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)