TJC H2 MATHS P2 Solutions

1 © TJC 2015 TJC/MA 9740/Preliminary Exam 2015 TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination 2015 Higher 2 Mathematics Paper 2 Section A: Pure Mathematics [40 marks] 1 Sketch, on an Argand diagram, the locus of the point representing the complex number z such that 5πarg( 3 i) . 6z   [2] Give a geometrical description of the locus of the point representing the complex number w such that i,wk where k is real. [1] (i) Given that the two loci intersect at exactly one point, show that k = a or kb where a and b are real constants to be determined. [3] (ii) In the case when k takes the value of a, find the complex number representing the point of intersection, in the form x + iy, where x and y are exact. [3] [Solution] Locus – half line with correct argument and excluded point, passes through origin The locus of w is a circle centered at (0, -1) and radius k units. (i) Two loci intersect exactly at one point: k CD or k AC In triangle ACD: π 3sin 62 3 CD CD   and 3AC , 3 2k or k 3 i.e. 3 2a , 3b (ii) Consider triangle BCD:  BCD = π π π 2 6 3 π 3 1 3cos 3 2 2 4BC k     , π 3 3 3sin 3 2 2 4BD k     At point D : 3 3 1, (1 )4 4 4xy    ,  complex number representing point D = 31 i44 B D –1 π 6 Re Im locus of z 5π 6 k A( 3, 1 ) C O locus of w

2 © TJC 2015 TJC/MA 9740/Preliminary Exam 2015 Alternatively, π11cos 32OD  complex number representing point D = 1 π π 1 3 1cos isin i2 6 6 2 2 2 31 i44                    2 (a) Given that the sequence 5, 11, 17, , x is arithmetic, solve the equation 5 11 17 2760 x     . [4] (b) Mr Tan set aside $80,000 for his two sons. On the first day of the year that his sons turned 7 and 17 years old, he deposited $ x into the younger son’s bank account and the remaining sum of money into the elder son’s bank account. Mr Tan adds a further $1000 into the younger son’s account on the first day of each subsequent year. The bank pays a compound interest at a rate of 2% per annum on the last day of each year. Each son will withdraw the full sum of money from his account (after interest had been added) on the last day of the year that he turns 21 years old. (i) Find the amount of money the elder son will withdraw in terms of x. [1] (ii) Show that the younger son will withdraw    15 14$ 1.02 51000 1.02 1x . [3] Find the valu e of x if Mr Tan wanted both sons to receive the same withdrawal amount, giving your answer to the nearest integer. [2] [Solution] 2(a) Let x be the nth term of the given AP. 5 6( 1)xn   61n --------- (1)  5 27602 n x ---------- (2) Sub (1) into (2): 23 2 2760 0nn   9230 or 3nn    (reject as n  ) 179x 2(b) (i) At the time of withdrawal, the amount in the elder son’s account =  51.02 80000 x (ii) Year Balance at the end of the year in younger son’s account 1(age 7) 1.02(x) 2(age 8) 1.02(1.02(x) + 1000) = 1.022 (x) + 1.02 (1000) 3(age 9) 1.02 (1.022 (x) + 1.02 (1000)+1000) =1.023 (x) + 1.022 (1000)+1.02(1000) … 15(age 21) 1.0215(x) + 1.0214 (1000)+ 1.0213 (1000)++ 1.02(1000) [B1] AEF

3 © TJC 2015 TJC/MA 9740/Preliminary Exam 2015 At the time of withdrawal, the amount in the younger son’s account     14 15 1.02 11.02 1.02 1000 1.02 1x         15 141.02 51000 1.02 1x   If they should receive the same amount of money at the time of withdrawal,       5 15 141.02 80000 1.02 51000 1.02 1xx       5 14 15 5 1.02 80000 51000 1.02 1 29401.851.02 1.02x   Therefore, $x = $29402 (correct to nearest integer) 3 (a) By considering a standard series expansion, find the general solution of the differential equation 23 d 1 d 1 d 1 d1. d 2! d 3! d ! d                             r y y y yx x x x r x [4] (b) An empty rectangular tank has vertical sides of depth H metres and a horizontal base of unit area. Water is pumped into the tank at a constant rate such that if no water flows out, the tank can be filled up in time T seconds. Water flows out at a rate which is proportional to the depth of water in the tank. At time t seconds, the depth of water in the tank is x metres. When the depth of water is 1 metre, it remains at this constant value. Show that d (1 )d x kxt  , where k is a constant in terms of H and T. [2] Find x in terms of t, H and T. [4] [Solution] 3(a) 23 d 1 d 1 d 1 d1. d 2! d 3! d ! d                             r y y y yx x x x r x d de y x i.e. d d de ln d y x yxx x   Integrate wrt x: 1ln d ln dy x x x x x x x    lnx x x C  

4 © TJC 2015 TJC/MA 9740/Preliminary Exam 2015 3(b) “In” rate: 1HH TT   “Out” rate: px dd dd V x H pxt t T   (Note 1Vx as horizontal base is of unit area) dd dd Vx tt When d1, 0d 0 xx t H pT Hp T       dd 1dd V x H H H xxt t T T T     i.e. Hk T 1d 1d xH x t T Integrate wrt t : 1 d 1 d1 ln 1 HxtxT Hx t CT       1 e e H tC Tx   1e H tTxA     When 0, 0 0 1 1 tx AA       1e H tTx    

5 © TJC 2015 TJC/MA 9740/Preliminary Exam 2015 4 The curve C has equation 2 2 a bxy b ax   where x and, a and b are constants such that 0 < a < b. (i) (a) Using an algebraic method, find the range of values of y in terms of a and b. [3] (b) Find the equation of the asymptote and the coordinates of the stationary point of C. [4] (c) Sketch C, indicating the intercept(s) on the axes. [1] (ii) Given that 2ba , find 2 2 d  a bx xb ax . [3] [Solution] (i) (a) 2 2 a bxy b ax    2( ) ( ) 0ay b x by a    Since x , discriminant = 0 – 4 (ay – b) (by – a)  0 (ay – b) (by – a)  0 by aQ since coefficient of 2x cannot be zero ab yba   Alternative 2 2 a bxy b ax    by + ayx2 = a + bx2 2 a byx ay b   Since x2  0, 0a by ay b    (a  by)(ay  b)  0 ab yba   (b)   22 2 2 22 2 b aa bx b a b a by b ax a b ax a a b ax          Equation of asymptote: by a         2 2 2 2 2222 2d 20d a b x b ay axx a b ax b ax        0, axy b   Coordinates of stationary point is 0, a b   . (c) x 0 by a 0, a b   y

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