2006 O Level Add Math 4018 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 NOT within the 4049 syllabus (i) xA B1 (ii) ( )n ' 16B = B1 (iii) CD = B1 [3] 2 (i) 2a= B1 (ii) 360 360120 3 120bb =  = =  B1 (iii) 1c=− B1 [3] 3 (i) ( ) 2 8 3 4yx − =− ( ) 3d ( 2) 8 3 4 3d y xx − = −  −  M1 ( ) 3 d 48 d 34 y x x =− − A1 When 2x= , gradient is 6− . A1 (ii) NOT within the 4049 syllabus E2 [5]

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 4 NOT within the 4049 syllabus (i) Unit vector of 22 33113 4 = 44 53 ( 4)    −=    −−+−     ij Unit vector of 22 44114 3 = 33 534    +=    +     ij M1 Attempt to find either unit vector s.o.i. 3 3 610 24 4 85 68 OP OP      = = =     − − −       = − ij   A1 4 4 1215 33 3 95 12 9 OQ OQ      = = =            = + ij   A1 (ii) 12 6 6 9 8 17PQ      = − =     −       M1 Find QP  o.e. 226 6 1717PQ = = +  M1 Apply distance formula o.e. 325 5 13 5 PQ  == =  A1 [6]

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 5 NOT within the 4049 syllabus (i) Let A and B represent the number of aircrafts for each type and the number of Economy, Business and First seats in each type of aircraft respectively. ( )5 8 4 10=A B1 300 60 40 150 50 20 120 40 0 100 0 0   =   B B1 (ii) Let C represent the total number of seats in each class. ( ) ( ) 300 60 40 150 50 205 8 4 10 4180 860 360120 40 0 100 0 0   ==    C B2 B1 for any two correct B1 for all correct (iii) Let D represent the percentage of empty seats for Economy, Business and First seats on that particular day. 0.05 0.1 0.2  = D B1 (iv) ( ) ( ) 0.05 4180 860 360 0.1 367 0.2  ==  CD 367 seats B1 [6]

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 6 62 6 6 1 6 2 3 63 2 662 2 (2) (2) 122 2 2 6 (2) ...3 2 64 96 60 ... x x x x xx −− −         − = + − + −                   + − +  = − + + B3 B1 for each simplified term Coefficient of 2 60( ) ( 96)(1)xk= + − M1 60 96 84 60 180 k k −= = M1 3k = A1 [6] 7 NOT within the 4049 syllabus (i) Minimum value of f 11=− M1 s.o.i. 2 2 1f (3) 9 3 11 533 1f ( 3) 9 3 11 89 3 = − − = − = − − − = M1  fR 11,89=− A1 (iia) 1 , 113  − B1 Minimum point B1 (iib) 1 ,113   B1 Maximum point B1 [7]

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 8 (a) lg ( 12) 1 lg (2 ) lg ( 12) lg (2 ) 1 12lg 1 2 xx xx x x + = + − + − − = + = − M1 Quotient low s.o.i. 12 102 x x + =− M1 Remove lg 12 20 10 11 8 8 11 xx x x + = − = = A1 (b) 2log 2a pa p = = M1 8 2 3 2 log log log 2 qb q b = = M1 Change of base low s.o.i 2 2 3 log 3 log 3 2 b q b qb q = = = M1 3p c a bq = = − A1 [7]

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 9 (i) ( ) 2 d 2( 3) (2 4) d 3 y x x x x + − −= + M1 Attempt to use quotient rule ( ) ( ) 2 2 2 6 2 4 3 10 3 xx x x + − += + = + A1 ( ) ( ) 2 2 10 d3 0 0 0 d3 yx xx +      + A1 Show that d d y x is increasing Since d 0d y x  and d 0d y x  , the curve has no turning points. AG Explanation given (ii) When 0y= , 2x= . Hence, P(2, 0). B1 s.o.i. When 2x= , ( ) 2 d 10 2 d5 23 y x == + M1 2 5 240 (2)55 y x c cc =+ = +  =− 40, 5Q − M1 Area 14(2)25 =  M1 Shoelace method o.e. 0.8= units2 A1 [8]

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 10 (i) ( ) 2f ( ) ( 1)( )x x x k x k= − − − M1 o.e. ( )( )( ) 2 f (2) 7 2 1 2 2 7 kk = − − − = M1 Remainder theorem s.o.i. 23 32 4 2 2 7 2 2 3 0 k k k k k k − − + = − − − = A1 AG (ii) Let 32f ( ) 2 2 3k k k k= − − − 32f (3) 3 2(3) 2(3) 3 0= − − − = Since f (3) 0= , by Factor Theorem, ( 3)k− is a factor of f ( )k . B1 Factor Theorem s.o.i. Attempt to long divide or compare coefficients M1 s.o.i. ( ) 2f ( ) ( 3) 1k k k k= − + + M1 ( ) 2 ( ) 0 ( 3) 1 0 fk k k k = − + + = 30k−= or 2 10kk+ + = For 2 10kk+ + = , Discriminant 21 4(1)(1) 3 0= − =−  Since discriminant < 0, 2 10kk+ + = has no real roots. M1 Hence, the only real value of k is 3k = . A1 AG [8]

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 11 (a) 2 2 2cot 1 tan 2 1 tantan 2 tan tan tan tan 2 0 xx xx xx xx =+ =+ =+ + − = M1 Obtain equation (tan 1)(tan 2) 0xx− + = M1 Factorise o.e. 1 tan 1 0 tan 1 45 x x  −= = = 1 tan 2 0 tan 2 tan (2) 63.43494882 x x x − += =− = = M1 1 1 2 2, 180 , 180 , 360 45 , 225 , 116.6 , 296.6 (1 d.p.) x    = + − − =     A2 A1 for any pair of correct answers (b) ( ) ( ) ( ) 6sin 2 1 5 0 6sin 2 1 5 5sin 2 1 6 y y y + + = + =− + =− M1 Make ( )sin 2 1y+ the subject 1 5sin 0.985110778336 − ==  M1 Basic angle s.o.i. 21 π , 2πy + = + − M1 Identify quadrants π 1 2π 1 ,22 1.56, 2.15 (3 s.f.) y + − − −= = A2 [10]

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 9 12 EITHER (i) When 0y= , ln 2x=− . ( )ln 2,0A − B1 o.e. When 0x= , 3y= . (0,3)B B1 (ii) 2d 2ed xy x −= M1 When 0x= , d 2d y x = . M1 Find gradient s.o.i. Gradient of normal 1 2=− Equation of normal is 1 32yx=− + . M1 o.e. When 0y= , 6x= . ( )6,0C A1 (iii) Area 0 2 ln 2 14 e d (6)(3) 2 x x− − = − + M2 02 ln 2 e49 2 x x − − = + + M1 Correct integration ( )1 4 ln 2 1 92= − − − + M1 Substitution 10.3= units2 (3 s.f.) A1 AG [11]

General Certificate of Education Ordinary Level 2006 Additional Mathematics 4018/01 Syllabus 4018 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 10 12 OR (i) x 15 20 25 30 lg y −0.82 −0.42 −0.02 0.37 Plotting of all points above P1 Line drawn on suitable axes L1 (ii) ( ) 10 lg lg 10 lg lg10 lg lg lg Ax Ax Ax yb yb yb y x b A − − − = =