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       