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