YJC H2 MATH P1 ANS

1 YISHUN JUNIOR COLLEGE 2015 JC2 PRELIMINARY EXAM PAPER 1 H2 MATHEMATICS SOLUTION Qn Solution 1 Let x, y, and z be the number of trays of blueberry, strawberry and chocolate cupcakes respectively. Time: 8 7 6 17 60 1020x y z     Amt: 0.6 0.6 0.8 96x y z   Price: 12 (1) 12 (0.9) 12 (0.8) 1572x y z   Using GC, 50, 50, 45x y z   2(a) If , a b 0 and , then 0ab implies that the two vectors a and b are perpendicular to each other i.e. ab . 2(b) If a lies in axisx and m a 0 , then m is parallel a , and hence is parallel to i . Since 1m then mi or i (just one will do) 2(c) Method 1: Let the diagonals BD and AC intersect at E. Given DE=EB -------(1) and AE=EC-----------(2) (*) (*) [from (1) and (2)] = AB AE EB DC DE EC EB AE AB    AB DC i.e. AB DC and //AB DC Therefore ABCD is a parallelogram. (Proven) Method 2: Using Ratio Theorem,     11 22OE OA OC OD OB    OA OC OD OB   OB OA OC OD   AB DC i.e. AB DC and //AB DC Therefore ABCD is a parallelogram. (Proven) 2(d) OP p x , OQ q y and OR r sxy Since ,P ,Q and R are collinear, PQ kPR for some k .    q p k r s p k r p ks      y x x y x x y ()k r p p  ----- (1) ks q ----- (2)

2 (1) (2) : r p p sq   rq pq ps  Therefore ps rq pq (Shown) 3 External volume =   2( 1) 1rh  Internal volume= 2 1000rh  2 1000h r   2( 1) 1 1000V r h    2 2 1000( 1) 1 1000Vr r       (Shown) 2 23 d 1000 20002 ( 1) 1 ( 1)d V rrr r r                  OR 2 2 2 3 2000 1000 d 2000 20002 2 2 d VV r r r r r r r r              For stationary ,V d 0d V r  2 2 3 2000 2000 2000( 1) 2 0r r r r         OR 23 2000 20002 2 0 r rr    i.e. 3 2000( 1) 2 0r r     OR    4 3 3 1000 1000 0 1 1000 0r r r r r          Since 3 20001 0 2r r    OR 31 0 1000rr     3 3 1000 10r    and 2 3 2 3 1000 1000 10 100h r        2 2 3 4 d 2000 60002 ( 1)d V rr r r               = 21.6105 > 0 when r = 3 10  OR 2 2 3 4 d 4000 60002 0 , 0d V rr r r      minimum for r    R OR r 3 10     3 10  3 10     d d V r negative 0 Positive Slope Therefore V is minimum when 3 10rh   cm.

3 4 2 2 dd dd yy xx  dd ddd yy xxx d d y yCx   , C d1 lnd x x C y Dy C y     , D eex D xC y A      , eD A  e xC y B  , 0B e xy C B  When 0x , 0y and d 1d y x  0 CB   i.e. CB and 1 0 1 CC    Therefore f ( ) 1 e xyx    4

4 5i 2 f ( )yx 5ii 1 f ( )y x 5iii f '( )yx 12y  2 30,  1,2 y x 0  1,0 y 2y 1x x 2y  320,  1 21,  320,  1 21, 0 1x  1,0 y x 0

5 6a  1 1 sin2x  d 1 1 cos ...d ... cos dd 2 2 x x      When 3 4x then 3 21 sin  and 1 2sin  1 6 When 1 4x then 1 21 sin  and 1 2sin  1 6  3 4 1 4 2 dx x xx  1 6 1 6 1 2 211 24 (1 sin ) 1 cos d2(1 sin ) (1 sin )              1 6 1 6 1 2 1 4 (1 sin )cos1 d2 (1 sin ) 2 (1 sin )            1 6 1 6 1 2 1 2 (1 sin )cos1 d2 (1 sin )(1 sin )          1 6 1 6 2 1 (1 sin )cos d2 1 sin          1 6 1 6 1 (1 sin )cos d2 cos       1 6 1 6 1 (1 sin ) d2       Therefore 3 4 1 4 2 dx x xx    1 6 1 6 1 cos2     1 1 1 1 6 6 6 6 1 cos cos2         1 6 6bi 3 3 3 11 33 000 e d e e d mm mx x xx x x x       3311 33 0 ee mxxx    3 3 01 1 1 3 3 3 e e emmm       331 9 1 3 e e mmm    ii Hence 3 0 e dxxx      331 9 lim 1 3 e e mm m m      1 9

6 OR As ,m 3e0 m  and 3 3 0e m m  then   331 9 11 3 e e 9 mmm    7a 11 1 ( 1) 1 [1 ( 1) ](1 ) a n a na n a na         1 1 ( 1)LHS 1 ( 1) 1 na n a n a na            11 [1 ( 1) ] 1 [1 ( 1) ] 1 na na a a n a na n a na           RHS  1 [1 ( 1) ] 1 N n a n a na    1 11 1 ( 1) 1 N n n a na     11 11 11 1 1 2 11 1 2 1 3 ... ... ... ... ... ... 11 1 ( 2) 1 ( 1) 11 1 ( 1) 1 a aa aa N a N a N a Na                       1 1 11 11 Na Na Na    1 Na Na  (Shown) Letting 1 2a then  1 [1 ( 1) ] 1 N n a n a na      11 1 22 11 2 1 ( 1) 1 N n nn     3 3 5 2 2 2 1 1 1 1 th term2 (1)( ) ( )(2) (2)( ) N      1 2 1 21 N N  1 2 11 1 N 

7 i.e. 3 3 5 2 2 2 1 1 1 th term(1)( ) ( )(2) (2)( ) N 1 2 22 1 N  As N  , 1 2 2 01 N  then 3 3 5 2 2 2 1 1 1 ...(1)( ) ( )(2) (2)( ) converges sum to infinity is 2 . 7b Let nP be the statement, 1 1 (2 ) 2 1( 2)! ( 2)! rnn r r rn    for n  When 1n , LHS 11(2 ) 2 (1 2)! 6  RHS 22 4 211 (1 2)! 6 6      LHS i.e. 1P is true Assume that kP is true for some k  i.e. 1 1 (2 ) 2 1( 2)! ( 2)! rkk r r rk    Show that 1kP  is also true i.e. 21 1 (2 ) 2 1( 2)! ( 3)! rkk r r rk    LHS 1 1 (2 ) ( 1)2 ( 2)! ( 3)! rkk r rk rk     112 ( 1)21 ( 2)! ( 3)! kk k kk      11( 3)2 ( 1)21 ( 3)! kkkk k     12 ( 3 1)1 ( 3)! k kk k      122 (2) 211 ( 3)! ( 3)! kk kk       RHS i.e. 1kP  is true Therefore by mathematical induction, nP is true for n  8a Let ka be the no. of marbles placed in the thk bag and A.P.: 12, ,..., na a a where 1 6a  and 1 6kkd a a    Consider [2(6) 6( 1)] 19222 n nSn    0 24.816n When 24n , 24 1800S 

8 Using GC, n nS . … 24 1800 less than 1922 25 1950 . … 24 bags contain 1800 marbles i.e. 122 marbles were left behind 8b Month Start ($) End ($) 1 (Feb) 1 10000A  1 1.015(10000)B  2 2 1.015(10000) 1200A    2 2 1.015 (10000) 1.015 1200B  3 2 3 1.015 (10000) 1.015(1200) 1200 A     32 3 1.015 (10000) 1.015 (1200) 1.015 1200 B   … … k 1 2 2 1.015 (10000) 1200(1 1.015 1.015 ... 1.015 ) k k k A           2 1.015 (10000) 1200 1.015 (1 1.015 ... 1.015 ) k k k B       Amount owed at end of month k is   11.015 11.015 (10000) 1200 1.015 1.015 1 k k kB      Final payment will be at the start of the month after 1200kB  From GC, 8 2345. 2052 10B  9 1162. 2070 10B  Amount of final payment $1162.70 , made on start of 10th month Final payment of $1162.70 (nearest cents) is paid on 1st November 2015. OR Amount owed at start of month k is 1 1 1.015 11.015 (10000) 1200 1.015 1 k k kA       Final payment is made when 0kA  From GC, 9 1145.52A  10 27.30A  Final payment of 27.30 1200 $1162.70   (nearest cents) is paid on 1st November 2015.

9 9i 332sin , cosxy  where 0 2  2d 6sin cosd x   , 2d 3sin cosd y   d cos 1 cotd 2sin 2 y x    At 0.25x , 32sin 0.25 sin 0.5   Hence 1 6 , d3 d2 y x  and   3 31 6 3cos 2y     Tangent: 3 3 3 1 8 2 4yx     31 322yx  Normal: 3 3 2 1 84 3 yx     2 5 3 243 yx 9ii At 0.25x , d d d d d d yy tt     211 66 13 sin cos 18   1 3 13 2 4 18         1 16 units/sec y is decreasing at 1 16 units/sec. 10i cos[ln(1 )]yx d1 sin[ln(1 )]d1 y xxx  d(1 ) sin[ln(1 )]d yxx x   (Show