NJC 8865 2022 Prelim Solutions vetted

* © NJC 2022 [Turn over NATIONAL JUNIOR COLLEGE SENIOR HIGH 2 Preliminary Examination H1 MATHEMATICS 8865/01 1 ( ) 2 2 2 2 2 3 1 2 2 1 0 ( 2 1) 4 0 4 4 1 4 0 4 8 1 0 mx mx m x mx mx x mm m m m mm − = − − − − + + = − + −  − + −  − +  31 2 m− or 31 2m+ 2 23e e 70xx+ 23e e 70 0xx− −  Let exu= , then 23 70 0uu− −  Consider 23 70uu−− = 0, then 1 1 4(3)( 70) 2(3)u  − −= u = 5 or u = 28 14 63− =− For 23 70 0uu− −  , 285 or 6uu − 28e 5 or e (reject as e 0)6 x x x −  x > ln5 3(i) Let $x, $y, $z be the amount invested in plan I, II and III respectively. x + y + z = 10000 … (1) y = 2x  2x − y = 0 … (2) 0.014(2) 0.035 8(50)( ) 356010000 zxy+ + = 0.028x + 0.035y + 0.04z = 3560 … (3) From GC, x = 20000, y = 40000, z = 40000 2( 8) 8 4(4)(1) 2(4) 8 4 3 3 182 m m − −  −= = = 

2 © NJC 2022 She deposits $20000 in plan I, $40000 in plan II and $40000 in plan III 3(ii) If she deposits in savings plans II and III, the total interest is 400000.035(40000) 8(50)( ) 300010000+= If she deposits $80000 in savings plan IV, the interest ranges from Min 0.037(80000) = $2960 to Max 0.039(80000) = $3120. The average interest is 3.8%, 0.038(80 000) = $3040 Tammy is advised to deposit in savings plan IV as she can earned $120 more against loss $40 over savings plans II and III. Further the average interest is $40 more than the plan II and III. 4 4 2 1 4 2 1 4 1 13d 619 d 19 6ln 11= 36 6ln 4 9 6ln141 111= 6ln 44 xx xxx xx x  − = − + = − −    − − − − −       −    Or 27.75 − 6ln4 24 31 1 13 1e d 3 d p x xx x + − =−   31 1 3 1 111e 6ln 434 p x+ −  = − ( ) 311 111e 1 6ln 434 p+ − = − By GC, p = 1.0275179 = 1. 0275 (4dp) Method 2 31 333e 1 18ln 4 4 p+ − = − e3p + 1 = 59.2967015 3p + 1 = ln(59.2967015) 3p = 4.082553681 − 1 p = 1. 027517894 = 1.0275 (4dp)

3 © NJC 2022 [Turn over 5(i) 2 5 2( 2) 1 22 12 2 xxy xx x − − −== −− =− − 5(ii) 5(iii) The exact area of the region bounded by the curves C1, C2, and the y-axis 1 1 0 1 1 0 1 1 0 25e 2 d 2 1e 2 2 d 2 1e d 2 x x x x xx xx xx − − − −= + − − = + − + − =+ −    11 0 e ln( 2 ) 1 e ln 2 x x−= − + − = − + − 5(iv) 2 5 1 222 xy xx −= = −−− ( ) 2 d1 d 2 y x x = − When x = 4, y = 1.5 and d1 d4 y x = . ( )11.5 4 4yx− = − 2 4 xy += Equation of tangent is 2 4 xy += or 0.25 0.5yx=+ O x y

4 © NJC 2022 5(v) Enter equation of tangent, 0.25 0.5yx=+ . Find intersection with C2 giving x = 6.0262534 From GC, x < 6.0262534 = 6.03 (3sf) 6(i) 32 2 7 8 20 d 3 14 8d C t t t C ttt = − + + = − + For stationary points on C, d 0d C t = 23 14 8 0tt − + = ( )( )3 2 4 0tt− − = 2 43t or t = = t 0.6 2 3 0.7 d d C t 0.68 0 −0.33 slope t 39. 4 41. d d C t -0.97 0 1.03 slope Or 2 2 d 6 14d C tt =− When 2 3t = , 2 2 d2 6( ) 14 10 0d3 C t = − =−  When 4t= , ( ) 2 2 d 6 4 14 10 0d C t = − =  2nd derivative test for either of the points.

5 © NJC 2022 [Turn over Therefore, C is a maximum when 2 3t = and minimum when 4t= . 6(ii) Area of the region ( ) 6 32 0 7 8 20 d 84 t t t t− + + =  The total production cost to manufacture the bottled drinks over a period of 6 months is $84 000. 6(iii) ( ) 0.25100 e 3.2CP −=+ d d P C = ( ) 0.25 0.25100 0.25e 0 25eCC−−− + =− When t = 6, from (i) 326 7(6) 8(6) 20 32C= − + + = d dC P = 0.25(32)25e−− = −25e−8 6(iv) d d d d d d P P C t C t= When t = 6, from (i), d d C t = 3(62) − 14(6) + 8 = 32 d d P t = (−25e−8)(32) = −800e−8 = 0.2683701023 The rate of decrease in profit when t = 6 is $268.37 per month. 7(i) Mean = 0.5(0.83 + 2.41) = 1.62. Let X be the height of the plants. If X is normally distributed then X ~ N(1.62, 1.52), then P(X ≤ 0) = 0.14007109. This means that about 14% of the large number of plants will height that is less than 0*. Hence, a normal distribution would not be a good model. 7(ii) E(X) = 1.62, Var(X) = 1.52. Let X be the sample mean height of the plants. Since the sample size is large, by Central Limit Theorem, 21.5~ N 1.62, 100X   approximately ( )P 1.2 1.6X = 0.444409753 = 0.444 (to 3 s.f.)

6 © NJC 2022 8(i) P(Accident) = ( ) ( ) ( )0.3 0.01 0.5 0.03 0.2 0.06 +  +  = 0.03 P(class H | accident) ( ) ( ) accident class H accident P P = 0.2 0.06 0.40.03 == 8(ii) P(all three drivers are of class H and exactly one has at least an accident) ( ) 2 0.2 0.94 0.2 0.06=    3 = 0.00127 (to 3sf) 9(ai) Method 1 Number of ways Ann and Alice separated from each other = 10! 11 P2 = 399168000 Method 2 Number of ways to bundle Ann and Alice as a unit and arrange the students = 11!2! = 79833600 Require ways = 12! − 11!2! = 399168000 9(aii) Number of ways to bundle 3 boys between Ann and Alice as a unit and arrange them = 7 3   3!2! Arrange the bundle and 7 others in a row = 8! Number of different sitting arrangements required = 7 3   3!2!8! = 16 934 400 (b) Number of ways Ann and Alice are team leaders and Ann’s team has 2 girls and Alice’s team has Case 1: 1 girl (i.e. only Alice), 3 7 2 4 1 3 0 4               = 105 Case 2: 3 girls (i.e. Alice and 2 other girls), 3 7 2 4 1 3 2 2               = 630 Number of ways Ann and Alice are team leaders and exactly one team has 2 girls = 2(105 + 630) = 1470

7 © NJC 2022 [Turn over 10(i) (ii) The equation of the least-squares regression line of p on s is p = 0.010853939s −3.827569368 p = 0.0109s −3.83 (3sf). The equation of the least-squares regression line of s on p is s = 646.8814847 + 79.44195692p s = 647 + 79.4p (3sf). (iii) The product moment correlation coefficient between s and p, r = 0.928578571 = 0.929 (3 s.f) As r is close to 1, there is strong positive linear correlation between s and p. (vi) If p = 70, s = 646.8814847 + 79.44195692(70) = 6207.818469 = 6210 (3sf) The value of the asset is US$ 6210 billion Alternative solution As profits depends on assets, use p on s p = 0.010853939s −3.827569368 70 = 0.010853939s −3.827569368 s = (70 + 3.827569368)/0.010853939 = 6801.914896 = 6800 (3sf) The estimate is not reliable as p = 70 is outside the data range [2.0, 65.8]. 11(i) The probability of a switch being faulty is a constant and the same for all switches. OR A switch being faulty is independent of any other switches being faulty. (ii) np = 0.54 p s 65.8 2 326.7 4914.7 (2488.8, 23.2) × p = 0.0109s −3.83

8 © NJC 2022 np(1 − p) = 0.5319 1 − p = 0.5319/0.54 = 0.985 p = 0.015 n = 0.54/0.015 = 36 (iii) X ~ B(36, 0.015) P(X ≤ 5) − P(X ≤ 1) or P(2 ≤ X ≤ 5) = 0.999984928 − 0.898541026 = 0.101443902 = 0.101 (3sf) (iv) P(X > 3) = 1 − P(X ≤ 3) = 0.002032816 (at least 5sf) Let Y be the number of days where more than 3 switches are found to be faulty out of 30 days Y ~ B(30, 0.002032816) P(Y ≤ 1) or P(Y = 0) + P(Y = 1) = 0.998269258 = 0.998 (3sf) (v) P(X > 3) = 0.002032816 The expected number of days = 0.002032816  30 = 0.060984478 = 0.0610 (3sf) 12(i) Unbiased estimate of the population mean, ( )250 25060 200 1 760250 253 or 60 3 3 xx −=+ = + =  Note: 253.3333333 is wrong as the value is not exact Unbiased estimate of the population variance, ( ) ( )( ) 2 22 2 2501 25060 1 60 1 200 3940013 80059 60 177 xsx  −= − − −  = − =   Note: 222.5988701 is wrong as the value is not exact (ii) Let µ be that the population mean time to repair cars. To test H0: µ = 250 against H1: µ > 250 at 5% level of significance. Under H0, since n = 60, is large, by Central Limit Theorem 39400~ N 250, (177)60X   approximately Formatted: Font: Not Italic, Complex Script Font: Not Italic Formatted: Font: Not Italic, Compl