AJC H2 MATH P2 SOLUTIONS

Anderson Junior College Preliminary Examination 2011 H2 Mathematics Paper 2 (Solutions) Section A: Pure Mathematics [40 marks] 1 (i) d 100 3d 97 , 0 u kut ku k      1 du 1 d97 1 ln 97 ln 97 97 e where 1 97 e (shown) kt kc kt tku ku t ck ku kt kc ku A A e uA k                  (ii) For the processing tank, when t = 0, u = 0. Hence, 1 97 0 97AAk    . Therefore,   1 97 97e ktu k  . Since u = 70 when t = 1, we have   170 97 97e 0.69220 (5 s.f. from GC)k kk    Amount of water in the processing container when t = 2 =   2(0.6922)1 97 97e0.6922  = 105.03 litres Amount of water pumped in when t =2 = 200 litres Hence, amount of water processed = 200 – 105.03 - 6 litres = 89.0 litres (3 s.f.)

(iii) As   0.692219 7, 97 97e 140.13 litres0.6922 0.6922 ttu      The minimum volume of the tank is 141 litres. 2 (a) (i) Method 1 (using differentiation) 1 2 1f' ( ) ( 1 ) s i n 1 x xx x     For 10 2x , f ' () 0 f ()x x is a strictly increasing function and therefore f is a 1 to 1 function. Since f is 1-1, the inverse function f -1 exist. Method 2 (using graphical test): Any horizontal line y = k intersects the graph of f at at most one point  f is a one-one function  the inverse function f -1 exist. ( Or : any horizontal line y =k, 01 4k   , intersects the graph of f at exactly one point  f is a one-one function  the inverse function f -1 exist) To find the value of f -1(1), solve f(x) = 1 1(1 ) s i n 1 1 1 (rejected since 0) or 0 xx xx x      1f( 1 ) 0  (ii) At 0, f '(0) 1 and 1xy  Equation of tangent to the curve y = f(x) at x = 0: y = x +1 (iii) At x = 1, we will use the symmetry properties of the curve of f and f -1 along y = x  Equation of tangent to the curve y = f-1(x) at x = 1: x = y +1  y = x – 1

2  (b)  (b) (i) Range of g = [-11,5] (ii) Different values of a will translate the curve 1= 1+y x along the x-axis by a units to give 1 = 1+y x a Domain of g g [-11, 5] h Range of hg Since domain of h = [-11, 5] and we need range of hg to be subset of (1, ), therefore [-11, 5] must fall in the interval (a, ) From the diagram, it can be observed that the greatest value of a is -12. 3  (i) 22 2 2 2 1 2 330 2( 3 )1 2 c o s c o s 7012                             22 2 22 22 2 2 322 1 4 2 70 52 2 9 2 20 2 1 9 2 11 40 16 0 (11 4)( 4) 0 44 as = (rejected as <0) 11                      y y x =a x 5-11 y=1 x -11 5 y=g(x)

(ii) Using ratio theorem, 42 1 0 31 1 30 3 344 4 21 7 OB OAOM                               10 10 44 1 1 3333 4 3 77 OC OM                     (iii) p represents the perpendicular distance of C from the line AB (or p is the height of the triangle ABC with AB as its base). 10 2 16 11 33 633 71 4 AC                  , 42 6 03 3 21 1 AB                 16 6 6 1 63 83 41 1 2 11 1 2 2 33 2 336 9 1 46 AC AB p AB                           3 1 1223 23 qO M pM C qp   4  (i) 943 34iz i zi    943 34iz i zi     2 34 9zi     34 3zi    A B M C O p q

(ii) max arg (z + i) 1 22 32sin2 35      2.65  arg 2.652 zi     (iii) *4ww 4ww - locus is the line x = 2 [1] ()zi w z i w   = distance between locus of z (circle) and the locus of points representing – iw . Locus of – iw is the line y = -2 Therefore least value of 3zi w A B  5  (i) c = 0 as the line l is on 2. [B1] (ii) Since the line l is on 2, l is perpendicular to the normal of 2. 1 50 11 51 0 5 1 (shown) b a ba ba            

(iii)   Method 1:     11 2 21 01 2 a a              is a vector  to  1      2 15 0 2 5 2 0 2 7 21 b ba a b a                           Method 2:       15 51 11 5 ba ab ab                is a vector  to  1   51 12 0 5 2 2 0 2 7 50 a ba b a b ab                  

Method 3:     12 25 1 01 5 2 b b               is a vector  to  1      21 1 0 2 5 20 27 52 1 aa b a b b                           Method 4:    11 2 21 01 2 a a              is a vector  to  1   12 25 1 01 5 2 b b               is a vector  to  1   22 11 25 2 k ab                        1a n d 2 52ka b         27ab    Using 51ba  from part (ii) and solving simultaneous equation, we get a = -1 and b = 4.   (iv) Possible answers are: 44 5 4 42 or 5 4 42 11         rr   Section B (Statistics) [60 marks] 6  There are 9 distinct letters: C, O (2), R (2), E, L, A, T, I, N (i) Treat all the letters as distinct then arrange them according to the restriction. This will be followed by taking the repeated letters into account. Number of arrangements 59 ! 6 27216002!2!  (ii) Case 1: All distinct

No. of ways = 9 6 60480P  Case 2: 1 pair of repeated letters No. of ways = 28 14 6! 504002!CC  Case 3: 2 pairs of repeated letters No. of ways = 7 2 6! 37802!2!C  Total no. of ways 60480 50400 3780 114660  7  (i) Let X be the number of lemon candies in a randomly selected packet of 20. (20,0.24)XB ( ) 20 0.24 4.8EX   ( ) 20 0.24 0.76 3.648Var X    Since 60 ( 50)n , by Central Limit Theorem, 3.6484.8, 60XN   . ( 5) 0.20865208PX  =0.209 (3sf)   (ii) The sample is biased, as only students are surveyed. Not everyone in the population has an equal chance of being surveyed. It will be difficult to get an exhaustive list of people of all age groups to do a proper stratification. (no sampling frame) Use Quota Sampling 8  (i)   (ii) From GC, the product moment correlation coefficient = 0.940 (3 sig fig) (iii) The points in the scatter plot fits closely to the curve Btx Ae (left) and the r value is close to 1. Thus this model is suitable. Transformation of Btx Ae into a linear model: ln lnx AB t  From GC, ln 1.71086 5.53, 0.0476AA B     (iv) If t is increased by 5, the increase in x is as follows:    ( 5) 5 5(0.0476) 1 (1.269) Bt o Bt B oo o xA e x Ae e x e x x    The rate of chirps is estimated to increase by 26.9% from the rate before the temperature is increased. The estimate is not reliable because we are extrapolating beyond the region where the

data is collected and analyzed. (v) The value of r remains the same since it is not affected by any translation or scaling. 9  2~ (45000, 2000 )AN 2~ (30000,1850 )BN (i)  122 ~ (30000, 22845000)ABB N  122 25000 0.85224 0.852PA BB   (3 sig fig) Assumption: The distributions of the lifespans of all televisions are independent of each other. (ii) Let W denotes the number of plasma televisions out of 50 with a life span of more than 30000 hours. ~( 5 0 , 0 . 5 )WB (14 22) (15 21)PW PW   (2 1 ) (1 4 )PW PW  0.16112 0.0013011 0.15982 0.160   (iii)