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Application Of Derivatives

Question
CBSEENMA12035351

Find the intervals in which the function f given by 
f (x) = sin x + cos x. 0 ≤ a ≤ 2 straight pi is increasing or decreasing.

Solution

          f (x) = sin x + cos x
therefore space space space straight f apostrophe left parenthesis straight x right parenthesis space equals space cos space straight x minus sin space straight x
Now,    straight f apostrophe left parenthesis straight x right parenthesis space equals space 0 space space space space space rightwards double arrow space space space space cosx minus sinx space equals space 0 space space space space rightwards double arrow space space space space sin space straight x space equals space cos space straight x space space space rightwards double arrow space space space tan space straight x space equals space 1
rightwards double arrow space space space space space straight x space equals space straight pi over 4 comma space space space space fraction numerator 5 straight pi over denominator 4 end fraction space space as space space space 0 less or equal than straight x less or equal than 2 straight pi
      The points space straight x space equals straight pi over 4 comma space space space space fraction numerator 5 straight pi over denominator 4 end fraction space divide the interval left parenthesis 0 comma space space 2 straight pi right parenthesis into three disjoint intervals open parentheses 0 comma space straight pi over 4 close parentheses comma space space open parentheses straight pi over 4 comma space fraction numerator 5 straight pi over denominator 4 end fraction close parentheses comma space space open parentheses fraction numerator 5 straight pi over denominator 4 end fraction comma space 2 straight pi close parentheses.
             In open parentheses 0 comma space straight pi over 4 close parentheses comma space space straight f apostrophe left parenthesis straight x right parenthesis space equals space square root of 2 open parentheses fraction numerator 1 over denominator square root of 2 end fraction cos space straight x space minus space fraction numerator 1 over denominator square root of 2 end fraction sin space straight x close parentheses space equals space square root of 2 open parentheses cosx space cos space straight pi over 4 minus sinx space sin space straight pi over 4 close parentheses
                                   equals space square root of 2 space cos space open parentheses straight x plus straight pi over 4 close parentheses greater than 0
                In open parentheses straight pi over 4 comma space space fraction numerator 5 straight pi over denominator 4 end fraction close parentheses comma space space straight f apostrophe left parenthesis straight x right parenthesis space less than space 0 space
               In  open parentheses fraction numerator 5 straight pi over denominator 4 end fraction comma space space 2 straight pi close parentheses,   straight f apostrophe left parenthesis straight x right parenthesis space greater than space 0
therefore space space space space space space space space straight f space is space increasing space in space the space intervals space open parentheses 0 comma space straight pi over 4 close parentheses space space and space space open parentheses fraction numerator 5 straight pi over denominator 4 end fraction comma space space space 2 straight pi close parentheses space and space straight f space is space decreasing space in space the space interval space open parentheses straight pi over 4 comma space fraction numerator 5 straight pi over denominator 4 end fraction close parentheses.          


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