2024 TPSS AMATH P2 PRELIM MS

Marking Scheme for 2024 4E5N A.Math Prelim Paper 2 [√ means follow through] Total Marks : 90 Setter:Mdm Ho Qn 1 (a) Let 3 8 3 2 ) (f 2 3 + − + =x x x x. By the Remainder Theorem, 12 3 8 3 2 3 ) 1 ( 8 ) 1 ( 3 ) 1 ( 2 ) 1 (f 23 = + + + − = + − − − + − = = − p p p p M1 A1 [2] (b) ) 1 2 (−x is a factor of ) (f x [ or (x+3) or (x-1) is a factor] 1or 2 1 ,3− =x M1 M1 [another quad expression] A1 [3 ans] [3] (c) 0 3 ) 1 ( 8 ) 1 ( 3 ) 1 2( 0 11 8 1) 3( 1) 2( 23 23 = + − − − + − = + − − + − x x x x x x Let 1− =x y . 2or 2 3 ,2 1or 2 1 ,3 1 1or 2 1 ,3 0 3 8 3 22 3 − = − = − − = = + − + x x y y y y Hence, 2or 2 3 ,2− =x . M1 A1 [2] 2 (a) $120 000 B1 [1] xo 2 1 1 2 0 1 2 = = = − x x x ( ) 0 ) 1 )( 3 )( 1 2 ( 03 2 ) 1 2 ( 0 3 8 3 2 2 2 3 = − + − = − + − = + − + x x x x x x x x x 3 0 3 − = = + x x or 1 0 1 = = − x x

(b) (i) 12 12 12 Value after 1 year 120000 90000 120000 0.75 12 ln 0.75 ln 0.75 0.023973512 0.02397 a a a e e e a a shown − − − = = = −= = =− = (ii) 70000 120000 7 12 70.02397 ln 12 22.49 23 months at at e e t t − − = = −= = ≈ (iii) 5 years = 60 months ( )60 0.02397 Value after 5 years 120000 $28482.55 e − = = Yes, since car dealer is paying more ($29000). or No, there is not much difference, so I would rather use the car. M1 M1 must show 0.0239735 M1 M1 A1 M1 A1 [2] [3] [2] 3 (a) (Shown) 4 26 2 1 2 2 2 3 2 2 30sin45sin30cos45cos ) 30 45cos(75cos −=            −            = ° ° − ° ° = ° + ° = ° M1[use 45+30] M1 [any one -surd form]] M1 [simplify to desired ans] [3] (b) RHS sin cos 1 sin cos cos sin cos sin cos sin sin 1 cos 1 sin cos cos sin LHS 22 = += +×+= + + = x x x x x x x x x x x x x x x x M1 use identities M1 [same denominator] M1 simplify [6]

13 cos sin 4cos 3sin 3cos 3sin 4cos 3sin 6sin cos 1tan 6 0.165148677 0.165, 3.31 xx x x xx xx xx x x α =+− +=− = = = = M1 cross multiply M1 [change to tan] A1 4 (a) 2196 , 2AB x BC x= −= ( ) ( ) 2 2 1 ( 2 ) 1962 3 1962 A xx x x x = +− = − B1 [ 2196AB x= − ] M1 formula with h=3x [2] (b) 2 2 2 2 2 22 2 2 2 d 3 2 3 196 d2 22 196 3 1962 196 3 196 2 196 3(98 ) 196 Ax x x x x x x x xx x x x  −= −+ − = − +− −  −−=  − −= − 2 2 d 0d 3(98 ) 0 196 98 9.90 A x x x x x = − = − = ≈ 1 2 22 22 2 2 13 196 ( 2 ) 3(98 )( )(196 ) ( 2 )d 2 d 196 xx x x xA xx − −−− − − − = − 0 d d 2 2 < x A , A is max. [or Using first derivative test] M1 M1 A1 M1 A1 [5]

(c) ( ) 2 3 98 196 982 147 cm A= − = M1 A1 [2] 5 (a) ( ) ( ) ( ) 4 2 2 32 2 8 4 2 2 2 2 2 4 2 2 2 2 523 5 2 3 + = + + = × + × + = + ++ x xx x xx x x x Let xy 2= . Since 0 2>x , 3− =x . M1 [ ( ) 2 82 x seen] M1 [ ( )32 2 x seen] M1 factorise A1 for both y A1 [5] (b) x y x a ba a ab a ababa 2 3 )(log2 ) (log3 ) (log log )]([log2 1 log loglog 2 22 2 3 2 2 3 23 += += = = M1 [change base 2] M1 [ 22log ( ) 3log ( )ab+ ] A1 [3] 3 2 2 8 12 8 1 1 8 0 1 8 3 − = = = = = = − − x y y y x x 4 2 4 0 4 − = − = = + x y yor 0 ) 4 )( 1 8 ( 0 4 31 8 4 32 8 2 2 = + − = − + + = + y y y y y y y

(c) 5 25 5 25 5 55 5 5 55 2 55 2 2 log 50 4log log (2 4) 2 loglog 50 4 log (2 4) log 5log 25 log 50 2log log 25(2 4) log 50 log 25(2 4) 50 50 100 20 ( 2)( 1) 0 2 or 1 (reject) yy y y yy yy yy yy yy y + = ++ + = ++ += + = + = + −−= − += = − M1[ 5 5 log log 25 y ] M1[2= 2 5log 5 ] M1 use product law A1 [4] 6 (a) 10cos sin 8 8sin 10cos 8sin (shown) AE ED ED AD θ θ θ θθ = = = = + Use trigo ratio M1 M1 [2] (b) 10cos 8sin cos( )Rθ θ θα+= − 22 1 8 10 164 or 12.8 8tan 10 38.66 38.7o R α − = + = =  = ≈ 10cos 8sin 164cos( 38.7)θθ θ+= − M1 M1 A1 [3] °θ A D C B 8 m 10 m E °θ

(c) 164 cos( 38.7) 12 12cos( 38.7) 164 θ θ −= −= basic 20.4∠= 38.7 20.4, 20.4θ −= − 59.1, 18.3 (1 decimal place)θ = M1 M1 A1 both [3] 7 (a) Show that ( ) ( ) 52 33{ 31 } ( 81 ) 31d xx x xdx −= − − . ( ) ( ) ( ) 5 3 2 53 3 { 3 1} 5 ( 31 ) 3 313 d xxdx xx x − = − +− ( ) 2 53 35 ( 31 ) 31xx x= −+− [ ] 23( 31 ) 5 31x xx= − +− [ ] 23(3 1) 8 1xx= −− B1 for 235 (3 1)3xx − seen B1 for ( )3 seen B1 for ( ) 5331x− seen M1 for factorization A1 for showing clear working leading to the desired expression [5] (b) ( ) ( ) 25 33( 81 ) 31 31x x dx x x c− − = −+∫ ( ) ( ) ( ) 2 25 3 338 31 31 31x x dx x dx x x c− − − = −+∫∫ ( ) ( ) ( ) 5 25 3 33 318 31 31 ( ) 5 (3)3 xxx d xxx c−− = −+ +∫ ( ) ( ) ( ) 5 25 3 33 318 31 31 ( ) 5 xxx d xxx c−− = −+ +∫ ( ) ( ) ( ) ( ) 2 55 3 33 5 3 1131 31 31 8 40 11 =( ) 3 18 40 xx d x xx x c x xc − = −+ −+ + −+ ∫ M1 M1 M1 ( ) 5 331() 5 (3)3 x− A1 [ 1/8 seen] A1 [1/40 seen] [5]

8 (a) (i) From the diagram, by similar triangles, 2 4 42 (shown)2 r h rh hr = = = (ii) From (a)(i), (shown) 12 4 3 1 2 3 1 3 1 3 2 2 2 h hh hh h r V π π π π =      =     = = M1 M1 M1 M1 [2] [2] (b) 2 3 4d d 12 hh V h V π π = = Given: 3d 2m /mind V t =− When 2=h , 2 d dd d dd d2 (2)4d d2 d d2 d 0.637m/min (correct to 3 sig. fig.) V Vh t ht h t h t h t π π π = × −= × −=× −= =− M1 B1 M1 M1 A1 [5] 9 (a) 06 622 2 = − − + +y x y x cm 4 16 )6( ) 3 ( 1Radius 22 = = − − − + = ) 3 ,1 ( is Centre−∴ M1A1 M1A1 [or A2] [4]

(b) M1 M1 A1 [3] (c) (i) Coordinates of centre C2 = (-1, -3). (ii) P(5,0) lies outside the circle C 2. B1 M1 A1 [1] [2] 10 (a) 2 1) 2sin( 0 1 ) 2sin(2 = + = − + π π x x 6 2 1sin 1 π α = = − 12 11,12 7,12,12 5 26,26,6,62 π π π π πππ ππ ππππ − − = + − + − = + x x x coordinate of A = 12 7π M1 M1 A1 [3] (b) x coordinate of B = 12 11π B1 [1] (c) [ ] [ ] [ ] [ ] 7 11 12 12 70 12 117 1212 70 12 2 Shaded area 2sin(2 ) 1 d 2sin(2 ) 1 d cos(2 ) cos(2 ) 2.69862 1 2.01377 ( 2.69862) 3.69862 0.684853 4.38 unit xx xx xx xx ππ π ππ π ππ ππ = − +− + +− = − − +− + − +− =−− − +− −− = + = ∫∫ M1 M1 M1 A1 [4] Total = 90 m 22(5 1) (3 0) 45 6.71 4 radius + ++ = = >= 0 ) 1 )( 7 ( 0 76 06 6) 1 ( 2 ) 1 ( 2 2 2 = + − = − − = − − − + + − k k k k kk (NA) 1or 7 − = =k k