SRJC H2 MATH P2 TEACHER S

1 2015 SRJC H2 Mathematics Prelim Paper 2 Section A: Pure Mathematics [40 marks] 1 Functions f and g are defined as below. 1f : , 2 1xx x  g : ln(2 ) , 2x x x x    (i) Sketch the graph of y = g(x) and state its exact range. [3] (ii) Determine whet her the composite function s fg and g f exist, justifying your answer. Find the range of the composite function if it exists. [4] (iii) On the same diagram as part (i), sketch the graph of y = g −1(x), indicating on your sketch the line in which the graph of y = g(x) must be reflected in order to obtain the graph of y = g−1(x). [2] Suggested Solution (i) Range of g, Rg = (2 + ln4, +∞) (ii) From GC, the range of f is (0, 1]. Since fRD g = (−∞, −2), the composite gf does not exist. Since Rg = (2 + ln4, +∞) ⊂ [2, ∞) = Df the composite fg exists. Direct Method for finding range of fg Range of fg = (0, 1 1 ln 4 ) y x (−2, 2+ln4) y x 1( 2, )1 ln 4  y = fg(x) = 1 ln(2 ) 1xx   Rfg y = g(x) (2+ln4, −2) y = x y = g−1(x)

2 Indirect Method for finding range of fg Range of fg = (0, 1 1 ln 4 ) 2 A tank contains water which is heated by an electric water heater working under the action of a thermosta t. When the water heater is first switched on, the temperature of the water is 35 C . The heater causes the temperature to increase at a rate Cr per minute, where r is a constant, until the water temperature hits 75 C .The heater then switches off. (i) Write down, in terms of r, the time taken for the temperature to increase from 35 C to 75 C . [1] The temperature of the water then immediately starts to decrease . The temperature of the water at time t minutes after the heater is switched off is C . It is known that the temperature of the water decreases at a variable rate ( 25) Ck  per minute, where k is a positive constant, until 35  . (ii) Write down a differential equation involving  and t, to represent the situation as the temperature is decreasing. [1] (iii) Given that when 55  , the temperature is decreasing at a rate of 5C per minute, find the total length of time for the temperature to increase from 35 C to 75 C and then decrease to 35 C , leaving your answer in exact form, i n terms of r. [7] y x y = f(x) 2+ln4 1 1 ln 4 Rg Rfg

3 Suggested Solution (i) Time taken 40 r mins (ii)  d 25d kt    (iii) 1 d d25 kt   ln 25 kt c   , where c is a constant, since  > 25oC  25 e .e kt c  25 e ktA  , where A is an arbitrary constant. When 55  , d 5dt   , 1 6k  When t = 0 and 75  , A = 50 1 625 50e t    When 35  , 1 635 25 50e t 1 6 1e 5 t  1 56ln 6ln 5t   Total length of time = 40 6ln 5r   mins

4 3 [It is given that the volume of a sphere of radius r is 34 π3 r and that the volume of a right square pyramid with a square base of length x and height h is 21 3 xh .] In the diagram below, a hemisphere of fixed radius a cm lies on the base of a right pyramid such that its curved surface is in contact with all four faces of the pyramid. The pyramid has a square base of length x cm and height y cm. Given that the volume of region inside the pyramid that is not part of the hemisphere is denoted by V, (i) show that V = 3 3 22 12 π33 4 ax a xa   . [3] (ii) use differentiation to find, in terms of a, the minimum value of V exactly, proving that it is a minimum. [7] Suggested Solution (i) Method 1 Consider the similar triangles ABC and ACD BC CD = AB AC y a = 2 21 2 1 2 xy x   y a = 221 4 1 2 xy x  x x y a D a 1 2 x y A B C

5 y a = 221 42 1 2 xy x  xy = a 22 4xy x2y2 = a2x2 + 4a2y2 y2 (x2 – 4a2) = a2x2 y2 = 22 22 4 ax xa y = 22 4 ax xa or y = 22 4 ax xa   (rejected x > 0) V = 2312 π33x y a = 3 3 22 12 π33 4 ax a xa   Method 2 Considering the area of ABC, we have 2 211 2 2 2 2 xx y y a              22 2 2 2 44 xx y y a          22 2 2 2 44 xx a y a  y = 22 22 224 4 a x ax xa xa   V = 2312 π33x y a = 3 3 22 12 π33 4 ax a xa   (ii) V = 3 3 22 12 π33 4 ax a xa   a 1 2 x y A B C

6 d d V x =     2 2 2 3 22 2 22 243 1 24 3 4 xx a x x xaa xa    =      2 2 2 4 3 22 2 431 3 4 x a x x a xa   =   4 2 2 3 22 2 1 2 12 3 4 x a xa xa   =     2 2 2 3 22 2 62 3 4 x x a a xa   Let d d V x = 0.      2 2 2 3 22 2 62 3 4 x x a a xa   = 0  x2 – 6a2 = 0 or x = 0 (Reject 0x )  x = 6 a or x = 6 a (Reject 0x ) Method 1 2 2 d d V x =   4 2 2 3 22 2 d 2 6 d3 4 x a xax xa      =            31 2 2 3 2 2 2 4 2 222 322 34 4 12 2 4 62 2 3 4 x a x a x x x a x a x a xa       When x = 6 a, 2 2 d d V x =           31 2 2 3 3 2 2 4 422 322 6 4 24 6 12 6 3 6 6 4 36 362 3 64 a a a a a a a a a a aa       = 6 6 2 48 3 0 38 aa a  = 43 a > 0

7 Method 2 x 6 a– 6 a 6 a+ d d V x -ve since     2 2 2 3 22 2 6 0 4 x x a xa    for x < 6 a 0 +ve since     2 2 2 3 22 2 6 0 4 x x a xa    for x > 6 a Slope \ – / Hence, V is a minimum value when x = 6 a. The minimum value of V = 4 3 22 1 6 6 2 π33 64 a a aa   = 3223 π3 a 

8 Im(v) Re(v) 0 –3 8 9 5 12  x 1 5 4 (a) The complex number z is given by iz re  , where r > 0 and 0 1 2 . (i) Given that w = ( 3 – i) z, find |w| in terms of r and arg w in terms of  . [2] (ii) For a fixed value of r, draw on the same Argand diagram the the loc i of z and w as  varies. [2] (iii) If r = 1.5 units, calculate (in terms of  ) the length of the locus of w for Im(w)  0 as  varies. [2] (b) Sketch on a single Argand diagram the set of points representing all complex numbers v satisfying both of the following inequalities: 5 8i 5v- - £ and 12 8i 12 10ivv- - ³ - - . Hence find (in radians) the least value of arg ( 5 3i)v-+ . [6] Suggested Solution (a) (i) 223 ( ) 3 1 2w riz r      arg( ) arg 3 arg( ) 6w i z      (ii) (iii) Length of the locus of w = 1 (2)( )6 )(3  units (b) 12+10i 12+8i = = Im Locus of z 2r r 0 Re Locus of w r 2r 6 

9 From the diagram, 2 2 21 =5 24xx   [M1] 24tan 0.3876 rad12   [M1] Therefore, the least value of  arg 5 3i 0.3876 1.18 rad2v      [A1] Section B: Statistics [60 marks] 5 A company with eight hundred employees wishes to find out how much time its employees take to travel to work. It is given that the employees go to wo rk either by car or by bus/train and that each of them takes the same form of transport to work every day. The following table shows the number of employees going to work by car and the number of employees going to work by bus/train. Car Bus/Train Men 165 260 Women 82 293 The company wants to take a random sample of 180 employees. (i) Explain what is meant, in this context, by the term “a random sample”. [1] (ii) Describe how a random stratified sample can be obtained. [2] (iii) Give a reason why quota sampling is not as suitable in this context compared to stratified sampling. [1] Suggested Solution (i) It means that every employee has an equal chance of being selected and that the selection of each employee is independent of one another. (ii) Randomly assign numbers to each employee in each stratum: Men taking car from 1 to 165; Men taking bus/train from 1 to 260; Women taking car from 1