RI Post+Prelim+Revision+P1%28LT1%29 Solutions

RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 4 Post-Prelim Revision Paper 1 (Source: Other JCs’ Prelim Qns) Source of Question: MI Prelim 9758/2024/01/Q1 1 (i) A quadratic curve passes through the point ( )1, 4−− and has its turning point at ( )2,5 . Find the equation of the curve. [4] (ii) Given instead that a cubic curve passes through the same point ( )1, 4−− and has the same turning point as stated in part (i). Explain whether it is possible to obtain a unique equation of the curve based on given information. [1] Solution: 1(i) METHOD 1 METHOD 2 - Preferred ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 Let . For point 1, 4 : 114 4 ---- (1) For point 2,5 : 225 4 2 5---- (2) d 2.d dAt turning point, when 2, 0. d 22 0 4 0 ---- (3) y ax bx c a bc abc a bc a bc y ax bx yx x ab ab = ++ −− − + −+= − −+= − + += + += = + = = += += ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 Let . Using turning point 2,5 : Clearly 2, 5 25 For point 1, 4 : 12 5 4 1 2 5 41 y ax h k hk y ax a a y x xx =−+ = = =−+ −− −− + = − =− = − − += + + From GC, 1, 4, 1a bc= −= = . Hence, equation of the curve is 2 4 1.yx x= −++ (ii) Not possible. For a general cubic equation 32y ax bx cx d= + ++ , there are 4 unknowns to solve for. But we can only form 3 equations from the given information. Therefore, we will obtain infinitely many solutions.

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 2 of 23 Source of Question: ACJC Prelim 9758/2024/02/Q1 2 The diagram shows part of the graph cos2 ,xyx= for 05 x≤≤ , which represents the path of a roller coaster. The horizontal distance travelled by the roller coaster is denoted by x units and its vertical distance travelled is denoted by y units. (a) Show that ( )cos 2d cos 2 2sin 2 lnd xyx x xxxx = − . [2] (b) At the point on the graph where x π= , find the rate at which the roller coaster is moving vertically when it is moving horizontally at a rate of 8 units per hour. [2] (c) Find the acute angle that the tangent to the graph where x π= makes with the horizontal. [1] Solution: 2(a) cos2 xyx= ( )ln cos 2 lny xx= ( )1 d cos 2 2sin 2 lnd yx xxyx x = +− ( )cos 2d cos 2 2sin 2 lnd xyx x xxxx = − (shown) (b) dd d dd d yy x tx t= × ( )cos 2 cos 2 2sin 2 ln 8x xx xxx =−×  ( )cos 2 cos 2 2sin 2 ln 8π ππ π ππ = −×  1 08 8 π π = −× = (c) Since d 1d x y x π= = , gradient of tangent is 1. Thus the angle that the tangent makes with the horizontal is or 454 π °.

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 3 of 23 Source of Question: NJC Prelim 9758/2024/01/Q3 3 Do not use a calculator to solve this question. (i) Solve the inequality 2 6 145 x xx − ≥+− . [3] (ii) Hence solve the inequality 2 2 6 145 xx xx − ≥+− . [2] Solution: 3(i) Method 1 ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) 2 222 22 22 2 2 6 145 56 4 5 4 5 , , 1 4 45 6 45 0 4 5 4 10 4 5 14 10 4 5 1 0 since 4 1 1 0 for all x xx x xx xx x x xx x xx xx x xx x xx xx − ≥+− ⇒ − +− ≥ +− ≠ − ≠ ⇒ +− − − +− ≥ ⇒ +− − −≥ ⇒ + − − −≥ ⇒ + −≤ − −≤< ∈  Method 2 ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) 2 222 22 22 2 2 6 145 6 45 45 45 6 45 0 4 5 4 10 4 5 14 10 4 5 1 0 since 4 1 1 0 for all x xx x xx xx xx x xx xx x xx x xx xx − ≥+− ⇒ − +− ≥ +− ⇒ +− − − +− ≥ ⇒ +− − −≥ ⇒ + − − −≥ ⇒ + −≤ − −≤< ∈  55Therefore, 1 since and 144 xxx− < < ≠− ≠ (ii) 2 2 6 145 xx xx − ≥+− Replace x with 1 y and we get 1 5 4−

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 4 of 23 2 2 2 22 2 11 66 61 11 45451145 y yy y y yyyy yyy    −−    −   ≥⇒ ≥⇒ ≥+− +−+−   5 51Therefore, 1, i.e. 1.44 y x−<< −<< So 51 1 0 or 0 14 4 or 15 xx xx −<< << <− > Source of Question: HCI Prelim 9758/2024/02/Q2 4 A sequence is defined by the recurrence relation 1 1 1 n n n uu u + += − , for 1n≥ . (a) State what happens to the sequence when 1 0u = . [1] It is now given that 1 2u = . (b) Find 2 345 6, , , and u uuu u . [2] (c) By observing the pattern in part (b), find 4 1 n r r u = ∑ in terms of n. [2] Solution: 4(a) 1 2 1 1 10 11 10 uu u + += = =−− 2 3 2 1 11 1 11 uu u + += =−− , which is undefined • The sequence ends at 2 1u = as the subsequent terms are undefined. OR • There are only two terms and terminate(ends) at the 2nd term.

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 5 of 23 (b) From GC, 2 3u =− , 3 1 2u =− , 4 1 3u = , 5 2u = , 6 3u =− 1 2u = 1 2 1 1 12 31 12 uu u + += = =−−− ( ) 2 3 2 1 13 1 1 13 2 uu u + −= = =−− −− 3 4 3 111 12 113 1 2 uu u −+= = =− −− 4 5 4 111 3 211 1 3 uu u ++= = =− − 5 6 5 1 12 31 12 uu u + += = =−−− (c) From observation, the sequence repeats with a period of 4. ( ) ( ) ( ) 4 1234 5 8 4 14 1 ... ... 11423 234 7 6 n r nn r u uuuu u u u u n n − = = +++ + ++ ++ +  = −+− + ×  =− ∑

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 6 of 23 Source of Question: ACJC Prelim 9758/2024/01/Q3 5 Elly started planking as an exercise and she continues the exercise every day to build her core muscles. If she meets her target duration, she increases the target duration of the exercise by an additional 4 seconds on the next day. On any day, she will stop her exercise once she meets her target duration for the day. However, Elly does not always meet her target. Each day when Elly misses her target, she decreases her target duration by 5% on the following day. On Day 1, Elly carries out 20 seconds of planking, and she hopes to reach her target of 2 minutes by the end of 30 days. (a) Assume that Elly met her targets for the first 11 days but missed her target duration from Day 12 to Day 15. Determine whether Elly will be able to reach her target of 2 minutes by the end of 30 days, if she met all her targets from Day 16 onwards. [3] Due to the difficulty level, Elly decides to restart the programme by increasing the target duration of the exercise by a% each day, regardless of whether she meets her target. (b) Find in terms of a, the total target duration Elly has completed by the end of 30 days if she carries out 20 seconds of planking on Day 1. [2] [You may assume that on any day, she will stop her exercise once she meets her target duration for that day.] (c) If the total target duration she has completed by the end of 30 days is at least 30 minutes, find, to the nearest integer, the least value of a. [1] Solution: 5(a) Let nU be the Elly’s targeted time of nth day ( )12 20 11 4 64U = += ( )13 64 0.95U = ( ) 41 15 64 0.95U − = ( ) 51 16 64 0.95U − = ( ) 4 17 64 0.95 4U = + ( ) ( )( ) 4 20 64 0.95 5 1 4U = +−  ( ) ( ) 4 30 64 0.95 14 4 108.1284U = += < 120 seconds Since the maximum time Elly can ach