0 votes
20 views
in Mathematics by (20 points)

Please log in or register to answer this question.

1 Answer

0 votes
by (30 points)
Calling the expression mm and focussing on just one period,

∂m∂x=cosxsinysin(x+y)+sinxsinycos(x+y)=0∂m∂x=cos⁡xsin⁡ysin⁡(x+y)+sin⁡xsin⁡ycos⁡(x+y)=0

and

∂m∂y=sinxcosysin(x+y)+sinxsinycos(x+y)=0∂m∂y=sin⁡xcos⁡ysin⁡(x+y)+sin⁡xsin⁡ycos⁡(x+y)=0

from which

cosxsinysin(x+y)=sinxcosysin(x+y)⇒sin(x+y)sin(x−y)=0cos⁡xsin⁡ysin⁡(x+y)=sin⁡xcos⁡ysin⁡(x+y)⇒sin⁡(x+y)sin⁡(x−y)=0.

Then x=yx=y. From this sinx=0sin⁡x=0 or \sin 3x=0 the latter giving x=π3x=π3 giving a maximum of 33√8338.

I leave it to you to determine the minimum and track through the 2 ignored situations.

No related questions found

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...