ASRJC 9758 2023 Prelim P2 Solution

[Turn over ANDERSON SERANGOON JUNIOR COLLEGE H2 MATHEMATICS JC2 Preliminary Examination Paper 2 (100 marks) 9758/02 3 hours Additional Material(s): List of Formulae (MF26) CANDIDATE NAME CLASS / READ THESE INSTRUCTIONS FIRST Write your name and class in the boxes above. Please write clearly and use capital letters. Write in dark blue or black pen. HB pencil may be used for graphs and diagrams only. Do not use staples, paper clips, glue or correction fluid. Answer all the questions and write your answers in this booklet. Do not tear out any part of this 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. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. All work must be handed in at the end of the examination. If you have used any additional paper, please insert them inside this 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 11 Total This document consists of 21 printed and 3 blank pages.

2 © ASRJC 2023 Section A: Pure Mathematics [40 marks] 1(a) Solution ( ) ( ) p a1 pa pa λλ λλ λ = +− = +− −= − b bb bb BP BAλ=       and common point B ∴the points A, B and P are collinear. (b) Solution Let F be the foot of perpendicular from O to AB ( ) for someOF µµ= +−a ba   0OF AB =      ( ) ( )[ ]0 µ+ − −=a ba ba  2 0µ+ −−=ab b a aa 2 2µ = − a ba OF   = ( ) 2 22+− + aa ba ab 22 2 += −a b ba

3 © ASRJC 2023 [Turn over 2 Solution Let 432P( ) 4 78z z z az bz=− + ++ Since 3 2i+ is a root of P( ) 0z = , P(3 2i) 0+= . ( ) ( ) ( ) ( ) 432 3 2i 4 3 2i 3 2i 3 2i 78 0ab+−++++++ = ( )( 119 120i) 4( 9 46i) (5 12i) 3 2i 78 0ab− + − −+ + + + + + = Equating real parts: 535ab+= ----- (1) Equating imaginary parts: 12 2 64ab+= ----- (2) Solving (1) and (2), we get 7a= and 10b=− Now 432P( ) 4 7 10 78zz z z z= −+−+ Since coefficients of P( )z are all real, 3 2i− is also a root of P( ) 0z = . A quadratic factor of P( )z ( )( )3 2i 3 2izz= −+ −− 2 6 13zz=−+ Consider ( )( ) 432 2 2P( ) 4 7 10 78 6 13z z z z z z z z cz d=− + − += −+ ++ Comparing constants: 78 13 6dd= ⇒= Comparing coefficient of z: 10 6 13 2d cc− = − + ⇒= Equating 2 2 60zz+ += , we have 22 2 4(1)(6) 2(1)z −± −= 2 4( 5) 2 −± −= 1 5i= −± Therefore, the other roots are 3 2i− , 1 5i−+ and 1 5i−− .

4 © ASRJC 2023 3 (a) Solution 2 22 11e sin d e sin e cos d22 x xx xx x xx= −∫∫ 2 2 dsin ed de cos d2 x x vux x u xvx = = = = 2 221 11 1e sin e cos e sin d2 22 2 x xx x x xx= −+  ∫ 2 2 dcos ed de sin d2 x x vux x u xvx = = = −= 2e sin dx xx∫ 22 2 1 111e sin e cos e sin d244 xx x x x xx C=−− + ∫ 25 e sin d4 x xx =∫ 22 1 11e sin e cos24 xx x xC−+ 2e sin dx xx =∫ 22 1214e sin e cos55 5 xx Cxx−+ 2e sin dx xx =∫ 21e (2sin cos )5 x x xc−+ where c = 14 5 C (Shown) (b) Solution Required volume 2 222 0 1e sin d (1) 34 x xx ππ πππ − = − ∫ pi or 2 2 222 0 4 4e sin d 1 d x xx x x πππ πππ π − = −−  ∫∫ 222 0 1e sin d 12 x xx π πππ −= −∫ 222 0 1e sin de 12 x xx π π π π= −∫ 222 0 11e (2sin cos )e 5 12 x xx π π π π= −− 02 1e (2sin cos ) e (2sin0 cos0)5e 2 2 12 π π π ππ π= −− −−

5 © ASRJC 2023 [Turn over ( ) 212e 15e 12 π π π π= +− ( ) 212e5 12 ππ π−=+− 4 (a) Solution ( ) 33Let e 1 1 e xxy −−= − −= − 3x≤ 3e1 3 ln(1 ) x y xy − = − = +− 1f : 3 ln(1 )xx− +− , 01 x≤< (b) Solution (c) Solution ( )g xx= 22axx−= 22 2 2 0 11 024 x xa xa +− =  + −−= 211 24xa= −± + 2211 11 or 24 24x ax a= −+ + = −− + (rejected 0x< ) y x

6 © ASRJC 2023 (d) Solution ( ) ( ] 2 g f R, D ,3 a= −∞ = −∞ Since gfRD⊆ for 01 a<< , therefore fg exist. ( ) ( ) 2f23 g fg, 1 e ,1aRaR −= − ∞  → = −

7 © ASRJC 2023 [Turn over 5 (a) Solution 2 2 3y ax x ax −= + , 0x≠ Since 2 2 0y x ≥ for all values of x and y, 3 0ax ax − ≥+ , xa≠− ( )(3 ) 0ax ax+ −≥ 3ax a−<≤ Since 0x≠ , 0ax∴− < < or 03 xa<≤ Alternative Solution ( ) 2 2 3a xxy ax −= + , 0x≠ Since 2 0 ,yy≥ ∀∈  ( ) 23 0a xx ax − ≥+ , 0x≠ ( ) 23 0x ax ax −−≥ + ( ) 23 0x ax xa − ≤+ 0ax∴− < < or 03 xa<≤ (b) Solution Let 1a= , we have 22(1 ) (3 )y xx x+= − . 22d2 (1 ) 2 (3 )d yy xy x xxx ++= −− When d 0d y x = , 22 2 (3 )y x xx= −− 263xx= − Substitute 22 63y xx= − in 22( ) (3 )ya x x a x+= − , 22(6 3 )(1 ) (3 )xx xx x− += − 2 2 3 23 3 2 66 3 3 3 30 ( 3) 0 x xxxx x xx xx +−−=− −= −= x -a 3a x – a 0 3a + – – +

8 © ASRJC 2023 0x= or 3x=± Since 0x≠ and 3ax a−<≤ , we have 3x= When 3x= , 2 6 3 3(3)y = − 63 9y= ±− The coordinates of the points are ( )3, 6 3 9 − and ( )3, 6 3 9−− (c) Solution When 1x= , 22(1 1) (1) (3 1)y += − 1y=± When 1x= , 1y= , 22d d12 (1 1) 1 2(3 1) 1d d2 yy xx ++= −−⇒ = Equation of normal is 1 2( 1)yx−= − − 23yx⇒= −+ When 1x= , 1y=− , 22d d12 (1 1) 1 2(3 1) 1d d2 yy xx− ++= −−⇒ = − Equation of normal is ( 1) 2( 1)yx−− = − 23yx⇒= − Solving 23yx= −+ and 23yx= − , we get 3 2x= and 0y= Therefore, the coordinates of N are 3 ,02  

9 © ASRJC 2023 [Turn over 6 12 18 24 30 100 x x x x O t x 20 40 80 60 (1, 97) (30, 27) x Section B: Probability and Statistics [60 marks] 6(a) [Solution] A student attempts to model the relationship between θ and t using two models: (A) a btθ = + or (B) dc tθ = + . (b) [Solution] For model A, r = – 0.909 For model B, r = 0.933 Model B is more appropriate since its r is closer to 1 and the scatter diagram shows that a curve fits the data points better than a straight line. (c) [Solution] 64.3607935.02439 tθ = + 64.435.0 tθ = + When 45t = , o 64.3607935.02439 45 36.5 C (3 sf) θ = + =

10 © ASRJC 2023 The prediction is not reliable as 45t = lies outside the data range of 1 30t≤≤ , hence extrapolation, and the linear relationship may not hold outside the data range. When 45t = , o 64.435.0 45 36.4 C (3 sf) θ = + = (d) [Solution] 64.36079θ 35.02439 θ 273.15Tt= += + 64.4308T t = +  7(a) [Solution] P(Score = 5) = ( ) ( ) 2 2 π 104 5 π 30 × = 4 45 P(Score = 3) = ( ) ( ) ( ) 22 2 π 20 π 104 5 π 30 −× = 4 15 P(Score = 1) = ( ) ( ) ( ) 22 2 π 30 π 204 5 π 30 −× = 4 9 (Shown) (b) [Solution] (b) P(X = 0) = 11 55      = 1 25 P(X = 1) = ( ) 2 14 4 259 9    +     = 152 405 P(X = 3) = ( ) 2 14 4 4 25 9 15 15    ++       = 56 135 P(X = 5) = ( ) 2 144 4 4 25 9 15 45 45    ++ +       = 344 2025 x 0 1 3 5 P(X = x) 1 25 152 405 56 135 344 2025