Northland Sec Prelim 2024 AM P1 MS

Mark Scheme for 2024 S4E5N Add Math Prelim Paper 1 1a 22 4 1xx+− 22[( 1) 1] 1x+ − − 22( 1) 3x+− B2,1 1− for each error 2 8xx− + − 2( ) 8xx− − − 2 11 824x − − − − 2 1 31 24x− − − B2,1 1− for each error Accept ( ) 2 0.5 7.75x− − − 1b Minimum point ( 1, 3)−− , Maximum point (0.5, 7.75)− B1√ Only need the y-values max of 2 8y x x=− + −  min of 22 4 1y x x= + − → curves will not intersect B1 Correct argument. Diagram may be used as long as all necessary working is shown [6] 2a Let 2e xu= → 19 14 8uu −+= 29 14 8 0uu+ − = M1 Form quadratic equation using substitution (9 4)( 2) 0uu− + = → 4 9u = or 2u =− (impossible) M1 Solving the quadratic and finding u 2 4e 9 x = → 2e 3 x = A1 Do not accept 0.666… [3] 3 2V r h= 2(26 3 20 5) ( 5 3) h− = − M1 Use of 2V r h= formula 2( 5 3) 8 2 15− = − B1 For correctly squaring 53− (26 3 20 5) 8 2 15 h −= − or (13 3 10 5) 4 15 h −= − (26 3 20 5) 8 2 15 8 2 15 8 2 15 −+  −+ M1 For rationalising denominator 208 3 52 45 160 5 40 75 64 4(15) + − − − DM1 Depend on previous method. Correct expansion of numerator 8 3 4 5 4 − 2 3 5 12 5− = − A1 A1 A1 unsimplified answer A1 correct simplified [6] 4 f '( ) 3cos 2sin 2 ( )x x x c=− − + B2,1 1− for each error 9 3cos( ) 2sin(2 ) c=− − + M1 Use gradient of 9 with x = 9 3 0 c= − + → 6c= f ( ) 3sin cos 2 6 ( )x x x x d=− + + + B2,1√ 1− for each error. √ from f '( )x 6 6 68 3sin( ) cos 2( ) 6( ) d  =− + + +

3 1 228 c=− + + + → 9c =− f ( ) 3sin cos 2 6 9x x x x =− + + + − A1 Use f ( )x of 8 with 3x = , c.a.o. [6] 5ai 3 − B1 5aii Principal value for 1cos x− lies between 0 and  inclusive B1 Accept 10 cos x − and 0  5b 2sin( ) 1 2sin( )A B A B+ = − − 2(sin cos cos sin ) 1 2(sin cos cos sin ) A B A B A B A B + = − − B1 Use of addition formula 4sin cos 1AB = 1 34(sin )( ) 1A = → 3 4sin A= M1 For sin Ak= 2243− M1 For finding adjacent 3tan 7 A= A1 c.a.o. [6] 6a coord of 4.5xC−= B1 1 2 (6 3) 6 h −  = → 4h= coord of 3 4 1yC− = − =− M1 For 3 height− or use of shoelace 3 ( 1) 8 3 4.5 3 ACm −−= =−− M1 For finding gradient of AC 78 10 7.5 3 DE km −= =−− M1 For DE ACmm = 1 3k = A1 6b 1 3 4.5 10 7.5 4.51 1 7 12 −− 11 33 1 (4.5) 10(7) 1(7.5) ( 10) (7.5) 7(4.5)2 + − − − − − M1 Correct use of shoelace method 20 units2 A1 [7] 7a d 2 d sec y x x= B1 1y= → 4x = M1 For finding x value dd d d d d yy x t x t= 2 d 4d0.12 sec ( ) x t − =  M1 Chain rule used correctly. Ignore − in d d y t d d 0.06x t =− units/second A1 a.e.f. 7b (120)(2) 240k == B1 PA k= → 1A kP −= M1 Make A the subject 2d dP A kP−=− A1 2d d 1240(120) 60 A P −=− =− A1 Accept 0.0166...− [8]