My question is " Separate [0,straight pi/2] into sub intervals in which f(x)=sin 3x is increasing or decreasing.

Asked by Navya Benny | 2nd Sep, 2015, 06:56: PM

Expert Answer:

C o n s i d e r space t h e space f u n c t i o n space f open parentheses x close parentheses equals sin 3 x D i f f e r e n t i a t i n g space t h e space a b o v e space f u n c t i o n space w i t h space r e s p e c t space t o space x comma space w e space h a v e comma f apostrophe open parentheses x close parentheses equals 3 c o s 3 x F o r space f open parentheses x close parentheses space t o space b e space i n c r e a s i n g comma space w e space h a v e comma f apostrophe open parentheses x close parentheses greater than 0 G i v e n space t h a t space x element of open parentheses 0 comma straight pi over 2 close parentheses rightwards double arrow 3 x element of open parentheses 0 comma fraction numerator 3 straight pi over denominator 2 end fraction close parentheses C o sin e space f u n c t i o n space i s space p o s i t i v e space i n space t h e space f i r s t space q u a d r a n t space a n d space n e g a t i v e space i n space t h e space s e c o n d space q u a d r a n t. c a s e space 1 colon W h e n space 3 x element of open parentheses 0 comma straight pi over 2 close parentheses rightwards double arrow c o s 3 x greater than 0 rightwards double arrow 3 c o s 3 x greater than 0 rightwards double arrow f apostrophe open parentheses x close parentheses greater than 0 space f o r space 0 less than 3 x less than straight pi over 2 rightwards double arrow f apostrophe open parentheses x close parentheses greater than 0 space f o r space 0 less than x less than straight pi over 6 therefore f open parentheses x close parentheses space i s space i n c r e a sin g space i n space t h e space i n t e r v a l space open parentheses 0 comma straight pi over 6 close parentheses c a s e space 2 colon W h e n space 3 x element of open parentheses straight pi over 2 comma fraction numerator 3 straight pi over denominator 2 end fraction close parentheses rightwards double arrow c o s 3 x less than 0 rightwards double arrow 3 c o s 3 x less than 0 rightwards double arrow f apostrophe open parentheses x close parentheses less than 0 space f o r space straight pi over 2 less than 3 x less than fraction numerator 3 straight pi over denominator 2 end fraction rightwards double arrow f apostrophe open parentheses x close parentheses less than 0 space f o r space straight pi over 6 less than x less than straight pi over 2 therefore f open parentheses x close parentheses space i s space d e c r e a s i n g space i n space t h e space i n t e r v a l space open parentheses straight pi over 6 comma straight pi over 2 close parentheses

Answered by Vimala Ramamurthy | 3rd Sep, 2015, 08:56: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.