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] 8a 6 3 48xy+= B1

Perpendicular height 2 2 1 2xx =−  or Area of triangle 1 ( )( )sin 602 xx= M1 For finding perpendicular height or area of equilateral triangle 13 ( )( )22V x x y=  or 23 4 xy  M1 For cross section area  height 2 23 3 (8 )(16 2 )42 xxV x x −= − = A1 Answer was given – so all working must be correct 8b 23 1 24 3 3V x x=− 2d3 d2 8 3 3V x xx=− B1 23 28 3 3 0xx−= M1 Sets d d V x to 0 and solves 3 23 (8 ) 0xx −= 0x= (rejected) or 16 3x= A1 a.e.f. 65.7V = A1 Accept 65.68 … [8] 9ai 2( 1)( 1)x x x+ − + B1 2( 1)( 1)x x x− + + B1 9aii 6 3 2 26 1 (6 ) 1− = − 33(6 1)(6 1)+− M1 For 22 ( )( )a b a b a b− = + − 22(6 1)(6 6 1)(6 1)(6 6 1)+ − + − + + (7)(31)(5)(43) A1 SR1 for 2 3 3 2 2 2 2(6 ) 1 (6 1)((6 ) (6 ) 1)− = − + + (35)(1333)= but A0 thereafter 9bi 321 1 1 2 2 22( ) 3( ) 11( ) 6− − − − − + M1 Use of remainder theorem or long division 10.5 A1 9bii 3 2 22 3 11 6 ( )(2 3)x x x Ax B x ax− − + = + + + or 2 32 2 + +3 2 3 11 6 x ax Ax B x x x+ − − + M1 Sets up ( ) ( )(factor)P x Ax B=+ or long division with factor as dividend 3 2 22 3 11 6 ( 2)(2 3)x x x x x ax− − + = + + + B1 For ( ) ( 2)Ax B x+ = + 2 2 234x ax x− = + or 11 3 2x x ax− = + 7a=− A1 ( 2)(2 1)( 3)x x x+ − − A1 [10] 10ai 12 1 3 12 1 2 r r rTx r x − +  =  B1 i.s.w. 10aii 3 12 12 4 1()r r r xxx−−= power of x 12 4 r=− B1 Accept 12 4 rx −

10aiii 12 4 0r−= or 2 12 11 10 33 12 12 11 ...12 22x x x xx       + + +             M1 Sets power to 0 or Correct expansion of first 3 terms 3 12 3 31 3 12 1 55 3 22Tx x − +  ==  A1 a.e.f. 10b 2(1 )(1 ) nxax++ ( 1) 2 2 2 2(1 )(1 ( ) ( ) ...) nnxxax n − + + + + B2 B1 for each 1 n n   = and ( 1) 2 2 n nn −= ( 1) 2 28(1 )(1 ...) nnnax x x − + + + + ( 1) 22 2 8 21 ... nnn anx x ax x − + + + + + ( 1) 22 2 8 21 ... nnn anx ax x x − + + + + + 1 12 na+= M1 Equate coeff. of x terms to 1 ( 1) 582 n n an− + =− M1 Equate coeff. of 2x terms to 5− 1 24(1 )( 1) 588 nnnn −− + =− 1 2( 1) 4 (1 ) 40n n n n− + − =− 22 4 2 40n n n n− + − =− 2 3 40 0nn− − = 24 2 42 0aa− − = 8n= or 5n=− (rejected) A1 7 2a= (rejected) or 3a=− 1 21 (8) 3a= − =− A1 8n= [10] 11 d d 2 10 y x x=− B1 2 10 4x− =− → 3x= M1 For finding x-coord of P 3x= → 23 10(3) 24 3y= − + = 1 4 PRm = B1 1 4 9 4 91 44 3 (3) 09 c c yx yx =+ = =+ = → =− or 30 1 34 x − − = → 9x=− M1 For finding x-coord of R 1 (9 3)(3) 182 += or 3 91 449 dxx − + M1 For finding area of triangle. Answer 18 may be implied 2 10 24 0xx− + = → 4x= B1 For finding x-coord of Q Integrate curve → 321 5 243 x x x −+ M1 A1 Knowing to integrate. A1 might be implied Limits 3 to 4 → 3 2 3 211 33(4) 5(4) 24(4) (3) 5(3) 24(3)   − + − − +    1 31= M1 Uses definite integral method on antiderivative 11 3318 1 19+= A1 [10]

12ai Plot all points correctly (allow  1 mm) M1 Straight line drawn through all points A1 Dep. on method 12aii Read off at 0t = (allow 0.005 ) M1 6e 403= → 403 thousands A1 Accept 403 000 12aiii Read off at ln 200 5.30= → 1995 8 2003+= M1 A1 c.a.o. 12b 21 2s ut at=+ 1 2 s at ut =+ → Plot graph of s t against t M1 A1 Divide throughout by t 1 2 a value obtained from gradient of graph DB1 Dep. on method u value obtained from y-intercept of graph DB1 Dep. on method Alternative Answer: 2 1 2 su att =+ → Plot graph of 2 s t against 1 t M1 A1 Divide throughout by 2t 1 2 a value obtained from y-intercept of graph DB1 Dep. on method u value obtained from gradient of graph DB1 Dep. on method [10] ln P t 5 10 15 20 0 3 3.5 4 4.5 5 5.5 6 6.5