SAJC H2 MATHS P1 Solutions

2017 Prelim Paper 1 Solutions 1 3 2 4 3 dd 4dd Vr Vr rtt     When V=20, 3 1 3 420 3 15 r r     When 1 315r   , d d V t  . 2 3 2 3 15 d4 d d d4 1 5 r t r t         Surface Area, 24 dd 8dd Ar A rrtt     When 1 315r   , 2 3d d4 1 5 r t      . 12 3 1 2 3 3 dd d dd d 15 = 8 41 5 2 15 cm /s AA r tr t            2 BC = BD + DC = tan tan3 4 hh x    BC = 3 tan tan4 1t a n t a n4 hh x x     

= 3( 1 t a n ) 31 t a n hh x x   3( 1 ) 31 hh x x   = 13 (1 )(1 )3 h hx x   = 23( 1 ) ( 2 )(1 )[1 ( 1) ) ...]32 ! h hx x x     = 23 (1 )[1 ...]3 h hx x x   = 23 (1 2 2 ...)3 h hx x   23 1 2 + 23hx x 3 (i)

(ii) 4 3 20 2 22 2 2 2 2 2 2 f( ) d 1 d cos=1 ( s i n ) d =s i n d 1c o s 2 d2 sin 2θ= θ22 22 4 a a a xx xbx a aba a ab ab ab ab ab                               4 (i)  2 2 ee axaxy bb   If 2 f( ) e xx  , then    2 fe ax ax  and so    ff fy xy a xy a x b     Hence the sequence of transformations are: 1. Scale by a factor of 1 a parallel to the x-axis, 2. Translate the resulting curve by b units in the negative y-direction.   (ii) y y=f(x) O (0,1-b) x

5 (i) Since uvw is perpendicular to  uvw ,   0 0        uvw v uvw uvw u uv vv w uw vwww uu Since 22 2 ,,  uvwuu vv ww , and ,,  vv ww wuu u v w uv , 22 2 20  uvw v w Since ,,uvw are unit vectors, 1,1, 1 uvw , 111 2 0 1 2 1cos 2 1cos 2         vw vw vw Hence, 60  5 (ii) O  V W y x O y=1/f(x) y=0

    2 Area of OVW 1 sin 602 13 1122 3 units4 OV OW      

5 (iii) Since u and vw are parallel, we have ,OV O OOU U W . Volume of OUVW     3 1 Area of OVW3 13 134 3 units12 OU   

6 (a) (i) Using integration by parts, ec o s dx nx x sin ees i n d x x nx nx xnn    O  U V W dec o s d ds i ned x x vun x x un x vxn     des i n d dc o sed x x vun x x un x vx n   

= s i n 1 ec o s ec o sed xx x nx nx nx xnn n n       = 2 sin 1 e cos 1ee c o s d x xx nx nx nx xnn n n      Rearranging, 2 11e c o s d x nx xn   = sin 1 e cose x x nx nx nn n     ec o s dx nx x = 2 2 sin 1 e cose1 x xnn x n x cnn n n              where c is a constant (ii) 2 ec o s dx nx x   = 22 2 sin 1 e cose1 x xnn x n x nn n n                  2 2 2 sin 2 1 cos 2 sin 1 cosee1 nn n n n nn n n n n n                                 For any positive integer n, 02sin n and 12cos n If n is odd, 0sin n and 1cos n 2 ec o s dx nx x   = 2 2 222 11e0 e 01 n nnn           2 2 1 ee1 n    (Ans) 6 (b) 216 xy x    2 2216 xy x   Hence volume required

222 0 drh y x         2 2 20 2 2 2 20 2 212 2 0 3 2 2d12 16 22 2d12 2 16 162 212 2 1 41 1 144 2 12 16 5 units288 x x x x x x x                             7 (i) i i i ii i22 2 i 2 e e e ee e e 2i sin 2 cos isin22 2i sin 2 11 cot22 i 2 11 cot i22 2 r rr                                        (ii) i 32ez 

(iii) 2 2 2 44 40(1 ) 1 22 24 011 ww ww ww ww         Let 2 1 wz w  , then 2 24 0zz  From (ii) the solutions are ii 332e or 2e zz      Since 2 1 2 (2 ) wz w zw z w wz z     2 zw z  Part (i) result can be used as i 32e z   , where 2r  with , 33    . 11 11 cot i or i cot22 6 22 6ww           13 13 i or i22 22ww  8 (i) Distance travelled per lap is in AP: a = 2(30) = 60, d = 2  3 = 6. Given total distance travelled > 3000 2 n [ 2(60) + (n  1)6] > 3000 3n2 + 57n  3000 > 0 (n + 42.52)(n  23.52) > 0 n < 42.52 or n > 23.52 Since n  Z , least n = 24

8 (ii)  1 012 2 22 1 1 3 Distance of the coach from just before the runner completes the th lap = 3 02 ( 3 )2 ( 3 )2 ( 3 ). . .2 ( 3 ) 30 2 1 3 3 .... 3 3130 2 31 30 (3 1) 29r r r r r rS                      1 1 1 11 1 1 Distance covered by the athlete after laps 23 23 5 8 23 5 8 3125 831 315 8 29 n r r nn r rr n r r n n n n n n                   When D = 8000m  315 8 From GC, 8.1254 8000 n n n    Hence the athlete has run 8 complete laps. The athlete has completed 7024 m Hence he still have 8000-7024=976 m On the 9th lap, the coach is 913 29 6590 m  from S. Hence the athlete would be 6590- 976 = 5614 m away from the coach once he finishes 8 km. 9 (i) d 2cosd x   , d 3s i nd y   d3 s i n 3 tand2 c o s 2 y x     When 6   , 2x , 11 2y  , d1 d2 y x  Equation of normal : 11 222yx   32 2yx  (ii) 12 s i n ( 1 ) 43 c o s ( 2 ) x y       Substitute equation (1) and (2) into 32 2yx 

343 c o s 2 ( 1 2 s i n ) 2 1 3c o s 4s i n2 8sin 2 3cos 1         At Point Q,   8sin 2 3 cos 1 (shown) Using GC: 2.847916 or 0.52359 (Reject, same as ,point )6 P  Hence, using GC coordinates of Q (0.42105, 2.3421) Q (0.421, 2.34) (iii) Req when 0.42105 0.42105 1 2sin sin 0.289475 0.29368 or 2.8479 ( at point Q) x         Required Area   0.42105 0.42105 22 0.4 12 26 0.29368 2 2105 d d 343 c o s 2 c o s d 2 d 2 8.9613 6.1911 2.7702 2.77 units (3 s.f.) yx yx x x             10 (i) d4 0 lnd x cxtx                      y                                                                                                                                                                                                                                                               x