NYJC 8865 2022 Prelim Solutions vetted

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 1 of 9 Q1 Suggested Answers For ( ) 23 4 2 0k x x k− + −  , ( )( ) ( )( ) 2 2 2 3 4 0 and 2 4 3 4 0 3 and 4 12 16 04 3 and 4 3 1 04 3 and 4 1 1 04 31 and or 144 k k k k k k k k k k k k k k k −  − − −   + −   − −   + −   −  Combining solution on a number line, 1k  Q2 Suggested Answers (i) 32y ax bx cx d= + + + At ( )0,2 , d = 2 (ii) 2d 32d y ax bx cx = + + Since ( )0,2 is a maximum point, c = 0 1195, 3 −− is a minimum point, ( ) ( ) 2 3 5 2 5 0 75 10 0 (1) a b c ab − + − + = − = −−−−−−−−− ( ) ( ) 32119 5 5 23 125125 25 (2) 3 ab ab − = − + − + − = −−−−−− Using GC, 2 3a=− , 5b=− Hence 322 523y x x=− − + Q3 Suggested Answers (i) 3 eln ln e 3ln(1 2 ) 3ln(1 2 ) (1 2 ) x x x x x x  = − − = − − − ( )3 1 2 d e dln 3ln(1 2 )dd (1 2 ) 12 32 1 2 16 122 x xxxx x x x xx −  = − − − −=−  − =+ −

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 2 of 9 (ii) ( ) 2e 1 e 1 e2 1 2 2 2 32d 94 12 d 2 12 9ln 2 12 9ln 2 12 9ln 1 2 12 9 14 2 12 5 xx x xx x x x x e e e ee ee − − −  − = − + = − + = − + − + + − = − + − = − −   Q4 Suggested Answers (i) 37 37 37 37 14 2e d 14 14ed 0 14 14e 1e ln1 3 7 3 7 x x x x yx y x x x − − − − =+ =− =− = =− = At 3 7x= , 337 7314 2e 6 2 87y − = + = + = Coordinates of turning point: 3 ,87   (ii) (iii) 44 d1 14 2e & 14 14e d yxy x −−=  = + = − Equation of the tangent: ( ) ( )( ) 4414 2e 14 14e 1yx −−− + = − − ( ) ( ) 4 4 4 44 14 14e 14 14e 14 2e 14 14e 16e yx yx − − − −− = − − + + + = − + (iv) Area required ( )( ) 1 3 7 4 4 0 114 2e d 1 16e 14 2e2 xxx − − −= + − + + ( ) 30,2e

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 3 of 9 137 2 0 2e7 7.16487 5.57 (3 s.f.) x x −= − − = Alternative: Area required ( ) 11 3 7 4 4 00 14 2e d 14 14e 16e d 5.57 (3 s.f.) xx x x x− − −= + − − + =  Q5 Suggested Answers (i) Volume of container, For maximum volume, (rejected) or r E.g. 16.65 E.g. 16.67 10.4615 > 0 0 −2.0948 < 0 Slope / − \ Hence gives a maximum volume of 29088.82 cm3 Alternative method When ,  gives maximum volume of 29088.82 cm3 (ii) Amount of discarded metal Alternative method Amount of discarded metal = ( )( ) 22 2 40rr  − ( ) 22 80 rr =− 100 4hr=− 2V r h= ( ) ( ) 2 23 100 4 4 25 rr rr   =− =− ( ) 2d 4 50 3d V rrr =− d 0d V r = ( ) ( ) 24 50 3 0 4 50 3 0r r r r − =  − = 0r= 50 3r = 50 3 −   50 3 50 3 +   d d V r 50 3r = ( ) 2 2 d 4 50 6d V rr =− 50 3r = 2 2 d 200d V r =− 50 3r = ( ) 2100 40 2 40 rh= − − ( ) 24000 2 40 100 4rr= − − − 2160 2rr =− ( ) 22 80 rr =−

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 4 of 9 (iii) ( ) 2e 1.05 2 80 e 1.05rrC D r r r r −−= + = − + From the sketch, the least cost to discard the waste metal from a container is $7.99. (iv) Maximum volume occurs when 50 16.667 6.49643r= =  Hence, the supervisor’s claim is false. (iv) When 50 3r = , 2 50 350 50 502 80 e 1.05 $17.503 3 3C  −      = − + =            Q6 Suggested Answers (i) Case 1: 5 boys, 0 girls: No. of ways 10 5 252C== Case 2: 4 boys, 1 girls: No. of ways ( )( ) 10 15 41 3150CC== Case 3: 3 boys, 2 girls: No. of ways ( )( ) 10 15 32 12600CC== Total no. of ways 252 3150 12600 16002 = + + = (ii) 5 2C ways to insert the Chairperson & Secretary such that there are exactly 2 students between them 2! ways to arrange between Chairperson & Secretary 4! ways to arrange the other 3 students and the Civics tutor Total no. of ways ( )( )( ) 5 2 2! 4! 480 C= = (ii) Alternative: Complement method ( )( )6! 5! 2! 480−= C r 5 20 (5, 9.58196) (20, 21.000001) (6.4964, 7.9895)

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 5 of 9 Q7 Suggested Answers (i) Let X be the number of ripe apples, out of n. ~ B(12, )Xp Var( ) 12 (1 ) 1.53X p p= − = 0.15p= (rejected as 0 5 1. p ) or 0.85p= (ii) ~ B(12,0.85)X P( 9) 1 P( 8) 0.90779 (5 s.f.) 0.908 (3 s.f.) X X  = −  = = (iii) Let Y be the number of boxes with at least 75% of the apples being ripe, out of 20. ~ B(20,0.90779)Y P(13< 18) P( 18) P( 13) 0.56062 (5 s.f.) 0.561 (3 s.f.) Y Y Y =  −  = = (iv) Let T be the number of apples that are not ripe, out of 240 apples. ~ B(240,0.15)T P( 30) 0.15995 (5 s.f.) 0.160 (3 s.f.) T  = = Alternative Let T be the number of apples that are ripe, out of 240 apples. ~ B(240,0.85)T P( 210) 1 P( 209) 0.15995 (5 s.f.) 0.160 (3 s.f.) T T  = −  = = Q8 Suggested Answers (i) P( )P( | ) 0.7 0.7 P( ) 0.35P( ) 0.5 0.7 ABBA A A =  =  = = P( ) P( ) P( ) P( ) 0.9 0.5 0.35 0.75B A B A A B=  − +  = − + = (ii) P( ' ')AB represents the probability that both A and B do not occur. P( ' ') 1 P( ) 1 0.9 0.1A B A B = −  = − = (iii) P( | ) 0.7 P( )B A B=  A and B are not independent. Alternative: P( )P( ) (0.5)(0.75) 0.375 P( )A B A B= =    A and B are not independent.

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 6 of 9 (iv) Q9 Suggested Answers (i) (ii) P(Ariana wins the match) (0.45)(0.9) (0.45)(0.1)(0.85) (0.55)(0.175 )(0.35) 0.47694 (5 s.f.) 0.477 (3 s.f.) = + + = = (iii) P(Ariana lost the second set|Ariana lost the match) P(Ariana lost the second set and lost the match) P(Ariana lost the match) (0.45)(0.1)(0.15) (0.55)(0.825) 1 0.47694 0.88039 (5 s.f.) 0.880 (3 s.f.) = += − = = A B 0.35 0.1 0.15 0.4 0.45 0.9 0.1 0.55 A A B 0.175 A B 0.825 B 0.85 0.15 A B 1st Set 2nd Set 3rd Set 0.35 0.65 A B

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 7 of 9 Q10 Suggested Answers (i) (ii) Using GC, the product moment correlation coefficient, r = 0.98762, i.e. 0.988 Since r is close to 1, it suggests that as the years of working experience of an employee, x, increases, the yearly salary, y, increases in a strong linear correlation. This is also consistent with the scatter diagram which shows that as the years of working experience increase, the yearly salary also increases in a linear trend (iii) Using GC, equation of the regression line is 8.8670 30.223yx=+ , i.e. 8.87 30.2yx=+ (iv) When 6x= , ( )8.8670 6 30.223 83.425y= + = Hence yearly salary of an employee with 6 years of working experience is $83425 Estimate is reliable as 6x= lies within the data range 1.3 10.5x and r is close to 1. (v) Value of r is not changed as scaling does not affect the relationship between the variables. Q11 Suggested Answers (i) Unbiased estimate of population mean, 141.68 35x = = 4.048 Unbiased estimate of population variance, ( ) 2 2 141.681 573.7434 35s  =−   = 0.0064518 = 0.00645 (ii) Let X be the random variable denoting the mass of a paperclip and µ the population mean mass. To test 0 1 :4 :4 H H   =  at 1% level of significance y x 1.3 10.5 121.9 43.5 (iii)

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 8 of 9 Under 0H , since n = 35 is large, by Central Limit Theorem, 0.0064518~ N 4, 35X   approximately ( )4 ~ N 0,1 0.0064518 35 XZ −= Using a one-tailed test, x _ = 4.048 gives p-value = 0.00020364 Since p-value < 0.01, we reject 0H and conclude that at 1% level of significance, there is sufficient evidence that the quality control officer’s claim that the mass of paperclip is understated is justified. (iii) 1% level of significance means that there is a probability of 0.01 of wrongly concluding the mean mass of paperclip is understated when it is not. (iv) Given ( ) 2~ N ,0.08X  To test 0 1 :4 :4 H H   =  a