Northland Sec Prelim 2024 AM P2 MS

Mark Scheme for 2024 S4E5N Add Math Prelim Paper 2 1 2 22 1 8 3 1( 1)(2 3) 2 3 x x Bx C A xx x x − − + −− + + =+ M1 Realising the form of the partial fractions 221 8 3 (2 3) ( 1)( )x x A x x Bx C− − = + + − + M1 Realising the need to eliminate denominator When 1x= , 221 8(1) 3(1) [2(1) 3]A− − = + 10 5 A−= → 2A=− When 0x= , 21 2[2(0) 3] (0 1)( ) C=− + + − 16 C=− − → 7C =− When 2x= , 221 8(2) 3(2) 2[2(2) 3] [2 7] B− − =− + + − 27 22 2 7 B− =− + − → 1B= 2 72 1 23 x x x −− − ++ A3, 2, 1 1− for each error inc. final answer [5] 2a 22 7 4 0x x c− − − = M1 Eliminates y or x 2 4b ac−= 2( 7) 4(2)( 4 ) c− − − − M1 Uses the discriminant 49 32 8 0 c++ 10.125c− A1 Accept 81 8c− smallest integer c 10=− A1 2b 2 2(2 5) 9y kx k x k= + − + 2 4b ac−= 2(4 10) 4( )(9 )k k k−− M1 Uses the discriminant 2216 80 100 36 0k k k− + −  220 80 100 0kk− − +  → 2 4 5 0kk+ −  ( 5)( 1) 0kk+ −  5k − or 1k  A2 A1 for each 0k  → 5k − B1 s.o.i. when 1k  rejected [8] 3a 1 cos sin cosec cot x x x x − −+ 1 11 c s st − −+ B1 cosec and cot all correct 2 1 1 c sc s − −+ 2 (1 ) 1 sc sc −→ −+ M1 Correct algebra 2 (1 )sc cc − −+ M1 Uses 22 1sc+= (1 ) (1 ) sc tcc − =− A1 All correct 3b tan 2 3x=− M1 Uses part (a) and replaces x with 2x 1tan (3) −= M1 For finding basic angle 2 180 , 360x = − − 54.2 ,144.2x=   A1 Accept 54.21… and 144.21… 3c 2 2 1 tt t= −

3 2t t t−= 32 0 ( 1) 0t t t t+ = → + = 0t = (not valid for 0 180x   ) B1 Must include and then reject 0t = or 2 1t =− not possible. No solution B1 Realises there is no solution to 2 1t =− Alternative Answer: Correct sketch of tanyx= and tan 2yx= B1 No point of intersection for 0 180x   and so no solution to the equation tan tan 2xx= B1 [9] 4a 22 2 22 log log 2log log 8 log 4 xx+= M1 Change of base 2 2 log 1log 32 xx+= B1 For ‘ 1 2= ’ 22 1 1log log 23xx+= 2 3log 8x→= M1 Making 2log x the subject 3 82x= → 3 8m= A1 4b B2 B1 for curvature of decreasing gradient observed B1 for graph close to asymptote observed with x-intercept 1 4c 2 11 5 4log ( ) 1x x − − = M1 Combine to single log 2 11 4 5x x − − = 2 11 5( 4)xx− = − → 3x= A1 If 3x= , 55log (2 11) log ( 5)x− = − or 55log ( 4) log ( 1)x− = − M1 Substitute into log expression Does not exist → No solutions A1 Correct argument and conclusion [10] 5a 2 f '( ) e (2 1)xxxx +=+ B1 Since 0x , 2 1 0x+ and 2 e0xx+  M1 Argues correctly f '( ) 0x  → increasing A1 f '( ) 0x  or 2 e (2 1) 0xx x+ + must be seen 5bi 22 2 2 6( 3)(2 ) (1) ( 3) ( 3) xxx x x x x ++− =+ + B2, 1 B1 for unsimplified 2 2 6 0( 3) xx x + =→+ 0x= , 6x=− M1 Sets to 0 and solves (0, 0) and ( 6, 12)−− A1 A1 SR1 answers not in coordinate form 5bii 3 18 ( 3)x+ B1 0x= 2 2 2d 0d 3 y x→ =  → minimum point DB1 Correct 2 2 d d y x value and conclusion x y

6x=− 2 2 2d 0d 3 y x→ =−  → maximum point DB1 Correct 2 2 d d y x value and conclusion [11] 6a 7 1 3 2 ( 6) 4PQm −== −− B1 2 ( 6) 71 22( , ) ( 2, 4)PQM +− += = − B1 Gradient of Perpendicular 4 3=− M1 Uses 12 1mm =− 44 ( ( 2))3yx− =− − − 44 33yx=− + A1 6b 44 33 24xx− + =− − M1 Realises the need to use sim eqns 8x=− ( 8,12)− A1 22( 8 2) (12 7)− − + − M1 Uses distance formula correctly with P or Q 125 A1 22( 8) ( 12) 125xy+ + − = B1√ √ for their radius and centre 6c ( 8 125,12)−− B2√ B1 for each, √ for their radius and centre [11] 7a 20.3(4 ) 1.2vt=− − + 0.6(4 )at=− B1 When 0v= → 2(4 ) 4t−= M1 Sets to 0 2t = , 6t = A1 0.6(4 2) 1.2a= − = m/s2 A1 7b 20.3(4 ) 1.2 ds t t= − − + M1 Realises need to integrate 30.1(4 ) 1.2 ( )s t t c= − + + A1 When 1t = , 2.5s= , 1.4c=− → 30.1(4 ) 1.2 1.4s t t= − + − Use 1t = and 3.9s= to find c When 0t = , 5s= m A1 7c Particle turned during 0t = to 7t = B1 When 7t = , it only gives distance from O B1 7d When 2t = , 1.8s= or When 6t = , 5s= OR 2 0 distance d 3.2vt== m or 6 2 d 3.2vt = m M1 For finding displacement at either time it turns When 7t = , 4.3s= OR 7 6 distance d 0.7vt== m M1 For finding displacement at ending time distance (5 1.8) (5 1.8) (5 4.3) 7.1= − + − + − = m A1 [12]

8a B3 B1 for two complete sine cycles with correct amplitude and intersections at x-axis B1 for one negative cosine cycle B1 for cosine cycle with correct amplitude and intersections at axis of rotation 8bi 5sin 2cosAB CE CF = − = − B1 5cos 2sinAD ED BF = + = + B1 Perimeter 5sin 2cos 5cos 2sin 7   = − + + + → 3cos 7sin 7++ B1 Answer was given – so all working must be correct 8bii 223 7 58R= + = B1 7tan 3 = → 66.8 = M1 For finding  58 cos( 66.8 )− A1 8biii 58 cos( 66.8 ) 7 13−  + = 6 58cos( 66.8 ) −  = 38.0 = M1 For finding  66.8 ,   − =− 28.8 , 104.8 (rejected) =   A1 Accept 28.78… 6.75 cm2 A1 [12] 9a 3 113 1 3 1 x xx =−++ B1 13 d ln(3 1)31 3 x x x x cx = − + ++ B3, 2, 1 1− for each error in the integration – needs +c 9b d1 ln(3 1) 3d 3 1 y xxxx= + +   + M1 B1 A1 Uses product formula. B1 ‘ 1 31x+ ’. A1 ‘ 3 ’ 3ln(3 1) 31 xx x= + + + 9c 33ln(3 1) d d 3 1 3 1 xxx x x xx+ + − ++ → 3ln(3 1) d 31 xx x x x+− + OR d3ln(3 1) d 3 1 yxx xx+ = − + → 3ln(3 1) d 31 xx x x x+− + M1 Attempts to use the result of part (b) x y 4 –1 – 2 0 1

3ln(3 1) d 31 xx x x x+− + → 1ln(3 1) 1 d 31x x x x+ − − + M1 Realises the need to use part (a) 44 1 0 3 0[ ln(3 1)] ln(3 1)x x x x + − − + A1 All correct 1 3[4ln13 0] 4 ln13− − − M1 Use definite integral formula on ln(3 1)xx + or antiderivative from part (a) 13 3 ln13 4− A1 For both values [12]