2024 ACJC H1 Prelim (solution with marker's report)

ACJC 2024 H1 Preliminary Exam 1 Let selling price of an electric car be $ e Let selling price of a gas-powered car be $ g Let selling price of a hybrid car be $ h. 4e 3g 172800 ...... (1) h 1.2g ...... (2) 3h 2g 2e 716880 ..... (3) −= = + + = By GC, Selling price of an electric car is $109 800. Some interpreted eqn (2) wrongly. Some students attempted to solve algebraically not knowing that GC can solve simultaneous equations involving 3 eqns and 3 unknowns. 2 Discriminant =D 22( 6) 4( 36)(0.5)kk= − − + 22 12 36 2( 36)k k k= − + − + 22 12 36 2 72k k k= − + − − 2( 12 36)kk=− + + 2( 6)k=− + Since coefficient of 2x = 2 36k + > 0 the curve has a minimum point. 22( 36) ( 6) 0.5 0+ + − + k x k x if 0D 2 2 ( 6) 0 ( 6) 0 − +   +  k k kR For 22( 36)ln[ ] ( 6) 0.5)+ + − +k x k x to be defined, 22( 36) ( 6) 00.5)+ + + −k x k x for all real x. Hence, D < 0 2 2 ( 6) 0 ( 6) 0 − +   +  k k ,6k R k − Most students considered the discriminant but did not proceed to solve the correct inequality. Those who arrived at this step: 2( 6) 0+k , didn’t conclude correctly. Students should note that since this inequality is always true for any values of k, hence k is the set of real numbers. Students should note that the calculation will result in 2( 6) 0+k and hence k can be any real numbers except - 6. 3a. ( ) 4 71 22 91 22 25 4 d = 25 4 d 50 8 c9 x x x x x x xx − + + = + +   Students should attempt to simply their answer and not leave it as: 91 2225 4 91 22 ++xx c 3b. 4 3 13 444 x xx − =− +−− Long division is an assumed knowledge for H1 students.

4 3 13 dx 4 dx44 4 13ln 4 x xx x x c − = − +−− =− − − +  4i 5ln( 1) 2 2yx x=− + + + − 2 d 1 5 d 1 ( 2) y x x x=− −+− To show no stationary points Method 1 For ln( 1)yx=− + to be defined 10x+ Hence 1 01x− + for all real x Also 2 5 0( 2)x− − for all real x 2 d 1 5 0d 1 ( 2) y x x x =− − +− for all real x. Hence curve C has no stationary points. Method 2 2 d 1 5 d 1 ( 2) y x x x=− −+− 2 2 ( 2) 5( 1) ( 1)( 2) xx xx − − − += +− 2 2 ( 4 4 ) 5 5 ( 1)( 2) x x x xx − + − − −= +− 2 2 ( 9) ( 1)( 2) xx xx − + += +− For stationary points d 0d y x = 2 90xx+ + = Discriminant = (-1)2-4(1)(9) = -35 <0 No solution. Hence curve C has no stationary points. Method 3 2 d 1 5 d 1 ( 2) y x x x=− −+− 2 2 ( 2) 5( 1) ( 1)( 2) xx xx − − − += +− Many students attempted to find d d y x and didn’t make any progress thereafter. They should show d 0.d y x

2 2 ( 4 4 ) 5 5 ( 1)( 2) x x x xx − + − − −= +− 2 2 ( 9) ( 1)( 2) xx xx − + += +− = 22 351 1 1 2 4 2 4 22 (( ) 9 ) (( ) ) ( 1)( 2) ( 1)( 2) xx x x x x − + + − − + +== + − + − <0 for all real x. Hence curve C has no stationary points 4ii When y = 0 , x = -0.222 or x = 11.5 Asymptote