PJC P1 MS

FYE P1 1       2 2 2 2 2 2 21 0 1 21 0 13 24 13Since 2 1 0 and 0, for all 24 1\{ }2 x xx x x x x x x                   2 2 2 2 2 2 44 0 1 2 01 114 4 1 0 11 1 xx xx x xx xx xx                           Replace 1 with :x x 11, 2 2xx x     2 (i) AB DC OB OA OC OD OD OD           b a c a b c (i) Area of Parallelogram ( ) () = = ABCD AB BC                  b a c b b c b b a c a b b c a b a c (proved) +   a b b c c a

(ii) Area of Triangle ACD = 11Area of Parallelogram 22 ABCD AB BC 1 2 AB BC = 2 +   a b b c c a = 1 (base) height2 AC  +   a b b c c a = height AC  Shortest distance from D to AC = +     a b b c c a ca (shown) 3 1 1dln 1 or d 1 d d 1 dd d 1 d 1 dd x x te tt x e x tt tt x x xx t t t t           When 0, When 1, 1 x t e xt  

  11 1 1 1 1 0 1 1 1 1 1 21 1 1tan d tan d = tan d = tan d (shown) = tan d 1 xx e e e e e e e x t t t t tt tt ttt t                        1 21 1 12 1 1 2 1 12 = tan d 2 1 1 = tan ln 12 11 = tan ( ) ln(1 ) 1 tan (1) ln(222 e e e t tt t t t t t t e e e                 12 12 1 ) = tan ( ) ln( 1 ) ln( 2) 4 = tan ( ) ln( 1 ) ln( 2) 4 = tan ( ) e e e e e e ee                         211ln 1 ln 24 2 2 e    4 Let nP be the statement 1 11sin( ) sin22cos 12sin 2 n r n r      , 1 ,2,3,...n When n = 1, LHS = 1 1 cos cos r r   RHS = 3 1 1sin sin 2cos sin2 2 2 cos112sin 2sin22         Thus, 1P is true. 1 2 dtan 1 d d1 1 dd 1 vut x u vtt t vt    

Assume that kP is true for some k, k = 1,2,3,… 1 11sin( ) sin22cos 12sin 2 k r k r      To prove that 1kP is also true, i.e. 1 1 31sin( ) sin22cos 12sin 2 k r k r        LHS 1 1 cos k r r    1 cos cos( 1) 11sin( ) sin22 cos( 1)12sin 2 1 1 1sin( ) 2cos( 1) sin sin2 2 2 12sin 2 1 1 1 1sin( ) sin( 1 ) sin( 1 ) sin2 2 2 2 12sin 2 1 3 1 1sin( ) sin( ) sin( ) sin2 2 2 2 12sin k r rk k k kk k k k k k k                                                2 31sin( ) sin22 12sin 2 k      = RHS Thus, 1kP is also true. Since 1P is true and kP is true  1kP is also true, then by mathematical induction, nP is true for 1 ,2,3,...n .

5 Let M be the center of the hemisphere. By Pythagoras Theorem, 22 PM n x      22 22 2 2 2 height diameter 2 2 2 2 2 cm (shown) A x nx x n x x n x            As x varies,  22 22 22d 2d 2 xxA nxx nx          2 2 2 2 2 2 22 dWhen 0, for stationary points,d 2 2 0 0 20 A x n x x n x x nx          2 2 2 (reject -ve) 2 nx nx   To show A maximum, x 2 n    2 n 2 n    d d A x + 0 - n x M P

22 2 2 2 Diameter of cylinder Height of cylinder 2 2 2 2 2 2 2 2 2 2 1 2 x nx n nn n n         2k  6. Solutions: (i) cos d 1 sind x t t x tt   3 sin 2 d 2cos 2d yt y tt   d d d d d d 2cos 2 1 sin y y t x t x t t    For turning point, d 0d y x 

2cos 2 01 sin cos 2 0 32 or 22 3 or 44 2When , cos , 3 sin 2 24 4 4 4 2 4 2The coordinates of the turning point is ,242 t t t t t t x y                                      3 3 3 2 3 3When , cos , 3 sin 2 44 4 4 2 4 4 23The coordinates of the turning point is ,424 t x y                            ii)   2cos 2d 2 2 2When , 2 d 1 1 0 1 sin 2 cos 2 2 2 Equation of tangent: 2 yt x x x                       iii)    2 4 2 4 22 d 3 sin 2 1 sin d 0.16584 0.166 (3 s.f.) (By G. C.) y x t t t             7.   2 22 AB r r r r 

    2 ( 2) 0 : 2 2 1 2 : 2 ( 2) 1 2 1 1 22 A r Br r A A r B B r r r r                   11 2 1 1 22 NN rr r r r r         11 13 11 24 11 35 11 46 ..... 11 2 11 11 11 2 1 1 11 2 1 2 3 1 1 2 1 2 NN NN NN NN NN                                                    1 2 2 2 2 2 2 2 2 ...2 1 3 2 4 3 5 3 1 2 1 1 2 N r r r N N N N N N N N              

                                 3 6 1 2 2 2 2 2 2 ...4 2 2 4 3 5 3 1 2 1 1 2 2 2 2 3 2 3 1 1 2 2 2 1 2 3 2 5 1 1 2 6 1 2 2 N r N r r r N N N N N N r r N N N N N N N N N N                                                       2 2 1 2 15 6 1 2 5 2 5 2 6 1 2 2 1 25 6 1 2 52 6 N N N N N N N N NN N N N NN N N N N                     1 1NN  Alternatively,                          3 6 1 1 2 2 2 2 2 2 ...4 2 2 4 3 5 3 1 2 1 1 22 23 3 1 1 2 2 1 3 5 2 1 61 N r N r r r N N N N N N rr NN N NN                                    66 22 lim4 2 4 2 5 2 1 lim 61 5 6 N N N rr r r r r N NN                 8. (i)

             d 20d d 20d d 20 , for some constant 0d 1d 20 d 1d dd20 d ln 20 When 0, 20 , ln(20 20) 0 ln 40 Hence, ln( 20) ln 40 F Ft F Ft F k F kt F kFt F t k tFt F kt C t F C C F kt                          When 10, 10 , ln 10 20 10 ln 40 ln10 ln 40 10 1 ln 104 11 ln 10 4 t F C k k k k                 1 10 1 10 1 10 11ln 20 ln ln 40 10 4 1ln 20 ln 40 4 120 40 4 140 20 (Shown)4 F F F F              