from Mathematics Continuity and Differentiability

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 Multiple Choice QuestionsShort Answer Type

571. If space straight y equals left parenthesis cos to the power of negative 1 end exponent space straight x right parenthesis squared comma space prove space that space left parenthesis 1 minus straight x squared right parenthesis space straight y subscript 2 minus straight x space straight y subscript 1 minus 2 equals 0.
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572. If space space space straight y equals left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis squared comma space prove space that space left parenthesis 1 minus straight x 2 right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 minus 2 equals 0. space
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573. If space space straight y equals open square brackets log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses close square brackets squared comma space prove space that space left parenthesis 1 space plus space straight x squared right parenthesis straight y subscript 2 space plus space straight x space straight y subscript 1 space equals space 2.
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574. If space space straight y equals sin left parenthesis straight m space sin to the power of negative 1 end exponent straight x right parenthesis comma space prove space that space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 plus straight m squared space straight y equals 0.
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575. If y = em sin1 x , prove that (1 - x2) y2 - x y1 = m2 y.
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576. If y = ea cos-1 x , show that
left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx minus straight a squared straight y equals 0.
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577. If space space straight f left parenthesis straight x right parenthesis equals vertical line straight x vertical line cubed comma space show space that space straight f " left parenthesis straight x right parenthesis space exists space for space all space real space straight x space and space find space it.


We space have space
straight f left parenthesis straight x right parenthesis equals open vertical bar straight x close vertical bar cubed equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell space space space straight x cubed end cell cell comma space straight x greater or equal than 0 end cell row cell negative straight x cubed end cell cell comma space straight x less than 0 end cell end table close space space space space space space space space space space space space space space space space space left square bracket because space open vertical bar straight x close vertical bar equals negative straight x space for space straight x less than 0 right square bracket
because space straight f apostrophe left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell space space 3 space straight x cubed end cell cell comma space straight x greater or equal than 0 end cell row cell negative 3 straight x cubed end cell cell comma space straight x less than 0 end cell end table close
space space space space space straight f apostrophe apostrophe left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell space space 6 space straight x cubed end cell cell comma space straight x greater or equal than 0 end cell row cell negative 6 straight x cubed end cell cell comma space straight x less than 0 end cell end table close
Hence space the space result
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578. If x = f(t), y = g (t) possess second order derivative for all t in (a, b) and f"(t) is inevitable then prove that

fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator straight f apostrophe left parenthesis straight t right parenthesis space straight g apostrophe apostrophe left parenthesis straight t right parenthesis minus straight g apostrophe left parenthesis straight t right parenthesis space straight f apostrophe apostrophe left parenthesis straight t right parenthesis over denominator left parenthesis straight f apostrophe left parenthesis straight t right parenthesis right parenthesis squared end fraction
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 Multiple Choice QuestionsLong Answer Type

579. If space left parenthesis straight x minus straight a right parenthesis squared space plus left parenthesis straight y minus straight b right parenthesis squared space equals straight c squared space comma space prove space that space space fraction numerator open square brackets 1 plus open parentheses begin display style dy over dx end style close parentheses squared close square brackets to the power of begin display style 3 over 2 end style end exponent over denominator begin display style fraction numerator straight d squared straight y over denominator dx squared end fraction end style end fraction space is space straight a space constant space independent space of space straight a space and space straight b.
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 Multiple Choice QuestionsShort Answer Type

580. Verify space Rolle apostrophe straight s space theorem space for space the space function
straight f left parenthesis straight x right parenthesis equals fraction numerator 8 straight x squared over denominator 3 end fraction minus 2 straight x comma space straight x element of open square brackets 0 comma 3 over 4 close square brackets
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