YIJC 9758 2024 Promo Solutions

1 ©YIJC 9758/01/JC1PE/2024 Solutions for 2024 JC1 H2 Mathematics Promotional Examination 1 (a) Let x, y, z be the original price ($) of a ticket for ‘Senior Citizen’, ‘Adult’ and ‘Child’ respectively. 5 12 6 2440 --- (1)x y z+ + = ( )15 0.8 10 5 2660 12 10 5 2660 --- (2) x y z x y z + + = + + = 8 10 12 2560 2 8 16 2560 --- (3) yxy xy + + =  += Using GC, x = 80, y = 120 and z = 100. The original price of a ticket for ‘Senior Citizen’, ‘Adult’ and ‘Children’ is $ 80, $ 120 and $ 100 respectively. (b) 2(80) 2(120) 3(100) 700++= Lee family spends $700. 2 ( ) 3 2 1 22 11 34 4 n rn nn r n r n rn rn =+ = + = + + =+   ( )( ) 3 2 1 3 1 4 nn rr r r n n == = − + ( ) ( ) ( ) 2 2 2 2224 2 1 144 n nn n n  = + − + + ( ) ( ) 22 2224 2 1 1n n n n n= + − + + ( ) ( )2 2 24 4 4 1 2 1 1n n n n n= + + − + + + ( )2215 14 4n n n= + +

2 ©YIJC 9758/01/JC1PE/2024 3 (a) ( ) 22 2 3 1 2x x x− + = − + Since ( ) 2 10x − for all x , ( ) 2 1 2 0 2x −+   . 2 23xx − + is always positive for all values of x . (b) ( )( ) 2 2 2 3 1 02 x x x xx − + − −− Since 2 23xx −+ is always positive, ( ) 2 1 02 x xx − −− . ( ) ( ) ( )( ) 2 1 02 1 012 x xx x xx − −− − +− 1 1 or 2xx −    (c) Replace x with x− in ( )( ) 2 2 2 3 1 02 x x x xx − + − −− 1 1 or 2 0 1 or 4 xx xx  −       1 - - + -

3 ©YIJC 9758/01/JC1PE/2024 4 (a) ( )ln cos lny x x= ( ) ( )1 d 1 sin ln cosd y x x xy x x = − + ( ) ( )d1 sin ln cosd y y x x xxx = − + ( )cos cos sin lnx xx x x x =−  OR ( )ln cos lny x x= (cos )ln (cos )lnd1 [( sin )(ln ) (cos )( )] e ed xx xx y y x x xxx = = − + (b) (i) 22dd 2 2 0dd yyx xy x y yxx    + − + =         ( )22 d22 d yx xy y xy x− = − 2 2 d2 d 2 y y xy x x xy −= − (ii) 2 2 d2 1d 2 y y xy x x xy −== − 22 22y xy x xy− = − 22yx = y x=  yx= 33 6 06 xx −= = Thus, no solution. yx= −  33 3 3 6 3 3 xx x x − − = =− =− 3 3y = Thus ( )333, 3− .

4 ©YIJC 9758/01/JC1PE/2024 5 (a)   213 sin 6 0 22 0 30xy x +  = 233 300 2xy x+= 233 300 2xy x=− 100 23 xy x=− 21 sin 602V x y= 23 100 4 23 xx x  =−   3125 3 8xx=− (Shown) (b) 2d3 25 3 0d8 V xx = − = 2 200 3 x = 4 200 3 x =  Since 0x  , 4 200 10.746 (5 sf) 3 x == . 2 24 d 3 3 200 04 43d V x x = − = −  Hence V is maximum when 4 200 3 x = . Alternatively: x 10.7 4 200 3 10.8 d d V x 0.368 0 −0.439 Slope / − \ Maximum V 3 3 44 200 1 20025 3 310 cm (3 sf)833    = − =        

5 ©YIJC 9758/01/JC1PE/2024 6 (a) (b)(i) ( )d 21d x att =− ( )d 4d y att =− d4 d 2 1 yt xt −= − At 1t =− , d4 d3 y x = − Gradient of normal is 3 4 ( ) ( )3 24y a x a− − = − 35 42y x a=− (b)(ii) ( ) ( )22 3512 42a t a t t a− = − − 224 8 3 3 10t t t− = − − 211 3 14 0tt − − = ( )( )11 14 1 0tt− + = 14 11t = or 1t =− (rej. original point) 214 14 42 11 11 121x a a = − = , 214 27112 11 121y a a  = − = −  42 271,121 121aa − y x O

6 ©YIJC 9758/01/JC1PE/2024 7 (a) ( ) ( ) ( ) ( )1 Let f e4 e4 ln 4 f ln 4 x x yx y y xy xx− = =+ =− =− =− ( (  1 1 ff ff D R 4, 4 e R D ,1 − − = = + = = − (b) 0.5y = cuts the graph of g twice (more than once), g is not one-one, 1g− does not exist. (c)   ( gf gf R 1,1 , D ,1 R D , hence the composite function fg exis ts. = − = −  (d) ( ) ( ) sin fg f sin e4 x xx = =+ ( )fg gDD π, π= = − (e) To find fgR : Let  gR 1,1=− be new domain of f : 1 fgR e 4, e 4−= + + 8 (a) Translation by 2 units in the negative x-direction. y x g y = 0.5  − y x y = 4

7 ©YIJC 9758/01/JC1PE/2024 Scaling by a scale factor of 1 3 , parallel to the x-axis. Translation by 1 unit in the positive y-direction. OR Scaling by a scale factor of 1 3 , parallel to the x-axis. Translation by 2 3 units in the negative x-direction. Translation by 1 unit in the positive y-direction. (b) (i) (b) (ii) x y O x = −3 x = 3 y = 2 (−a, b) (a, b) (0, c) y x O (a, 0) y = 0 x = − x = 

8 ©YIJC 9758/01/JC1PE/2024 9 (a) 2 1 cos6sin 3 d d 2 11 sin 62 12 C     −= = − +  9 (b) ( ) ( ) ( ) ( ) ( ) 22 22 2 22 2 2 2 2 1ln d ln 2ln d22 1 ln ln d2 11 ln ln d2 2 2 11 ln ln2 2 4 xxx x x x x x x x x x x x xxx x x x x xx x x x C = −  =− = − −   = − + +     9 (c) ( ) ( ) ( ) ( ) ( ) 3 12 3 23 2 3 11 31 3 2 1 11 33 22 1 6 2 1 621 211 16 2 1 2(3) 1 2(1) 13 1 53 1 dd 3 x xxx x x x − =− −  −=    = − − − =−    9 (d) d ed xu ux == When 0x = , 1u = When ln 3x = , 3u = ( ) ( ) ln 3 3 33 22 01 3 2 1 31 1 11 e1 e 1 1 11 1 = tan 3 tan 3 1 tan 1 31 34 31 12 dd d x x xu u u uu u uu   − −− =++ =− +  − = − − − = − − + = − −   10 (a) (i) 12 423 xz y +−= = − =

9 ©YIJC 9758/01/JC1PE/2024 12 4 23 x y z    = − + =+ =− 12 4 1 , 23 r  −        = +        −    (ii) Plane 1p : 1 4 5 0 5 4 x z r  − =   = − Let  be the acute angle between 1l and 1p . sin = 21 10 34 21 10 34               −−                  −−    2 12 14 17 += 14 14 17 = 65.2 = (iii) Substituting line 1l equation into 1p equation: 1 2 1 4 1 0 5 2 3 4  −           +  =           −−      1 2 1 4 0 5 2 3 4    −+       +  =      −−    1 2 8 12 5− + − + = 1 = 1 2 1 Required position vector 4 (1) 1 5 2 3 1 −           = + =          −−      (b) (i) Let F be foot of perpendicular from R to 2l . R(−  ) (  −) F

10 ©YIJC 9758/01/JC1PE/2024 1 1 2 2 2 0 2 1 3 AR −−           =−=          −      Method 1: Direct formula RF = 22 03 31 2 3 1 −       −           − = 9 8 6 181 14 14    = = 3.60 units (3 s.f.) Method 2: Use length of projection AF = 22 03 31 2 3 1 −       −           − = 1 14 AR = 222 3 13+= RF = 22 13 1 14   − 181 14= =3.60 units (3 s.f.) Method 3 (Foot of perpendicular): (Not recommended) r 12 2 3 fo some 11 OF         + −       −   =  22 03 31 RF OF OR         = − = + −       −    2 2 2 2 3 0 3 3 0 1 3 1 1 4 4 9 3 0 1 14 RF                        − = + − − =              −        +  + − + =  −  = 2 2 21 4 26 3 26 1 3 3 434 141RF   − = = + +  = 3.60 units (3 s.f.)