SAJC 9758 2023 Prelim P2 Solution

2023 H2 Math Prelims Paper 2 Solutions 1 No. Solutions 1 Substitute i into the equationzkk= + ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 22 22 2 2 2 3 i (5 i) i 0 3 1i ( 5i ) 1i 0 3 1 2i i (5 5i 5i+i ) 0 3 (2i) (4 6i) 0 ----------(*) 4 6 6i0 kk kk kk kk kk k kk α α α α α + −+ + += + − + ++= ++ − ++ += − + += −+ − = Comparing real and imaginary parts: 40kα −= ----(1) ( ) 26 6 0 (2) 6 10 1 or 0 (rejected since is non-zero) kk kk kk k − = −−−− −= = = Substitute 1k = into (1) 4α = Let the other root be w ( )( ) 23 (5 i) 4 3 1 iz z zwz− + + = − −− Comparing constants: ( ) ( ) ( )( ) 4 3( )( 1 ) 4 31 i 41 i 31 i 1 i 22 i33 wi w = − −− = + −= +− = − 2(i) 2 36 6 PQR πππθ π θ −−+  = ∠ − = Using Sine Rule, P Q R 2 3 π 6 π θ+ a

2023 H2 Math Prelims Paper 2 Solutions 2 sin sin3 6 2 PQ a π π θ =  − sin 3 sin cos sin66 2 13cos sin2 2 cos 3 3 2 3 co sis n a PQ a a π ππ θθ θθ θθ − − =   −  = = 2(ii) ( ) 2 12 222 2 2 2 3sin 13 2 13 2 ( 1)( 1) 3 cos 3 3 3 33 1 ( 1) 3 32 2! 2 13 2 713 23 aPQ a a a a a θθ θ θ θθ θθθθ θθθ θθ − − −− −−   −−+ − −−+ −−   ++ = ≈ =   = + = + ++  ≈   ++   2(iii) 2 2 713 2 3s 33 cosf( ) 3 cos inLet 3 sin 713 2 3 sin 3 sin 1 cos 1 cos aa a θθ θθθ θθ θθ θθ θθ ++ − −  −=  −= ++ − − 0f( .) 05θ <∴

2023 H2 Math Prelims Paper 2 Solutions 3 From GC, 0.190( 4 sf )3θ< < 3(a) ( ) ( ) 2 22 2 22 2 2 2 2 2 2 1d d tan , secdt1 1 sec t dt tan tan 1 1 sec t dttan (sec ) sec dttan cos dtsin (cos )(sint) dt 1 , where is an arbitrary constantsin 1 , xx xt t xx tt tt t t t t t CCt x Cx − = = + = + = = = = = −+ += −+ ∫ ∫ ∫ ∫ ∫ ∫ Alternatively, 2 cos dt cosec cot dsin = cosec t t ttt tC = −+ ∫∫ (b)(i) ( ) 3 23d cos 3 sind x xxx =− f( )y θ= 0.05y = 0 0.194 x 1

2023 H2 Math Prelims Paper 2 Solutions 4 (ii) 53 3 23 3 32 3 2 3 3 3 23 3 33 sin d ' sin 1cos cos d ' 3 cos33 1cos 3 cos d33 1cos sin33 x xx ux v x x x x x xx u x v x x x x xx x x xC = = = −+ = = − = −+ = − ++ ∫ ∫ ∫ 4(i) (i) 2 90xy x y+− = Differentiate with respect x, ( ) dd2 2 19 0dd d 29 21d d 21 d 92 yyxy xx y xyx yy xx + +− = −= −− += − (ii) (ii) Let d d yG x= . 21 92G y x + −∴= Diff wrt x, ( ) ( )( ) ( ) 2 d2 92 2d d 12d 92 y xyxG x x = − − +− − When 3x= , 0 39 0 1 29 6yy xy y yx+= +− = ⇒ − ⇒ = d 2(1) 1 1d 9 2(3) y x + −∴ = = Hence, when 3x= ( ) ( ) ( ) 2 d d 2(1) 9 2(3) 2(1) 1 ( 2) 0.02 9 2(3) dd dd 0.022 or sf)67(375 x t GG tx ⋅ − − +−= × − = = Therefore, required rate is 2 75 units/s.

2023 H2 Math Prelims Paper 2 Solutions 5 (or 0.0267 units/s) 5(i) (ii) (iii) ( ) 11L.H.S. = ! ( 1)! 11 ( 1)! = R.H.S. (Shown)( 1)! rr r r r r − + +−= + =+ ( ) 1 1 123 ...2! 3! 4! ( 1)! = (1)( 1)! 11 = (2)! ( 1)! 11 = 1! 2! 11 + 2! 3! 11 +( 1)! ! 11 + ( )! 1 ! = n n r n r nS n r r rr nn nn = = = ++ ++ + −−−+  − −−− + − − • • • −− − + ∑ ∑ ( ) 11 1!n− + 1As , 0 ( 1)! 1 Since the limit 1 is unique and finite, the series converges. n n n S →∞ → + ∴→ S∞ =1

2023 H2 Math Prelims Paper 2 Solutions 6 (iv) 2 2 2 ( 1)Let ( 2)! Let ( 1) ! 0 and 0 for all . (1) ( 1) ( 1 )! ( 2)! ( 2) ( 1) 4 1 0 for 1( 2)! ( 2)! This implies that 0 for all . (2) r r rr rr rr ra r rb r ab r rrba rr rr r r rrr ba r + + −= + = + ≥ ≥ ∈ −−− −−= − ++ +−− −= = >≥ ++ − ≥ ∈ −−−   Hence by using the comparison test, since 1 ( 1) !r r r ∞ = +∑ converges then ( ) 2 1 1 ( 2)!r r r ∞ = − +∑ also converges. 6(i) The possible values of X are 2, 3, 4 and 5 54 5P( 2) P( ) 8 7 14X RR= = =×= 53 4 5P( 3) P( or ) 2! 8 7 6 14X RBR BRR= = =×××= P( 4) P( ,the first 3 balls can be in any order but last one must be ) 5323 !4 3 8762 !51 4 X RBBR R = = =××× ×= P( 5) P( ,the first 4 balls can be in any order but last one must be ) 5 3 2 1 4! 4 1 8 7 6 5 3! 4 14 X RBBBR R = = =×××× ×=

2023 H2 Math Prelims Paper 2 Solutions 7 6(ii) all E( ) P( ) 5531234514 14 14 14 3 x X xXx= = = ×+ ×+ ×+ × = ∑ [ ] [ ] 22 22222 Var( ) E( ) E( ) 55312345 314 14 14 14 69 6977 XX X= − =×+×+×+×− = −= 6(iii) Expected profit = Expected Loss $ $2 P( 2)+$3 P( 3)+$4 P( 4) $5 P( 5) 5531234514 14 14 14 3 yX X X X y y = ×= ×= ×= + ×= = ×+ ×+ ×+ × = Or $ $1 E( ) 3 yX y = × = 7(i) P(toy chosen is either a Triangle or a Star given not Yellow) ( ) ( ) P(Triangle or Star and not Yellow) P(not Yellow) 423 531 18 340 (40 1 3 4 2) 30 5 40 = ++ + ++ = = =−−−− 7(ii)a) P(both toys chosen are purple and different shapes) P(Purple Square, Purple Triangle) + P(Purple Star, Purple Triangle) P(Purple Square, Purple Star) 23 13 21= 22240 39 40 39 40 39 11 780 = + ××+ ××+ ×× = 7b) Let A and B be Ben’s two favourite combinations. Let ,1 5AAnn +∈ ≤≤ and ,1 5BBnn +∈ ≤≤ be the number of A and number of B respectively.

2023 H2 Math Prelims Paper 2 Solutions 8 1240 39 39 ABnn× ×= 20ABnn = ∴the possible cases given that 1,5 ABnn≤≤ are: 5, 4 4, 5 AB AB nn nn = = = = Thus, possible A and B are: Yellow Triangle, Green Star Yellow Triangle, Red Square Green Triangle, Red Square Green Triangle, Green Star 8(i) Let X be the random variable denoting the Calculus score of a randomly selected student and 1 µ be the population mean. Test 01H : 52µ = against 11H : 52µ ≠ at 5% level of significance 2Unbiased estimate of the population variance s 2 sample variance1 30= 1529 6750 29 n n= ×− × = Under 0H , since 30n= is large, by Central Limit Theorem, 6750~ N 52, (29)(30)X   approximately. Test statistic ( )52 ~ N 0,1 6750 870 XZ −= approximately. Using a 2-tailed z-test, reject 0H if p-value 0.05≤ Using GC, the test statistic value 46x= and calc 2.1541z =− gives p-value = 0.0312 < 0.05 We reject 0H and conclude that there is sufficient evidence at the 5% level of significance that Professor’s claim about the mean score is not valid.

2023 H2 Math Prelims Paper 2 Solutions 9 8(ii) Let Y be the random variable denoting the Statistics score by a randomly selected student and µ be the population mean. Test 0H : 48µ = against 1H : 48µ > at 5% level of significance Under 0H , since n is large, by Central Limit Theorem, 213~ N 48, 30Y   approximately. Test statistic ( )2 48 ~ N 0,1 13 30 YZ −= approximately. Carry out 1-tailed z-test at the 5% level of significance. Since Professor B has understated the score, we will reject 0H . For 0H to be rejected , 2 1.6449 48 1.6449 13 30 48 3.9041 51.904 52.0 (Round in) [Accept 51.9] calcz k k k kk ≥ − ≥ −≤ ≥ ⇒≥ ≥ 8(iii) Yes. Since it is not known whether the Calculus score and Statistics score by a randomly selected student is normally distributed, it is important that the sample size is at least 30 in order to use Central Limit Theorem so that the mean Calculus score and the mean Statstics score by students are approximately normal. 1.6449 0.05

2023 H2 Math Prelims Paper 2 Solutions 10 9(i) The probability that a patient has diabetes mellitus is a constant at 1 15 for each patient. A patient having diabetes mellitus is independent of any other patient having diabetes mellitus. (ii) Let Y be the random variable “number of patients having diabetes mellitus out of 49. 1~ B 49, 15Y   Required Probability ( ) ( )P 3 P the 50th patient is the fourth patient who has diabetesY= = 10.2284556 0.015215= ×= (iii) 6 10Required Probability 100 100 100 0.06 0.001 p p =+× = + (iv) 100 – 125 mg/dL | not diagnosed with diabetes mellitus) 100 – 125 mg/dL not diagnosed with diabe tes mellitus) not diagnosed with diabetes mell P( P( P( 10 1100 100 1 (0.06 0.001 ) 0.1 0.001 0. itus) p p p = ×−= −+ −= ∩ 100 94 0.001 940 p pp −=−− (v) Let W be th