PH 2022 AMath Prelim Paper 1 Anskey

2022 PHSS Prelim AMATH Paper 1 Solutions Question Solution 1 Sum of length of the 2 parallel sides ( )2 12 5 2 32 18 2 + = − 24 10 2 32 18 2 32 18 2 32 18 2 ++= −+ 768 432 2 320 2 360 1024 648 + + += − 1128 752 2 376 += ( )3 2 2=+ cm 2 3 1 0yx− + = ---------- (1) 11x y+= ---------- (2) From (1), 31xy=+ ---------- (3) Subst. (3) into (2), 13 1 1y y+ + = 23 2 1 0yy+ − = 1 or 13y=−  y-coordinates of A and B are 1 3 and −1. 3i 21 254 xx− − + ( ) 21 854 xx=− + + ( ) ( )( ) 2221 8 4 4 54 xx=− + + − + ( ) 21 494 x=− + +

Qn Solution 3ii stationary value of y = 9 corresponding value of x = −4 4 ( ) 4 2 94 4532 x dxxxx  + −−  ( ) 4 49 3 2 45x dx x −= − + − ( ) ( )( ) ( ) 3 9 3 2 4ln 4 5 3 3 4 xx c − −−= + + − ( ) ( )3 1 ln 4 5 32 xc x =− + − + − 5 ( )( ) 32 2 15 19 116 6 5 1 9 x x x xx + + − −+ ( )( ) 2 2 22 19 213 5 1 9 xx xx −+=+ −+ ( )( ) 2 22 22 19 21 5 1 95 1 9 x x A Bx C xxxx − + + =+ −+−+ ( ) ( )( ) 2222 19 21 9 5 1x x A x Bx C x− + = + + + − Subst 1 5x= , 452 226 25 25 A= A = 2 Subst x = 0, 21 = 18−C C = −3 Subst. x = 1, 24 = 20 + (B − 3)(4) B = 4 ( )( ) 32 22 15 19 116 6 2 4 3 3 5 1 95 1 9 x x x x xxxx + + − − = + + −+−+

Qn Solution 6a By remainder thm, subst. x = −2, ( ) ( ) 32 2 2 3 2 2 30 36 m− + − − + = m = −5 6b ( )( ) 3 2 22 3 29 30 3x x x x x q Ax B+ − + = + + + Compare coeff. of x3: A = 2 Compare coeff. of x2: B = −3 Compare coeff. of x: q = −10 ( )( ) 3 2 22 3 29 30 3 10 2 3x x x x x x+ − + = + − − = ( )( )( )5 2 2 3x x x+ − − Alternative Solution: 23x− 2 3x x q++ 322 3 29 30x x x+ − + 322 6 2x x qx− − − 23 29 2 30x x qx− − − + 23 9 3x x q+ + + 20 2 3 30x qx q− − + + Since 2 3x x q++ is a factor of 322 3 29 30x x x+ − + , 20 2 3 30 0x qx q− − + + = −20 − 2q = 0 and 3q + 30 = 0  q = −10 ( )( ) 3 2 22 3 29 30 3 10 2 3x x x x x x+ − + = + − − = ( )( )( )5 2 2 3x x x+ − −

Qn Solution 7a Period 10 2 2 −= = 4 p 3 42 = + 2 = 8 Alternative Solution: Midpoint of AC 2 10 , 9.52 += ( )6 , 9.5= P 6 10 2 += = 8 7b amplitude 9.5 0.5 2 −= = 4.5 24b = b = 2 c 9.5 0.5 2 += = 5  equation of the curve is 4.5cos 52 xy=− + . 8a Since CQR is similar to CAB, 280 50 80 QR x −= 2550 8QR x =− m 22 550 8A x x =−  24 550 8xx=− m2 (shown)

Qn Solution 8b 35100 2 dA xxdx =− For stationary value, 0dA dx = 35100 0 2xx−= ( ) 25 40 02 xx −= 0 (rej.), 6.3245 or 6.3245 (rej.)x=−  stationary value of A = 1000 m2 (3 sf) 2 2 2 15100 2 dA xdx =− When 6.3245x= , 2 2 199.99 0dA dx =−  the stationary value of A is a maximum. 9i gradient of PQ = tan 135 = −1 (shown) 9ii Subst. x = 2, y = 0 into y = −x + c c = 2  coordinates of P is (0, 2). Let coordinate of Q be (a, b). ( )2, 2, 022 ab + = 22 a = and 2 02 b+ = a = 4, b = − 2  coordinates of Q is (4, −2).

Qn Solution 9iii Since PM ⊥ MR, gradient of MR = 1 0 132 r− =−− r = −5 9iv area of PQR 0 3 4 01 2 5 2 22 −= −− = ( )1 6 8 6 202 + + + = 20 units2 10a LHS sin coscos 1 sin1cos     =+ + sin cos 1 cos sin  =++ ( ) 22sin cos cos sin 1 cos     ++= + ( ) 1 cos sin 1 cos   += + 1 sin= = cosec  =RHS (proved) 10b tan 2 cot 2 4sec2 1   + =−+ cosec 2 = −4 sin 2 = 1 4− basic  = 14.477 2 = 194.477, 345.523  = 97.2, 172.8 (1 d.p)

Qn Solution 11a LHS = QRT = SQP (s in alt. segment) = QSR (alt. s; PQ // SR) = RHS (shown) 11b RTS = PTQ (vert. opp. s) RST = PQT (alt. s; PQ // SR)  RTS is similar to PTQ (AA similarity test) PQ PT RS RT= 1 3 PQ PT RS PR = 3PQ PT RS PR= PQ × PR = RS × 3 PT 12a 3 4 3 24 9 2 3x x x x +− =  6 2 4 3 2 32 3 2 2 3 x x x x =   2 823 9 xx= 812 9 x = 8ln12 ln 9x = x = −0.0474 (3sf) 12b ( ) ( )3 3log 6 3 log 2 0xx− − − = ( ) ( )3 3 3 log 2log 6 3 0 log 3 xx −− − = ( ) ( ) 2 33log 6 3 log 2xx− = − ( ) 2 6 3 2xx− = − 26 3 4 4x x x− = − + 2 20xx− − = x = 2 (rej.) or −1 13ai 2 270r = r = 9.2705 = 9.27 cm (3 sf)

Qn Solution 13aii A = 2r 2dA rdr = 10 2 (9.2705) dr dt= 0.172dr dt = cm/s (3sf)  rate of change of radius is 0.172 cm/s. 13b Subst. t = 2 , M = 8, k = 56 56 31M t= + ( )( ) ( ) 2 56 1 3 1 3dM tdt − = − + ( ) 2 168 31t =− + when t = 5, ( )( ) 2 168 3 5 1 dM dt =− + = 21 32− g/min  rate of decrease of M is 21 32 g/min.

Qn Solution 14a 14yx=+ ( ) ( ) 1 2 1 1 4 42 dy xdx − =+ 2 14 x = + gradient of tangent at A = 2 3 22 314 x = + 1 4 9x+= x = 2 When x = 2, y = 3 Coordinates of A is (2, 3). subst.x = 2, y = 3 into 3 2y x c=− + c = 6  coordinates of B is (0, 6). 14b Shaded area ( ) ( ) 12 2 0 1 6 3 2 1 42 x dx=  +  − +  ( ) 23 2 0 19 1 46 x= − + ( )( ) ( )( ) 33 22119 1 4 2 1 4 166 = − + − + = 14 3 units2 Alternative solution: Shaded area ( ) 23 1 11 6 3 242 y dy −= +  −  3 3 1 11 343 yy = − +  ( ) 31 1 1 3 3 1 34 3 3 = − − + + = 14 3 units2