SAJC P1 MS

2 [Turn Over 1 Using sine rule, 2 12 222 2 2 sin sin6 3 sin 6 sin 3 sin 6 sin cos cos sin33 1 2 31cos sin22 1 31 2 1 1 233 1 1 ... 223 3 3 15 1 633 1 1 5 or33 6 3 BC AC x BC AC x xx xx x x xx x x x x x x xx                                               23 1 5 3 (shown)3 3 18xx Hence, a = , b = - , c = (Ans). 2 (i) 1 1 1 kk kuu k    2 3 1 1 1 12 11 1 2 3 1 1 5 4 3 1 2 3 3 2 1 ( 1) 2 k k kk ukk k k k uk k k k k k uk k k kk u              A B C π 6 π 3 x+

3 [Turn Over (ii) Let P(n) be the statement “   123  n nS n n for all positive integer 1n ” When n = 1, L.H.S = 11 2Su R.H.S =   1 1 1 1 2 23    P(1) is true Assume P(k) is true for some 1k ,   S 1 23  k k kk To prove P(k+1) is true:   1 1 233    k kS k k       1 1 1 .. 2 = 1 23 2 ( 1) = 1 232 k kk k L H S S Su kk k k u k k k k kk k u k                            21= 1 2 232 = 1 1 23 3= 1 23 1 = 2 3 . .3 kkk kk k kk k kk k k k R H S              P( 1) is true k Since P(1) is true and P(k) is true implies P(k+1) is true, by Mathematical Induction, P(n) is true for all n  . 3(i) 2 23 , , 02x x x xx     2 23 02xx x- - ³ Method 1:Using GC to sketch the graphs of 2 23 2y x x x= - - . From the graph: 0 or 2xx<³ Method 2: 2 23 2y x x x= - - 0 2

4 [Turn Over 2 32 23 0, 02 2 3 4 02 x x x x xx x - - ³ ¹ -- ³ Multiplying by 2x2, ( ) 322 3 4 0x x x - - ³ Sketch the graph of ( ) 322 3 4y x x x= - - using GC: From the graph: 0 or 2xx<³ Method 3: Analytical method 2 32 23 0, 02 2 3 4 02 x x x x xx x - - ³ ¹ -- ³ Multiplying by 2x2, ( ) 32 2 22 151 4 16 2 3 4 0 ( 2)(2 2) 0 ( 2) 0 since 2 2 2( ) 0 x x x x x x x x x x x x - - ³ - + + ³ - ³ + + = + + > 0 or 2xx<³ (ii) 2 1 23 d2 a x x xx 2 22 12 2 3 2 3 dd22 a x x x x x xxx                23 2 3 2 12 332ln 2ln3 4 3 4 a x x x x xx               328 1 3 3 82ln 2 3 2ln1 2ln 2ln 2 33 3 4 3 4 3 aa a                         321 3 12ln 2 2ln 2ln 212 3 4 3 aa a       32 314ln 2 2ln 3 4 4 aaa      32 2 1 2 3 3 1 d2 3 4 4 a aax x xx      3 2 3 23 1 3 14ln 2 2ln 3 4 4 3 4 4 a a a aa        ln 24 = ln a2  a = 22 = 4. 0 2 32(2 3 4)y x x x= - -

5 [Turn Over 4 (i) | 3| 3iz 3| || | 3iz i   | 3 | 3zi   3arg 3 4z i      arg( (3 3 )) 4zi     arg( ( 3 3 )) 4zi      (ii) (a) Min |z - (3+3i)|=PT sin64 PT  32PT Max possible |z - (3+3i)|=QT = 6 units, but Q is not to be