RVHS 2026 9758-02 Prelims MS

1 ©RIVER VALLEY HIGH SCHOOL 9758/02/2026 Solutions and Comments for H2 Mathematics P2 2026 Section A: Pure Mathematics [40 marks] 1 The curve C has parametric equations 4xt t=+ , 2y t=− , 02 t . (a) Show 2 d2 d 4 y x t = − . Hence find the value of d d y x as 0t → . [3] (b) The equation of an asymptote of C is of the form y kx= , where k is a real constant. Find the value of k. [1] (c) Sketch C, indicating clearly the coordinates of the endpoint of the curve and the equations of any asymptotes. [2] 1 Solution [6] Parametric curve (a) 22 42, d 4 d 21 , dd x t y tt xy tt tt = + = − = − = 2 2 2 2 2 2 d d d d d d 2 41 2 4 2 4 y y x x t t t t t t t t = = − = = − − As 0t → , d 2 1 d 0 4 2 y x → = −− Generally well done. More working should be presented in trying to arrive at 2 2 4t − since it is a ‘shown question’. (b) From (a), since d1 d2 y x →− (when 0,tx→→  ), then the asymptote of C should be given by 1 ,2y x k= −   . Hence, 1 2k =− . Majority of the students who obtained 1 2k =− did not provide the correct reason.

2 ©RIVER VALLEY HIGH SCHOOL 9758/02/2026 (c) Poorly attempted. Common mistakes include no open circle at (4, –1), solid line drawn for the asymptote. C x y O

3 ©RIVER VALLEY HIGH SCHOOL 9758/02/2026 2 A start up company is launching a project to capture airborne dust particles using two different filtration models A and B, with its efficiency measured over time. Model A is programmed such that the amount of dust particles captured in its first month of operation is 8 units and increases by a fixed amount of d units each month. In its first 12 months of operation, Model A captured a total of 360 units of dust particles. (a) Find the amount of dust particles captured by Model A in its 20th month of operation . [3] Model B uses a depleting chemical catalyst to capture dust particles. The amount of dust particles captured in the first month is a units. For each subsequent month, the amount of dust particles captured is r times that of the previous month. Over its lifespan, Model B can capture a theoretical maximum of 1000 units of dust particles. In the first two months, Model B captured a total of 640 units of dust particles. (b) Find the values of a and r. [4] The startup company will decommission Model B once its total amount of dust particles captured reaches at least 99% of its maximum capacity over its lifespan. (c) Find the minimum number of months that Model B will be deployed before it is decommissioned. [3] 2 Solution [10] AP/GP (a) ( )( ) ( )( ) 12 212 12360 2 8 112 360 96 66 66 264 4 nS a n d d d d d = + − =+ =+ = = ( ) 20 19 8 19 4 84 u a d=+ =+ = Model A captured 84 units of dust particles in its 20th month of operation. Generally, well done. Though there were occasional mistakes. Common mistakes: - Wrong formula usage: like ( )( )12 12 112 8Sd =+ . - Expanding incorrectly like: ( ) 16 16 11 196dd+ = + . - Interpreting that we require the total particles captured, rather than just the amount in the 29th month. - Interpreting 360 as the amount collected in the 12th month only. (b) ( ) ( )2 10001 1000 1 ---(1) 1 640 ---(2) aS r ar S a ar a r  == − =− = + = + = Substitute (1) into (2): Generally well -done. There were occasional mistakes too. Common mistakes: - Unable to interpret the “theoretical maximum” information, resulting in incorrect equations like ( )1 10001 nar r − =− .

4 ©RIVER VALLEY HIGH SCHOOL 9758/02/2026 ( ) ( )( ) 2 2 1 640 1000 1 1 640 6401 1000 0.36 0.6 0.6 (rej 0, amount of dust particles can not be ve) ar rr r r r rr += − + = −= = = =  − ( )1000 1 0.6 400 a a =− = - Writing infinite geometric sum incorrectly, like 1S ar  = − . - Writing ( ) 2 1 1raa ar r −+= − instead of ( ) 2 1 1a r r − − . - Accepting 3 5r =− because it is ‘decreasing’! (But this means there is negative amount of dust! In fact , 01 r would implying the amount of dust is decreasing in this context.) (c) Let n be the number of months that Model B will be deployed before it is decommissioned. ( ) 0.99 1 0.6 0.991 0.6 1 0.6 1 0.6 0.99 n n n SS a a  −  −−  − Though this question was well attempted. Many students did not arrive at the correct final answer. Method 1: 0.6 0.01 ln 0.6 ln 0.01 ln 0.01 ln 0.6 9.015 n n n n     Method 2: By GC, when 9n = , 1 0.6 0.9899 0.99n− =  when 10n = , 1 0.6 0.994 0.99n− =  It was heartening to see that students realise ln 0.6 0 and changed the inequality sign. Reminder: Students cannot simply state ‘By GC’ without showing further working! Please reminder to state the table or sketch the graph from the GC! Model B will be deployed for a minimum of 10 months before it is decommissioned. There were a significant number of students who arrived at 9 months instead due to incorrect rounding or misreading from their table.

5 ©RIVER VALLEY HIGH SCHOOL 9758/02/2026 3 Solutions [9] Complex Numbers (a)(i) iz represents the point C, hence iwz= . ( )i 1 iz z z+ = + represents the point B, hence ( )1ivz=+ . When attempted, this was well done. Students are able to identify w more successfully as compared to v. (a)(ii) 22 18 3 2v OB OA OC=+ == = . ( ) ( ) .a π π 11πarg 4 7 8rg 2BOA zv  + + == = Students who saw most success appreciated the geometry of the problem. For example, they realise v was just the length of the diagonal of the square, i.e. OB. A minority of students tried to use modulus 3 The point A on the Argand diagram represent s the complex number z, where 3z = and ( ) πarg 7z = . Points B and C are on the diagram such that OABC is a square. (a) (i) The points B and C represent complex numbers v and w respectively. Find v and w in terms of z. [2] (ii) State the exact values of v and ( )arg v . [2] Points D and E represent the complex numbers i*z− and ( )2 Re z respectively. (b) (i) On the same Argand diagram in the Printed Answer Booklet, plot the points D and E. Show clearly the geometrical relationship between the points C and D, and between points A and E. [2] (ii) Show that the complex number representing DB  is i i *zzz++ . [1] (iii) Show that BD is perpendicular to CE. [2] O A B C Re Im ×

6 ©RIVER VALLEY HIGH SCHOOL 9758/02/2026 properties, but were unable to apply argument properties. These are not in syllabus anymor e, so students are discouraged to apply them as it might blind them from another solution. A common mistake was deriving w and ( )arg w instead. Several students also had misconceptions about the modulus brackets, usually splitting the sum incorrectly like i i 3 3iz z z z+ = + = + . (b)(i) ( )( )i * i *zz− = − . Consider the point representing *z , then rotate it 90 clockwise about O. This question proved challenging to students. Although some students marked the points correctly, many attempts did not show the required geometrical relations between the points. This meant indicating equal lengths, angles, special angles, or parallel lines, etc … Students had far more success in labelling E as compared to D. (b)(i) Alternative ( ) ( ) ( )i * * i * * * iz z z− = − = , so i*z− and iz are conjugates. Students did notice that D was the reflection of C. However, they did not managed to capture the full geometrical description of the reflection. They labelled lengths OC and OD are equal, but that is not sufficient to indicate a reflection. It is important that CD is perp