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Continuity And Differentiability

Question
CBSEENMA12035499
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Verify Rolle's Theorem for the function straight f left parenthesis straight x right parenthesis space equals space cos squared space straight x space in space left square bracket 0 comma space straight pi right square bracket

Solution

Here space straight f left parenthesis straight x right parenthesis space equals space cos 2 space straight x space left parenthesis straight i right parenthesis space We space know space that space cos space straight x space is space continuous space in space left square bracket 0 comma space straight pi right square bracket space Now space cos 2 straight x comma space being space th space product space of space two space continuous space functions space cos space straight x space and space cos space straight x comma space is space continuous space in space left square bracket 0 comma space straight pi right square bracket space left parenthesis ii right parenthesis space straight f apostrophe left parenthesis straight x right parenthesis space equals space minus 2 space cos space straight x space sin space straight x comma space Which space exists space in space left parenthesis 0 comma space straight pi right parenthesis space therefore space straight f left parenthesis straight x right parenthesis space is space derivable space in space left parenthesis 0 comma space straight pi right parenthesis space left parenthesis iii right parenthesis space straight f left parenthesis 0 right parenthesis space equals space cos 2 space 0 space equals space 1 comma space straight f left parenthesis straight pi right parenthesis space equals space cos 2 space straight pi space equals space left parenthesis negative 1 right parenthesis 2 space equals space 1 space therefore space straight f left parenthesis straight x right parenthesis space satisfies space all space the space conditions space of space Rolle apostrophe straight s space theorem space therefore space there space must space exist space at space least space one space value space straight c space of space straight x space such space that space straight f apostrophe left parenthesis straight c right parenthesis space equals space 0 space where space 0 space less than space straight c space less than space straight pi. Now space space space space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space minus 2 space cos space straight c space sin space straight c equals 0 therefore space space space space cos space straight c equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space sin space straight c not equal to 0 space in space left parenthesis 0 comma space straight pi right parenthesis right square bracket therefore space space space space straight c equals straight pi over 2 element of space left parenthesis 0 comma space straight pi right parenthesis therefore space Rolle apostrophe straight s space theorem space is space verified.

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