If f(x) is a polynomial satisfying  f(x2+1) = {f(x)2}+1 and f(0) =0 ; then f’(0) = ?

Asked by Malavika Umesh | 3rd May, 2015, 03:18: PM

Expert Answer:

T h e space c o n d i t i o n space f open parentheses x squared plus 1 close parentheses equals f left parenthesis x squared right parenthesis plus 1 space c a n space b e space s a t i s f i e d space o n l y space b y space a space l i n e a r space p o l y n o m i a l. left parenthesis T h e space a b o v e space s t a t e m e n t space c a n space b e space p r o v e d space b y space t a k i n g space a space p o l y n o m i a l space o f space h i g h e r space d e g r e e space a n d space t r y i n g space t o space f i t space i t space i n t o space t h e space g i v e n space c o n d i t i o n right parenthesis L e t space t h e space l i n e a r space p o l y n o m i a l space b e f left parenthesis x right parenthesis equals K x plus C G i v e n space t h a t space f left parenthesis 0 right parenthesis space equals space 0 space rightwards double arrow space C equals 0 S o space f open parentheses x close parentheses equals K x N o w space u sin g space t h e space c o n d i t i o n space f open parentheses x squared plus 1 close parentheses equals f left parenthesis x squared right parenthesis plus 1 comma space w e space g e t K open parentheses x squared plus 1 close parentheses equals K x squared plus 1 T h e space a b o v e space e q u a t i o n space i s space s a t i s f i e d space w h e n space K equals 1. H e n c e comma space t h e space p o l y n o m i a l space b e c o m e s space f open parentheses x close parentheses equals x H e n c e space f apostrophe open parentheses x close parentheses equals 1 space f o r space a l l space apostrophe x apostrophe. H e n c e space f apostrophe open parentheses 0 close parentheses equals 1

Answered by satyajit samal | 5th May, 2015, 11:38: AM