RVHS P2 MS

RV 2012 Yr 6 H2 MA Prelim Paper 2 Question 1 [7 Marks] 32 32 2 2 2 d (1 ) 0 (1)d ddLet dd Subs. into (1): d 0d d ( 1) (shown)d yx x y xx uyu xy x y xx u y y x y xx u ux x u xx                     3 3 3 2 3 3 3 3 3 1 dd1 ln | 1| 3 1 e where e e1 e 1 1 (shown) and f 3 x C x x u x xu xuC u A A uA Ay x x x             (i) 1A (ii) 1A

2 Question 2 [10 Marks] (a) 2 dtan e sec e d xx yyy x   2 2 2 2 2 1 1 de (1 e ) 1cot cot sec d1 tan cot d (cos ec 1) d cot e tan (e ) xx xx x y y y y y yy yy y y C C                   (b) 2 1exy  When 0x , ey and when 1x , 2ey . 2 1exy   2 ln 1xy Required volume 22π(1) e J where 2e e π (ln 1)dJ y y 2e e π (ln 1)d J yy      222 eee ee e π (ln ) 1dy y y y      2e e 2 2 2 π (ln ) 2 π e ln e eln e 2e 2e π e y y y      Required volume   22 2 π(1) e πe πe =πe e 1 J   Note that ln y d ln 1 d lny y y y y y y c    

3 Question 3 [11 Marks] ai aii aiii b 2 2 2 2 2 2 1 1 ( 1) 14 AByx x y x y         A: Scaling parallel to the y-axis with a scale factor of 1 2 . B: Translation of 1 unit in the positive direction of the y- axis. Alternatively 2 2 2 2 2 2 ( 2) (2 2)1 1 14 4 4 ABy y yx x x         Note that 2 2 (2 2) 14 yx  is equivalent to

4 22 ( 1) 1xy   . A: Translation of 2 units in the positive direction of the y-axis. B: Scaling parallel to the y-axis with a scale factor of 1 2 . Note: Scale factors must be positive numbers. Question 4 [11 Marks] i 3 (0) 8 8e iz  3 (0 2 ) (2 )8e 8ei k i kz  2 32e , 0, 1 ki zk    ii 2 3 1 2e i z  , 2 2z  , 2 3 3 2e i z  Note that the points 1Z , 2Z and 3Z lie on the circumference of a circle with centre O and radius 2. The points are equally spaced out separated by an angle of 2 3 radians. 1Z and 3Z represent complex numbers which are conjugates to each other. Thus 1Z and 3Z lie on the same vertical line. iii Since 1zz satisfy the equation 23| | | |z z z z   with 1 2 1 3| | | | 12 2 3z z z z     , thus locus of P passes through the point 1Z . Note 1 2 1 3| | | | 12z z z z    can be easily found using GC. Re Im Z1 Z2 Z3 O

5 iv Locus of P: 23| | | |z z z z   , is the perpendicular bisector of the line joining 2Z and 3Z . It passes through the origin and 1Z . Locus of Q: 1| | 2 3wz , is a circle with centre at 1Z and radius 23 . It passes through 2Z and 3Z . v Let V be the point of intersection of the loci of P and Q in the first quadrant. 2 12 3Z OZ  1 2 3VOZ    2 3 1 2e 1 3 i zi    1 i 3 ( i )V v x y     where 12 3cos 2 3 3 2x      , 32 3sin 2 3 3 2y       . 1 i 3 ( 3 3i) ( 1 3) i( 3 3)v           Alternatively 11 2 3 2v Z V Z O    1 2 3arg( )v VOZ    (Refer to above working.)   32 3 2 e i v     2 3 2 cos 2 3 2 sin33vi               V Z1  O Z2 Re Im Z1 Z2 Z3 O P Q V 23

6     132 3 2 2 3 222vi       ( 1 3) i( 3 3)v       Question 5 [4 Marks] i The age group may be classified as 21 – 30, 31 – 40, 41 – 50 and last group with age more than 50. For each age group, interview 25 people which add up to 100. ii The sample selected is biased and not representative of the district. iii Stratified random sampling Question 6 [7 Marks] No. of combinations = 2 3 3 4 72    No. of ways 4[72 (1 2 2 3)] 2 54        Alternatively No. of ways 2 3 3 4 542222                           No. of ways = 82 1 255 Alternatively, number. of ways = 8 1 8 255 r r    Note: 8 0   is not counted as the customer must choose at least 1 ingradient.

7 Question 7 [8 Marks] a Let the sample of bivariate data be ( , )iixy where 1,2, ...in . Let ()i i ie y a bx   be the vertical deviation between the point ( , )iixy and the line y a bx . The line y a bx is the least square regression line for the sample of bivariate data if the sum of the squares of the vertical deviation   2 1 n i i e   is the minimum. b i From GC, r = 0.96953 0.970 Although the value of r is close to 1, it does not indicate that the linear model is suitable as the scatter diagram may show a curvilinear relationship between the points. b ii Calculate the r values for both models and the model with a higher absolute value of r closer to 1 is a better model. For lny a b x , value of r = 0.995> 0.970 lny a b x = 1.4595375 + 4.2820506 lnx Thus y = 1.46 + 4.30 lnx is a better model. b iii From (ii), y 1.4595375 + 4.2820506 lnx Sub 6.4y , then 3.17x . Since 0.995 1r , the regression line of ln x on y and the regression line of y on ln x , it is reasonable to use the regression line of y on ln x for the estimate. –(1) In addition, 6.4y is within the input data range of 2.2 9.9y , the estimate is an interpolation. –(2) Hence, the estimate is reliable. Note: Both (1) and (2) should be stated to justify the reliability of the estimate. Simply quoting 0.995 1r in (1) is not acceptable.

8 Question 8 [8 Marks] i Let S and L be the r.v. denoting the amount of oil in a Small and Regular bottle respectively. S ~ N 545, 202   and   2~ N 1020, 50R       P 0.97 1 P 0.97 P 0.03 925.96 R R R           Greatest 925 ii   2 1 2 3 ~ N 3 545, 3 20S S S      2 12 ~ N 2 1020, 2 50RR   Let  1 2 3 1 2 1 5T S S S R R       1 2 3 1 2 22~ N 3 545 2 1020, 3 20 +2 50 S S S R R         2 11~ 3675, 620055TN    ~ 735, 248TN  P 730 0.375T  iii S ~ N 545, 202   P(bottle with the least content holds more than 560 ml) = P(all 3 bottles holds more than 548 ml of oil)   3 = P 547S = 0.0974 (3.s.f) iv Suppose   2~ N 1020, 50R  0 1080 0.885PR    OR  1080 0.115PR  There is a probability of 0.115 that the Regular bottle of cooking oil will not fall within the bottle, hence using a normal distribution is not appropriate. Alternatively, Suppose   2~ N 1020, 50R , then 1020 and 50     3 3 930 1230 0.997P R P R           There is a probability of 0.997 that the amount of cooking oil in a Regular bottle falls within the range 930 ml to 1230 ml. A wide range of this set of values exceeds the capacity of the bottle i.e. 1080 ml, hence using a normal distribution is not appropriate.

9 Question 9 [11 Marks] i 00 10 H: H:     ii It means that there is a probability of 0.025 that the test concludes that the mean time spent by a technician of Open Network to install a fibre optic cable unit in a household unit is more than 0 minutes when in fact it is not true. iii 2 sample variance1 ns n   22 12 4.511s  Under 0H ,  X 11oTt sn  Using GC, invT(0.975,11) gives 2.200985 0H is not rejected if 2.200985calt  0 0 0 0 21.3 2.200985 12 (4.5) 11 12 (2.200985)(4.5)21.3 11 18.31370 18.4 (3s.f.)          The time spent by a technician of Open Network to install a fibre optic cable unit in a household unit, is assumed to be normally distributed. Note: In this case, X is defined in the question. Therefore the statement “X is normally distributed” is acceptable. Candidates are to ensure that X is properly defin