PJC P2 MS

2012 J2 FYE P2 1 (i)     291 2 1 11 2 1 cos3 2 13Since is small, h 2 sin 2 2                (ii) (iii) 22    (iv)           1 ' 1 13Let ln ln 1 3 ln 22 1 d 3 d 1 3 2 1 d 1 7 when 0, 3 2 d 2 2 2 170 0 Verified 22 n n n n nn y y n n y n n y y n nyn ny f f                                          11 13 1 3 22 1 1 .... 1 3 ...22 1 1322 1 7 1 7 ,2 2 2 2 n nn n n n n n n n n nn nn ab                          

2 (i)  4 1 4 22 zi z i zz              13 13arg 2 arg 2 arg 24 3 4 4z z z           ` (ii) z nearest to origin is complex represented by A. Method 1 Using triangle ADE, 1cos 2 24 2 DE DEAD       , 1sin 2 24 2 AE AEAD       Therefore  2 i 2 2z   Method 2 DC = 222 2 2 2 =>AC= 2 2 2 Using triangle ABC,   1sin 2 2 2 2 24 2 AB ABAC          1cos 2 2 2 2 2 24 2 BC BC OB BC OCAC            Therefore  2 i 2 2z  

3 Total volume of revolution of four rectangles = 222 2 1 1 1 1 5674 1 2111 444                           2 2 2 2 4 4 4 4 4 9 10 11 12                             4 2 1 4 8r r     Total volume of revolution of n rectangles =       22 2 2 2 2 2 1 1 1 1 12 1 2 2 1 2 2 2 21 1 1 1 n r n n n n n n n n n n nrnn                                                               (i) Exact volume of solid formed     22 1 21 1 1 d 1 61 xx x       Since estimated volume < exact volume of solid formed     22 11 11 6622 nn rr n nn r n r        (ii)     2211 1lim lim 6622 nn nn rr nn n r n r         4(a) By long division, 2 222 411144 12 x xx x       Hence, 1A , 1B , 1 2C  . The sequence of transformations is: (i) Scaling parallel to x-axis by a factor of 2 (ii) Reflection about the x-axis (iii) Translation of 1 unit in the direction of the y-axis

(b) (i) (ii) 5   1 1 (i) f 1 From graph, there exists no horizontal line that cuts the graph at 2 more points f is 1-1 f exists for any , (ii) Since 1, Let , 1 1 1 f ( ) 1 xx xx xyx x xy y x yx y xx x                                f x               1 (2) (2012) (2013) (2013) f ( ) f f f f f f f 0 xx x x x x x x             x y 0 f ( ) yx  1 ,9 2 4y x y 0 f ( 1) yx 1 1 4y From sketch, any line, y = k will cut the graph at most once.

      22(iii)g( ) 3 6 2 3 1 1 min. value = 1 R = 1, Since \ and fg does NOT exists g f x x x x D              => R D R 1g f g             2 2 3 6 2(iv) fg 3 6 2 2fg 0 2 2 2 2 4 from iii 12 xxx xx                      6 (i) Stratified Sampling (since the service provider should have complete list and relevant details of his subscribers) (ii) (1) Divide the subscribers into mutually exclusive strata like age, gender, locations etc (2) Simple random samples are drawn separately from each stratum with appropriate proportion. Eg. If there are 60% male subscribers and 40% female subscribers => out of n subscribers to be selected, simple random sample of 0.6n male subscribers and 0.4n female subscribers need to be formed. (3) Simple random samples put together to form a complete sample of subscribers to obtain feedback (i) 69 531, 441 (ii) 9 3 6! 75602! 2! 2!C   (iii) Case1 : 3 odd 3 evens 4 3 3 3 5 4 32000C    Case 2 : 2 odd 4 evens 5 2 4 2 5 4 64,000C    Required answer = 32,000 + 64,000 = 96,000

8(a)       Let represent the number of times occu rs per day. ~ Po 8 E( ) Var 8 8~ N 8, approx (by Central limit theorem,si nce sample size =50 is large)50 P 7 0.0062096799 0.006 XA X X X X X          (b)         Let represent the number of times occu rs per day. ~ Po 0.5 ~ N 0.5, 0.5 approx (since is large, henc e is 0.5) P P 0.5 (continuity correction) 0.5 YA Ym Y m m m m Y m Y m           9 (i) Method 1 1( ) ( ) 1 3a d a a d a         1P2 3nd a (Shown) (ii) Method 2       12 1 3 1 3a a d a d a d a d             1P2 3nd a d   (Shown) 1 14(1 )3 45 1 15 p p   1st 2nd 3rd C’ C C’ C C’ C a – d a + d a p 1 – p 0.1 0.12

(iii) P(3rd yr | C’) = 0.6 (3rd yr ') 0.6( ') 1 (0.12) 133 0.61 1 1 1 81(0.1) (0.12)3 3 15 3 PC PC d d dd                    Proportion of 3rd year students = ad 1 13 40 3 81 81 (iv) P(3rd year | did not complete) = 0.6 P(3rd year) = 40 81  0.6  The two events are not independent. 10. (i) Each phone is selected to purchase (by customers) independently from one another. (OR the probability that a camera phone is purchased is constant for each phone purchased). (ii) Let C’ denote the number of non-camera phones purchased =>  ' ~ B ,0.05CN  ' 5 0.1PC      1 P ' 5 0.1 P ' 5 0.9CC      From GC, When N = 63,  ' 5 0.90551 0.9PC    When N = 64,  ' 5 0.89990 0.9PC    Therefore, largest number is 63. (iii) Let W denote the number of non-camera phones purchased out of 180 phones monitored. => W ~ B(180,0.02) Since 180 is large, 3.6 5n np   ~ Po(3.6)W approx.. (less than 171 camera phones) =P(at least 10 non-camera phones) =P(W 10) =1 P(W 9) 0.004024267 0.004 P    ….. 13 81d

11 (i) From GC, the product moment correlation coefficient for the data is 0.950. (3 s.f.) (ii) The product moment correlation figure may suggest a strong linear relationship, but the scatter plot suggests otherwise. (iii) C : lnt a b s , since the scatter plot has a shape similar to curve y=lnx. (iv) Model C should have a higher positive value of product moment correlation coefficient than (ii)’s model (v) Use regression line of Y=lns on X=t. From GC, Y= 0.4837235603+0.019805927X When t=5, Y=0.5827532 => 0.5827532e e 1.790962527 1.79Ys    Not reliable as this is an extrapolation since t=5 is out of the data range for TV. 12 (i) 2 Unbiased estimate for the mean 1997.610 Unbiased estimate for the variance = 67. 37777778 67.4 xx s      (ii) Test 01: 2000 : 2000H vs H Level of significance : 0.5% Since n is small and 2 is unknown, we use the t-test. Test statistic: "claimed value" 1997.6 2000 0.92468.208396785 10 xt s n    From GC, p-value is 0.18965 > 0.005 Since the p-value is _ more___ than the level of significance, we do not reject 0H . There is __insufficient_______ evidence, at the 0. 5% level, to conclude that the battery’s lifespan less than 2000 hours. Assumption : The lifespan of the battery is normally distributed. Therefore Tom should recommend the battery to his classmates.

(iii) The 0.5% level of significance is the probability of rejecting the fact that the mean life span of the battery is 2000 hrs when it is true. (iv) To have a different conclusion => to reject Ho Use z-test 1997.6 2000 (0.005)5 28.779 z test invnorm n n      13(i)       22 2 ~ (0.9,0.005 ), ~ (0.91,0.005 ) 0 ~ ( 0.01, 2 0.005 ) Req. prob = 0 0.0786496525 0.0786 ps p s p s ps ps D N D N P D D P D D D D N P D D            (ii) The pencil’s and sharperner hole’s diameters are independe