2007 AJC Prelims Paper 2 solutions

2007 H2 Mathematics Prelim Paper 2 solutions No. Solutions 1a () 32 2 13 3 1( l n ) 19 ( l n ) dy dx x x xx ⎛⎞== ⎜⎟+ +⎝⎠ 1b 2 22 33 2 2, 2 1 (2 1)(2 1) 22 22 dx dyttdt t dt dy t t ttdx t t −=+ = − ⎛⎞ −=− = ⎜⎟ −−⎝⎠ At intersection, 3 2 22 0 1 2 131 3 tt x p −=⇒= =+= = 2 23 2 (1 ) (1rr r r+= ++− − ) ))() ( )( ([]∑ ∑ == − + + += + n r r n r r r r rr 11 11 2 2 2 32 () ()[]∑ = + − − + + = n r rrr r r r 1 1 2 1 2 2 1 () ()[]∑ = −+ − − + + = n r rrr r r r 1 11 2 1 2 2 21 where []∑ = − − + + = n r r f r f r f 1 ) 1 ( 2 ) ( ) 1( rr r f2 ) (= (2) (1) 2 (0) (3) (2) 2 (1) (4) (3) 2 (2) .............................. f ff f ff f ff =+− ++− ++− + ................................................. (2 ) (3 ) 2 (4 (1 ) (2 ) 2 (3 ) () ( 1 ) 2 ( 2 ) (1 ) ( ) 2 (1 ) fn fn fn fn fn fn fn fn fn fn fn fn + +− +− − − +− +− − − ++− − − ++ + − − ) ) (1 ) 2 ( ) ( 1 ) 2 ( 0f nf n f f=+ + −− () 112 22 2nnnn +=+ + − ()22 22 2n nn=+ + − ( )24 2 2n n=+ − 2 () 121 2 nn +=+ − Page 1 This study source was downloaded by 100000851604598 from CourseHero.com on 11-29-2022 21:19:05 GMT -06:00 https://www.coursehero.com/file/5568602/2007-AJC-Prelims-Paper-2-solutions/

Page 2 No. Solutions 1 2 2 ) 3 2 (− = ∑ + r n n r r = ()∑ = + n n r rr 2 2 3 22 1 () () ⎥⎦ ⎤ ⎢⎣ ⎡ + − += ∑∑ = − = n r n r rr rr 2 1 1 1 2 3 2 2 3 22 1 () ( ) ()[] 2 2 1 1 2 2 2 1 42 1 1 2 + + − − − + =+ nn nn ()() () 21 11 41 2 21 2 24 1 22 1 nn nn nn nn − −+ =+ −− ⎡ ⎤=+ − +⎣ ⎦ 3a ∠ AOB = ∠ AOD = 90° O B = 2 O A , O D = O A 3a Minimum |z - w| = EC = 3 2 a 3b 1arg arg zz ⎛⎞ =−⎜⎟⎝⎠ 1sin 22 a a B E D O AC α y 2zi aa−= ww i a= + x 6 παα== ⇒ = arg arg arg26 26az a π ππ⎛⎞ ⎛+ − ≤≤+ +⎜⎟ ⎜⎝⎠ ⎝ π ⎞ ⎟⎠ ⇒ 2arg arg arg33az aπ π+ ≤≤+ ⇒ 21arg arg arg33aa z π π⎛⎞− −≤ ≤ − −⎜⎟⎝⎠ 4a 22 45xy y−−− = () () ()() 22 2 22 22 450 24 5 0 21 0 21 xyy xy xy ⎡⎤−+ + =⎣⎦ ⎡⎤−+− + =⎣ −+ − = −− − = 0 ⎦ 2xy ⇒ A translation of – 2 units in the y – direction. 4bi) This study source was downloaded by 100000851604598 from CourseHero.com on 11-29-2022 21:19:05 GMT -06:00 https://www.coursehero.com/file/5568602/2007-AJC-Prelims-Paper-2-solutions/

Page 3 No. Solutions 4bii) 5i 1Π : 5 41 3 ⎛⎞ ⎜⎟•− =⎜⎟⎜⎟⎝⎠ r y ( )2y fx= − 5 Distance of A from 1Π 65 24 63 15 50 50 ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟•−⎜⎟ ⎜⎟⎜⎟ ⎜⎟−⎝⎠ ⎝⎠=− 41 5 1 1 5050 50 =−= 15 4 50 50 > A and O are on the same side of ⇒ 1Π 5ii Vector parallel to =2Π 56 22 0 86 1 2 −⎛⎞ ⎛⎞⎛⎞ ⎜⎟ ⎜⎟⎜−=⎜⎟ ⎜⎟⎜⎜⎟ ⎜⎟⎜⎟−−−⎝⎠ ⎝⎠⎝⎠ ⎟ ⎟ 6 1 ⎞ ⎟ ⎟⎟⎠ Normal vector of is 2Π 11 4 2 02 1 2 2 21 0 2 −⎛ ⎞ ⎛⎞ ⎛⎞⎛ ⎜ ⎟ ⎜⎟ ⎜⎟⎜×= − = −⎜ ⎟ ⎜⎟ ⎜⎟⎜⎜ ⎟ ⎜⎟ ⎜⎟⎜−− − −⎝ ⎠ ⎝⎠ ⎝⎠⎝ Angle between and 1Π 2Π x ()2, 2 5− 2y= 2− 2y=−()2, 2− x y 32− ( )'y fx= This study source was downloaded by 100000851604598 from CourseHero.com on 11-29-2022 21:19:05 GMT -06:00 https://www.coursehero.com/file/5568602/2007-AJC-Prelims-Paper-2-solutions/

Page 4 No. Solutions 11 52 46 31 31cos cos 46.8 50 41 50 41 −− ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟−• −⎜⎟ ⎜⎟⎜⎟ ⎜⎟ −⎝⎠ ⎝⎠== × = ° 5iii Equation of is 2Π 26 2 62 6 16 ⎛⎞⎛⎞ ⎛⎞ ⎜⎟⎜⎟ ⎜⎟ 1 • −= • −⎜⎟⎜⎟ ⎜⎟⎜⎟⎜⎟ ⎜⎟−−−⎝⎠⎝⎠ ⎝⎠ r i.e. 2 66 1 ⎛⎞ ⎜⎟• −=⎜⎟⎜⎟−⎝⎠ r 1Π : ------ c 5431xyz−+= 5 62Π : ------ d 26xy z−− = 3Π : 8x ya zb++= ------ e For line of intersection of and 1Π 2Π 54 3 1 5 261 6A −⎡⎤=⎢⎥ −−⎣⎦ rref 10 1 3 011 / 20A ⎡⎤= ⎢⎣⎦ 3xz+= ⎥ 3x z−⇒= 1 02yz+= 1 2y z⇒= − 3 32 1 02 02 zx yz z z μ −⎛⎞ −⎛⎞ ⎛⎞ ⎛ ⎞⎜⎟⎜⎟ ⎜⎟ ⎜ ⎟⎜⎟=− = + −⎜⎟ ⎜⎟ ⎜ ⎟⎜⎟⎜⎟ ⎜⎟ ⎜ ⎟⎜⎟⎝⎠ ⎝⎠ ⎝ ⎠⎝⎠ 1 1 r 32 0 02 μ −⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟=+ −⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ Let , two points on the common line are (3,0,0), and (1 , −1 , 2) 0,1 μ= Substitute into 8x ya zb++= 0 3ab b= ⇒ = : 38 ( 0 )++ & 18 2 2 7ab a b−+ =⇒ −= 21 0 5 aa=⇒ = Alternatively Since the common line lies in , 3Π 32 1 01 8 02 b a μ ⎡⎤ −⎛⎞ ⎛ ⎞⎛⎞ ⎢⎥⎜⎟ ⎜ ⎟⎜⎟+− =⎢⎥⎜⎟ ⎜ ⎟⎜⎟⎜⎟ ⎜ ⎟⎜⎟⎢⎥⎝⎠ ⎝ ⎠⎝⎠⎣⎦ for all μ 3 - 2μ – 8μ + 2 μa = b This study source was downloaded by 100000851604598 from CourseHero.com on 11-29-2022 21:19:05 GMT -06:00 https://www.coursehero.com/file/5568602/2007-AJC-Prelims-Paper-2-solutions/

Page 5 No. Solutions 3- μ (10 – 2a) = b b = 3, a = 5 6i E E I I I G L T M S Number of arrangements 6 5 5!5! 72003!2!C=× × = 6ii All letters different: A pair of identical letters: 7 3 210p = 6 1C 3! 23 62!×× = Three I’s : 1 Total number of ways = 210+36+1= 247 7a Required Probability = () 'PS M∩ = P (M) – ()PS M∩ = 0.18 – 0.16 X 0.65 = 0.076 7b Required Probability = P (S / M’ ) (' ) (' ) 0.16 0.35 10 . 1 8 0.06829... 0.0683 PS M PM ∩= ×= − = = 7c ( ) () ( ) ( ) 0.16 0.18 0.16 0.65 0.236 PS M PS PM PS M ∪ =+ − ∩ =+−× = 8a let X = number of red pens in sample of 10 ()10,0.35XB Required probability = () 51 ( 5 )PX P X>= − ≤ = 0.0949 S (60 or over) S’ (less than 60) M(myopia) M’ 0.16 0.84 0.65 0.35 P(M) = 0.18 C C CC CArrange consonants first This study source was downloaded by 100000851604598 from CourseHero.com on 11-29-2022 21:19:05 GMT -06:00 https://www.coursehero.com/file/5568602/2007-AJC-Prelims-Paper-2-solutions/

Page 6 No. Solutions 8b P( at most n more pens) > 0.98 () () () () () () () 21 0.35 0.65 0.35 0.65 0.35 .... 0.65 0.35 0.98 n− ++ + + > ()()0.35 1 0.65 0.9810 . 6 5 n − >− ()0.65 1 0.98 n <− () () ln 0.02 ln 0.65n> n > 9.08 least n = 10 9a Let A = number of calls Alice receives in 30 minutes Let B = number of calls Brenda receives in 30 minutes (0.5AP o ) and ()1.8BP o ∴ ()2.3AB P o+ required probability = () 3 5PA B+<⎡⎤⎣⎦ = () 3 4PA B+≤⎡⎤ ⎣⎦ = 0.91625 3 = 0.769 9b required probability = () 1/ 5PA A B≤+ < = () ( ) () ( ) () 0. 0 4 1. 0 3 5 PA P B PA P B PA B =≤ ≤ + =≤ ≤ +< = 0.60653 0.96359 0.30327 0.89129 0.91625 ×+× = 0.933 9c Let W= number of calls Alice received in one day ()8WP o () ( 65PW PW<= ≤ ) ) = 0.191 Let C = number of days out of 60 days, with less than 6 calls per day ()60,0.19124CB∼ since np > 5 and n(1-p) >5, approximately (11.4744, 9.28CN () ( )10 20 9.5 19.5PC P C≤< = << = 0.737 10ai 18.7 80 79.06520x −=+ = 2 2 21( 1 8 . 7 )[102.5 ] 4.4745 (2.11530)19 20 xS −=− = = 1:8 0 , :8oHH 0μ μ=< If Ho is true, the test statistics is 80 ~( 1 9 )2.1153 20 XTt −= . This study source was downloaded by 100000851604598 from CourseHero.com on 11-29-2022 21:19:05 GMT -06:00 https://www.coursehero.com/file/5568602/2007-AJC-Prelims-Paper-2-solutions/

Page 7 No. Solutions We perform a one tailed t-test at level of significance and reject H o 5% if p < 0.05. Use GC with 80, 20, 79.065, (2.11530) xnx Sμ === = we have p = 0.03138 As p = 0.03138 < 0.05, we reject Ho at level of significance. 5% And conclude that there is significant evidence that the manufacturer’s claim is justified at 5% level of significance. Assume that the weights (X) follow a normal distribution. 10aii The probability of concluding that the manufacturer is justified in his claim when actually he is not justified is 0.05 σ2 is known to be 10kg, ~( 8 0 , 0 . 5 )XN or 80 ~( 0 , 1 ) 0