Verify Roll's Theorem for the function :  from Mathematics Con

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 Multiple Choice QuestionsShort Answer Type

591. Verify Roll's Theorem for the function :f(x) = x (x - 3)2 in 0 ≤ x ≤ 3
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592. Verify Roll's Theorem for the function :f(x) = (x2 - 1) (x - 2) in [-1, 2]
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593. Verify Roll�s Theorem for the function :f{x) = e1 - x2 in [-1, 1]
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594. Verify Roll�s Theorem for the function :f(x) = log (x2 + 2) - log 3 in [-1, 1]
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595. Verify the truth of Rolle's Theorem for the function
straight f left parenthesis straight x right parenthesis equals space sin space 3 straight x comma space straight x element of open square brackets 0 comma space straight pi over 3 close square brackets
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596. Verify Rolle's Theorem for the function f(x) = sin2 x in [0, straight pi].
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597. 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
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598. 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
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599. Verify Roll's Theorem for the function : straight f left parenthesis straight x right parenthesis equals space sin space 2 space straight x space or space cos space 2 open parentheses straight x minus straight pi over 4 close parentheses space in space open square brackets 0 comma space straight pi over 2 close square brackets


Here space straight f left parenthesis straight x right parenthesis equals space sin space 2 space straight x
left parenthesis straight a right parenthesis space Since space sine space curve space is space continuous space
therefore space space straight f left parenthesis straight x right parenthesis space is space straight a space continuous space function space in space straight f open square brackets 0 comma space straight pi over 2 close square brackets
left parenthesis straight b right parenthesis space straight f apostrophe left parenthesis 0 right parenthesis equals space 2 space cos space 2 straight x comma space which space exixts space in space open parentheses 0 comma space straight pi over 2 close parentheses space rightwards double arrow space straight f left parenthesis straight x right parenthesis space is space derivable space in space open parentheses 0 comma space straight pi over 2 close parentheses
left parenthesis straight c right parenthesis space space straight f left parenthesis 0 right parenthesis equals space sin space 0 equals 0
therefore space space straight f open parentheses straight pi over 2 close parentheses equals sin space straight pi equals 0
therefore space space straight f left parenthesis 0 right parenthesis space equals straight f open parentheses straight pi over 2 close parentheses
therefore space 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.
therefore space 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 c element of space open parentheses 0 comma space straight pi over 2 close parentheses
Now space straight f apostrophe left parenthesis straight c right parenthesis equals 2 space cos space 2 straight c
therefore space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space space 2 space cos space 2 straight c equals 0 comma space straight i. straight e. comma space cos space 2 straight c equals 0
straight i. straight e. comma space space 2 straight c equals straight pi over 2 space or space straight c equals straight pi over 4 space lies space in space open parentheses 0 comma space straight pi over 2 close parentheses
therefore space space rolle apostrophe straight s space Theorem space is space verified.
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600. Verify Roll's Theorem for the function : straight f left parenthesis straight x right parenthesis equals cos space straight x plus sin space straight x space in space left square bracket 0 comma space 2 straight pi right square bracket
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