YIJC 8865 2022 Prelim Solutions vetted

Qn Solutions 1 For ( )2 10 22 knx k x − + + + −   , Discriminant < 0 and 10a=−  2 40b ac− and 10a=−  ( ) ( ) 2 2 2 1 4 1 0 22 2 1 2 2 0 4 1 2 0 knk k k k n k k n + − − −   + + + −  + + −  Let 2 4 1 2 0knk + + − = , then ( )24 4 4 1 2 2 2 4 1 2 2 32 nk n n − − −= =− − + =−  +   Therefore 2 3 2 2 3 2n k n− − +  − + + . Qn Solutions 2 Let $x, $y and $z be the cost of a jigsaw puzzle, a spinning top and a Rubik’s cube respectively. 4 2 5 264.5 34 4 2 49 4 2 4 40 9 3xz x y z xz x y z x y z + + = = + = + + − −  = = Using GC, 30, 16, 22.5x y z= = = The cost of a spinning top is $16.00 −2 − √3 + 2𝑛 −2 + √3 + 2𝑛

Qn Solutions 3(a) ( ) ( ) ( ) ( ) 1 3 2 3 3 32 2 3 23 2 d 2 d 21dd 1 12 1 32 31 xxx x xx xx − − −  =− − = − − − =− (b) ( ) ( )( ) 2 2 2 2 2 4ln ln 4 ln 1 e1e d ln 4 ln 1 ed 4 2e 4 1 e 1 2e 1e x xxx xxx x xx x xx  = − − − −− =+ − =+ − (c) 3 62 2 1 3 2 6 2 1 3 62 1 04 4 11 e d 1 e d 11ln e 2 1 1 1ln 3 e ln1 1 e3 2 2 71 ln 3 e62 x x x xxx xxx x x − −− − +− = + − = − +    = − + − − +       = + −  

Qn Solutions 4(a) (b) ( )ln 3 7y x= + d3 d 3 7 y xx= + At 7 2x= , d 3 6 7d 35 37 2 y x ==  + , 7 35ln 3 7 ln22y     = + =        Equation of tangent to the curve: 35 6 7ln 2 35 2yx    − = −       3535 35ln 6 21 2 3535 6 35ln 21 2 yx yx − = − − = −  (c) Numerical value of area under curve ( ) 1 1 ln 3 7 d 3.8269 (4 .d.p) xx − =+ =  𝑦 𝑥 𝑥 = − 7 3 𝑦 = ln⁡(3𝑥 + 7) (0, ln 7) (−2,0) 𝑂

Qn Solutions 5(a) (i) Let P m be the perimeter of the amusement park and B m be the width of the square. 2224 2 Bx Bx = = 4 2 4 2 10 5 2 2 2 P x x y y x x = + + = = − − ( ) ( ) ( ) 2 2 22 2 22 2 5 2 2 2 2 10 4 2 6 10 4 2 6 A xy x x x x x x x x xx =− = − − − = − − = − + 5(a)ii d 10 8 2 12 0d 10 5 8 2 12 4 2 6 A xxx x = − − = == ++ x 5 4 2 6 −   + = 0.4288 5 4 2 6+ 5 4 2 6 +   + = 0.4290 Sign of d d A x 0.0030818 0 − 0.001581 slope Therefore 5 4 2 6 x= + gives maximum area. 5b(i) When 1, 1.5tC==

1.5 3 3e 3e 1.5 1e 2 1ln 2 1ln ln 22 k k k k k − − − =− = = −= =− = (ii) ln 23 3e tC −=− d 0.25993 0.260 (3 s.f)d C t == (iii) ln 23 3e tC −=− ,3tC→ → The maintenance cost approaches $3000 in a long run. (iv) 5 ln 2 0 3 3e d 10.8t t−−= The total maintenance cost over the 5 months is $10800 Qn Solutions 6(a) 8 12 42 8 12 51 8 6 Case 1 : 4 girls , 2 boys No. of ways C C 4620 Case 2 : 5 girls , 1 boy No. of ways C C 672 Case 3 : 6 girls No. of ways C 28 Total no. of ways 4620 672 28 5320 =  = =  = == = + + = (b) Total no. of ways 3! 3! 2 72=   = Qn Solutions 7(a) P( ) P( ) P( ) 0.65 0.4 0.25 A B A A B  = −  = − =

(b) P( ) 8P( | ) P( ) 9 0.4 8 P( ) 0.45P( ) 9 P( ) 0.55 ABAB B BB B  ==  =  = = (c) P( ) 1 [P( ) P( )] 1 0.95 0.05 A B A B B   = −  + = − = P( ) 0.25 P( ) P( ) 0.65 0.55 0.3575A B A B =   =  = Events A and B are not independent. Qn Solutions 8(a) 8(b) P(a LED light is damaged) 0.25 0.02 0.75 0.04 0.035=  +  = 8(c) P(From B | is damaged) P(From B and is damaged) P(is damaged) 0.75 0.04 0.8570.035 = == Required Probabilty = 20.035 0.0( ) (1 ) 3 0.0 5535 03−  = 0.75 0.25 A B otherwise damaged damaged otherwise 0.04 0.02 0.96 0.98

Qn Solutions 9(a) Let X be the random variable denoting the number of electrical components that are faulty out of a box of 20. ~ 20, 100B pX   Since ( )P 2 0.2X == , 2 18 2 18 20 1 0.22 100 100 190 1 0.2100 100 pp pp      −=            −=       Using GC, since 10p , 0.052929(to 5 s.f.)100 5.29(to 3 s.f.) p p = = 9(b) Given 5p= , ( )~ 20,B 0.05X . ( )P 3 0.984 (to 3 s.f.)X = 9(c) ( ) ( )P 3 1 P 3 0.015902 (to 5 s.f.) XX = −  = Let Y be the random variable denoting the number of rejected boxes out of 5. ( )~ 5,0.B 015902Y ( ) ( )P 2 P 1 0.998 (to 3 s.f.) YY =  = 9(d) ( ) ( ) E 20 0.05 1 Var 20 0.05 0.95 0.95 X X =  = =   = Let 1 2 40 ... 40 XX XX+ + += , since 40n= is large, by Central Limit Theorem, 0.95~ N 1, 40X   approximately. ( )P 1 0.500 (to 3 s.f.)X =

Qn 10(a) (b) Using GC, product moment correlation coefficient, 0.989 (3 s.f)r= There is a strong positive linear correlation between the monthly advertising expenditure and the monthly sales of the bubble tea. As the monthly advertising expenditure increases, the monthly sales of the bubble tea also increases. (c) 7.6145 3.2356 7.61 3.24 (3 s.f.) yx yx =+ =+ (d) When 3.8, 7.6145(3.8) 3.2356 =32.1707 =32.2 (3.s.f) x y = =+ The monthly sales of the bubble tea is $32200. Since r = 0.989 is close to +1 and x= 3.8 is within the data range of x, the estimation is reliable. Qn 11(a) Let X be the mass of a randomly chosen tyre (in kilograms) Unbiased estimate of population mean, 128 15 12.4450x −= + = Unbiased estimate of population variance, x y 1 5 10 40 𝑦 = 7.61𝑥 + 3.24

( ) ( ) 2 2 1281 53335.1 0.151 3sf49 50 350s  −= − = =  (b) 0 1 H : 12.6 H : 12.6   =  Test at 5% significance level Under 0H , since 50n= is large, by Central Limit Theorem, 53N 12.6, 50 350X     approximately, where 2 53 350s = is a good estimate of 2 Using GC, the test statistics 11.44x= gives 2.9074calcz =− and p-value = 0.0018224 0.00182 (3 sf)  0.05. Since value 0.0018224 0.05p− =  , we reject 0H and conclude there is sufficient evidence, at 5% level of significance, that the mean mass of tyres is less than 12.6 kg. Hence, the claim of the group of customers is not valid at 5% level of significance. (c) Let Y be the mass of a randomly chosen tyre (in kilograms) from the new shipment. 0 1 H : 12.6 H : 12.6    = Test at 5% significance level Under 0H , since 75n= is large, by Central Limit Theorem, 0.5N 12.6, 75Y   , approximately. Since there is sufficient evidence that the population mean mass of tyres from this shipment differs from 12.6 kg, we reject 0H .

12.6 1.95996 0.5 75 12.6 0.160030 12.44 (2 d.p) k k k − − − −  or 12.6 1.95996 0.5 75 12.6 0.160030 12.76 (2 d.p) k k k −  −  Qn Solutions 12(a) ( ) 2~ N 16,3L ( )P 17 0.36944 (to 5 s.f.)L= ( ) 3 Required probability P 1 0.050 7 (to4 3 s.f.) L = = (b) p is more than the answer in part (i). The cases found in (i) are subsets of the cases found in (ii). (c) ( ) 2N 15,~ 1.5M Let 1 2 3 2A M M M L= + + − ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 E 3 15 2 16 13 Var 3 1.5 2 3 N 13 42. , 42.7 75 5~ A A A = − = =+ = ( ) ( ) ( ) 1 2 3 1 2 3 P P P 0.977 (to 3 s. 2 2 ) 0 0 f. M M L M M L M M A + +  + = += − =  (d) Let ( ) ( )1 2 3 1 21.05 1.1T L L L M M= + + + + ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 E 1.05 3 16 1.1 2 15 83.4 Var 1.05 3 3 1.1 2 35.2125 1.5 N 83.4,35.2125~T T T =  +  = = += ( )P 83 0.473 (to 3 s.f.)T =