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 at   2 2, ppe  is   22 1 2ppy e e x p p    Equation of normal at 12,C e   is  1 2y e xe   At  ,0Ax : 2 2 2 2 1 1 2 122 exx e e e       At  0,By : 21 1 1 222 ey e y ee e e        0,1 0 x y (0, 1)

5 22 2 12 1 1 2OA ee eeOB e  : 1:OA OB e 8 2 2 232x a ay x a x a x a     Asymptotes: , x a y x a     2 2 2 2 2 3 2 2 1x a a dy ay x a x a x a dx xa         Since   222 0 & 0 a x a x     ,   2 2 21 1 0dy a dx xa      2 2 232x a ay x a x a x a     has no stationary points. (shown) Axes intercepts:  3,0a ,  3,0a ,  0,3a     2 2 2f1 ayx xa     From the graph of  fyx  , f is increasing for xa OR     2 3 4f0 0 ax xa xa xa       xa 1y x y x=a y=x+a 0 (0,3a)  3,0a  3,0a 22 3xay xa  

6 9 : or 2kk     or  , 2  10 (i) 11 a b a b aba b a b a b     (ii)     2500 1 2500 1 2500 1 1 4 3 4 1 4 3 4 1 4 3 4 1 1 4 1 4 34 5 1 4 r r r rr rr rr rr                     1 9   5 13 9 9997 9993 10001 9997 10001 1 4            (iii) 2500 2501 01 2500 1 11 4 1 4 5 4 3 4 1 11 4 3 4 1 10001 10003 10001 1 1 4 10001 10003 10000 1 1 4 10001 10003 124.75 10001 10003 24 rr r r r r r rr                        2 1 −2 2 O 6  −2 R 3 Re(z) Im(z) 3  3,1

7 11 (i) 2 acOM  2 2 2 2 2 cos 60 2 0.5 2 0.5 2 3 ()4 a c cLength of Projection c a c c c c a c c c a c c Note: a cc c c c c c Shown          11 (ii) 2 1 4 1 sin 604 3 8 3 8 3 8 a c ac c c Note: a c c Area of OMC k         11 (iii) 5 2 5 2 sin 605 2 53 4 53 4 53 4 c a a ca a ca= a ca a c OD Shortest OMC OD t        

8 12 2 2 4 diff. w.r.t. 2 yw t t dyt ytdw dt dt t          23 2 2 3 4 4 2 2 3 2 3 2 3 2 2 2 2 2 22 2 22 2 2 2 2 1 1 2 1 1 2 11 1 2 2 2 11ln ln 222 1 ln22 ln 2 2 2 1 t t dwt w t wt wdt dyt yt y y ydtt t t t t t t dyt t y t y yt ytdt dy yydt dy dt yy dy dtyy dy dtyy y y t c y tcy y tby y Aey y Ae                               2 2 2 2 22 2 2 1 12 1 t t t t t Ae Aey Ae Aew t Ae      For A=0 is the x-axis. x y A>0

9 13 Common ratio r = tan . For S to exist, 1r  , i.e. 1 tan 1 44          For  in this range, 23 11 tan tan tan 1 tan 1 3 3 1 tan 2 21 tan 33 2tan 1 33 13tan 33 1 3 3 3tan 3 3 3 3 23tan 6 1tan 3 6                                Hence 46     . 14(a)           2 2 2 22 22 22 2 2 2 2 3 2 Re 32 34 3 6 9 0 12 3 0 3 11 4 0 3 1 14 12 zz x y x x y x x x y x x y xy x y                      Axes intercepts:  1,0 ,  3,0 Asymptotes:  31yx  (-1, 0) (3, 0) x y 3( 1)yx  3( 1)yx 22( 1) 14 12 xy 

10 14(b)  2 2 3wi    4 2arg 3 w w    224 cos sin 33 nn nnwi              2 is real sin 0 3 3 , even, or 3 , 2 n nw n m m m n k k                5050 50 50 100 100 100 100 * 100 100 100 1004 cos sin 4 cos sin3 3 3 3 22 2 sin 3 322 2 23 2 3 ww ii i i i k                                                    