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 8 4 1 2 2 x x x x x x x x x x x − + − = − − = − = − = = = = = 2 113 or 0 == yy Coordinates:      − 2 113 , 2 1 and 0) (2, 10ii Using First Derivative Test, x −2 2 +2 d d y x + 0 + (2, 0) is a point of inflexion. x 1 2 − 1 2 1 2 +

6 d d y x ̶ 0 + 1 1 , 13 2 2   −    is a minimum point 11i 210 3 d d 2 −+= = tt t sv 11ii a = 6 t + 4 ctt tav ++= =  43 d 2 When t = 1, 4v = . 3c = − 343 2 −+=∴ ttv cttt tvs +−+= =  32 d 23 When t = 0, 6s = 6c = 3 2 2 3 6 s t t t ∴ = + − + 11iii 3 2 3 2 2 2 3 6 5 2 4 3 2 0 (3 2)( 1) 0 Q P s s t t t t t t t t t t = + − + = + − + + − = − + = 2 3t = or 1 (reject) t = − 2At 3t = , vP = positive vQ = positive Therefore, P and Q are travelling in same direction 12ai 21 10 2 +− xx

7 ( ) ( ) 2 2 2 10 5 5 21 x x = − + − − − + ( ) 2 5 4 x= − − Since ( ) 2 5 0 x − ≥ , ( ) 2 5 4 4 x − − ≥ − Thus 21 10 2 +− xx can never be smaller than 4− 12aii (5, 4) − Minimum 12aiii 021 10 2 =+− xx ( 7)( 3) 0 x x − − = 7 or 3 x x = =

8 12aiv 12b 2 2 16 0 xx a b + − = Discriminant 2 2 1 1 4(6) a b     = − −         2 2 1 24 a b = + Since 2 1 0a > and 2 24 0b > , Thus, 2 2 1 24 0a b + ≠ and there are no values of a and b for which the equation has equal roots. 13i 3 ) 1(1 ) 2(4 of Gradient = −− −−=PQ 3 7 21 (5, -4) x y

9 Gradient of perpendicular bisector of PQ = 3 1− Midpoint 1 1 2 4 ,2 2 − + − +   =     ( )0,1 = 1 ( ) 3y x c = − + 11 (0) 3 c= − + 1c = Equation of perpendicular bisector of PQ is 13 1 +− = xy . 13ii At x-axis, y = 0. 3 13 1 13 10 = = +−= x x x Coordinates of R are (3, 0) 13iii Since PS is parallel to x-axis and P is the point ) 2 , 1( −− , y-coordinate of S is ̶ 2. 2 3 9 3 7 7 3 x x x − = − = = Coordinates of S are      − 2 ,3 7 . 13iv 2 73 1 1 3 1Area of 32 0 4 2 2 0 40 units 3 PQRS −= − − = Can infer from previous working that y-intercept is 1