RVHS P1 MS

RV 2012 Yr 6 H2 MA Prelim Paper 1 Question 1 [7 Marks] Let 4 3 2f ( ) 2 14z z z z az b     Consider 4 3 22 14 0z z z az b     ---- (1) Sub 1 2iz into (1), using GC ( 7 24i) 2( 11 2i) 14( 3 4i) (1 2i) 0 ab            ( 7 22 42 ) ( 24 4 56 2 )i 0a b a           ( 27 ) (36 2 )i 0 0ia b a       Comparing the real and imaginary coefficients 27 0 ----(2) 36 2 0 ----(3) ab a       Solving (2) and (3), 18a and 45b Therefore 4 3 2f ( ) 2 14 18 45z z z z z     Using GC to solve 4 3 2f ( ) 2 14 18 45 0z z z z z      , i2z  , i2z  , 3z , 3z Replace z with iz in (1), we obtain 4 3 22 14 18 45 0z iz z iz     i 1 2iz , i 1 2iz , i 3iz , i 3iz i2z  , i2z  , 3z , 3z Alternatively, since all the coefficients of the polynomial f ( )z are real  1 2iz and 1 2iz are roots of f ( ) 0z      2(1 2i) (1 2i) 2 5z z z z       is a quadratic factor of f ( )z . Let 2z qz r be the other quadratic factor of f ( )z . 4 3 22 14z z z az b     22 25z z z qz r        Comparing coefficient of 3z : 22 q    0q Comparing coefficient of 2z : 14 5r  9r Therefore f ( )z 22 2 5 9z z z        4 3 2f ( ) 2 14 18 45z z z z z     18a and 45b f ( ) 0z   1 2iz , 1 2iz , 3iz , 3iz

2 Question 2 [8 Marks] 9 ( 3)( 3) 0 x x xx x    3 0 or 3xx    4 34 3 3422 3 2 2 9 d 99 dd 9ln | | 9ln | |22 999ln 3 9ln 8 9ln 4 9ln 32 2 2 18ln 3 1 9ln(4 ) 2 n n n xx x x x x xxx xxxx nn nn I                                              As 0n , ln(4 )n  I  Question 3 [9 Marks] i OAQB is a parallelogram OA BQ OA OQ OB   11 02 22 OQ          2 2 4 OQ     ii  a c c is the projection vector of a onto b . iii |a| < |b| 22 ( 1) 4 1 4 4pp      2 20 ( 1)( 2) 0 12 pp pp p          But p > 0, therefore 02 p . 3 –3 0

3 iv 2 1 2 1 3 1 1 2 . 1 2 1 . 3 3 3 0 0 2 2 2 2 4 0                                                                                  Thus ab and ab are perpendicular. Note: ab and ab are the diagonals of the parallelogram with a and b as the adjacent sides. The parallelogram with a and b as the adjacent sides must be a rhombus. | a b| is the area of a rhombus formed by the vectors a and b. OR | a b| is the area of the rhombus OAQB. Question 4 [9 Marks]   21cos xy  --- (1)            2 1 1 1cos2d d x xx y --- (2) Squaring both sides, we get   2 212 1 cos4 d d x x x y       yx yx 4d d)1( 2 2      . (shown) yx yx 4d d)1( 2 2      Differentiate with respect to x: x y x yxx y x yx d d4d d)2(d d d d)2()1( 2 2 2 2          --- (3) Substitute x = 0 into (1), (2) and (3):   420cos 22 21       y                 )1(22 01 10cos2d d 2 1 x y   2d d4d d)2()01( 2 2 2 2 2  x y x y  ...!2 2)(4 2 2  xxy  ...4 2 2  xxy 

4 i Equation of tangent: 4 2  xy . ii 1 2 2cos 1 x x   1 2 2cos 1 x x    = 22 1 11 cos2 xx x    = 2 1 2 )1(d d   xx y =      ...2 11...)2( 2xx  x2 Question 5 [10 Marks] i Amount of water at the: End of 1st day = 80(0.8) End of 2nd day = (80(0.8)+40)(0.8) = 80(0.8)2 + 40(0.8) = 83.2 cm3 ii End of 2nd day = 80(0.8)2 + 40(0.8) End of 3rd day = (80(0.8)2 + 40(0.8)+40)(0.8) = 80(0.8)3 + 40(0.8)2 + 40(0.8)  End of nth day = 80(0.8)n + 40[0.8 + 0.82 + 0.83 + … + 0.8n–1] = 80(0.8)n + 40        8.01 )8.01(8.0 1n = 80(0.8)n + 160  1)8.0(1  n = 80(0.8)n + 160 – 160 1)8.0( n = 80(0.8)n + 160 – 200 n)8.0( = 160 – 120(0.8)n So, the amount of water at the end of the nth day is  n)8.0(120160 cm3 (shown) iii Since the maximum capacity of the glass is 180 cm3, we will find the least n value such that the amount of water at the end of nth day is more than 140 cm3. 160 – 120(0.8)n + 40 > 180 160 – 120(0.8)n > 140

5 Method 1 (Algebraic approach): 160 – 120(0.8)n > 140 (0.8)n < 6 1 n > )8.0ln( )ln(6 1 n > 8.0296…  least value of n is 9. At the end of the 9 th day, amount of water is more than 140 cm3. So, the day when overflowing happens is the 10th day. Method 2 (Graphical approach): 160 – 120(0.8)n > 140 20 – 120(0.8)n > 0 Plot the graph y = 20 – 120(0.8)n From the graph, n > 8.0296…  least value of n is 9. So, the day when overflowing happens is the 10th day. iv As n , (0.8)n 0 160 – 120(0.8)n 160 Amount of water at the end of any day will not exceed 160cm3, so the minimum capacity of the glass to b e used is (160+40)cm3 = 200cm3.

6 Question 6 [10 marks] (i) 2f : 2 3, 1.x x x x    A horizontal line y = k where 40 k   cuts the graph of y = f(x) twice, thus f is not one-to-one. Therefore f −1 does not exist. Quoting a specific line eg 3y  is acceptable. (ii) For f −1 to exist, largest domain is ( , 1].  Largest value of 1a . 2 2 1 Let 2 3. 2 3 0 2 4 4( 3 ) 2 2 4 16 2 14 Since 1, 1 4 f ( ) 1 4, 4 y x x x x y yx y y x x y x x x                            (iii) y = f(x), y = f −1(x) and y = x intersect at the same point. 2 2 f ( ) 23 30 1 13 2 1 13Since 1, . 2 xx x x x xx x xx           x −3 1 (−1,−4) y

7 (iv) Rg = (0,2) Df = ( ,1] Rg  Df Thus, fg does not exist. (v) For fg to exist, Rg  Df. Let Rg = (0,1] Thus Dg = [1.2,3) gf[1.2,3) (0,1] ( 3,0]     Question 7 [11 Marks] i 3 23132 2   x xxy yxyxx 323132 2  0)323()13(2 2  yxyx The equation above has no real roots when 042  acb 0)323)(2(4)13( 2  yy 01522  yy 0)5)(3(  yy 53  y So, C cannot lie between –3 and 5. ii 3 23132 2   x xxy = 3 272  xx The asymptotes are 27yx and 3x .

8 iii iv 2 22 2 3 20122)3( kx xxx         2 22 2 3 323132)3( kx xxxx           222 1)3( kyx  Add a circle with centre (–3, 1) and radius k To have a positive root, we first find the distance between (–3, 1) and      3 23,0 .

9 3 481 3 203 2 2      So, range of values of k: 2 2 481 3k   481 481or33kk  . Question 8 [11 Marks] (i) ttt xtx tansecd dsec  tt yty 2secd dtan  t x t y x y d d d d d d  = tt t tansec sec2 = t t tan sec = tsin 1 = tcosec (ii) At point  sec , tanP  , t  d cosecd y x  Equation of tangent at P: tan (cosec ) ( sec )yx      (cosec ) tan (cosec ) (sec )yx       sin 1(cosec ) cos sin cosyx       2sin 1(cosec ) sin cosyx    2cos(cosec ) sin cosyx   cos(cosec ) sinyx   (cosec ) cotyx 

10 (iii) When 0y , cot cos sin coscosec sinx      So, coordinates of  cos , 0A  When 0x , coty  So, coordinates of  0, cotB  Area of triangle AOB = 1 cos cot2  When 6   , Area = 6cot6cos2 1   = 32 3 2 1  = 4 3 units2 (iv)  cos , 0A   0, cotB  So, mid-point of AB = cos cot,22     cos cot,22xy    11sec , tan22xy   Since 22tan 1 sec Then 22 11 144yx  22 11 144xy   [Note: 0 