EJC 8865 2022 Prelim Solutions vetted

Error! Reference source not found. EJC Q1 For ( ) 2 1 2 1kx k x k+ − + + always negative, Discriminant < 0 and 0k  . ( ) 2 2 22 2 40 1 4 (2 1) 0 2 1 8 4 0 7 6 1 0 (7 1)( 1) 0 b ac k k k k k k k kk kk − − − +  − + − −  − − +  − +  11 or 7kk−  and 0k  Combining solutions, 1k − Set of values of k is { : 1 }kk − Q2 (i) Let the mass (in kg) of arabica, robusta and liberica coffee beans brought in by Starluck Coffee on International Coffee Day be x, y and z respectively. 550 ............(1)x y z+ + = 4 1.5 3.5 1685 ............(2)x y z+ + = ( ) ( ) ( )4 4 44.00 1.50 70 3.50 13155 9 5 23.2 2.8 245 13153 26 1560 ............(3)3 x y x x y x xy + + − =  + + − = += Using the GC, 240, 180, 130x y z= = = (ii) Profit ( ) ( ) ( )1315 240 3.70 180 1.10 130 3.20 187= − − − =− Starluck Coffee made a loss of $187 on International Coffee Day -1 1/7 -1 1/7 0

Q3 (a) 52 3 421 33 2 d2 3d 225 3 2 −  −     = − +        xxx xx xx (b) ( ) ( ) 1 0 11 2 0 11 2 0 11 22 1 d 32 3 2 d 32 ( 2)(0.5) (3 2) (3 0) [1 3] 31 x x xx x − − =−  −=− =− − − −  =− − =−   Q4 (a) 20 px−= 2x p= (b) At A, y=0 3 3 0 3 ln(2 ) ln(2 ) 3 2e 2e px px px x p = − − −= −= −= x-coordinate of A is 32e p − . (c) d d 2 2 y p p x px px −=− =−− At point A,

Error! Reference source not found. 3 3 3 d d 22 22 yp x ep p p e p e =  −−  = −+ = Equation of tangent at A: 3 3 20 peyx ep  −− = −   3 33 33 2 21 peyx ee pyx ee −=− = + − Q5 (i) (ii) y = 1 x = 0 (-9, 0) O x y = 1 x = 0 (-9, 0) O

From the GC, the point of intersection is ( )1.5,7 Area   3 2 30 2 3 2 92 4 d 1 d 8.25 9ln 338.25 9ln 9ln 22 36.75 9ln 9ln 2 26.75 9ln 3 a a x x x x xx aa aa aa = + + + = + + = + + − − = + + − = + +  Alternative (using area of trapezium) ( )   3 2 3 2 1 3 9 4 7 1 d22 8.25 9ln 338.25 9ln 9ln 22 36.75 9ln 9ln 2 26.75 9ln 3 a a xx xx aa aa aa = + + + = + + = + + − − = + + − = + +  Q6 (i) 0.5(0) 1 When 0, 112e 0.4(0) ee − = = − − = t C 1 e million dollars (ii)

Error! Reference source not found. 0.5 1 0.5 1 d 2e (0.5) 0.4 0d e 0.4 0.5 1 ln 0.4 0.16742 0.167 0.36515 0.365 − − = − = = −= =  = t t C t t t t C Method 1: Using 1st derivative test t 0.167- 0.167 0.167+ d d C t E.g. use 0.166=t 0.5(0.166) 1 4 d 2e (0.5) 0.4d 2.84 10 0 − − =− =−   C t 0 E.g. use 0.168=t 0.5(0.168) 1 4 d 2e (0.5) 0.4d 1.16 10 0 − − =− =   C t Slope of tangent \ - / Value is a minimum Method 1: Using 2nd derivative test 2 0.5 1 2 2 2 d e (0.5)d dwhen 0.16742, 0.20000 0d Value is a minimum −= = =   tC t Ct t (iii) (iv) (0.167, 0.733) (0, 1/e) C (millions) t (years) O

3 0.5 1 1 12e 0.4 e d 1.8330 1.83− −− = t t t The total cost incurred between 1st Jan 2023 to 1st Jan 2025 is 1.83 million dollars. (v) Using GC, when 5 ,12=t d 2.53453 2.53d P t = Rate of increase of total profit is 2.53 million dollars per year. Q7 (i) Number of ways = 4 6 8 2 3 5 6720C C C  = (ii) Number of ways to arrange 17 units ( )4 5 7 1 pair of sisters 17!+ + + = Number of ways to arrange within the group of sisters 2!= Probability 17! 2! 1 18! 9 == (iii) ( ) ( ) P 4 students from School of Engineering are all separated | 2 sisters are next to each other P 4 students from School of Engineering are all separated 2 sisters are next to each other P 2 sisters are ne = ( )xt to each other Number of ways to arrange all students except those from the School of Engineering ( )5 7 1 pair of sisters 13!+ + = Number of ways to arrange within the group of sisters 2!= Number of ways to slot the students from the School of Engineering 14 4 4!C= Probability 14 413! 2! 4! 14318! 1 340 9 C   == Alternatively, Probability 14 413! 2! 14318! 1 340 9 P == Q8

Error! Reference source not found. (a) ( )P '|AB represents the conditional probability that event A do not occur given that event B has already occurred. (b) ( )P '| 0.8AB = ( ) ( ) ( ) P' 0.8P( ) P ' 1.6 P( ) P( ) P ' 0.728 1.6 0.28 (Shown) AB B A B p A B A A B pp p  = =  = +  =+ = (c) ( ) ( ) ( )P P P 'A B B A B = −  2 1.6 0.4 0.112 pp p =− = = Since ( )P 0.112 0AB =  , events A and B are not mutually exclusive. (d) Q9 (a) P(student is in Year 1 and takes up sports) 320 16 1100 55== (b) P(student is in Year 2  takes up sports) 285 154 61 320 41 1100 55 + + +== Alternatively, P(student is in Year 2  takes up sports) =P(student in Year 2) + P(takes up sports) − P( in Year 2 and takes up sports) A B 0.112 0.448 0.168 0.272

500 605 285 1100 1100 1100 41 55 = + − = (c) P(student in Year 2  does not take up performing arts) 346 173 800 400== (=0.4325) (d) Required Probability 3 2 195 194 905C 1100 1099 1098    =        = 0.0774 Q10 (i) Let X represent the number of rotten avocados in a box of 16 avocados. ~ 16, 100 pXB   Using mean np= 16 3.52100 22 (Shown) p p  = = (ii) ( )~ 16,0.22XB ( ) ( )3 1 3 0.48143 0.481 P X P X = −  = = (iii) Let Y represent the number of boxes with more than 3 rotten avocados, out of 15. ( )~ 15,0.48143YB ( ) ( )54 0.078234 0.0782 P Y P Y =  = = (iv) Probability ( ) 6 1 0.48143 0.019446 0.0194 =− = = Q11 (i) Let X denote the mass of one oatmeal cookie, X ~ N (40, 1.2 2) in grams. Let Y denote the mass of one white chip cookie, Y ~ N (45, 0.4 2) in grams.

Error! Reference source not found. ( )( ) ( )1.02 40 40.8 0.25249 0.252 P X P X =  = = (ii) ( ) ( ) 2 44.9 44.9 0.40129 0.16103 0.161 P Y P Y   = = = (iii) Let A = ( )1 2 3 60.04 ...X X X X+ + + + ( ) ( ) ( ) ( ) 2 2 2 2 2 E( ) 0.04 40 40 40 ... 40 9.6 Var 0.04 1.2 1.2 1.2 ... 1.2 0.013824 ~ N 9.6, 0.013824 A A A = + + + + = = + + + + = Let B = ( )1 2 3 100.06 ...Y Y Y Y+ + + + ( ) ( ) ( ) ( ) 2 2 2 2 2 E( ) 0.06 45 45 45 ... 45 27 Var 0.06 0.4 0.4 0.4 ... 0.4 0.00576 ~ N 27, 0.00576 B A B = + + + + = = + + + + = ( )~ N 36.6, 0.019584AB+ ( )36.5 0.23743 0.237 P A B+  = = (iv) ( ) ( ) 2 1 2 80 E( ) 10 Var 2.5 1 ...80 T T T T T T = = = + + + Since 80n= is large, by Central Limit Theorem, 22.5~ N 10, 80T    approximately. ( ) ( )P 0 10 0.5 P 10 10.5 0.46318 0.463 TT −  =   = = Q12 (i) Using GC, unbiased estimate of the population mean 37.15=x unbiased estimate of the population standard deviation 4.0734=s unbiased estimate of the population variance 22 4.0734 16.593 16.6= = s (ii)

The probability of any student being selected for the sample is the same and the selection of any student is independent of the selection of other students. (iii) Let X denote the studying time, in hours, of one randomly chosen student Let  be the population’s mean studying time. 0 1 H : 38 agains Test t H : 38   =  at 10% Level of significance Under 0H , since n = 40 is large, by Central Limit Theorem, 24.0734~ N 38, 40X   approximately. Using a one-tailed test, 37.15=x gives value 0.093459 0.0935p− =  Since value 0.0935 0.1p− =  , we reject 0H . Hence, there is sufficient evidence at 10% level of significance to conclude that the mean studying times of the students are less than 38 hours/ management has overstated the mean studying times of the students. The teacher’s suspicion is valid. (iv) Sample mean, 38.4=x 0 1 H : 38 agains Test t H : 38   =  at 10% Level of significance Under 0H , since n = 90 is large, by Central Limit Theorem, 2 ~ N 38, 90X   approximately. ( )~ N 0,1XZ n   −= critical value 0 critical value Rejection Region Rejection Region