2008 O Level A Math 4038 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2023 Mathematics 4038/01 Syllabus 4038 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (i) Angle ABX 180 120 60= − =  (adj. angles on a st. line) 3sin 60 4 42 AX AX =  =  M1 3 2 s.o.i. 23AX = cm A1 (ii) By Pythagoras’ Theorem, ( ) 224 2 3 2 cmBX = − = M1 23tan 4ACB= M1 1 3tan 2ACB − =  (Shown) A1 AG [5] 2 ( ) ( ) 3 20 2 9 27 1 3 3 3 yx y x =  =  ( ) ( ) 11 34 22 8 2 16 2 2 2 2 2 xy x y =    =        M1 Attempt to change to a common base in either equation 32 0233 320 2 3 (1)4 xy xy xy + = += =− −− 193 2222 193 22 6 9 (2) yx yx yx − = −= − = −− M1 Either equation obtained through equating powers s.o.i. Substitute (1) into (1) 369 4yy − − = M1 Substitution or elimination s.o.i. 27 94 4 3 y y = = A1 o.e. 1x=− A1 [5] 3 NOT within the 4049 syllabus E5 [5]

General Certificate of Education Ordinary Level 2023 Mathematics 4038/01 Syllabus 4038 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 4 (i) ( ) 3 3 2d1 ln 3 lnd x x x x xxx =+  M1 Apply the product rule correctly 22 3 lnx x x=+ A1 (ii) 2 2 3 13 ln d lnx x x x x x c+ = + M1 Use result in (i) 2 3 2 1 3 3 12 3 ln d ln d ln 3 x x x x x c x x xx x c c = + − = + − +  M1 Correct integration 33 2 ln d ln 39 xxx x x x d = − + , where 12d c c=+ A1 Final answer with arbitrary constant [5] 5 (i) Let 8 46 ( 5)( 1) 5 1 x A B x x x x − =+− + − + M1 8 46 ( 1) ( 5)x A x B x− = + + − When 1x=− , 54 6 9 B B − =− = When 5x= , 66 1 A A −= =− M1 Find A or B correctly 8 46 9 1 ( 5)( 1) 1 5 x x x x x − =−− + + − A1 (ii) 22 d 9 1 d ( 1) ( 5) y x x x=− + +− M2 M1 for each term At 2x= , d8 d9 y x =− . A1 [6]

General Certificate of Education Ordinary Level 2023 Mathematics 4038/01 Syllabus 4038 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 6 (i) 12 seconds B1 c.a.o. (ii) Distance 12 2 0 16d 2t t t=− M1 Integral seen 12 23 0 13 6tt=− M1 Correct integration 23 13(12) (12) 1446= − = m A1 (iii) d 6d vat t= = − M1 22 m /s− A1 c.a.o. [6] 7 2 2 d (2 cos )(cos ) sin d (2 cos ) y x x x xx −−= − M2 M1 Attempt to use quotient rule M1 for any correct differentiation 22 2 2cos cos sin (2 cos ) x x x x −−= − M1 Simplify 22 2 d 2cos cos sin00d (2 cos ) y x x x xx −−=  = − M1 s.o.i. ( ) 222cos cos sin 0 2cos 1 0 x x x x − + = −= M1 Convert using 22sin cos 1xx+= 1 πcos 26xx=  = A1 [6]

General Certificate of Education Ordinary Level 2023 Mathematics 4038/01 Syllab