stgss_Prelim_2021_4E_AMath_P1_Solution

1 2021 AM 4E Prelim P1 Marking Scheme Solutions: 1i 21 p− obtained 21 p− − 1ii 21 p p − − 2 098 09412 4912 2 2 2 >−++ >−+−+ +>++ kxx kxxx kxxx For the equation to be always positive, no real roo ts 7 28 4 0428 0436 64 0)9 )( 1 ( 4) 8 ( 04 2 2 −< −< <+ <+− <−− <− k k k k k ac b 3 2 2 2 2 2 2 2 2 2 2 2 d (2 1)6e 6e d (2 1) 12 e 6e 6e (2 1) 12 e (2 1) 4 3e (2 1) 4 2 1 x x x x x x x y x x x x x x x x x xy x + − = + + − = + = + = + = + 4=∴ k 4i 2πA r = d 2πd A rr = 2 1 d d 2πd d 2π 5 3 30 π cm s A r rt t − = × = × × = 4ii 2π 4πr = r = 2

2 2 1 d d 2πd d 2π 2 3 12 π cm s A r rt t − = × = × × = 5i Amplitude = 2 Period = π 5ii Minimum value = ̶ 1 Maximum value = 3 5iii 6i When P = 3000, t = 0, ( ) (shown) 2000 5000 3000 e5000 3000 0 − = =− += A A A k 6ii kt P e2000 5000 −= When P = 3800, t = 10, ( ) fig.) sig. (3 0511 . 0 10 5 3ln 5 3ln ln 2000 1200 e2000 5000 3800 10 10 10 −= ÷     =      = = −= k k e e k k k 3 -1 y x π

3 6iii 0.0511 5000 2000e tP −= − 0.0511 Since e 0 t− > , 0.0511 2000e 0 t−− < . 5000 0.0511 2000e 5000 t−− < Thus the number of fishes can never reach 5000. 7i Let BAE x ∠ = (tangent-chord theorem) ACB x ∠ = (alt. s, // ) CAD x CE DA ∠ = ∠ Thus BAE CAD ∠ = ∠ 7ii In triangles BAE and DAC , [from part (i)] BAE CAD ∠ = ∠ 180 (opp. s of cyclic quad) CDA ABC ∠ = ° − ∠ ∠ o o 180 (180 ) ( on a str. line) CDA ABE s ∠ = − − ∠ ∠ ABE = ∠ Hence, triangles BAE and DAC are similar. (AA similarity test): Encouraged to write. Not necessary to penalize) 7iii [from part (i)] ACB x ∠ = (base of isos. triangle) BAE BEA s ∠ = ∠ ∠ ( similar to ] DCA BAE DAC = ∠ ∆ ∆ x= ACB DCA x ∠ = ∠ = ∵ , the line AC bisects the angle BCD . 8 6 2 2 (1 ) 1 6 15 ax ax a x + = + + + … 6 2 2 2 2 (1 )(1 ) (1 )(1 6 15 ) 1 (6 ) (15 6 ) bx ax bx ax a x a b x a ab x + + = + + + + = + + + + + … … 6 a + b = 0

4 2 2 21 15 6 4 75 2 4 a ab a ab + = − + = − Sub 6b a = − into 2 75 2 4a ab + = − 2 2 2 75 12 4 77 4 1 2 a a a a − = − − = − = ± When (reject) 3 ,2 1 − == ba When 3 ,2 1 =− = ba 9a 2 3 0 (4 )(8 ) 1 2 2 x y x y + = = 2 3 0 x y + = 3 2 y x = − ─ (1) ( ) ) 2 ( 2 1 3 13 55 5 15125 2 1 3 3 3 −−=− = =÷ −− xy xy xy Sub. (1) into (2), 14 3 2 1 3 12 0 2 1 3 12 = = =−+ x x xx 7 1− =∴ y 9b ( ) ( ) 2 2 π 12+4 2 π π 2 1 V r h h = = −

5 ( ) 2 12 4 2 2 1 h += − 12 4 2 3 2 2 h += − ( )( ) ( ) ( ) 12 4 2 3 2 2 3 2 2 3 2 2 h + + = − + 36 24 2 12 2 16 9 8 h + + + = − 52 36 2 h = + 10i 3) 42 ( −= xxy 2 3 d ( )3(2 4) (2) (2 4) (1) d y x x x x = − + − 2 3 2 2 6 (2 4) (2 4) 0 (2 4) (8 4) 0 (2 4) 0 or (8 4) 0 2 4