2012 NYJC H2 Math Prelim P2 solutions

1 2012 NYJC H2 Math Preliminary exam Paper 2 solutions 1 (a) ( )( )1 i i i i 6 11i zw a b b ab a = + − − = − − − + = − 6 1 11 a b ab − = + = ( )1 6 11 a a ⇒ + − = 2 6 10 0 a a − − = 6 36 40 6 76 3 19 2 2 a ± + ± = = = ± Since a > 0, 3 19 a = + , 3 19 b = − + (b) 22, arg 3u u π= = − 35, arg 4v v π= = 2 5 4 v u = 2 3 4 25 arg arg 2 arg 4 3 12 v v u u π π π   = − = + =     2 1arg 12 v u π  ∴ =     2 5 cos isin 4 12 12 n n v n n u π π       = +             Since 2 n v u       is imaginary, cos 0 12 nπ = 3 5 , , ,... 12 2 2 2 nπ π π π ⇒ = Since Im part of 2 n v u       is negative, sin 0 12 nπ < 3 7 , ,... 12 2 2 nπ π π ⇒ = Hence, smallest n = 18 2 (i) Clearly f is 1-1 on the domain ( )1,1 − since every horizontal line y = k ( k ∈ /Rbb) cuts the graph at most once. So f –1 exists. Alternatively, Since 2 n v u       is imaginary and negative, 2 , 12 2 n k k π π π= − + ∈ /Zbb 6 24 n k = − + To get smallest positive n, let k =1. Hence, smallest n = 18.

2 Let y = x 2 – 4 x. Rearranging, we have x2 – 4 x – y = 0. Solving for x, we obtain x = ( ) ( ) 2 444 2 y−−−± = 2 4 y± + . Since 1x < , x = 2 4 y− + as 2 4 2 y+ + ≥ . So f –1 (y) = x = 2 4 y− + . Hence, f –1 : x → 2 4 x− + , x ∈ /Rbb, 3 5 x− < < . (ii) Since R f = ( )3,5 − ⊂ /Rbb\{–3} = D g, gf exists. (iii) ( ) ( ) ( ) 2 2 1gf g 4 1 4 3 x x x x x = − = + − + = 2 11 4 3 x x + − + (iv) ( ) ( ) ( ) 2 2 2 2 2 4 4 2 h 4 3 4 3 x x x x x x x − − ′ = − = − + − + Given 1 1 2 2 2 x x − < < ⇒ − < − < 2 4 2 6 x⇒ < − < . Since ( ) 224 2 0 4 3 0 x and x x − > − + > , ( )h 0 x′ > and so h is increasing on D h. Note that ( ) 1 9 h 1 1 1 4 3 8 − = + = + + and h( x) → ∞ as x → 1-. Hence h 9R , or (1.125, ) 8   = ∞ ∞     . x 5 1 -1 y -3 y = f( x) O

3 3 2 2 1 4x y + = . Let V be the volume of the silo. 2V x y π= 21( ) 4 y y π= − 31= ( ) 4 y y π − km 3 21= ( 3 ) 0 4 1 1 (reject ) 2 3 2 3 dV ydy y π − = ⇒ = − 2 2 3 16 0 when . 2 3 1 1 1 1 3 maximum ( ) or km 4 12 36 2 3 12 3 d V y y dy V π π π π = − < = = − = Let Vd be the volume of the dome. Vd = 32 3 rπ ⇒ 22ddV dr rdt dt π= ⇒ 0.75 = 2 32 ( ) 4rπ π= ⇒ r2 = 0.5 ⇒ r =1/ √ 2 km A = 2 π r2 4dA dr rdt dt π= 1 3 = 4 ( ) 42 π π 2 -1 3 2 = km s 2 4 (a) ( ) ( ) 1 1 1 0 Area = 2 ln 2 ln 2 d or 2 ln 2 2 ln 2 d R x x x x − − − − − ∫ ∫

4 ( ) 1 0 = 2 ln 2 2 ln 2 d since 0 x x x − − > ∫ ( ) 11 0 0 2ln 2 2 ln 2 d 2 xx x x x     = − − +     −  ∫ 1 0 22ln 2 2 0 1 d 2 xx   = − + − +   −  ∫ ( ) ( ) 1 0 2ln 2 2 2ln 2 2ln 2 2 1 2 ln 2 2 2ln 2 x x   = − − − −   = − − + = − (b) Volume needed 1 2 0 d y x π= ∫ When x = 0, ( ) 2 3 0 1 1 t t = + ⇒ = − When x = 1, ( ) 2 3 1 1 1 1 t t = + ⇒ + = ± 2 or 0 (rejected out of range) t = − ( ) 2 31x t = + ( ) 1 3 d 2 = 1 d 3 x tt − + Thus, volume needed ( ) ( ) 2 122 3 1 2ln 1 d 3t t t π − − −   = +   ∫ = 1.80 units 3 5 Either stratified sampling or quota sampling method can be accepted as appropriate methods. Description - Stratified– i) states that sampling frame is obtained listing t he participants according to the types of educational institutions that they come from ii)states that using simple random sampling, 35, 25 , 15 and 5 participants are selected

5 from each of the strata (primary, secondary, JC, poly) respectively Advantage – participants from all the 4 institution s are represented proportionately Responses can be analysed by the institutions that participants come from Disadvantage – relatively inconvenient to carry out Quota – States/sets the quota to be selected from each inst itution type, such that they add up to 80 in total (e.g. 30 primary, 20 secondary, 20 JC, 10 poly) Selects from a list the participants for the survey , in accordance with the quota set above. Advantage – easy to choose the 80 participants Disadvantage – non-random method, so results obtain ed may be biased (e.g. the list may cluster participants from the same school) 6 (i) 6 5 2 4 75 C C × = ways (ii) Number of ways if at least one of the sisters are included = number of ways without restriction – number of ways if none of the sisters is included = 11 8 6 6 434 C C − = Or 3 8 3 8 3 8 1 5 2 4 3 3 434 C C C C C C × + × + × = (iii) Select a man to be between the 2 sisters and group the 3 of them as one unit and arrange 4 units round a table 3 1 3! 2 36 C × × = (iv)First arrange the other 4 persons round the tab le. There are 4 ways to insert the sisters. 3! 4 24 × = or 4 2 2! 2! 24 C × × = 7 (i)By G.C., we obtain a scatter plot of the diagr am as follows:

6 (ii)Although from the scatter diagram, it seems tha t there is a negative linear relationship between y and x within the given range. However, we note that the trade-in value of the car cannot continue to decrea se linearly to become a negative value eventually. Thus, a linear model may not be ap propriate. (iii)Under a quadratic model, the trade-in value of the car would eventually increase over time, which would not make sense. (iv)By G.C., we transform the y values as follows: Thus, we have ln 0.103 4.196 y x = − + . When 5.5 x = , 37.631 y = . Thus, the trade-in value is $37631. 8 (i) P(first red bead is obtained on or before t he 5th draw) = 2 3 4 2 3 2 3 2 3 2 3 2 .5 5 5 5 5 5 5 5 5             + + + +                         or 4 0 3 2 5 5 r r =             ∑ = 0.922 Or 1 – P(no red on first 5 draws) = 5 31 0.922 5   − =     (ii) P(obtaining a first green bead on the 8 th draw given that no green bead has been obtained after 5 draws) = P(red on 6th and 7 th draws and green on 8 th draw)= 2 2 3 5 5             = 0.096 (iii) P(exactly r draws are required for beads of both colours to be obtained) = 1 1 2 3 3 2 5 5 5 5 r r − −         +               

7 = 2 2 2 6 3 6 5 25 5 25 r r − −         +                = 2 2 6 2 3 25 5 5 r r − −         +                  , where r = 2,3,4,… (iv) P(first obtaining beads of different colours after 5 or more draws) 3 3 4 4 6 2 3 2 3 ... 25 5 5 5 5                 = + + + +                                       ( ) ( ) 3 3 32 5 5 32 5 5 6 25 1 1 0.155     = +     − −       = Or P(first obtaining beads of different colours after 5 or more draws) = P(obtaining same colour in t