EJC 9758 2023 Prelim P2 Solution

2023 EJC Prelim Paper 2 Solutions Q1 4 5i 7 (1) (1 i) 8 30 (2) zw zw += −+= By (2), 4 16(1 i) 60(1 i) (3)zw++= + (3) (1) :− (16 11i) 53 60iw+= + 4iw= + By (1): 4 5i(4 i) 7z+ += 3 5iz= − Q2 (a) Consider the following cases: 2 7 3 13xx x− += − ( )( ) ( ) 2 2 6 10 0 6 6 4 1 10 21 6 76 6 2 19 22 3 19 xx x −−= ±− − = ±±= = = ± ( ) 2 7 3 13xx x− −+=− ( )( ) ( ) 2 2 8 16 0 8 8 4 1 16 21 4 xx x −+= ±− = = The roots are 3 19± and 4. (b) From the graph, the solution is 3 19 4 or 4 3 19xx− << <<+

Q3 (a) secyz x= dd sec tan secdd yz zxx xxx= + Sub in the DE, ( )dπ sec tan sec sec 3 π tan 0d dπ sec tan π sec 3 sec π sec tan 0d dπ sec 3 sec 0d dπ sec 3 secd d3 (shown)d π zzxx xzx x x zzxx xzxzxx x z xzxx z x zxx zz x  + + −= + +− = += =− −= 3 π 3 π 13dd π 3ln π e e where e xC x C zxz z xC z zA A − + − −= −= + =± = =± ∫∫ 3 π 3 π esec cos e x x y Ax y xA − − = = When π 3x= and 2y= , 3 π π3π2cos e3 e A A − = = 3 π 31 π cos e e 3cos e where 1 and π x x yx yx a b − −  ∴=  = = =−

(b) 31 πe cos x y x − = For vertical asymptotes, consider cos 0x= The asymptotes closest to y-axis are π 2x=− and π 2x= Q4 (a) 33 Ax y xy+−= Differentiating with respect to x, 22 0dd3 d3 d yyx xy xy x+− − = ( ) 223 d 3dyy yxx x =−− + (*) When d 0d y x = , 20 3 yx= −+ 23yx= (shown) (b) Sub 23yx= into 33 Ax y xy+−= , ( ) ( ) 332 2 33 Ax x xx+−= 3 6327 3xAxx+ −= 63 027 2 Ax x− − = ( ) ( ) ( ) ( ) 2 3 2 4(272 ) 2 27x A−−− ± − = − ( ) 3 22 2 27 1 27 Ax ±= + 3 71 27 12 Ax ±= + 33 11 127 1 27 13 27AxA= ±+ +± =

For more than one stationary point, Discriminant 1 027A+> So 1 27A>− (c) From (*), ( ) 223 d 3dyy yxx x =−− + Differentiate with respect to x , ( ) 2 2 2 13 d dd d ddd 6 6d y yy yxx xx xy x y =−  +−+−  When d 0d y x = , 23yx= , so we have ( ) 222 2 d 63 d 3 yxx x x − =−  ( ) 2 4 2 d27 6 d yxx x x − =− 2 24 3 d6 6 d 27 27 1 yx x xx x −= − −− = When 3 71 27 12 Ax += + , 2 23 d6 6 0 d 27 1 1 27 y xx A =− + = −< − so this is a maximum point. When 3 71 27 12 Ax −= + , 2 23 d66 0 d 27 1 1 27 y x Ax = =− + > − so this is a minimum point. Q5 (a) Since p and q are perpendicular, a direction vector parallel to q is 2 0 1   − .

Normal to q 22 2 1 0 2 4 22 11 4 2 −         = × −= −= −        −−     Equation of q: 1 51 r2 3 2 7 2 42 227xyz    = −=    ++=  (b) 2 10xz−= 227xyz++= Solve using GC: 15 2 51 4 xz yz zz = + = − = 52 115 404 x yz z      =+−       Let 1 4 zµ = , equation of m is 52 1 5 , 04 µµ      = +− ∈       r  (c) Let F be the foot of perpendicular from A to q. Method 1 21 : 1 2 , 62 AFl αα      = +∈    −   r  Since F lies on AFl , 2 12 62 OF α α α + = +−+   for some α∈  Since F also lies on q, (2 ) 2(1 2 ) 2( 6 2 ) 7αα α+ + + + −+ = 89 7 5 3 α α −+ = = 11/ 3 13 / 3 8/3 OF  ∴= −   5/3 10 / 3 5 10 / 3 k AF  = = =   

Method 2: Since p and q are perpendicular, the foot of perpendicular F lies on m. 52 15 4 OF ω ω ω + = −   for some ω∈  32 64 AF OF OA ω ω ω + =−= −  +       Since AF is perpendicular to m, 32 2 5 50 64 4 ω ω ω +     − ⋅− =    +   6 4 25 24 16 0 45 30 2 3 ωω ω ω ω ++ ++ = =− =− 11/ 3 13 / 3 8/3 OF  ∴= −   5/3 10 / 3 5 10 / 3 k AF  = = =    (d) The other point lying on p with shortest distance 5 units from q is A’, the reflection of A about m. By Ratio Theorem, ' 2 OA OAOF +=     11/ 3 2 16 / 3 ' 2 2 13 / 3 1 23 / 3 8/3 6 2/3 OA OF OA        = −= − =       −−         Both the lines must be parallel to m. ∴Equation of lines: 22 1 5 , 64 αα    = +− ∈  −  r  and 16 / 3 2 23 / 3 5 , 2/3 4 ββ      = +− ∈       r  Q6 (a) Let P( )AC x∩= . 1P( ') 5AC x∩= −

3P( ' ) 10ABC x∩∩ = − 11 3 1P( ' ') 2 5 5 10 10ABC x x  ∩∩ = −− − =−     13 1 0 and 0 and 05 10 10x xx−≥ −≥ − ≥ 1 31 and and5 10 10xxx≤≤≥ Hence, 11 10 5 x≤≤ . Greatest value of 1P( ) 5AC∩= Least value of 1P( ) 10AC∩= (b) (c) 1P( ) 5A =

1 1 31212 5 60( )1PB C −= ′∩= − 31 3 49P( ) 60 10 60C =+= For A and C to be independent, 1 49 49P( ) P( ) P( ) 5 60 300AC A C∩= × = ×= Q7 (a) ( )E1X Y µ+ = + ( )Var 1 2 3YX + =+= ( ) 2 ~N 1 , 3YX µ++ ( ) 4P .0 043XY+<≤> Using graph on GC, 11.1013 2. 013µ<−< 21.10 .10µ<−< (b) ( ) ( )12E 2 2 10 20nXX YnX n+ +… = =+ −− − ( ) ( )2 12Var 2 2 2 8nXX Yn X n−= +++ =… ++ ( )12 82 ~ N 20,nX X YX nn+ +… −−++ ( )12P 2 0.0310nXX X Y− ≥ <+ +…+ From GC, n ( )12 0P2 1nXX X Y++ −…+ ≥ 20 0.0294 21 0.0473 So largest n is 20 (c) We need to assume that X and Y are independent.

Q8 Solution (a) Whether or not a vase is defective is independent of other vases being defective. The probability that a vase is defective is a constant. (b) ( ) ( ) ( ) ( ) ~ B 30, 0.04 P 2 1P 2 0.11690 0.117 3 f X XX s >= − ≤ = = (c) Let Y be the no. of days where more than 2 defective vases are found, out of 5 days. ( ) ( ) ~ B 5, 0.11690 P 1 0.893 Y Y ≤= (d) Let V be the total no. of defective vases found in 5 days. ( )~ B 150, 0.04V ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 123 3 12 3 Required probability 4!P 4 P 1 P 0 2! 3! 4!P 3 P 2 P 0 2! 3! P5 0.0056980 0.03500.16280 XX X XX X V = = = ××  + = = = ×× = = = = Q9 (a) No. of results 32 5 5!C= × (b) No. of results no. of resultsno. of results with traits from without restrictions only 1 domain  = −   8 524165120 5! 4C= − ×× 24138240= (c) Case 1: Domain repeated is Cognition No. of results   pick two slots for the Cognitionchoose 1 trait traits among thechoose 2 each from the no of wayslast 3 and placeCognition non-Cognition to permute thethem theretraits domain remainin 8 83 3 21 2 ( ) 2! 3!CC C= × × ××     g 3 traits 516096= 24165120=

Case 2: Domain repeated is not Cognition No of results     Pick 1 trait Place No of ways3 possible ways Pick 2 traits from each of Cognition to permuto pick "repeating" from "repeating" the r emaining trait in last domain domain domains three slots 3 8 83 3 11 21 ( ) 4!C C CC= × ×××  te remaining 4 traits 3096576= Total no. of results 516096 3096576= + 3612672= (d) A: event that Rapport traits are consecutive B: event that Planning traits are consecutive Method 1 ( ) ( ) ( )( ) ' () () () ( ' )n A B nE n A B nE nA nB A∪ = − ∪= − + ∩ Case 1: R consecutive No. of results 4! 2! 48=×= Case 2: P consecutive, R not consecutive No. of results 3 22! 2! 2! 24C=× ××= Total no. of results no. of results no. of results with without restrictions R consecutive no. of results with P consecutive, R not   = −     − 5! 48 24=−− 48= Alternative 1: ( ) ( ) ( )( )( ) ' ( ) ( ) ( ') ( ')n AB n E n AB n E n AB n BA n AB∪ = − ∪= − ∩+ ∩+ ∩ Complement cases are P consecutive but R not, R consecutive but P not and (R consecutive and P consecutive). Case 1: P consecutive, R not consecutive No. of results 3 22! 2! 2! 24C=× ××= Case 2: R consecutive, P not consecutive No. of results same as case 1 24= = A B We want the shaded region.