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 = o