VJC H2 MATH P1 ANS

1 2015 VJC JC2 Prelim Paper 1 Solutions Q1) Let nP be the statement: ( 1)nw an n   , n  . LHS of 11Pw a  (given) RHS of 1 (1) (1 1)Pa a    1 is true .P Assume kP is true for some k  i.e. ( 1)kw ak k   We want to show 1kP  is true i.e. 1 ( 1)kw a k k    LHS of 11kkP w   2 1 ( 1) ( 1) ( 1) = RHS of 1 [( 1) +1] 1= ( 1)[ ( 1)]+1 ( 1)( 1) 1 k k ak ak kk P kwk k ak kk kk k k k a                  1is true is true kkPP  Since we have shown that (1) 1P is true and (2) 1is true is true. kkPP   By mathematical induction, nP is true for all positive integers n. Q2) Sub (1,1) and (2,2) into h( )yx . 1 ----- (1) 8 4 2 2 -----(2) a b c d a b c d         Since (2,2) is also the stationary point, h '(2) 0 . i.e. 12 4 0 ----- (3)a b c   Using the GC, 11 24 35 24 2 ad bd cd     0 1 1 3 5 2 4 2 4 02 ab c dd d            6{ : 2 or 0} 5d d d     2 6 5 0    

2 Q3) 2 2322 ax bx d kyx xx     By observation, 2a    22 2 3 2x bx d x x k       Compare coefficents of x : 3 4 1bb    22 2 x x dy x   Given: 6d  Asymptotes: 2 3, 2y x x   Axial intercepts: when 0x , 2 dy When 0y , 220x x d     1 1 4 2 1 1 8 44 d dx   The coordinates are 0, 2 d  , 1 1 8 ,04 d  , 1 1 8 ,04 d  . Let 1 1 8 4 d  and 1 1 8 4 d  Q4(i) dLet e 7 . So, = e 7. d d = 0 e 7 0d ln 7 min ln 7 xx x yyx x y x x          Q4(ii) Let   1g1x   g1 e 7 1x x x     From the GC, 3.13x  ,0  ,0  20, d O 2x 23yx x y x y e7xyx 1y 3.13 y x e7xyx

3 Q4(iii) Q5(i)   1 fy x Q5(ii)                 0 0 2 f ' d 2 f 2 0 f 0 2 f 5 2 4 3 aa x x x x aa a a a a                              Q6(i) n Amount at end of year n 1 1.08(1000) 2  1000 1.081.08 1000   21000 1.08 1.08 3 :       2 23 1000 10001.08 1.08 1000 1.08 1000 1.08 1.08 1.08      n   21000 1.08 1.08 1.08 n   1x  ,0a  f'yx  ,0a x y 1y xa 0y  1,0   1 fy x  1 4, aa x  1 50, a y  ln 7,7 7ln 7  7 7ln 7,7 7ln 7 0.169 0.169 x y  gyx   1ggyx   7 7ln 7,ln 7   1gyx 

4 Amount at the end of year 2040     2 26 26 1000 (1.08) 1.08 ... 1.08 1.08 1 (1.08) 1000 1 1.08 86351 (to nearest dollar)           Q6(ii) terms 1000 1080 1160 ... 86351n n S        2 2(1000) ( 1)(80) 863512 40 960 86351 0 59.987 (N.A.) or 35.987 Least number of years that he still needs to save = 36 n nSn nn nn             The year at which Mr Woo’s savings in this savings plan will first exceed $86351 2015 36 1 2050    Q7  Rate of salt flowing into tank per minute is 12 0.125 1.5 kg 12Rate of salt flowing out per minute is 0 .03400 dTherefore, 1.5 0.03 .d qq q qt    d 1.5 0.03d q qt  0.03 0.03 1 d 1 d1.5 0.03 1 ln 1.5 0.030.03 1.5 0.03 e 1.5 0.03 e t t qtq q t C qA qB           When 0, 100, 1.5 0.03(100) 1.5 t q B B        0.03 0.03 1.5 0.03 1.5e 1.6 kg per litre = 0.16 400 = 64 kg of sal t in the tank Thus 1.5 0.03 64 1.5e 42.4 min t t q t         (3 s.f)

5     0.03 0.03 0.03 0.03 1.5 1 e 50 1 e When is large, e 0 Thus, the amount of salt in the tank decreases to 50kg. t t t q q t       Q8(i)       2 22 22 22 LHS ( i ) * (i ) 2 i * 2 i * 2i xy x y xy x y xy x y xy               2 2 22 22 RHS ( i )* i i 2 i 2i xy xy x y xy x y xy       Q8(ii) Let iz x y , where ,xy  2 22 2 4 2 2 2 2 22 ( i ) 1 4 i 1 (1) 24 2(2) 12(1) 1 12 0 ( 4)( 3) 0 0 4 2, 2 i or 2 i xy xy xy y x x x x x xx x x x x y z                                 … Ö3 Ö 3 - (2) Ö3 Ö3 Ö 3 Ö 3 Q8(iii) 2 4 16 i 4(1 4 i)w    Ö 3 Ö 3 * both sides:   2 * 4(1 4 i)w  Ö3 using (i) :   2 * 4(1 4 i)w  Ö3 using (ii) : * 4 (2 i ) or 4 ( 2 i) 4 2 i or 4 2 i 4 2 i or 4 2 i w w              Ö Ö 3 Ö Ö 3 Ö 3 Ö 3 Ö 3 Ö 3   0.0350 1 e tq  100 t q 50q

6 Q8(iv) 1 2 i z  Ö3 and 2 2iz   Ö3 Given:   2arg z  .                12 2 2 arg arg 2 3i 2 3i arg 4 4 3i 3 arg 1 4 3i arg arg 1 arg zz z z                   Alternative Given:   2arg z  (where 0  )      1 2 1 2arg arg arg 22 z z z z         Q9(i) Q9(ii) d 0.1d x t  , d ?d y t  at 61 e2x   61e sin e 26      d e cos e sin cos sin d e cos e sin cos sin y x             d d d d d d y y x t x t       cos sind 66 0.1d cos sin66 3 122 0.1 3 122 31 10 3 1 y t                            e sin e cosy x      x y 2e  1 2e   O

7 Q9(iii) d cos sin d cos sin y x     At point P , tangent // y-axis d is undefinedd cos sin 0 tan 1 4 2 2 y x                 „ 44 1e sin e 4 2 x       Equation of tangent at point P is 41 e 2 x   Q9(iv) Coordinates of P: 4411e , e 22   2 2 411e e e22OP         Note that P lies on the line yx and OP  OQ, then Q lies on the line of yx . 44 44 2 2 4 e sin e cos tan 1 4 1e sin e 4 2 1e cos e 4 2 11e e e22 x y OQ                      Area of   1 2POQ OP OQ 4411 ee22          units2

8 Q10(i) 1 2 3 ...nnv u u u u     23 2 3 4 3 4 5 21 11 12 2 1! 2! 3! 2 2! 3! 4! 2 3! 4! 5! . . 2 ( 2)! ( 1)! ! 2 ( 1)! ! ( 1)! 2 ! ( 1)! ( 2)! n n n n n n n n n A A A A A A A A A A A A n n n A A A n n n A A A n n n                        2 2 1 1 2 2 1 2 22 2! 2! ( 1)! ( 1)! ( 2)! (shown)2 ( 1)! ( 2)! n n n nn A A A A AA n n n A A AA nn                   Q10(ii) 12 2 1 7( 1)! ( 2)! nnN nN n n AAvN n n               2 2 2 2 2 2 1 22 1 2 1 72 11 ( 1) 72 7 1 1 7 (1 7 ) 2 1 7 7 1 1 7 . . 7 126 7 1 49 1 1 2 6 7 7 N nN n N n N n N N N N N AAN AANN NA AN NA AN NA AN                                                    11 1N NN   As N  , 1 0N  , so 111 N . Also, 1 0 7N  . Hence as N  , Series 2 7 26 AA   12 2 1 7( 1)! ( 2)! nnN nN n n AAvN n n             converges. Limit = 2 7 26 AA .

9 Q11) Let M be a point on plane  . 01 0 , 2 13 1 OM MP                   , 2 2 22 1 2 . 2 11 1 5 5 4 1 ( 5) 10 25 2 a a aa a a a a                             Q11ii) 1 24 : 1 3 32 lr                    ,  2 41 : 0 1 50 lr                   ,  2 4 4 1 3 , 3 2 5 OA OB                       24 31 22 14 . 1 0, . 3 0 02 1 2 7 , 1 7 29 Solving, 22 5, 99 2 14 99 2 22, 39 37 9 AB AB AB OA OB                                               5        Q11(ii) 4 1 2 0 1 . 2 8 2 2 5 13 5 0 1                                     Hence 2l is in plane  .

10 Alternative 2 41 : 0 1 50 lr  