CJC Prelims 2025 H2 Math P2 Solutions

9758/02/J2PRELIM/2025 [Turn Over CATHOLIC JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 JC2 Preliminary Examination CANDIDATE NAME CLASS INDEX NUMBER MATHEMATICS 9758/02 Paper 2 17 Sep 2025 3 hours Additional Materials: Printed Answer Booklet List of Formulae (MF27) READ THESE INSTRUCTIONS FIRST Answer all the questions. Write your answers on the Printed Answer Booklet. Follow the instructions on the front cover of the 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 present the mathematical steps using mathematical notations and not calculator commands. You must show all necessary working clearly. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 43 printed pages.

2 9758/02/J2PRELIM/2025 Section A: Pure Mathematics [40 marks] Solution: Q1 (a) (b) y x O x = 3 (1, 0) y x O y =2 (2, 0) (3, 4) x = 1

3 9758/02/J2PRELIM/2025 [Turn Over (c) ( ) ( ) ( ) f translate 1 unit in positive -axis direc tion f1 reflect in -axis f1 yx x yx x yx = ↓ = − ↓ = −− ( ) ( ) ( ) f translate 1 unit in positive -axis direc tion f1 reflect in -axis f1 yx x yx x yx = ↓ = − ↓ = −− y x O y = −2 (−1, 0) (3, 0) (4, −4) (−2, −4) x = 0 x = 2 (1, 0)

4 9758/02/J2PRELIM/2025 Solution: Q2 (a) (b) For gf to exist, fgRD⊆ , )(fR 0,= ∞ )(gD 0,= ∞ Since fgRD⊆ , gf exists (shown) ( ) ( ) ( ) ( ) 2 2 2 4gf g 4 1ln 1 4 4 4ln 1 4 x x x x   =  −      = +    −   −= +  ( ) ( )gf fD D ,4 4,= = −∞ ∪ ∞ OR { }gf fDD 4 \= =  x y y = 0 x = 4 y = f (x) O

5 9758/02/J2PRELIM/2025 [Turn Over Method : Sketch ( )gfyx= for given domain ( )gfR 0,= ∞ Method : Mapping ( ) ( ) ( ) ( ) fg f f gfD ,4 4, R 0, R 0,= −∞ ∪ ∞  → = ∞  → = ∞ (c) Largest value of k is 4. (d) ( ) ( ) ( ) 2 2 2 4 4 4 2 24 rej 4 . 4r 4 4 o y x xx yy y x x y x = − −= ± = +< −= = − )1 ffD R (0,− = = ∞ 1f: , , 0 24 x x xx− →∈− > x y O 4 y = gf (x) x y y = 0 y = g (x) O

6 9758/02/J2PRELIM/2025 Solution: Q3 (a) By Pythagoras Theorem, 2 22 22 a xh h ax = + = − 2 22 2 3 2 3 2 V xh xa x = = − ( ) ( ) 2 42 2 2 24 6 3 4 3 (shown)4 V xa x V ax x = − = − (b) Method : ( ) 2 24 63 4V ax x= − Differentiate implicitly with respect to x: ( ) 23 5d32 46d4 VV ax xx = − At stationary point, d 0d V x = ( ) 23 5 32 2 4 60 22 3 0 ax x xa x −= −= Since 0x≠ , 22230ax−= 22 2 3xa= 23 22 2 66 6 6 3 32 2max 43 3 34 8 4 9 27 34 4 27 9 3 V aa a aa a a a   = −    = −  =  = = h x a x h a

7 9758/02/J2PRELIM/2025 [Turn Over Method : ( ) ( ) 2 24 6 1 24 6 2 3 4 3 2 V ax x V ax x = − = − ( ) ( ) ( )( ) 1 24 6 23 5 2 1 23 5 24 6 2 d 31 46d 22 3 464 V ax x ax xx ax x ax x − − =−−  = −− At stationary point, d 0d V x = ( )( ) ( ) 1 23 5 24 6 2 23 5 32 2 3 46 04 4 60 22 3 0 ax x ax x ax x xa x − − −= −= −= Since 0x≠ , 22230ax−= 22 2 3xa= 23 22 2 66 6 6 3 3 32 2max 43 3 34 8 4 9 27 34 4 27 9 units3 V aa a aa a a a   = −    = −  =  = =

8 9758/02/J2PRELIM/2025 (c) Total surface area, 6 area of isosceles triangles area of hex agonA= ×+ ( ) 22 22 22 2 22 2 2 2 22 22 22 22 1166 2 22 2 333 44 2 12 2 3233 3 43 3 43 25 2133 36 32 533 9 3 5 3 units xxA xa xx xxxa x aa a a a aa aa aa a  =××× − +××× −   = −+ = −× + × = ×+ × = + = + a x

9 9758/02/J2PRELIM/2025 [Turn Over Solution: Q4 (a) The student’s claim may not be true as p may not be real. (b) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) 432 432 2 14 33 26 0 2 3i 1 4 3i 3 3 3i 2 6 3i 0 2 8 6i 8 6i 14 3 i 8 6i 33 8 6i 78 26i 0 2 28 96i 14 18 26i 264 198i 78 26i 0 56 192i 252 364i 264 198i 78 26i 0 10 0 10 z z z zp p p p p p p − + − += +− ++ +− + + = + +− + ++ +−−+ = + − + + + −− += + − − + + −− += −+= = (c) Method : Since the coefficients of the polynomial are all real, 3i+ is a root,3i− is also a root. ( ) ( ) ( ) ( ) ( ) 2 2 2 2 3i 3i 3i 3i 3i 6 91 6 10 zz zz z zz zz −+ −−     = −− −+      = −− = − ++ =−+ ∴ ( )( ) 432 2 22 14 33 26 6 10 2 2 1z z z zp z z z z− + − += −+ − + Method : Since the coefficients of the polynomial are all real, 3i+ is a root,3i− is also a root. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 432 2 2 2 22 22 22 2 14 33 26 10 3 i 3 i 2 3 i 3 i2 3 i2 6 912 6 10 2 z z z z z z z Az B z z z Az B z z Az B z z z Az B z z z Az B − + − +=−+ −− + +      = −− −+ + +      =−− ++ = − ++ + + = −+ ++ Comparing constants: 10 10 1BB= ⇒= Comparing coefficients of z : ( )26 6 1 10 2 AA−= − + ⇒= − ( ) ( ) ( ) 2 2 432 43 2 32 32 2 2 2 21 6 10 2 14 33 26 10 2 12 20 2 13 26 2 12 20 6 10 6 10 zz zz z z z z zzz zz z zz z zz zz −+ −+ − + − + −−+ −+ − −− + − −+ − −+

10 9758/02/J2PRELIM/2025 ∴ ( )( ) 432 2 22 14 33 26 6 10 2 2 1z z z zp z z z z− + − += −+ − + ( ) 2 i 2 4 42 2 2 21 4 2i 1 4 2 0 2 4 4 1 2 zz z − = ± += = = ± = − ±− ± (d) Trapezium Area of qualilateral ( )( )1 1 2 2.5 3.752= += units2 (or 15 4 units2 ) Re Im O (3, 1) (3, −1) (½ , ½) (½ , ½) 2 units 1 unit 2.5 units