AJC_H2_MATH_P2_soln

Anderson Junior College Preliminary Examination 2015 H2 Mathematics Paper 2 (9740/02) 1a (i) 2 2 2d 2(e ) ( 2e )d xxw Ax      2 22 4e (e ) x x A    22 22 d4 LHS = e = RHS (verified)d (e ) x x w wx A    (ii) 2 d2 d e x y x A  2 2 d e xyx A  2 2 1 2 e d 1e x x A xA A   21 ln 1 e , xABA   where B is an arbitrary constant. 1b Let S be the amount of salt (in grams) at time t minutes.   d rate of salt entering tank rate of salt leaving tankd 2(5) 5100 200 g/min20 S t S S    1 20 1 20 1 20 1 d 1 200 d 20 1 d d 200 1ln 200 20 200 e 200 e t t S St StS S t c SA SA            When 0,t 50S  50 200 A 150A 1 20200 150e t S     When concentration is 1 g/litre, 100.S  1 20100 200 150e t    ⇒ 1 20e 2 3 t  ∴ 20ln(2 3) 8.11 min (3 s.f.)t  2 5 5 ( 2 ) 2()55 33 5 5 5 5 32 0 32 32 2 , 0, 1, 2 2 , 2 , 2 , 2 , 2 ki k i i i i i i z ze z e k e e e e e                   2i Method 1 45 51 3 2 5 2 * 2 n in nn i n in z e ez e          

For 1 2 * n z z    to be real and positive, smallest n = 5 Method 2     1 12 2 12 arg arg( ) arg( *)* arg( ) arg( ) 3 55 4 5 n z n z zz n z z n n            For 1 2 * n z z    to be real and positive, 4 2 , 5 n kk  5 , 2n k k , so smallest 5n 2iii Let the complex number represented by A’ be x+iy BA rotates 90o about B to get BA’: 5( ) 2 ( ) 2 2 iiix iy e i e e          1 5( 2) 2 ( 2) i x iy i e         since ei = -1 2 2cos 2 sin 2)55x iy i i      Real part = 22 2 sin 2 2sin55i     3(i) 4 6 2 1 2 0 2 2 1 4 0 4 2 AB                                         and 3 6 3 2 0 2 6 0 6 AC                              Normal to plane ABC = 1 3 2 1 2 0 2 6 1 AB                              Equation of plane ABC is 2 2 6 r 0 0 0 12 1 1 0                               5 2 i Ae   2 i Be   Re(z) Im(z) 2 A’

3(ii) Since S lies on the perpendicular from (3,-1, 4) to plane 1x y z   , 31 1 1 41 OS                       (Note S lies along perpendicular from D to p1. Line SD: r = 1 + pOD n ) S lies on plane ABC  32 1 0 12 41                        6 2 4 12 2 3         11 331 25 1 1 3341 14 3 OS         