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Continuity And Differentiability

Question
CBSEENMA12035503
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Verify Roll's Theorem for the function :straight f left parenthesis straight x right parenthesis equals space sin space straight x minus 1 space in space open square brackets straight pi over 2 comma space fraction numerator 5 straight pi over denominator 2 end fraction close square brackets

Solution
Let space straight f left parenthesis straight x right parenthesis equals space sin space straight x minus 1 left parenthesis straight a right parenthesis space Since space sin space straight x space and space 1 space are space both space continuous space functions space in space open square brackets straight pi over 2 comma space fraction numerator 5 straight pi over denominator 2 end fraction close square brackets therefore space sin space straight x minus 1 space is space continuous space in space space open square brackets straight pi over 2 comma space fraction numerator 5 straight pi over denominator 2 end fraction close square brackets space rightwards double arrow space straight f space is space continuous space in space space open square brackets straight pi over 2 comma space fraction numerator 5 straight pi over denominator 2 end fraction close square brackets left parenthesis straight b right parenthesis space straight f apostrophe left parenthesis straight x right parenthesis equals space cos space straight x comma space which space is space exists space in space open parentheses straight pi over 2 comma space fraction numerator 5 straight pi over denominator 2 end fraction close parentheses space rightwards double arrow space straight f space is space derivable space in space open parentheses straight pi over 2 comma space fraction numerator 5 straight pi over denominator 2 end fraction close parentheses left parenthesis straight c right parenthesis space straight f open parentheses straight pi over 2 close parentheses equals space sin space straight pi over 2 minus 1 equals 1 minus 1 equals 0 space space space straight f open parentheses fraction numerator 5 straight pi over denominator 2 end fraction close parentheses equals sin space fraction numerator 5 straight pi over denominator 2 end fraction minus 1 equals sin open parentheses 2 space straight pi plus straight pi over 2 close parentheses minus 1 equals sin straight pi over 2 minus 1 equals 1 minus 1 equals 0 therefore space straight f open parentheses straight pi over 2 close parentheses equals space straight f open parentheses fraction numerator 5 straight pi over denominator 2 end fraction close parentheses therefore space straight f space satisfies space all space the space conditions space of space Rolle apostrophe straight s space Theorem space in space space open square brackets straight pi over 2 comma space fraction numerator 5 straight pi over denominator 2 end fraction close square brackets therefore space there space must space exists 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 equals 0 space where space straight pi over 2 less than straight c less than fraction numerator 5 straight pi over denominator 2 end fraction Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space cos space straight c equals 0 therefore space straight c equals fraction numerator 3 straight pi over denominator 2 end fraction element of open parentheses straight pi over 2 comma space fraction numerator 5 straight pi over denominator 2 end fraction close parentheses space space rightwards double arrow space Rolle apostrophe straight s space Theorem space is space verified.

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