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 Asymptotes x = -1 and x = 2. ( 1) 5ln 2 2=− + + +−y x x Students should note how the asymptotes are obtained: 1 0 1 2 0 2 + =  =− − =  = x x x x 4iii 5 3 21ln( 1) d 6.29 2 xxx x +− + + = − Some students aren’t aware that numerical area implies that the integral can be evaluated using the GC. 5i When 23 4yk= 2 2 23 4 k k x=− 22 1 4xk= 1 2xk= d 2d y xx =− d1 2( )d2 y kkx =− =− 23 ()42 ky k k x− =− − 25 4y k kx=− To find tangent, students should find 1) The coordinates of the point ( , ) and 2) The gradient in terms of k. -1 2 x y

5ii Area of 231 3 5 9( )( )2 4 4 2 32 k k k kAFB = − = Area AFD = 2 22() k k k x dx− = 2 3 2[] 3− k kxkx 3 3 3 3 3 5()3 2 24 24 k k k kk= − − − = Area bounded by the curve C, the tangent to the curve at point A and the x-axis = 3 3 35 9 7 24 32 96 k k k−= Most students identified the correct region but made mistakes in the limits of the integrals. Students should firstly compute the x-intercepts of the curve and the tangent. 6i Method 1 0.2 2( 4) 9tPe=− − + 0.2 0.2d 2( 4)(0.2 )d =− − ttP eet For stationary points d 0d P t = 0.2 0.2d 2( 4)(0.2 ) 0d =− − = ttP eet 0.2( 4) 0−=te 0.2 4te = 0.2 ln 4t = 5ln 4t = Method 2 0.2 2( 4) 9tPe=− − + 0.4 0.2( 16 8 ) 9ttee=− + − + 0.4 0.2 78ttee=− − + 0.4 0.2d 0.4 1.6d ttP eet =− + For stationary points d 0d P t = Many students solved d 0d P t = using the GC, instead of doing differentiation as stated in the question. O C x y A k B F D

0.4 0.20.4 1.6ttee = 0.2 4te = 0.2 ln 4t = 5ln 4t = To justify whether it is a maximum or minimum point Method 1 2 0.4 0.2 2 d 0.4 1.6d ttP eet =− + When 5ln 4t = , 2 2 d 1.28d P t =− At 5ln 4t = , the stationary point is a maximum point Method 2 5ln 4t = =6.93 to 3 significant figures t 6.5 5ln4 7 dp dt 0.485 0 -0.0895 / \ At 5ln 4t = , the stationary point is a maximum point. For all maxima/minima questions, students are expected to determine the nature of the point unless otherwise stated in the question. 6ii Students have to indicate the endpoints of all curve sketching questions. 6iii 0.2 2( 4) 9 dtet− − + = 0.4 0.2 8 7) dtte e t= − + − 0.4 0.22.5 40 7tte e t C=− + − + 9 0.2 2 0 ( 4) 9 dtet− − + = 0.4 0.2 9 0[ 2.5 40 7 ]tte e t− + − 0.4(9) 0.2(9)2.5 40 7(9) 2.5 40ee=− + − + − = 49.9903= 50 (to the nearest integer) Or using GC 9 0.2 2 0 ( 4) 9 dtet− − + = 49.9903= 50 (to the nearest integer) Over a span of 9 months Mr Lim made a total profit of $50000. Most students tried to expand but made careless mistakes. Students have to note that: ( ) ( ) ( ) ( ) ( ) 20.2 0.2 0.2 0.2 0.2 0.4 . + = == t t t t t t e e e ee Some students aren’t aware that they can evaluate the integral using GC to get its numerical value to check their answers. 0 P t 0 9 (9,4.80)

6iv 200 375( 20 75 )3 2 5At t= − + + + 2 2 d 200 375(2)( 20 )d 3 (2 5) 200 750( 20 )3 (2 5) A tt t = − − + = − − + When t = 3, 2 d 200 750( 20 )d 3 11 A t = − − d 1746d A t =− (to the nearest integer) When t = 3, the remaining budget is decreasing at the rate of $1746 per month Many students evaluated the derivative using GC despite the question asking them to use differentiation. 7a ~ B(30, )100 18 30 18100 60 pX np p p =  = =  P( 20) 1 ( 19) 0.29147 0.291 (3 sf)X P X = −  =  Students have to note that the probability was given as p%=p/100. Many wrote p incorrectly as 0.6. Hence, they were penalized. Students to note that they have to show this step before they embarked on their GC calculations: P( 20) 1 ( 19) = − X P X 7b Let be the random variable representing the number of elderly patients with diabetes. ~ B(40, )100 P(9 20) 0.25 P( 20) P( 8) 0.25 Using GC, 17.367 or 56.512 Since 50, 17.4 (3 s.f) Y kY Y YY kk k k   =  −  = ==  = Many students interpreted the question wrongly. As GC only supports binompdf or binomcdf, students need to show this step: P(9 20) 0.25 P( 20) P( 8) 0.25   =  −  = Y YY

8a P( )P( | ) 0.7 P( ) 17P( ) 0.7 3 30 P( ) P( ) P( ) P( ) 3 1 7 1P( ) 5 3 30 2 ABAB B AB A B A B A B A ==  = =   = + −   = − + = 8b Method 1 Required answer P( ) P( ) P( ) P( ) 1 7 1 7 2 30 3 30 11 30 A A B B A B= −  + −  = − + − = Method 2 Required answer P( ) P( ) 37 5 30 11 30 A B A B=  −  =− = Students are encouraged to use the venn diagram to indicate the required region before they start their calculations. There is no need to apply any new formulas to do this question. 8c Method 1 7P( | ) 10 1P( ) 2 Not independent. AB A = = Method 2 7P( ) 30 1P( ).P( ) 6 Not independent. AB AB = = Students have to write down the value of both probabilities and make comparison. A B

9a But