DHS 9758 2023 Prelim P2 Solution

1 DHS 2023 Year 6 H2 Mathematics Preliminary Examination 2023 H2 Math Prelim solutions Qn Suggested Solution 1(a) Area of n rectangles under the above curve between 0 and 1,xx= = ( ) 1 0 1 1112 1 1f 0 f f ... f 1 f n r n n nn nn n n r nn − = −   = +++ +        = ∑ Since area of n rectangles is an underestimate of / less than the actual area under curve 1 0 f( ) d ,xx=∫ 1 1 00 1 f f( ) d (shown) n r r xxnn − = ∴< ∑ ∫ (b) Required expression 1 1 f n r r nn= = ∑ (c)(i) Lower limit 9 0 29 0 1 f10 10 1 110 10 1.285 r r r r = = =  = + = ∑ ∑ (c)(ii) From GC : 1 2 0 +1 d 1.3333xx =∫ Difference from 1 2 0 +1 dxx∫ (a) lower limit = 1.3333 1.285 0.0483 3.62%1.3333 1.3333 − = = (b) upper limit = 1.385 1.3333 0.0517 3.88%1.3333 1.3333 − = = The lower limit is a better estimate since it has a smaller % / absolute difference from the exact value. O … 3 n n n y x f( )yx= Upper limit 10 1 210 1 1 f10 10 1 110 10 1.385 r r r r = = =  = + = ∑ ∑

Qn Suggested Solution 2a(i) (A) c = 5 The sequence initially decreases and subsequently alternates , and converges to 2. (B) c = 2 It is a constant with a value of 2. a(ii) ( ) 1 21 32 3 0.5 3 0.5 3 0.5 3 0.5 3 0.5 1.5 0.25 uc uuc uu c c = = −= − = −= −−= + ( ) ( ) 32Given 2 5 2 1.5 0.25 5 3 0.5 3 0.5 15 2.5 2 18 9 uu cc cc c c =− + = −− + = −+ = = b(i) Given 3 21 12 ,v uu+= − ( ) 12 212 11 2 , 2 2 2 3 0.5 2 262 2 vv uu p cc pc cp + −+= − += − − +=− += b(ii) Given 12 , 2,v pv= = ( ) ( ) ( ) 3 12 4 23 5 34 2 12 2 12 1 2 1 2(2) 2 1 1 2 4 2 1 22 1 2 4 1 6 5 6 5 77 12 v vv p p v vv p p v vv p p p p p = + −= +−= + = + −= + + −= + = + −= + + + −= + += ∴=  Qn Suggested Solution

3(a) ( ) 2 22 2 2 22 2 2 2 20 2 2 100 20 20 100 5 (shown)20 xyh yh h yy yh hy = +  −= +  = +−+ = + = + 2 2 20 2 20 2 5 20 10 10 xy h h = − = −+  = − (b) Volume of prism, ( ) ( )( ) ( ) ( ) 2 2 23 43 2 1 2 1 10 20 22 10 100 1010 1000 100 1010 1 10 100 1000 (shown)10 V hxz hhh h hh h h hh hh h h = =−−  = −− = −−+ = −− + ( ) 32d1 4 30 200 1000d 10 V hh hh = −−+ For max.volume, 32d 0 4 30 200 1000 0d V hh hh = ⇒− − + = From GC : 6.40 or 10 (reject 0 10) 3.9039 hh h = − << ∴=  2 2 2 d1 (12 60 200)d 10 V hhh = −− ( ) ( ) 2 2 2 When 3.9039, d1 (12 3.9039 60 3.9039 200) 25.1 0d 10 h V h = = − − = −< 3.9039h∴= gives maximum volume 2 20 20 2 xy xy += = −

Max volume ( ) ( ) ( )( ) 324 3 1 3.9039 10 3.9039 100 3.9039 1000 3.903910 201.71 202cm (3 sf) =−− + = =

Qn Suggested Solution 4(a) Since the lines AB and AC meet, 12 4 44 6 92 ab λµ λµ λµ ++     += +    ++   . 23λµ−= --- (1) abλµ= --- (2) 23λµ−= --- (3) Solving (1) & (3) gives 1λ = & 1µ =− Hence (2) gives ab=− ⇒ 0ab+= (shown) (b) 21 2 3 12 2 ab ab ba −   ×= −   −  Since the normal of plane ABC is parallel to 3 1 3 −   , 29ab−= --- (1) 29ba−= − --- (2) Solving (1) & (2) gives 3a= & 3b=− Equation of plane ABC: r 313 1 4 1 19 3 63 −−       = =           ⇒ 3 3 19xy z−++ = (c) Since 1λ = & 1µ =− and 3a= & 3b=− , the coordinates of A are (3, 7, 7). Let the acute angle between lines AB and AC be θ. 21 33 12 5cos 14491194 θ      −      = = ++ ++  ⇒ 69.1θ = ° or 1.21 rad

(d) r 19 9 3 1 1 3 9 919 19 + −   = + ++  =  Distance of plane ABC from the origin is 19 units. Hence the two required planes: r 3 10 3 −  =  & r 3 1 2 19 19 38 3 −  = =  5(a) 1 7k = (b) Let G be absolute difference of two scores. Probability Distribution of G: g 0 1 3 4 ( )P Gg= 11 77 22 77 44 77 21 49   +    +    +   = 242 77 16 49      = 122 77 4 49      = 142 77 8 49      = ( ) 16 4 8E1 3 4 49 49 49 60 49 G =++ = E(2 ) 0Gm−> 2E(G) 0 m−> 602049 m −> 1200 2.4549m∴< < = (c) Tim’s winnings is based on ( )E G , which is the long term average score. He may still lose for some of the 2 games, but in the long run, he makes a profit.

Qn Suggested Solution 6(a) (i) Number of ways 5 12 9 1 51 4276800C PC=××= (ii) Case (1) : 3 letters + 4 digits 59 34 181440PP=×= Case (2): 4 letters + 3 digits 59 43 60480PP=×= Total Number of ways 181440 60480 241920 = + = (b) All possible ways with 3 identical letters 2 13 14 7! 12012003!CC= × ×= All letters and no digit Number of ways 24 14 7! 16803!CC= × ×= Since alphanumeric requires at least 1 digit, ∴ Using complement, # ways with 3 identical letters and at least 1 digit ( ) 2 13 4 1 44 7! 3! 1199520 C CC= ×× − = “e” or “s” Choose from 4 letters & 9 digits “e” or “s” All other 4 letters chosen

Qn Suggested Solution 7(a) Probability 12 33 51...96 96 96 16 8 54 27 =++ = = (b) Probability 12. 196 8 8 27 = = (c) To win $6 in total for 3 games, each game he must win $2. Probability 12 33 51. . . 3!96 96 96 540 157464 5 or 0.00343 (3 s.f.)1458 =××× = = (d) The participant should end the game by taking the first option because if he proceeds to throw the die, there is only a one-sixth chance that he will take home a higher amount. OR 2 31(0.02) (0.5) (2) 0.596 66 + += For the throw of die, the expected factor is 0.59 which is less than 1. This means that the participant is unlikely to take home a higher amount if he were to proceed to throw the die. 8(a) For a player to reach point B, there must be 5 right steps and 3 up steps in total. Let 1R be the number of right steps taken by a player out of 5 to move from X to Y. ( )1 ~ B 5,Rp ( )1 5 32 3 32 P3 10 R C pq pq = = × = Let 2R be the number of right steps taken by a player out of 3 to move from Y to B. ( )2 ~ B 3,Rp

( )2 32 2 2 P2 3 R C pq pq = = × = Required probability 32 2 53 10 3 30 (shown) pq pq pq = × = (b) ( ) 53 41~ B 15,30 ~ B 15,0.07864355WW     ⇒       ( ) ( ) P5 1P 4 0.0046167 0.00462 (3 s.f.) W W ≥ = −≤ = = ( c ) ( )~ B 15,0.078643W ( ) ( ) ( ) ( ) E 15 0.078643 1.1796 Var 1.1796 1 0.078643 1.0869 W W = = = −= Since sample size = 40 is large, by Central Limit Theorem, 1 40... 1.0869~ 1.1796, 40 40 WWWN ++ =  approximately. ( )P 1 0.138 (3 s.f.)W ≤= 9(a) Let F and G be the mass of a Fuji and Gala apple respectively ( ) 2~ N 205,9F , ( ) 2~ N 180,6G ( ) ( ) 22~ N 205 180, 9 6 ~ N 25, 117FG FG− − + ⇒− ( ) ( )P P0 0.98960 0.990 (3 s.f.) F G FG> = −> = = (b) Required probability ( ) ( ) ( ) 2 2 3!P 203 P 185 2! 0.58793 0.013134 3 0.013620 0.0136 (3 s.f.) FF= > ×<× = ×× = = (c) Let A denotes the mass of an assorted packet of ten apples. ( ) ( )1 2 1 2 10... ...nnA FF F GG G −= + ++ + + ++ ( ) ( ) ( ) ( )E E 10 EA nF n G= +−

( )205 10 180 25 1800nn n= +− = + ( ) ( ) ( ) ( )Var Var 10 VarAn F n G= +− ( )( ) 229 10 6 45 360nn n=+− = + ( )~ N 25 1800, 45 360 An n ++ ( )( ) ( ) 229 ~ N 25 1800 9(205), 45 360 9 9 9 ~ N 25 45, 45 6921 AF n n AF n n ∴− + − + + ⇒− − + ( )P 9 28 0.5AF−>≥ Method 1: Standardisation ( )28 25 45P 0.5 45 6921 nZ n −− >≥ + ( )28 25 45 0n− −≤ 2.93n⇒≥ Least n in an assorted packet is 3. Method 2: Graphic Calculator From GC, ( )P 9 28 0.5AF−>≥ When n = 2, ( )P 9 28 0.3918 0.5AF−>= < When n = 3, ( )P 9 28 0.5095 0.5AF−>= > Least n in an assorted packet is 3. (d) The mass of every apple is independent of one another. Qn Suggested Solution 10( a) Let X be the waiting time for a customer, in minutes. Using GC, unbiased est