2016 O Level A Math 4047 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2016 Additional Mathematics 4047/01 Syllabus 4047 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (i) 2 12 2 (1) 2 6 4 (2) yx y x x = − −−− = − − −−− Substitute (1) into (2) ( ) 2 2 2 2 6 4 12 2 2 4 16 0 2 2 8 0 2( 2)( 4) 0 x x x xx xx xx − − = − − − = − − = + − = M1 Factorisation o.e. 2 or 4xx=− = When 2x=− , 16y= . When 4x= , 4y= . M1 Attempt to find values of y (–2, 16) and (4, 4) A1 (ii) 2 2 2 4 12 2 2 (2 ) 16 0 x kx x x k x − − = − + − − = Discriminant 2 2 (2 ) 4(2)( 16) (2 ) 128 k k = − − − = − + M1 Attempt to find discriminant Since 2(2 ) 0k− , then 2(2 ) 128 128 0k− +   . Hence, discriminant 0 for all values of k, the line intersects the curve at two distinct points. (Shown) A1 AG Must justify clearly that D 0 [5] 2 ( ) 2 6 2 16 4 3 16 4 3 6 2 12 2 h h + = + += ++ M1 Expand ( ) 2 62+ correctly 16 4 3 8 4 3 8 4 3 8 4 3 h +−= +− M1 Multiply by conjugate s.o.i. 128 32 3 64 3 16(3) 64 16(3) +−−= − M1 Expand 80 32 3 16 5 2 3 −= =− A2 A1 for each term Deduct A1 for wrong form [5]

General Certificate of Education Ordinary Level 2016 Additional Mathematics 4047/01 Syllabus 4047 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 3 (ai) 190 sin 90 x−−    or 1ππ sin22 x−−   B1 o.e. (aii) 10 cos 180x−   or 10 cos πx− B1 o.e. (b) 1a=− B1 6b= B1 2c= B1 [5] 4 (i) B3 B1 for correct shape for each graph B1 for relative scale of graphs (i.e. 2 256yx= is much wider than 22yx= ) (ii) ( ) ( ) 22 4 3 3 2 256 4 256 0 4 64 0 0 or 64 8 xx xx xx xx = −= −= = = = M1 8yx= A1 [5] 22yx= 2 256yx= y x O

General Certificate of Education Ordinary Level 2016 Additional Mathematics 4047/01 Syllabus 4047 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 5 2 22 2 4 31 2 19 266 x x x x x x x + − − =++ − + − M1 Use of long division o.e. Let 2 2 19 6 2 3 x A B x x x x − =++ − − + M1 Partial fractions 2 19 ( 3) ( 2)x A x B x− = + + − When 2x= , 15 5 3AA− =  =− M1 FTA When 3x=− , 25 5 5 BB− =−  = M1 FTA 2 2 2 4 31 3 5 26 2 3 xx x x x x +− = − ++ − − + A1 [5] 6 (i) Since 4x= is the line of symmetry, then 6 42 p+ = . 6 8 2 (Shown)pp+ =  = B1 Working shown clearly (ii) The equation of the curve is ( 6)( 2)y a x x= − − . When 4x= , 6y= . 36 4 ( 0) 2a a a= −  =  B1 ( ) 2 3 ( 6)( 2)2 3 6 2 122 y x x x x x = − − = − − + M1 Expand 23 12 182 xx= − + 12b=− A1 18c= A1 (iii) 06 q B2 B1 for correct use of inequality B1 for correct limits used [7]

General Certificate of Education Ordinary Level 2016 Additional Mathematics 4047/01 Syllabus 4047 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 7 (i) 5m (100 10 ) m OP t OQ t = =− B1 Both seen By Pythagoras’ Theorem, 22(5 ) (100 10 )s t t= + − M1 s.o.i. ( ) 22 2 2 25 10000 2000 100 125 2000 10000 125 16 80 (Shown) t t t tt tt = + − + = − + = − + A1 AG (ii) ( ) 1 2 2d1 125 16 80 (2 16)d2 s t t tt − = − +   − M1 Attempt to use chain rule 22 125( 8) 5 5( 8) 16 80 16 80 tt t t t t −−== − + − + A1 o.e. (iii) 2 d 125( 8)00d 16 80 8 yt x tt t −=  = −+ = When 8t = , 44.72135955 44.7s== (3 s.f.) B1 By first derivative test, x 7.9 8 8.1 d d y x –0.279 0 2.279 Sketch of tangent Sketch of curve Thus, 44.7s= is the least distance. B1 Justify using 1st or 2nd derivative test For 1st derivative test, values of d d y x must be calculated [7]

General Certificate of Education Ordinary Level 2016 Additional Mathematics 4047/01 Syllabus 4047 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 8 (i) 3 452 3 45 (1) 22y x y x=− +  =− + −−− Gradient of AB 2 3= M1 s.o.i. 2 3 2 226 ( 2)33 2 22 (2)33 y x c cc yx =+ = − +  = = + −−− M1 Find equation of AB s.o.i. Substitute (1) into (2) 3 45 2 22 2 2 3 3xx− + = + M1 Substitution or elimination s.o.i. 13 91 66 7 12 x x y = = = M1 Find x or y s.o.i. B (7, 12) A1 (ii) When 0y= , 15x= . C (15, 0) B1 s.o.i. 15 2 0 6 13, , 32 2 2M −+   ==       M1 Let the coordinates of D be (p, q). 7 13 622 p p+ =  = 12 362 q q+ =  =− D (6, –6) A1 [8]

General Certificate of Education Ordinary Level 2016 Additional Mathematics 4047/01 Syllabus 4047 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 9 (i) 3 d 32 2d y xxx=− + M1 o.e. 3 d 32 0 2 0d y xxx= − + = M1 s.o.i. 4 4 2 32 0 16 2 x x x − + = = = M1 Solve for x When 2x= , 6y=− . When 2x=− , 6y=− . (2, –6) and (–2, –6) A2 A1 for each point (ii) 2 24 d 96 2d y xx =− − M1 FT from (i) When 2x= , 2 24 d 96 2 8 0d2 y x =− − =−  . Hence, (2, –6) is a maximum point. A1 Must calculate the value of 2 2 d d y x When 2x=− , 2 24 d 96 2 8 0d ( 2) y x =− − =− − . Hence, (–2, –6) is a maximum point. A1 [8]

General Certificate of Education Ordinary Level 2016 Additional Mathematics 4047/01 Syllabus 4047 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 10 (i) 2 24 1 ln 2d ln d x x xx x x x x  −  = M2 M1 Attempt to use quotient rule M1 for correct differentiation 4 4 4 3 3 2 ln 2 ln 1 2ln (Shown) x x x x x x x x x x x x −= = − = − A1 AG (ii) 13 3 2 1 2ln ln dxx xcx x x− = + M1 Knows to use result from (i) 3 132 2 122 ln ln2 d d ln 2 xx x c x xxx xx ccx − − − = + − = + − + −   M1 Integration of 3x− s.o.i. 3 2 2 ln ln 1d, 24 xx xdx x x =− − + where 12d c c=+ A1 (iii) 3 2 2 ln ln 1f ( ) d 24 xxx x d x x x= =− − + 3f (1) 4= 13 144 dd− + =  = M1 Attempt to find arbitrary constant 22 ln 1f ( ) 1 24 xx xx=− − + A1 o.e. [8]

General Certificate of Education Ordinary Level 2016 Additional Mathematics 4047/01 Syllabus 4047 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 11 (a) 2 2 (sec tan ) 1 sin cos cos    − =− M1 Change tan to sin cos   s.o.i. 2 1 sin cos   −= M1 Join fraction 2 2 2 2 (1 sin ) cos (1 sin ) 1 sin     −= −= − M1 Use of 22sin cos 1+= s.o.i. 2(1 sin ) (1 sin )(1 sin ) 1 sin 1 sin     −= +− −= + A1 AG (bi) π 30k = B1 NOT within the 4049 syllabus. (bii) 1 40 80sin 1sin 2 1 πsin 26 kt kt  − = = ==  M1 Find basic angle s.o.i. π π π π π, π , π , 2π30 6 6 6 6 5, 25, 35, 55 t t = − + + = M1 40 minutes A1 [8]

General Certificate of Education Ordinary Level 2016 Additional Mathematics 4047/01 Syllabus 4047 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 9 12 (i)   π π 6 6 00 33f ( ) sin cos 244x dx x k x=  + = ( ) 3sin cos 2 sin 0 cos 2(0)6 6 4 1 1 3 1 (Shown)2 2 4 2 kk k k k  + − + = + − =  =− B1 AG Show substitutions clearly (ii) d cos 2 1f ( ) sind 2 2 xxx x = − − M1 Knows to differentiate f ( ) cos sin 2x x