ACJC_Solutions_Paper_1

1 Anglo-Chinese Junior College H2 Mathematics 9740 Qn Paper 1 Solution 1(a)       2 2 2 2 2 21 1 2 4 2 4 7 2 4 7 4 7 12ln 4 72 23 1 2 2ln 4 7 tan2 33 xx dx dxx x x x x x x x dx x xx x C                     (b)         33 22 00 0 3 2 0 5 2 0 5 2 5 2 22 33 20 3 22 35 22 035 4 15 a aa a a a x a xx a xdx x dx a x dx ax a a                    2     2 22 2 2 2 2 23 2 5 2 2 5 2 23 02 5 2 12 0(2 1)( 2) 1 2 0, (2 1)( 2) 0 0.5 2 , 0 0.5 2 xx x x x x xx xx x xx Since x xx x or x Since x x or x                       3 When 0y , 21 and xe (1,0) (e2,0) 0 y x y=f(x)

2 When 0y , it cuts at 2 points on the curve. Therefore it is not a one to one function. Thus, 1f does not exist. ae       22 2 2 1 1 1 1 ln ln 2ln ln ln 1 1 ln 1 1 (n.a) or yy y x x y x x yx xy x e e               1 1 1f : , ( ,1] xx e x     ghR = ( 0,3 ]e 4 (i) Let n be the number of years after 2013. Tom’s pay   30000 1 1500n   Jerry’s pay   1 25000 1.05 n       1 30000 1 1500 25000 1.05 n n     From the table, 17n . Hence the first year in which Jerry’s pay is higher than Tom’s is 2029. (ii) Tom’s total income     2 30000 1 15002 n n   Jerry’s pay 1.05 125000 1.05 1 n        1.05 12 30000 1 1500 250002 1.05 1 nn n     From the table, 26n . Hence the first year in which Jerry’s pay is higher than Tom’s is 2038.

3 5 2 2 2 2 A r B rhC A r Bh rC        2 2 2 2 2 2 V r h A r Br rC r A r BC                2 2 2 12022 1 3 02 gives max 3 dV r rB A r Bdr C C A r BC ArV B           2Cost of base 3 3 rB A BB A        6 1 2 3 4 1 1 3 114 2 2 1 114 3 332 2 4 1 312 9 994 6 10 1 914 27 27 2710 18 28 1 27110 3 13 n n n u u u u u                     Let P(n) be the statement 1 1 3 13 n n nu    for all 0n . When 0n , LHS = 0 1 4u  RHS = 1 1 1 31 3 11 3 4 1 3      LHS = RHS  P(0) is true.

4 Assume that P(k) is true for some 0k , i.e., 1 1 3 13 k k ku    . To prove P(k + 1) is true, i.e., 1 3 13 k k ku    ,   1 1 1 1 1 1 11 1 11 1 LHS 3 21 33 13 by assumption 32113 3 13 2 3 1 3 13 3 2 3 1 3 3 1 3 3 3 RHS13 k k k k k k k k k kk k k kk k k k k u u u                                Since P(0) is true and P(k) is true  P(k+1) is true, by the Principle of Mathematical Induction, we conclude that P(0), P(1), P(2), P(3), … are all true. Hence P(n) is true for all integers 0n . 7 2 2 , tx t y e  2 2 xt dx dt   2 2 2 t t ye dy tedt     2 tdy tedx   Equation of normal