TMJC 8865 2022 Prelim Solutions vetted

Page 1 of 11 2022 H1 MATH (8865/01) JC 2 PRELIMINARY EXAMINATION SOLUTIONS Qn Solution 1 System of Linear Equations Let x, y and z be the original selling price for a waffle, a scoop of ice cream and a cookie in dollars. 2 (0.6) (3 3) 20.2 1.2 3 23.2x y z x y z+ − + =  + + = --- (1) 0.6 (2 3) 2 12.6 0.6 2 2 15.6x y z x y z+ − + =  + + = --- (2) 3x(0.6)+(6y – 3(3)) + 5z = 36.8  1.8 6 5 45.8x y z+ + = --- (3) Using GC, 6, 5, 1x y z= = = the original selling price for a waffle is $6. Qn Solution 2 Techniques of Differentiation & Techniques of Integration (a) ( ) ( ) ( )( ) ( ) ( ) 2 2 3 3 d1 d 2 1 3 d1 13d2 1 2 1 3 32 3 13 x x xx x x − −   + =+  = − + =− + (b) 1 12 1 12 1 12 1 e d .45 1eln 4 5 + 15 2 1 ln 4 5 2e 5 x x x xx xC xC + + +  + − =− − + =− − +  , where C is an arbitrary constant

Page 2 of 11 Qn Solution 3 Equations and Inequalities (a) 2 2 2 2 (2 ) 3 5 2 (2 ) 3 5 0 2 (5 ) 5 0 x k x k x x k x x k x k x k + − − =− − + − + − + = + − + − = Since 22 (2 )y x k x k= + − − and 35yx=− − intersect at two distinct points, Discriminant 0 ( ) ( )( ) 2 2 2 (5 ) 4 2 (5 ) 0 25 10 40 8 0 2 15 0 5 3 0 3 or 5 kk k k k kk kk kk − − −  − + − +  − −  − +  −  Set of values =  : 3 or 5k k k −  (b)(i) (b)(ii) 21 ln( 1) 11 21 1 ln( 1)1 x xx x xx + − − − +  + −− Range of x values: 1 5.97x (3 s.f) x y x y

Page 3 of 11 Qn Solution 4 Applications of Differentiation (i) ( ) ( ) ( ) 2 2 2 ln 4 7 d 2 4 d 47 24 0 2, ln 3 47 y x x yx x xx x xy xx = − + −= −+ − =  = = −+ Coordinates of stationary point: ( )2,ln 3 (ii) (iii) When d40, ln 7, d7 yxy x= = =− Equation of tangent: ( )4ln 7 0 7yx− =− − 4 ln 77yx=− + (iv) 4 ln 77yx=− + When 0, ln 7xy== and 70, ln 74yx== ( )7 has coordinates ln 7,0 and has coordina tes 0,ln 74PQ   ( ) ( ) 2 21 7 7Area of triangle ln 7 ln 7 ln 7 units2 4 8OPQ ==  𝑦 (0, ln 7) (2, ln 3) 𝑂 𝑥 ( ) 2ln 4 7y x x= − +

Page 4 of 11 Qn Solution 5 Cross topical – curve sketching, differentiation, integration (i) ( )0.04 4 195 174e 46.727 A − =− = sales = $46727 (nearest dollar) (ii) Using GC, d 5.0539973d A t = Rate of change of weekly sales is $5054 per week (nearest dollar) (iii), (iv) (iv) ( ) ( ) 2 0.04 0 2 0.04 0 3 2 0.04 0 3 2 0.04 3 2 0.04 0.25 9 30 195 174e d 0.25 9 165 174e d 0.25 9 174 165 e3 2 0.04 0.25 9 174 174 165 e3 2 0.04 0.04 19 165 4350e 435012 2 k t k t k t k k t t t t t t t t t k k k k k k − − − − − − + + − − = − + − + −= + − − − = + − − − −  =− + − − +   The integral represents the increase in total sales in the first k weeks due to the advertising campaign. (v) For an eight-week advertising campaign, 8,k = increase in total sales ( ) ( ) ( ) ( )32 0.04 819 8 8 165 8 4350e 435012 2 − =− + − − + 116.585= thousands > $100 000 which is more than the total cost incurred for the campaign. Hence, the company should release the cereal on the market with the advertising campaign. t A (0,21) (0,30) 20.25 9 30y t t=− + + (15,108.75) o

Page 5 of 11 Qn Solution 6 Normal Distribution ( )~ 0, 10P 0.748, 10 0.66821 0.66821 10 (1 1 ) ZZ N     − = − =− − = −−− ( ) ( ) ( ) P 18 0.748 0.725 P 18 0.023 18P0 ~ 0, 1.023, 18 1.9954 1.9954 18 (2) ZN X X Z       = − = − = − = + = −−− Solving equations (1) and (2), µ = 12.0 ,  = 3.00

Page 6 of 11 Qn Solution 7 Probability (i) ( ) ( ) ( ) ( ) 1Since P | and P 2 and 0,2 P | P A B p A p p A B A = =   Hence A and B are not independent events. (ii) ( ) ( ) ( ) ( ) 2 1P| 2 P 1 P2 1P 2 A B p AB pB A B p =  = = ( )P 0.6AB= ( ) ( ) ( )P +P P 0.6A B A B −  = 212 + 0.6 2p p p−= 26 1.2pp−= 2 6 1.2 0pp− + = Using GC, 0.207151 or 5.79284 (reject since 0 0.5)p p p= =   0.207p= (iii) ( ) ( ) ( ) 2 P ' ' 1 P 1 1 0.2071512 = 0.979 A B A B = −  =−

Page 7 of 11 Qn Solution 8 Binomial Distribution (i) The probability that any randomly chosen strawberry is bruised is constant at 2 0.02.100 = Whether a randomly chosen strawberry is bruised is independent of any other strawberries. (ii) The probability that a strawberry is bruised may not be the same across all strawberries due to the nature of the harvesting process. (iii) ( )B 20, 0.02X ( )P 0.272(3 s .1 .f )X = = (iv) ( ) ( )P 1 P 0.059899 0.0592 9 (3 s.f1 .)XX =−  == (v) Let Y be the number of punnets in a crate of n punnets that have at least 2 bruised strawberries. ( )0.05 899B, 9Yn ( )P0 1 .2Y  From GC, when 48,n= ( )P 0.20928 2,1 0.Y = when 49,n= ( )P 0.19983 2.1 0.Y = Hence, least 49.n=

Page 8 of 11 Qn Solution 9 Probability (Tree Diagram (i) Let R be the event Philip got a rare item from the loot box Let C be the event Philip got a common item from the loot box (ii) P(at least 1 common item from the loot boxes) ( ) ( )( ) ( ) 1P 1 0.4 0.24 0.144 0.986 3 s.f. RRR=− =− = (iii) P(at least 1 rare item | common item in the 2nd loot box) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) nd PP P common item on 2 loot box 0.4 0.76 0.6 0.48 0.676 0.4 0.76 0.6 0.48 0.842 3 s.f. RC CCR+= += + = Or ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) nd P P P P common item on 2 loot box 0.4 0.76 0.312 0.4 0.76 0.688 0.6 0.48 0.676 0.4 0.76 0.6 0.48 0.842 3 s.f. RCR RCC CCR++= ++= + = Qn Solution R C 0.144 0.856 R C 0.312 0.688 R C 0.312 0.688 R C 0.676 0.324 R C 0.24 0.76 R C 0.52 0.48 R C 0.4 0.6 1st loot box 2nd loot box 3rd loot box

Page 9 of 11 10 Correlation and Regression (i) Using GC, 5.2875x = Let k be Candidate H’s weight (in kg). 70.3 58.7 90.3 66.7 68.5 55.3 62.3 8 472.1 8 ky k + + + + + + += += Substitute ( ),xy into the regression line y on x, ( ) 101.68 6.1332 472.1 101.68 6.1332 5.28758 81.9 kg (3 s.f.) yx k k =− + =− = (ii) (iii) Using GC, 0.908 (3 s.f.)r=− Since 0.908r=− is close to 1− , there is a strong negative linear correlation between x and y. As the hours of sleep per day increases, the weight of the person decreases. (iv) Using GC, 98.870 5.7766 (5 s.f.) 98.9 5.78 (3 s.f.) yx yx =− =− (v) For 7x= , ( )98.870 5.7766 7 58.4 kg (3 s.f.)y= − = Since 7x= is within data range and r is close to 1− , it indicates a strong negative linear correlation between x and y. Hence, the estimate is reliable. (vi) There are other factors like exercising and dieting which contribute to weight loss. Hence, it is incorrect to claim that sleeping more causes weight loss. y x x x x x x x x x 90.3 55.3 3 8.2

Page 10 of 11 Qn Solution 11 Hypothesis Testing (i) An unbiased estimate for population mean is 17550 58530x == . An unbiased estimate for population variance is ( ) 2 2 1 17550 8325010350000 or 2870 to 3 s.f.29 30 29s = − = (ii) Let  denotes the population mean mass of cakes (in grams). 0 1 H : 600 H : 600   =  Under 0H , since 30n= is large, by Central Limit Theorem, 83250 29~ N 600, 30X      approximately. Using GC, value 0.062587p−= Since 0H is not rejected, value 100p −   0.062587 100 6.2587 : 0 6.25          (iii) 0 1 H : 600 H : 600   =  At 5% significance level, reject 0H if value 1.95996 or value 1.95996 600 6001.95996 or 1.95996 2000 2000 40 40 586.14 or 613.859 586 or 614 zz xx xx xx − − −  −− −     