ASRJC 2026 Prelim P2 Ans

[Turn Over ANDERSON SERANGOON JUNIOR COLLEGE H2 MATHEMATICS JC2 Prelim Paper 2 (100 marks) 9758/02 31 Aug 2026 3 hours Additional Materials: Printed Answer Booklet List of Formulae (MF 27) CANDIDATE NAME CLASS / READ THESE INSTRUCTIONS FIRST Write your name and class in the boxes on the Printed Answer Booklet. Please use a dark blue or black pen and write clearly . An HB pencil may be used for graphs and diagrams only. Do not use staples, paper clips, glue, or correction fluid. Answer all the questions. Write your answers on the Printed Answer Booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you must p resent the mathematical steps using mathematical notations and not calculator commands. You must show all necessary workings clearly. All work must be handed in at the end of the examination. If you have used any additional paper, please insert them inside the Printed Answer Booklet. The number of marks is given in brackets [ ] at the end of each question or part question. Question number Marks 1 2 3 4 5 6 7 8 9 10 Total This document consists of 13 printed pages and 3 blank pages.

2 Section A: Pure Mathematics [40 marks] 1 Solution Since w∈  z⇒ is real. Also 0w ≥ 5 0522wz − −⇒= ≥ Hence 52z −≥ . Let iwxy= + , 0x< and 0y> ( ) is in the 2nd quadrantw So 22 3xy+= ( )22 91xy⇒+=  Also, 1tan 2.411865yxπ − −= − 0.894427yx∴= − So from 1, 22 0.8 9xx⇒+ = ( )5 0xx= −<  And 2y= Hence 5 2iw= −+ 2 Solution (a) Area = ( )2 2 6 4cosec 3 cot cosec d 2cot 4cosec 7 x xx x xx π π − −+∫ cosec d cot cosec d ux u xxx = ⇒= − When ,26 ,12 xu xu π π = = = = ( ) ( ) ( ) 22 2 2 66 4cosec 3 cot cosec 4cosec 3 cot cosec dd 2cot 4cosec 7 2 cosec 1 4cosec 7 x xx x xxxx xx xx ππ ππ −−∴= −+ −− +∫∫ 1 22 43 d 2 45 u u uu −=− −+∫ 2 21 43 d 2 45 u u uu −= −+∫ 22 2211 44 1dd 2 45 2 45 u uu uu uu −= + −+ −+∫∫ 222 1 21 11ln 2 4 5 d52 2 2uu uuu= −+ + −+∫

3 [Turn Over ( ) 2 21 11ln 5 ln 3 d32 1 2 uu=−+ −+∫ 2 1 1 11 1ln 5 ln 3 tan2 33 22 x−       −  =−+         151 2ln tan33 6 − = +  (b) 13 11 45 6 4yyxx   =− − − <−      25 0 13 1d45 5 yy yπ  = −− −  ∫ 9.62= units3 3 Solution (a) r = 10 2 3 0 , 11 λλ −   +∈  −   OR r = 82 3 0 , 01 λλ −     +∈        (b) Let B be the foot of perpendicular from A to l2 Since B lies on l2, 41 1 0 for some 04 OB λλ  → −     = +∈        4 16 1 03 0 49 2 2 94 AB OB OA λ λ λ  →  →  → = − −     = +−       −− = −−+ Since AB  → is perpendicular to l2, 1 00 4 AB  → − •= 21 2 00 94 4 λ λ −− −     −•=    −+  

4 2 36 16 0 2 λλ λ +− + = = 4 12 1 2 0 = 1 0 48 OB  → −       = +          Coordinates of B is (2, 1, 8) (c) Using GC, equation of the line: r = 72 1 1 2 , 3 01 µµ    +− ∈        (d) 1 :p r 1 23 2  •= and 2 :p r 2 15 2  •=− Since S lies on line l2, 41 1 0 for some 04 OS λλ  → −     = +∈        C(3, 0, 0) lies on p1 and D(0, 5, 0) lies on p2. Since S is equidistant from p1 and p2, 12 12 CS n DS n nn  →  → •• =  222 2 2 2 4 31 4 0 2 10 2 15 1 40 2 40 2 1 2 2 2 1 ( 2) λλ λλ −  −               −• −•               −        = + + + +− 1 14 2 12 41 424 2 λλ λλ −−         • =−•         −     3 7 4 10λλ+= − 3 7 (4 10 )λλ+ = ±− 17 or 17 3λλ= = 4 1 67 1 11When , 1 0 1717 17 1704 4OSλ  → −         = = +=              415 7 71When , 1 0 33 33 0 4 28OSλ  → −         = = +=             

5 [Turn Over (e) Let θ be the acute angle between between the line l2 and the cursed boundary p2. 2 2 222 2 12 01 42sin (1 ) 0 4 2 1 (2 ) θ −     •     −  = − + + + +− 10sin 3 17 θ = 53.9θ = o The wizard escapes the Cursed Forest. 4 Solution (i) Total time, T = run swimTT + = 221 11 4 10 ww−+ +       28 4 10 1ww= −+ + ( ) ( ) 1 2 2d1 4 10 1 2d2T www −= −+ +  2 104 1 w w = −+ + For shortest time, d 0dTw = 2 104 1 w w ⇒= + 221 5 ww⇒ += ( ) ( ) 2241 5 ww⇒+= 221 4w⇒= 2 4 21w⇒= 2 since 0 21 ww⇒= ≥ Method 1 w 0.436 2 21 0.437 d d T w 0.00336− 0 0.00434 sketch Method 2

6 2 2 2 22 210 1 10 d 21 d1 www T w ww +−  += + ( ) ( ) 22 32 2 32 2 10(1 ) 10 1 10 0 is a minimum 1 ww w T x +−= + = >⇒ + Hence, when w = 2 21 , Paul would take the shortest time. (ii) (iii) When w = 2 21 , 248 4 10 1 2121T = − ++ 8 2 21= + 17.165= Shortest time taken by Paul 17 minutes (iv) tan yxθ = 2 2 dd d ddsec d yxxy tt tx θθ −= Since d 0.01dxt = and d1d4yt = , ( ) 2 22 10.01d 254sec d 100xy xytx xθθ −−  += = When 1x= , 4t = , 1.04y⇒= 1.04tan 1.04 1θ⇒== So ( ) ( )( )2 21 25 1.04d1 1.04 d 100 1tθ ++= T w O

7 [Turn Over d 0.12971 0.130 rad/mindtθ⇒= = Section B: Probability and Statistics [60 marks] 5 Solution (a) Base Multiplier $10 $20 $50 x 1 $10 $20 $50 x 2 $20 $40 $100 P(X = 10) = P($10, x 1) = 2 6 12 49 9 81 27×= = P(X = 20) = P($10, x 2) + P($20, x 1) 2 3 3 6 24 89 9 9 9 81 27=×+×= = P(X = 40) = P($20, x 2) 33 199 9=×= P(X = 50) = P($50, x 1) 46 89 9 27=×= P(X = 100) = P($50, x 2) 4 3 12 49 9 81 27=×= = x 10 20 40 50 100 ( )P Xx= 4 27 827 19 827 427 (b) E(X) 4 818 410 20 40 50 10027 27 9 27 27=×+×+× +×+× 1120 or 41.527= E(X2) 222 2 24 818 410 20 40 50 10027 27 9 27 27=×+×+× +×+ × 7600 or 2533.333= Var(X) = E(X2) – [E(X)]2 27600 1120 3 27 = −  = 812.62 (c) P( 57 30)X −> P( 57 30) P( 57 30)XX= − > + − <− P( 87) P( 27)XX= >+ < P( 100) P( 10) P( 20)X XX==+= += 448 27 27 27=++

8 16 27= 6 Solution (i) Let X be the daily screen time of a Singapore teenagers. ( )22 200 1.5199s = = 2.2613 = 2.26 (3 s.f.) (ii) Since n = 200 is large, by Central Limit Theorem, ( )1 2 200... ~ N 8.5(200), 2.2613(200)YX X X= + ++ approximately ( )i.e. ~ N 1700, 452.26 approximatelyY ( )P 1640 1730Y<< = 0.918 (3 s.f.) (iii) Let µ denote the population mean daily screen time of a Singapore teenagers. H0: µ = 8.5 H1: µ > 8.5 Right-tailed test at 5% level of significance Under H0, 21.2452~ N 8.5, 200X  p-value = 0.0305 (to 3 s.f.) < 0.05 Since p-value < 0.05, we reject H0 and conclude there is sufficient evidence that the Ministry of Education suspects the mean daily screen time of teenagers exceeds 8.5 hours at 5% level of significance. OR Since p-value < 0.05, we reject H0 and conclude there is sufficient evidence to support the Ministry of Education’s suspicion at 5% leve