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 Syllabus 4038 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 8 (i) ( ) sin 3 sin sin 2 sin sin 2 cos cos 2 sin sin xx x x x x x x x x + = + + = + + M1 Compound angle formula s.o.i. ( ) 222sin cos sin 2cos 1 sinx x x x x= + − + M1 Double angle formula s.o.i. 22 2 2sin cos 2sin cos sin sin 4sin cos (Shown) x x x x x x xx = + − + = A1 AG (ii) ( ) 22 22 2 4sin cos 2cos 4sin cos 2cos 0 cos 4sin 2 0 x x x x x x xx = −= −= M1 Factorise o.e. 2cos 0 cos 0 π 2 x x x = = = 4sin 2 0 1sin 2 π 5π,66 x x x −= = = A2 A1 for any value A1 for all values [6] 9 Let Ann and Betty’s age be x and y years respectively. 22 2 6( ) (1)x y x y− = − −−− M1 5( ) (2)x y x y+ = − −−− M1 From (2) 55 36 3 (3)2 x y x y xy xy + = − = = −−− Substitute (3) into (1) 2 233 2622y y y y   − = −       M1 Substitution s.o.i. 2 2 29 1 1 2 6 3 04 2 4 1 304 0 (reject) 12 y y y y y yy y or y − =  − =  −= == M1 Must reject When 12y= , 18x= . Ann’s age is 18 years and Betty’s age is 12 years. A2 Conclude [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) 5 10 (i) Discriminant 0 25 4 (2) 0a− M1 s.o.i. 25 8 0 258 25 8 a aa −    M1 o.e. Smallest integer 4a= A1 (ii) Discriminant 0 2 4( 5)( 2) 0b − − −  M1 s.o.i. ( )( ) 2 40 0 40 40 0 2 10 2 10 b bb b − + −  −   M1 o.e. Smallest integer 6b=− A1 [6] 11 (i) ( ) ( ) 71 1 7 7 2 7 77 rr r r r r r r T x kx r k x x k xrr −− + − − − =    ==       M1 Attempt to simplify the general term o.e. For 3x , 7 2 3 2rr− =  = . For x , 7 2 1 3rr− =  = . M1 Find values of r 2377 23kk   =       M1 Form equation ( ) 2321 35 35 21 0, then 0 3 5 kk k k k k = =   =  A1 o.e. (ii) For 5x , 7 2 5 1rr− =  = . For 7x , 7 2 7 0rr− =  = . M1 Find values of r Coefficient of 5 21 5x = Coefficient of 7 1x = M1 Either one s.o.i. Coefficient of 7 211 ( 5) 20 5x = + − =− A1 [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) 6 12 (i) ( )lg lg lg lg lg lg x x x y kb y kb kb kxb = = =+ =+ M1 Attempt to take lg on both sides to rewrite equation Gradient 1.3 0.8 1 0 11 22 −= =−− M1 s.o.i. 1 22 1lg 22 10 b b − =− = M1 Convert back to index form 0.9006280202 0.90 (2 s.f.)b== A1 1.3 lg 1.3 10 k k = = M1 Convert back to index form 19.95262315 20 (2 s.f.)k == A1 (ii) 81 1.3 2210 10y −=   M1 FT from (i) 8.637014255 8.64 (3 s.f.)y== A1 [8]

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) 7 13 (i) 2(5 ) 2 6 360 16 2 360 180 8 x h x xh hx + + = += =− M1 Expression for h in terms of x s.o.i. Height of triangle QRS 22(5 ) (3 )xx=− M1 Use Pythagoras’ Theorem s.o.i. 22 2 25 9 16 4 xx xx =− == M1 2 2 1(6 ) (6 )(4 )2 6 (180 8 ) 12 1080 36 (Shown) A h x x x x x x xx =+ = − + =− A1 AG (ii) d 1080 72d A xx =− M1 d 0 1080 72 0d A xx =  − = M1 s.o.i. 72 1080 15 x x = = M1 When 15x= , 21080(15) 36(15) 8100A= − = A1 (iii) 2 2 d 72 0d A x =−  Hence, this value of A is a maximum value. B1 [9] END OF MARKING SCHEME

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) 8 BLANK PAGE