marks Guidance 1 ( ) ( ) ( ) 2 22 2 3 8 212 1 2 x x A B C xxx x x −− = + ++−+ − + ( )( ) ( ) ( ) 222 3 8 2 1 1 2x x A x x B x C x− − = + − + − + + Let x =1, 9C = −9 C = −1 Let x = −2, 6 = −3B B = −2 Let x = 0, −6 = −2A A = 3 ( ) ( ) ( ) 2 22 2 3 8 3 2 1 212 1 2 xx xxx x x −− = − −+−+ − + 5 M1 M1 B1, B1 A1 Correct partial fractions, accept ( ) 2 2 Ax B x + + Comparing numerator For any two correct A, B or C All correct with partial fractions given

Page 4 of 18 4049/01/PRELIM/4E/2026/MARKING SCHEME 2 ( )3 9 9log 2log 3 log 4xx− = + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 2 3 9 9 39 3 3 3 3 3 33 2 33 2 3 2 2 2 log log 3 log 4 log 1 log 4 log 4log 1 log 9 log 4log 1 2 2log 2 log 4 log log 4 2 log 2 4 94 9 36 9 36 0 12 3 0 12 or 3 (rej.) xx xx xx xx xx xx x x x x xx xx xx xx − = + − = + +−= +−= − = + − + = =+ =+ =+ − − = − + = = =− 5 M1 M1 M1 M1 A1 Realises 92log 3 1= or combine with log term Change of base formula into a common base Getting rid of log Solving quadratic equation

Page 5 of 18 4049/01/PRELIM/4E/2026/MARKING SCHEME 3 ( ) ( ) 2 2 2 22 2 2 2 1 sin 2 tan 1cos 1 2sin cosLHS = cos cos sin 2sin cos cos tan 2 tan 1 tan 1 RHS x xx xx x x x x x x xx x − =− − +−= = − + =− = Alternative: ( ) 2 2 2 2 2 1 2sin cosLHS = cos 1 2 tancos sec 2 tan tan 2 tan 1 tan 1 RHS xx x xx xx xx x − =− =− = − + =− = 4 M1 M1 M1 AG1 M1 M1 M1 AG1 Double angle sin 2x Identity 22cos sin 1xx+= Divide to get in terms of tan x Double angle sin 2x Get sec2 x Identity 22sec 1 tanxx=+

Page 6 of 18 4049/01/PRELIM/4E/2026/MARKING SCHEME Question Answer Marks Partial Marks Guidance 4 ( ) ( ) ( ) 1 1 60 1.1 150 1.1 2.5 1 lg1.1 lg 2.5 lg 2.51 lg1.1 10.61 n n n n n − − = = −= −= = Therefore, Week 11. 4 M1 M1 M1 A1 Of the form ( ) 1 60 1.1 n− or ( )60 1.1 n Taking log or changing to log form 1.1log 2.5 Making n the subject and obtaining the value (either n = 9.61 or 10.61 is fine at this stage)

Page 7 of 18 4049/01/PRELIM/4E/2026/MARKING SCHEME Question Answer Marks Partial Marks Guidance 5 ( ) ( ) ( )( ) 22 22 2 2 2 2 3 subst into 19, 3 3 19 3 6 9 19 3 10 0 5 2 0 5 or 2 yx x xy y x x x x x x x x x xx xx xx =− − + = − − + − = − + + − + = − − = − + = = =− 4 M1 M1 M1 A1 Subst. to get equation in one variable Simplify into general form Solving quadratic equation

Page 8 of 18 4049/01/PRELIM/4E/2026/MARKING SCHEME Question Answer Marks Partial Marks Guidance 6(a) Let x be the side of the square. ( ) 222 2 2 2 28 2 4 8 4 8 6 2 8 6 4 2 xx x x x + = + = + + =+ =+ 3 M1 M1 A1 Pythagoras’ theorem expansion 6(b) ( ) 21 3 110 6 2 6 4 23 30 18 2 6 4 2 6 4 2 6 4 2 180 120 2 108 2 144 4 36 12 2 4 9 3 2 V x h h h = + = + +−= +− −+−= −= =− 4 M1 M1 M1 A1 Volume formula Rationalise denominator Expansion

Page 9 of 18 4049/01/PRELIM/4E/2026/MARKING SCHEME Question Answer Marks Partial Marks Guidance 7(a) ( ) 3 3 3d e 3 e ed x x xxxx =+ 2 B1, B1 Differentiating 3e x correctly, All correct 7(b) 3 3 3 3 3 3 3 33 1 33 2 3 e + e d e 3 e d e e d e3 e d e 3 eee d 39 x x x x x x x xx xx nx x x x c x x x x c x x x c c xx x c =+ = − + = − + + = − +     4 M1 M1 M1 A1 Using reverse of differention Making 33 e dxxx the subject Integrate 3e x

Page 10 of 18 4049/01/PRELIM/4E/2026/MARKING SCHEME Question Answer Marks Partial Marks Guidance 8(a) 2 2 2 2 2 2 3 4 432 4 432 432 4 432 4 432 4 x xh xh x xh x V x h xx x xx += =− −= =  −=  −= 3 B1 M1 AG1 Surface area h in terms of x 8(b) 2 2 2 2 d 432 3 d4 432 3 04 3 432 144 12 (rej. -ve) Vx x x x x x −= − = = = = 2 2 d3 18 0d2 Vx x =− =−  Therefore, stationary value is maximum. 4 B1 B1 M1 A1 Correct differentiation x = 12 and derivative = 0 valid test (or 1st derivative test) Conclusion