DHS H2 MATHS P1 Solution

1 2014 Year 6 Prelim Examination Paper 1 Suggested Solution Qn Suggested Solution 1(i) m : weight of mackerel in kg s : weight of salmon in kg t : weight of tuna in kg 800 7 21 39 20300 5 23 49 23900 ms t mst mst    Using GC, 200, 250, 350.mst Therefore the fisherman has 250 kg of salmon. (ii) m : weight of mackerel in kg s : weight of salmon in kg t : weight of tuna in kg 600 7 21 39 20300 5 23 49 23900 ms t mst mst    Using GC, 460, 990, 70.ms t   Since the weight of all fishes must be non-negative, the fisherman’s claim is not possible. Or Since the weight of salmon and tuna is more than 600kg, the fisherman’s claim is not possible.

2 Qn Suggested Solution 2(a) 2(b) 2 2(1 ) (2 ) 14 x y   Making y the subject of formula: 2(1 )21 4 xy   Let the volume of solid generated when the curve 2(1 )21 4 xy   is rotated about x-axis from 2 to 3xx be V1. V1 = 3 2 2 π dyx 2 23 2 (1 )π 21 d 4 x x     = 21.593 Let the volume of solid generated when the curve 2(1 )21 4 xy   is rotated about x-axis from 2 to 3xx be V2. V2 = 3 2 2 π dyx 2 23 2 (1 )π 21 d 4 x x     = 6.1573 Volume of required solid = V 1 – V2 = 15.4 (to 3sf) 15, 2   15, 2  O y x x = 2

3 Qn Suggested Solution 3(i) 2 2 2 2 2 2 2 e1 32 Differentiate with respect to , de3 4d Differentiate again with respect to , ddee 4dd dd 4e (shown)dd y y yy y xx x y xx x yy xx yy xx          3(ii) 2 2 2 dd 4edd yyy xx   Differentiating again with respect to x, 23 23 dd d d24 edd d d yy yy y x xx x         2 2 3 300 0 3 3 3 dd when 0, 0, 3, 5, dd de3 3 e3 5 e 0 d d 18d yyxy xx y x y x          23 2 351 8 503 3 3 2! 3! 2yx x xx x x        3(iii) 21 3 2 1.0302 1.51 (reject) or 0.01 xx xx           22 3 22 3 5ln 1 3 2 3 3 2 Using 0.01, 5ln 1 3 0.01 2 0.01 3 0.01 0.01 3 0.01 2 ln 1.0302 0.0298 (4 d.p.) yx x x x x x           Qn Suggested Solution 4(i) A C B 2 1

4 Qn Suggested Solution 1 12 0 23021 133 3 3 p p OC            ab  Area of triangle OAC = 1 2 OA OC     = 211 1123 33 pp           = 0012 63 3 366 2 1 pp pp             10 3 4(ii) Triangle OAD, triangle ADE and triangle OAC have the same height and base and thus they have the same area. Area of trapezium OAED = 1031 03   4(iii) 2 2 22 2 1 10 30cos135 10 1 1 2 10 10 2 10 10 (reject 10 since 0) p p p p pp p pp               ab

D C O 1 1 1 A E

5 Qn Suggested Solution 5(i)  πi 6 πi i6 πi 6 3i2 e 23 i 22 e e 4e 4 πarg 66 2 wz r r wr w                    Useful screenshots: 5(ii) Remark: Locus of z could also be drawn along the positive Im-axis as values of include π 2 . 5(iii) 222 * 2 * 16 822 2 Since 0 2, 01 6 .2 wwr rzz r r w z      Qn Suggested Solution 6  2d 30d  yxy x yx .........(1) Given ux y : dd dduy x yx x 8 B  A 0 2 6   Im Re Locus of z Locus of w

Su d dx   or  6(i) G 1 3 H U 6(ii) y i.e Qn Su 7(i) 7(ii) ubstitute int 2d 3d u ux 2 1 3duu  1 3 xu r 3u x  1 3y xx iven 11, 3     1 1 3 1(3  C ence, y sing GC : has no turn e. particular uggested So  ln ,0 0,1 to (1): d d u x 3 dx x C ( C 1 x C  1 x C (*)   6)  CC 1 3( 2 ) xx ning point w r solution is olution O ln ,  0,1  y O y  ln , y 230u C arbitrary ) 6 when 0C 2 1 3y x y = ,0 e xy  x x exy  0 y  constant) 0 , = 0  As hence, hori x y y       ,e 0 izontal asym i x   s 6

7 Qn Suggested Solution 7(iii) 7(iv)  ee e e1 1ee ee xx x x      i.e. e  Since 1 1e1e  , from the graphs of eexy=  and 1 eexy  will intersect 3 times. Thus there will be 3 solutions for   ee e e1xx  . Qn Suggested Solution 8(a) 2sec ( ) d tan( ) tan( ) d tan( ) ln sec( ) xx a x x xa xax x xa xa C          OR: tan( ) ln cos ( )x xa xa C  8(b) 22 2 11 2 2dd22 2 22 1 ln( 2 2)2 xx x xxx xx x xC        O x y 0y  1y  lnx  10,1    1 exy  

8 Qn Suggested Solution 8(b) (i)   2 21 22 2211 22 2211 2 221 11 11 4 d22 13 dd22 22 13 dd22 (1 ) 1 1 ln 2 2 3 tan ( 1)2 1 ln 2 ln1 3 tan 1 tan 02 13ln 2 24                          x xxx x x xxx xx x xxxx x xx x 8(b) (ii) Note that  22 11 :22 11    xx xx x     22 1 2221 21 22 1 1 d22 11 dd 11 11 12 d (by symmetry) 11 1=2 ln( 2 2) ln 2 22 p p p p p p x xxx xx x x xx x x x xx pp                  +  1

9 Qn Suggested Solution 9(a) 1 1log (log ) nn kn kn yy x kx k       1 1 log log log log (a constant, where is the common ratio) kn kn n k n k xx x x rr      Since the difference between any two consecutive terms is a constant,{}ny is an arithmetic sequence. 9b(i) 1.01(20000 ) 20000 1.01 0.01(20000) 198.02 x x x    9b(ii) No. of payme nts Amount owed after each payment in the middle of the month 1 20 000 x 2 1.01(20 000 ) 1.01(20 000) 1.01 xx x x    3    22 1.01 1.01(20 000) 1.01 1.01 20 000 1.01 1.01 x xx x xx      … n 11 2 1 1.01 (20 000) (1.01 1.01 ... 1.01 1) 1.01 11.01 (20 000) 1.01 1 nn n n n x x          For the loan to be paid in full after the n th payment, 1 1.01 11.01 (20 000) 0 1.01 1 n n x   1 1 1.01 11.01 (20 000) 0.01 200(1.01 ) (shown) 1.01 1 n n n n x x      Alternatively No. of payments Amount owed at the end of each month 1 1.01(20 000 ) x

10 Qn Suggested Solution 2   22 1.01 1.01(20 000 ) 1.01 (20 000) 1.01 1.01 xx x x    … n1 11 2 1 1 1.01 (20 000) (1.01 1.01 ... 1.01) 1.01(1.01 1)1.01 (20 000) 1.01 1 nn n n n x x            For the loan to be paid in full after the n th payment, then 1 1 1.01(1.01 1)1.01 (20 000) 0 1.01 1 n n xx     1 1 1 1 1 1.01(1.01 1)1.01 (20 000) 0.01 1.01(1.01 1) 0.01 1.01 (20 000)0.01 1.01 1 1.01 (20 000)0.01 n n n n n n x x x x           1200(1.01 ) (shown) 1.01 1 n nx    For the loan to be fully paid in 3 years (n = 36 months), 36 1 36 200(1.01 ) 1.01 1 x    657.709x Hence, for Thomas to fully pay up the loan in exactly 3 years, he should be paying a monthly amount of $657.71