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 31