ACJC 9758 2023 Prelim P1 Solution

1 2023 H2 Maths Prelim Paper 1 Marking Scheme Qn Solutions 1 ( ) 32f x x ax bx c=+ ++ 32 1 abc− =+++ ( ) 2f' 3 2x x ax b=++ 032 ab= ++ ( ) 1 fy x= has a vertical asymptote at 5x= implies that ( )fyx= has an x-intercept at 5x= . 0 125 25 5 a bc= + ++ 33 23 25 5 125 abc ab a bc ++= −  += −  + += − By GC, ( ) 32f 5 7 35xx x x=− +− . [ 5, 7, 35a bc= −== − ] 2 * 2 4i ----(1) 2 3i ----(2) wz zw + = −+ += From (1): 2 4i *wz= −+ − Substituting into (2): ( )2 3i 2 4i *zz+= −+ − 3i * 14 6izz+ = −− Let iza b= + , ( ) ( ) i 3i 3 14 6i 3 3 i 14 6i ab a b a b ab ++ += −− +++= − − Comparing real and imaginary parts: 3 14 ---(3) 3 6 ---(4) ab ab += − += − Solving simultaneously with (4) 3(3)− , we have 8 36 4.5bb− = ⇒= − Hence 0.5a=− , and 0.5 4.5iz= −− 2 4i *wz= −+ − : 2 4i 0.5 4.5i 1.5 0.5iw= −+ + − = − −

2 3(i) Method 1: Algebraic Manipulation 1 0 d 1 x x x+∫ 1 0 ( 1) 1 d 11 x x xx += − ++∫ 11 00 11d d 1 xx x x = +− +∫∫ ( ) ( ) 131 22 0 2 1213 xx= +− + ( ) ( ) ( ) ( ) 3 1 31 2 2 22 222 22 1 2133   =−−−     42 222 233   = − −−     42 233 2(2 2)3 = − = − Method 2: Integration by Parts Let ux= d1 d 1 v x x = + d 1d u x = 21vx= + 1 0 d 1 x x x+∫ 1 1 00 2 1 2 1dxx x x= +− + ∫ ( ) 13 21 0 0 212 12 3 xxx  += +−  ( ) ( ) 33 22 4220 2 1 3 = −− −   42 233= − ( ) 2 223= −

3 Method 3: Using Substitution (Change of Variable) Let 1ux= + . Then d 1d u x = . When 0, 1xu= = When 1, 2xu= = 1 0 d 1 x x x+∫ 2 1 1du u u −=∫ 2 1 1 du u uu = −∫ 112 22 1 duuu − = −∫ 231 22 1 2 23 uu= −  2 221 2 213   = −− −   42 233 2(2 2)3 = − = − 3(ii) Area of 1st rectangle: 2 11 11 1 1 1 1 1 1f 11 1 11 nn nn n n n n n nn n nn n      = = = =  +++  + Area of 2nd rectangle: 2 22 121 1 1 2 1 2f 22 2 21 nn nn n n n n n nn n nn n      = = = =  +++  + Area of nth rectangle: 2 11 1 1 1 21 nn n nnnnfnnn n n n nn nn n n n nn n      = = = =  ++  +

4 Total area of all the rectangles: 11 2 3fff f n n nnn n    = +++ +        11 2 3 123 2 n nn n n n n = + + ++ +++  All the rectangles above the curve enclosed the region bounded by the curve and the x-axis and include a small portion above the curve. Thus, the total area of all the rectangles is more than the area bounded by the curve and x-axis. Hence, ( ) 11 2 3 2 223123 2 n nn n n n n  + + ++ > − +++  . 4(i) ⋅pq is the length of projection of q onto p. OR ⋅pq is the length of projection of q onto the line with direction vector p. 4(ii) ( )OM = ⋅pqp   (vector projection, since M lies on OP.) 2OM∴= p   (shown) OR Since M lies on line OP, for some OM λλ= p   . Since M is the foot of the perpendicular from Q to the line OP, 0QM OP⋅=     ( ) ( ) 2 0 0 2 2(as is a unit vector) OM OQ OP λ λ λ λ − ⋅= − ⋅= ⋅=⋅ = = pqp pp qp p p       2OM∴= p   (shown) OR Since 2⋅=pq and ⋅pq is the length of projection of q onto p, M lies on OP produced, and 2OM OP= . 2OM∴= p   (shown)

5 4(iii) 2OM = p   ( )OR k= = −r pq   , OQ= q   2QM = −pq   ( ) ( )1QR k k k= − − = +−−pq q p q   ( ) ( )( ) ( ) ( ) ( ) Area of triangle 1 2 1 212 1 2 2 (since )2 1 222 1 22 2 2 RQM QM QR kk kk kk k k = × = − × +−− = −− ×− × ×=×= = −− ×+ × = −− × += × pq p q pq qp pp qq 0 pq pq pq pq     2 2 ka += When 2k =− , points R, Q and M are collinear. 5(i) 22d e 2ed xx xx −− =− ( ) ( ) 22 22 22 22 2 32 2 2 2 2 ed e d 11e e (2 ) d22 1 e ed2 11 ee22 1 e12 xx xx xx xx x x x xx x x xx x xx xc xc −− −− −− −− − =  =− −−   = −+ = − −+ = − ++ ⌠⌡ ∫∫ ∫

6 5(ii) 2 e xzy −= ( ) 2 22 2 2 2 dd edd d2e e d d 2edd d e2dd e x xx x x x z yxx yxy x z xyyz x xyxx − −− − − = = −+ + = = + For 3d 2d y xy xx−= , ( ) 2 2 2 22 2 22 22 3 3 32 2 22 de 22d d ed 11ed e e 22 11e ee 22 11 1 e 1e22 2 x x x xx x xx xx z xy xy xx z xx zx x x c yx c yx c x c − − −− − −−  + −= = = = − −+ = − −+ = − −+ = − ++ ∫ 6(i) 4 33 4 0z kz k z k−+ −= ( )( ) 22zkzk z k− +=− ( )( ) 4 33 4 222zk zk z k zkzc z d− + −= − ++ Comparing coefficients of 04 2 2:z k kd d k−= − ⇒= 13 2:z k kc c k= − ⇒= − ( )( ) 4 33 4 222 2zk zk z k zkzk z k∴− + −= − −+ For 22z kz k−+ , 22 43 i2 22 kk kk kz ±−= = ± The other two roots of the equation are 3 3 i22 kkz = + (shown) and 4 3 i22 kkz = − (as ( )3Im 0z > and ( )4Im 0z < ).

7 6(ii) 22 3 3 22 kkzk = +=    ( ) ( ) 11 3 3 /2 πarg tan tan 3/2 3 k z k −−= = = 6(iii) ( ) ( )3 3arg arg arg 1 i1i π 3π 34 5π 12 n z nz n n  = − −+ −+ = − =− For 3 1i n z −+ to be purely imaginary, ( ) ( ) 5ππ 2 1 , where is an integer12 2 6 215 nk k nk −= + = −+ For the two smallest positive values of n, consider ( ) ( ) 63: 6 1 6 5 68 : 16 1 185 kn kn = − = − −+ = = − = −−+= 7(i) The graphs of ( )fyx= and ( ) 1fyx −= are the reflection s of each other about the line yx= . y (0,3) (-4,0) y = 4 x y = x x = 4 (3,0) (0,-4) ( ) 1fyx −=

8 7(ii) Rg =[ 4, )−∞ From graph, Rfg =[0,4) 7(iii) ( ) ( ) 1g f0 4a −= =− 7(iv) ( ) 2 2 2 -1 Let 6 5 65 34 34 3 4 (rej 4 as 4) h() 3 4 yx x yx x yx xy x y yx xx =++ =++ = +− += ± + =− − + + + <− = −− + ( )1hD 3,− =−∞ 8(i) At π 2t = , 2 π ππsin 12 22 πsin 12 x y = −= − = = π 1, 12A − 8(ii) Area = ( ) π 12 0 1 πd 11 22 Cyx − −− ∫ ( ) ( )( ) π π 21 22 0 0 π 22 0 dd sin d d sin 1 cos d C xyx t t t t tt − =  = − ⌠⌡∫ ∫ π 222 0 ππ 22 3 0 0 ππ 22 3 00 sin sin cos d 1 cos 2 1 d sin23 11 1 sin 2 sin22 3 1 π 1 π1 22 3 4 3 t t tt t tt tt t = − − = −   = −−     = −=− ⌠⌡ ∫ Therefore area = π1 π1 1 43 42 6   −−−=    

9 8(iii) ( ) 2 1 1 sin , sin sin sin 0 πsin for 0 2 sin x t ty t ty t ty t x yy − − = −= = ≥ = ≤≤ ⇒= −  8(iv) V olume = ( ) ( ) 2 1 2 0 1 ππ 11π d32 Cxy −− ∫ (the volume of region bounded by the line and curve gives a cone) ( ) 2 1 21 0 1 ππ 1 π sin d32 0.25261 0.253 (3 s.f.) y yy−= −− − = ≈ ⌠ ⌡ 9(a) Let f () g ()Dx x= − . ( )dd f () g () f () g ()dd D xx x xxx ′′= −= − At xc= , distance is maximum, i.e. d 0 f () g () 0 f () g ()d xc D cc c cx = ′′ ′′= ⇒−= ⇒= (shown) From the diagram, at xc= , f( )yx= is concave downwards, i.e. f() 0c′′ < g( )yx= is concave upwards, i.e. g() 0c′′ > Therefore at xc= , 2 2 d f() g() 0d D ccx ′′ ′′=−< . Hence the distance is a maximum. 9(b)(i) O x y 50 72

10 9(b)(ii ) From (a), we know that the maximum will occur when A ' () B ' ()cc= , and c will satisfy the equation A ' () B ' ()xx= . ( ) ( )( ) ( ) 32 3 2 dd 0.16 12 5000log 9 10000dd 50000.48 24 9 ln3 xx xxx xx x − + = +− − += + Therefore 50000.48, 24, ln3P QR= −= = . Using GC to solve the equation gives ( ) ( ) ( ) ( ) 11.906 5 s.f. 11.9 3 s.f. 46.296 5 s.f. 46.3 3 s.f. x x = = = = 9(b)(ii i) ( ) ( ) ( ) ( ) A 11.906 B 11.906 2404.8 A 46.296 B 46.296 1580.9 −= − −= Therefore the furthest distance between them is 2404.8m (or 2400m) and Ben is in front of Andy. 10(i) ( ) ( ) ( ) 2 2 2 4 6 11 22 2 2 4 6 11 4 6 2 11 2 0 xxy x yx x x x yx y − −−= + += − −− ++ + + = ( ) ( )( ) ( )( ) 2 2 2 2 D0 6 2 4 4 11 2 0 36 24 4 176 32 0 4 8 140 0 2 35 0 7 50 57 yy yy y yy yy yy y < + − +< ++−−< −− < −−< − +< −< <