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 included. Therefore 3 2 | 3 3 | 6zi    since Q is not included. (Ans) (b) Max arg( (3 3 ))zi occurs at point Q (not included) and min arg( (3 3 ))zi occurs at point P. 3 arg( (3 3 ))4 zi     (Ans)

6 [Turn Over 5 (i) (ii) Differentiating w.r.t. x : 2(1 ) ''( ) 2 '( ) 0x f x xf x+ + = Differentiating w.r.t x: 2 2 (1 ) '''( ) 2 ''( ) 2 ''( ) 2 '( ) 0 (1 ) '''( ) 4 ''( ) 2 '( ) 0 x f x xf x xf x f x x f x xf x f x + + + + = + + + = When 0 '(0) 1 ''(0) 0 '''(0) 2 '(0) 2 x f f ff = = = = - = - Hence 13 3 2tan ( ) .... 3! 1 3 x x x xx - = - + »- (iii)

7 [Turn Over (iv) (v) The approximation in (iii) is better than that in (iv) because the value of x substituted in (iii) is closer to zero as compared to the value of x substituted in (iv). 6 (i) 2x at 3y at 2dx atdt  23dy atdt  233 22 dy at tdx at When 225 4x a at , 5 2t  When 3125 8y a at , 5 2t  . 5 2t  for 25 125,48aa  Note that 5 2t  does not give the correct point. When 5 2t  , gradient of tangent 15 4  , Eqn of tangent : 125 15 25 8 4 4y a x a            16 250 60 375y a x a   60 16 125 (1)x y a   (ii) Subst 2x at and 3y at into (1): 2360 16 125at at a 3216 60 125 0tt       4 5 2 5 2 5 0t t t    55 42t or  Note that 5 2t  is rejected. When 5 ,4t  2 5 25 4 16x a a  , 3 5 125 4 64y a a , Hence the coordinates of the point where the tangent meets the curve again is at 25 125,16 64aa 

8 [Turn Over (iii) 3 2 dy tdx  , gradient of normal 2 3t  Eqn of normal : 2 210 32y x a t      27 (2)3 ayx tt    Subst 2x at and 3y at into (2):   32 27 3 aat at tt  423 2 21 0tt       223 7 3 0tt    2 7 3t or 2 3t  (rejected) 7 3t  When 7 3t  , 7 3xa , 3 27 3ya  , When 7 3t  , 7 3xa , 3 27 3ya   , 3 277 ,33aa     and 3 277 ,33aa     7 2 2 ( 1)(4 ) (2 3) ( 1) dy x x a x ax dx x       2 22 2 0, ( 1)(4 ) (2 3) 0 4 4 2 3 0 2 4 ( 3) 0 dy dx x x a x ax x x ax a x ax x x a                   For stationary points to exist: 24 4(2)( 3) 0 1 (since 1) a aa      For 1f to exist, there should not be any stationary point in the given domain, hence 1a . Therefore, the set of values: { : 1 }.aa  (i) 22 3 3() 1 xxfx x  

9 [Turn Over 51: (0,3.5) :[ 0.25,6) :[ 3, ) 7 g g fgD R R     (ii) 1 2 h( ) f(f h( )) 2 3 3 1 xx x xx ee e    Therefore, 22 3 3h : , , 0 1 xx x eex x x e   8(a) 22ax a O x P 22 2 2 2 2 2 2 4 23 23 22 22 2 () d2 2 4d d 2d dd 6dd A x a x A x a x a x x AA a x xx AA a x xx AAA a xxx           For max A, d 0d A x  y =g(x) 1 4- 1 6 3.5 y x 1 4- -1 6 -3 y=f(x) y x 51 7 R Q

10 [Turn Over 22 2 2 ( 2 ) 0 Since 0, 2 ( 0) 2 dWhen , 0, d2 x a x axx axx aAx x     22 22 2 2 2 d 6 2 0d2 dHence 0d AaA a ax A x      2 ax gives a max A Perimeter of OPQR = 2x + 2 22ax When 2 ax , Perimeter of OPQR 2 222 22 2 2 2 2 4 4 4 2 aa a a a a a x OP          (b) C 6 cm P  D x cm Let PD = x cm and CPD  rads at any time t. Given: 1d 2 cmsd x t  , To find d dt  when x = 6 3 cm 2 6tan 6cot d 6cosd x x x ec      

11 [Turn Over 2 2 2 d d d d d d d 2 1 sind 6cos 3 1When 6 3, tan 63 d 1 1 sin rad/s = 0.0833 rad/s (3 s.f.)d 3 6 12 xx tt t ec x t                  9 (i) ; ; ; ; Hence the vector equation of line PR is 00 04 53                    r , where  . (ii) = (iii) l : , mÎ ¡ Equate: 0 0 0 0 0 4 1 3 5 3 8 1                                          2 3  (iv) | m | = | m | is the perpendicular/shortest distance from point X to line AB. Area of triangle AXB = (3 s.f.) 2 1 1y x  y x O 3 3/(n+1) 6/(n+1) 3n/(n+1)