YIJC 9758 2023 Prelim P2 Solution

1 ©YIJC 9758/02/Prelim Exam/2023 2023 JC2 H2MA Preliminary Examination P2 Solutions Solution 1a 41 0.754h −= = From the diagram, Total area of shaded rectangles < Area under the curve from 1x= to 4x= [ ] 4 1 f (1 ) Area of n nh h A = ∴+<∑ b [ ] 3 0 f (1 ) n nh h = +∑ or [ ] 4 1 f (0.25 ) n nh h = +∑ or [ ] 4 1 f (4 ) n nh h = −∑ or [ ] 4 1 f (1 ( 1) ) n n hh = +−∑ c By GC, [ ] [ ] 4 1 4 1 Lower Bound 0.75f (1 0.75 ) 0.75( 2ln(1 0.75 1) 7) 13.0 n n n n = = = + = − + ++ = ∑ ∑ y x 1 4 O h

2 ©YIJC 9758/02/Prelim Exam/2023 [ ] [ ] 3 0 3 0 Upper Bound 0.75f (1 0.75 ) 0.75( 2ln(1 0.75 1) 7) 14.4 n n n n = = = + = − + ++ = ∑ ∑ d (For students’ understanding only) Since translation by b units in the positive y-direction will result in an increase in the area by 4 rectangles (represented by c) with each length 0.75 units and breadth b units, 4 0.75 3 cb cb ∴=× = . Alternatively, [ ] [ ] [ ] [ ] 44 11 44 11 4 1 g(1 ) f (1 ) f (1 ) f (1 ) 4 nn nn n nh h nh b h nh h bh nh h bh = = = = = + = ++ = ++ = ++ ∑∑ ∑∑ ∑ 4 (0.75) 3 cb cb ∴= = 2a Method 1 3i 3 3 9 2 3− = += ( ) 1 32 πarg 3i 3 π tan 33 − −= − =  1i 11 2−= += ( ) 1 1 πarg 1 i tan 14 − −= − = −  y x 1 4 O b 0.75

3 ©YIJC 9758/02/Prelim Exam/2023 23 6 2 z = = 2π π 11πarg( ) 3 4 12z = −− = Method 2 (NOT recommended) ( )( ) ( )( ) ( ) ( ) 3i 3 1i 3i 3 1 i 1i1i 3 3 3 3i 2 z −= − −+ = −+ −−+− = ( ) ( ) 22 33 3 3 22 3963 3963 44 6 z   −− −  = +     ++ +−= + = ( ) ( ) 1 33 11πarg( ) π tan 1233 z −  −= −=  + b ( )( ) ( )( ) ( ) ( ) 2 3i 3 i 3i 3 i ii 3 3 3 3i 1 z k k kk kk k −= − −+ = −+ − −+ − = + ( )Im 0zz z ∗= ⇒= 2 3 1 3 0k k− + = 3 3k =

4 ©YIJC 9758/02/Prelim Exam/2023 3a ( )( ) ( ) ( ) ( ) ( ) ( ) 2 22 2 2 11ln 1 ln 11ln ln 1 ln 1 ln ln 1 2 ln l hnn o1 sw r rr rr r r rr r rr  −−=   −+=   = −+ +− = −− + + b ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1ln 1 ln 1 2ln ln 1 ln1 2ln2 ln3 ln2 2ln3 ln4 ln3 2ln4 ln5 ln4 2ln5 ln6 ln 3 2ln 2 ln 1 ln 2 2ln 1 ln ln 1 2ln ln 1 ln2 ln ln 1 1ln 2 n r n n r S r r rr n nn n nn n nn nn n n = = = −  = −− + + = −+ +− + +− + +− + + + −− −+ − + −− −+ + −− + + = −−+ + +=  ∑ ∑  c 1 11ln ln2 22 n nS nn + = = +   As n→∞ , 1 2 0n → . Hence, 11 1lim ln ln22 2n S n ∞ →∞ = +=  . Considering 0.05nSS ∞−≤ , using the GC, n nSS ∞− 19 0.0513 (> 0.05) 20 0.0488 (≤0.05) 21 0.0465 (≤0.05) Smallest n = 20. 4ai 0vu×=  • Either 0v= • Or 0 (and given 0)vu≠≠  , v and u are parallel (accept , where 0v ku k= ≠ ) Combining both cases, , where v ku k= ∈ 

5 ©YIJC 9758/02/Prelim Exam/2023 aii 1 2, 2 11 11 22 3144 22 1 1Alternative answer : 2 3 2 v n n  = −−    = −=−  ++  −−   = −−−    bi ( ) 2243 33 1Area of 142 12 3 1423 AB b a A C ab a ab ABC AB AC ba a b → → →→ = − = − −= − ∆= ×= −× − =        223 3 2833 Since 0, 0 and , 7 283 7 283 12 (Shown) ba bb aa ab aa bb ba ab ab ab ab ×− ×− ×+ ×= ×= ×= ×= −× − ×= ×= ×=             bii Since 5, 3= =ab 12 sin 12 15sin 12 4sin 5 a b a b AOB AOB AOB ×= ⇒ ∠ = ∠= ∠=    AOB∠ is obtuse 2 cos 0 4cos 1 5 3 5 AOB AOB ⇒ ∠< ∠ = −−  =−

6 ©YIJC 9758/02/Prelim Exam/2023 5a 5cos sin6 11 55sin sin cossin66 6 xx AB xπ ππ π= = − − 1 1 1cos 3 2 2 sin2 A x B x −−  = ( ) 1 11 co 322 ins sx B x A + = 3 in 1 cos sx AB x+ = (Shown) b 2113 1 2 x B x A ≈ −+ 2 1113 2xx − ≈+ −  22 21113 3 22xx xx ≈− − + −   2 2113 3 2xx x≈− + + 2713 2xx≈− + c ( )ln ln 1 ppx q q x q  += +   ln ln 1 pqx q =++  ln pqx q=++  Comparing, ln 1q= and 3p q =− eq= and 3ep=− 6a ( ) ( )P 1P 1 0.82 0.18 BBAA ′ ′∪= − = = ∩ − ( ) ( ) ( ) PP P A BAB B= ∩∣ ( ) 0.180.4 P B= ( )P 0.45B =

7 ©YIJC 9758/02/Prelim Exam/2023 b ( ) ( )P 1P 1 0.45 0.55 BB = − = − = ′ A BB′′∩⊆ ( )P 0.55BA ′∩≤ c B & C independent. Hence ( ) ( ) ( )P PP 0.4 0.1 0.0 5 5 4 BB C C=∩ × = = A & C are mutually exclusive. Hence ( ) ( ) ( ) ( ) P PP P 0.45 0.18 0.045 0.225 A B C AB B BC ′ = −− =−− = ′∩∩ ∩∩ 7a Group the 2 red discs and the blue disc as one unit. Together with the remaining 6 green discs, there are 7 units. Required no. of arrangements 7! 3! 6! 2! 21 = × = bi ( ) ( ) ( ) 27 03 9 3 27 12 9 3 27 21 9 3 5 765 5P 0 or 12 9 8 7 12 1 276 1P 1 or 32 987 2 1 217 1P 2 or 312 9 8 7 12 511 or 1 12 2 12 CCR C CCR C CCR C ×= = = ××= ×== = ×××= ×= = = ×××= − −= Probability distribution of R: r 0 1 2 ( )P Rr= 5 12 1 2 1 12 bii r 0 1 2 ( )P Rr= 5 12 1 2 1 12 Change in points 0(9) 3( 3) 9 +− =− 1(9) 2( 3) 3 +− = 2(9) 1( 3) 15 +− = A B C

8 ©YIJC 9758/02/Prelim Exam/2023 Expected change in points ( ) ( ) ( ) ( ) ( ) ( ) 9 P( 0) 3 P( 1) 15 P( 2) 51 19 3 1512 2 12 1 RR R= −= + = + =    = −++       =− Alternative solution ( ) 5 1 12E0 12 12 2 12 3R  = ++ =   Expected change in points ( ) ( )9 E 3 E3RR=× −× − ( )22(9) 3 333 = +− −  1=− 8a Unbiased estimate of population mean, 198.5 50035 505.671 506 (3 s.f.) x = + ≈ ≈ Unbiased estimate of population variance, ( )2 2 198.51 718835 1 35 178.3006 178 (3s.f.) s  = −−  ≈ ≈ b Ho : 500µ = H1 : 500µ ≠ Test at 2% significance level. Under Ho,, since the sample size n = 35 is large, by Central Limit Theorem, 178.3006~ N 500, 35X  approximately. Using GC, p-value = 0.011986 < 0.02 We reject Ho and conclude that there is sufficient evidence, at the 2% level of significance, that the mean mass of the packets of scallops is not 500 grams. c ( ) 22 40 11.739 140.4 s = = 00H: µµ= (claim) 10H: µµ<

9 ©YIJC 9758/02/Prelim Exam/2023 Level of significance: 5% Under 0H , s ince the sample size n = 40 is large, by Central Limit Theorem, 0 140.4~ N , 40X µ  approximately. Since 0H is not rejected, 0 0 510 1.64485 140.440 0 513.08 (2 d.p.) µ µ − >− << 9a ( ) ( ) 2 2 2 Area of inner circleP Obtaining 'hit' Area of outer circle 225 15 r r π π = = = bi Let X be the random variable denoting the number of ‘hits’ out of 8 throws. Then 2 ~ B 8, 225 rX   ( ) ( ) ( ) 782 22 8 7 0.08 P 7 P 8 0.08 0.08225 P6 1225 225 XX rC X rr ≤ = += ≤  >   −+  ≤   Using GC, ( )11.3439 6 s.f.r≤ Therefore, ( )11.3 3 s.f.r≤ bii 26 0.16225 = Let Y be the random variable denoting the number of ‘hits’ out of 4 throws. Then ( )~ B 4, 0.16Y

10 ©YIJC 9758/02/Prelim Exam/2023 Required probability ( ) ( ).P 061 1Y ×= = 0.060693 0.0607 (3 s.f.) = = Alternative solution Required probability ( )( ) ( )30.16 0.84 0 6 4 .1×= = 0.0607 (3 s.f.) biii Let W be the random variable denoting the number of ‘hits’ out of n throws. Then ( )~ B , 0.16Wn ( ) ( ) ( ) 1 0.7 1 P 0 0.7 P 03 P 0. W W W ≥≥ −= ≥ = ≤ By GC, n ( )P0W = 6 0.3513 0.3> 7 0.2951 0.3< 8 0.2479 0.3< Therefore, the least value of n is 7 . Alternative solution ( ) ( ) 1 0.7P 1 P 0 0.7 W W ≥≥ −= ≥ ( )P 0 0.3W = ≤ ( ) ( ) 0 0.16 0.84 0.30 nn ≤ ( )0.84 0.3 n ≤ ln 0.3 ln 0.84 6.9054 n n ≥ ≥ Therefore, the least value of n is 7 . biv ( )~ B 8, 0.16X Required probability ( )( )8 0.16 8 16P2 0.XX= × >×< ∣ ( )P 2.56 1.28XX= >< ∣