IJC JC2 H2 Maths 2012 Solutions Paper 2

1 IJC/2012/8864/BT/Marking Scheme Solutions for 2012 H2 Maths Prelim 2 Paper 2 1(i) 1(ii) 2(i) 1 4 1 22 2 2 1 d 1 4 u uu u ⋅ −   −    ∫ 1 4 1 22 2 2 2 1 d 1 4 u uuu u = ⋅ − −∫ 1 4 1 2 2 2 1 d 1 4 u u = − −∫ ( ) 1 1 4 1 2 2 1 sin 2 2 u−  = −   1 1 1 1 1 sin sin 2 2 2 − −       = − −             1 2 6 4 24 π π π   = − − =     ( )f 2 y x ′= − 1x = 1x = − O y x 4− 4 y x O ( )fy x = 1y = − 2x = 2x = − 2−

IJC/2012/9740/02/Prelim 2/Solutions 2 2(ii) Area = ( ) 4 22 2 1 1 1 2 2 d 2 4 2 4 x x x   +    − ∫ 21 unit 4 24 π= + 3(i) 1, , . n n n x l x l +→ ∞ → → ( ) ( ) 2 2 10 3 10 3 3 10 0 2 5 0 l l l l l l l l = − = − + − = − + = Solving, 2 or 5 (rejected) l l = = − 3(ii) ( ) 2 2 2 1 10 3 2 6 3 3( ) n n n n x l x x l x + − = − − = − = − 3(iii) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 1 1 1 1 1 Since 0 3( ) 0 0 0 (since 0) n n n n n n n n n n x l l x x l l x x l x l x l x l x l x l + + + + + + + < − > − = − − > − + > − > + > >

IJC/2012/9740/02/Prelim 2/Solutions 3 4 Volume of cone V 21 3 r h π= , where 2r θ= and 2 2 16 4 h π θ = − . ( ) 2 2 2 1 2 16 4 3V π θ π θ = − 4 2 2 4 4 4 3V π θ π θ = − 2 4 6 8 43V π π θ θ = − Differentiating with respect to θ, ( ) ( ) 1 2 4 6 2 3 5 2d 8 1 4 16 6 d 3 2 V π π θ θ π θ θ θ −   = − −       When d 0d V θ = , 2 3 5 16 6 0 π θ θ − = ( ) 3 2 2 16 6 0 θ π θ − = 2 2 16 6 π θ = 2 2 8 3 πθ = 2 3 2 2 28 8 8 43 3 3 V π π π π     = −         6 6 8 256 512 3 9 27 V π π π= − 48 256 3 27 V π= 4128 9 3 V π= 4128 3 27 V π= 128 , 27 p q = = θ 28 3 π − 28 3 π 28 3 π + gradient + ve 0 - ve Shape / _ \ Therefore V is a maximum when 28 3 πθ = (verified).

IJC/2012/9740/02/Prelim 2/Solutions 4 5 12 0 12 8 , 4 , 4 0 7 7 OB OD DB             = = =             −      /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp Equation of line DB: 12 12 8 4 0 7 r λ         = +         −    12 4 5 OE     =       /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp Let N be the foot of perpendicular from E to line DB . 12 12 12 8 4 , 4 4 7 5 7 k k ON k EN k k k +        = + = +         − − −     /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp . 0 12 12 4 4 . 4 0 5 7 7 144 16 16 35 49 0 51 209 1896 1 1468 209 357 dEN k k k k k k k ON =         + =         − − −     + + + + = = −     =       /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp The foot of perpendicular from E to DB is 1896 1468 357 , , 209 209 209       . (ii) 12 4 5 12 12 10 8 4 4 15 0 5 12 10 15 12 nOBE OE OB OE     =       −             × = × = −                   −    =       /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp

IJC/2012/9740/02/Prelim 2/Solutions 5 Let θ be the angle between DB and OBE. 1 1 12 10 4 . 15 7 12 144 sin sin 27.4 209 469 209 469 o θ − −   −                      −         = = =               (iii) 12 12 0 , 4 2 7 DE DB         = =         − −     /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp Length of projection 12 12 0 . 4 2 7 158 10.929 10.9 209 209                 − −     = = = = units (iv) As DB and AC are skew lines, they are not co -planar. 6(i) Obtain a list of households according to the addres ses. Select one household at random from the first 10 households on the list. Thereafte r select every 10 th household on the list. (ii) Stratified Sampling Obtain a list of all households and divide the hous eholds according to the different types of housing. Select a random sample from each type of housing such that the sample size is proportional to the relative size of each type of housing. One advantage is that the sample obtained is a bett er representative of the population than systematic sampling. Quota Sampling Divide the households according to the different ty pes of housing and set a quota for each type of housing. The interviewer interviews the households such that the quota is met. One advantage is that no sampling frame is needed.

IJC/2012/9740/02/Prelim 2/Solutions 6 7 Let X be the number of sixes obtained out of 25 throws o f the biased die. Then X ~ B(25, p). Given that std. dev. of X is 1.5. So, Var( X) = 1.5 2. 2 2 2 25 (1 ) 1.5 0.09 0.09 0 p p p p p p ∴ − = ⇒ − = ⇒ − + = From GC, 1 9 or 10 10 p = Since 1 1 , (shown) 6 10 p p < ∴ = 1Hence ~ B 25, . 10 X       Required probability = P(6 10) P( 9) P( 5) X X X ≤ < = ≤ − ≤ ≈ 0.033321 = 0.0333 (3 s.f.) Let Y be the number of sixes obtained out of 40 throws of the biased die. 1~ B 40, . 10 Y       From GC, P( 3) 0.20032 P( 4) 0.20589 P( 5) 0.16471 X X X = = = = = = Hence, most likely number of sixes obtained is 4. 8(a)(i) Let X be the number of genuine call-outs in a 2-week period. Then X ~ Po(4). P( 6) P( 5) 0.785 (3 s.f.) X X < = ≤ = (ii) Let T be the total number of call-outs in a 6-week period. Then T ~ Po(6(2+0.5)), i.e., T ~ Po(15). Since λ =15 (> 10), T ~ N(15, 15) approximately. P( T > 19) = P( T > 19.5) (with continuity correction) ≈ 0.12264 = 0.123 (3 s.f.)

IJC/2012/9740/02/Prelim 2/Solutions 7 8(b) Let W be the no. of genuine call-outs in a week at station B. W ~ Po( m). P( 1) 0.08 P( 0) P( 1) 0.08 e e 0.08 e (1 ) 0.08 m m m W W W m m − − − ≤ = ⇒ = + = = ⇒ + = ⇒ + = From GC, m = 4.17 (3 s.f.) 9(i) Unbiased estimate of population mean, ( )12 12 13 xx −∑= + 6.09 12 13 = + = 12.4685 = 12.5 (3 s.f.) Unbiased estimate of population variance, 2 2 1 6.09 20.853 12 13 s   = −     = 1.50 (3 s.f.) 9(ii) 0 0 :H =µ µ 1 0 :H >µ µ Significance level: 5% Under 0H , 0 (12) ~XT t S n µ−= From GC, P( 1.7823) 0.95 T < = If 0H is not rejected, ⇒ 012.4685 1.7823 1.50 13 µ− < ⇒ 0 11.863 µ > ⇒ 0 11.9 µ > (3 s.f.) Thus, set of values of 0µ for the null hypothesis not rejected is { }0 0 : 11.9 µ µ ∈ > /Rbb . 1.7823 test statistic t(12)

IJC/2012/9740/02/Prelim 2/Solutions 8 9(iii) 0 : 12 H µ = 1 : 12 H µ > Significance level: α % From GC, p-value = 0.09649 Significant evidence that the modified petrol does improve mileage in cars ⇒ reject 0H ⇒ p-value ≤ 100 α ⇒ α 9.649 ≥ Thus least significance level = 9.65% (3 s.f.) Assumption: X is normally distributed. OR The mileage of the particular model of car is normally distributed. 10(i) P(exactly two girls are chosen) = 15 14 10 9 4! 25 24 23 22 2!2! × × × × = 189 506 (or 0.373518 ≈ 0.374 ) Alt P(exactly two girls are chosen) = 15 10 2 2 25 4 4! 4! C C C × × = 189 506 (or 0.373518 ≈ 0.374 ) 10(ii) P(Chairperson and Vice-Chairperson are of opposite sex) = 15 10 225 24 ×