EJC H2 2020 Prelim P2 solution

Section A: Pure Mathematics [40 marks] 1 Suggested solution Suggested solution a) 2 The curve 1C has equation 2 3 2 3 1 x xy x    . The curve 2C has equation     2 2 2 41 1 yx b    , where 0b . (i) Sketch 1C , stating the equations of any asymptotes and the coordinates of any turning points and points where the curve crosses the axes. [4] y x O y x O  0, 6 (1, 3) (2.5, 0) (0, 1/6)

Page 2 of 16 (ii) Hence, find the range of values of b such that there is no point of intersection between 1C and 2C . Using the maximum value of b found, sketch 2C on the same diagram as part (i). [3] Suggested solution (i) 23 2 3 4 3 11 1 x xy x x x       (ii) 2C is a horizontal hyperbola with center 1,4 . Asymptotes:  1 4y b x   For no point of intersection between 1C and 2C , the asymptote of hyperbola must be as steep or less steep than the oblique asymptote of 1C . Note that the gradient of oblique asymptote of 1C is 3. 0 3 b   For 3b , Asymptotes:  3 1 4 3 1, 3 7y x x x       3 A curve C has parametric equations 1 cos ,x   sin 1,y     for 0 2   . 3 1y x  1x x  0, 3 0.155, 2.93   2.15,10.9 y O  0, 3  0,4  2,4 3 1y x  x y O 1x  2.15,10.9 3 7y x 

2020 JC2 H2 Mathematics Preliminary Examination Page 3 of 16 (i) The point P on C has parameter p. Show that the normal to C at P crosses the y-axis at a point Q with coordinates 0, 1p . [5] (ii) The normal to C at P crosses the x-axis at point R. Given that point S is the midpoint of QR, find a cartesian equation of the curve traced by S as p varies. [3] Suggested Solution (i) d sind x   d cos 1d y    d cos 1 d sin y x    Gradient of normal = sin cos 1    Equation of normal at point P  sinsin 1 1 cos cos 1 sin sin sin 1cos 1 sin 1cos 1 py p p x p p py x p p pp py x p p                    When 0, 1x y p   (Shown) (ii) Since equation of normal is sin 1cos 1 py x p p       when 0 1 1 cos sin y p px p    Midpoint of QR   1 1 cos 2sin p px p   1 2 py  Cartesian equation of the curve traced by S     1 cos 2 1 sin 2 1 y yx y      4 Vectors a and b are such that the modulus of a is 2 and b is a unit vector perpendicular to a. (i) State the value of (2 3 )a a b and explain your answer briefly. [1] (ii) Find the numerical area of the parallelogram with adjacent sides defined by 2a b and 5 7a b . [3] A vector c is such that 2 3  b c a b .

Page 4 of 16 (iii) Show that 6 c b a , where  is a constant. [2] (iv) State the geometrical meaning of b c and find the possible values of  if 5b c . [2] Suggested solution (i) (2 3 ) 0 a a b since 2 3a b is a vector that is perpendicular to 2a , and hence it is also perpendicular to a. (ii) Area of the parallelogram (2 ) (5 7 ) 10( ) 7( ) 5( ) 14( ) a b a b a a b b b a a b             10( ) 7( ) 19( ) 19 0 0 b a b a       19 sin 90b a  19(1)(2)(1) 38   (iii) 2 3 ( ) (2 3 ) ( ) 6( ) ( 6 )                  b c a b b c a b 0 b c b a 0 b c a 0  6c a is parallel to b and so 6  c a b , where  is a constant. 6  c b a (iv) b c c b  is the length of projection of c onto b. 2 ( 6 ) ( ) 6( ) 6(0)            b c b b a b b b a b     Since 5b c , 5  . 5 On 1 Jan 2019, Emma takes up a study loan of $30000 from a bank that offers two repayment plans. Pla n A charges a monthly instalment of $600 at the beginning of every month and increases by $50 for every subsequent month. There is no interest charged; instead, the bank charges a one -time administrative fee of $2500 which is to be added to the study loan. (i) (a) Show that, after the nth payment, she would have paid a total amount of $  225 575n n . [2] (b) How many payments will it take for Emma to fully repay her loan? [2] Plan B charges a fixed monthly instalment of $ x which is to be paid at the beginning of each month, starting from 1 Jan 2019. There is no interest charged for the first month only. Thereafter, interest is charged at the end of each month on the outs tanding amount at 1% per month, (i.e. interest will be first charged on 28 February 2019.)

2020 JC2 H2 Mathematics Preliminary Examination Page 5 of 16 (ii) (a) Show that the amount she owes after the fourth payment is         2 2 1.01 30000 1.01 2 1.01 x x x   . [2] (b) Use the formula for the sum of a geometric progression to find an expression, in terms of n and x, the amount she owes after the nth payment, for 2n . [2] (c) If Emma intends to repay the loan fully after 32 payments, find the least value of x. [2] (iii) Using the value of x in part (ii)(c), determine which plan will be cheaper for Emma. [1] Suggested solution (i)(a) Total amount paid = 600 + 650 + 700 + ….       2 600, 50 2 600 1 502 600 25 25 25 575 (Shown) n a d nS n n n n n             Total paid after nth payment   2$ 25 575n n  (i)(b) To complete fully repay her study loan, 2 Amount paid 30000 2500 25 575 32500 0 From GC, 49.345 or 26.345 n n n n        It will take Emma 27 payments to fully repay her study loan. (ii)(a) Mth Amount owed at s tart of the month Amount owed at e nd of the month Jan (n=1) 30000 x 30000 x Feb (n=2) 30000 2 x    30000 1.01 2 1.01x Mar (n=3)    30000 1.01 2 1.01x x        2 230000 1.01 2 1.01 1.01 x x   Apr       2 230000 1.01 2 1.01 1.01 x x x            3 3 2 30000 1.01 2 1.01 1.01 1.01 x x x    … … … n         2 2 3 4 30000 1.01 2 1.01 1.01 1.01 ... n n n n x x x x         Amount owed after 4 payments       2 230000 1.01 2 1.01 1.01x x x    (ii)(b) After nth payments, amount owed

Page 6 of 16                               2 2 3 4 2 2 3 4 5 0 2 2 2 2 2 2 30000 1.01 2 1.01 1.01 1.01 ... 30000 1.01 2 1.01 1.01 1.01 1.01 ... 1.01 1.01 130000 1.01 2 1.01 1.01 1 30000 1.01 2 1.01 100 1.01 1 n n n n n n n n n n n n n n n x x x x x x x x x x                                             Alternative After nth payments, amount owed                           2 2 3 4 1 2 2 3 4 1 3 2 2 2 2 30000 1.01 2 1.01 1.01 1.01 ... (1.01) 30000 1.01 2 1.01 1.01 1.01 ... (1.01) 1.01 1.01 1 30000 1.01 2 1.01 1.01 1 30000 1.01 2 1.01 101 1. n n n n n n n n n n n n n x x x x x x x x x x x x x x x                                               301 1n x   (ii)(c) For Emma to clear the debt, the amount that she owed after nth payments 0 When 32n ,                         2 2 2 32 2 32 2 32 2 30 30 30 30 30 30 30000 1.01 2 1.01 100 1.01 1 0 30000 1.01 2 1.01 100 1.01 1 0 30000 1.01 2 1.01 100 1.01 1 0 30000 1.01 2 1.01 100 1.01 1 1078.837557 n n nx x x x x x x x                        Least value of x is $1078.84 (iii) Under plan A, total amount repaid 30000 2500 32500   Under plan B, total amount repaid 1078.84 32 34522.88   Therefore, it is cheaper for Emma to take up Plan A

2020 JC2 H2 Mathematics Preliminary Examination Page 7 of 16 Section B: Probability and Statistics [60 marks] 6 Nine letter tiles that spell the word PRINCIPAL are used. (i) Find the number of ways to arrange the nine tiles in a row such that the vowels are together. [2] (ii) The tiles are shuffled and placed in a row randomly. Find the probability that the tiles spell PRINCIPAL. [2] (iii) Eight of the nine tiles are taken and arranged in a circle. Find the number of ways that this can be done . [3] Suggest