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

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 Multiple Choice QuestionsShort Answer Type

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601. Verify Roll's Theorem for the function :straight f left parenthesis straight x right parenthesis equals space 4 space sin space straight x space in space left square bracket 0 comma space straight pi right square bracket.


straight f left parenthesis straight x right parenthesis space equals space 4 space sin space straight x space
space left parenthesis straight a right parenthesis space space space space Since space sine space curve space is space continuous space space
therefore space straight f left parenthesis straight x right parenthesis space is space straight a space continuous space function space in space left square bracket 0 comma space straight pi right square bracket space space
left parenthesis straight b right parenthesis space space space space straight f apostrophe left parenthesis straight x right parenthesis space equals space 4 space cos space straight x comma space which space exists space in space left parenthesis 0 comma space straight pi right parenthesis space 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
space left parenthesis straight c right parenthesis space space space space straight f left parenthesis 0 right parenthesis space equals space 4 space sin space 0 space equals space 0 space comma space straight f left parenthesis straight pi right parenthesis space equals space 4 space sin space straight pi space equals space 0 space space
therefore space straight f left parenthesis 0 right parenthesis space equals space straight f left parenthesis straight pi right parenthesis space
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 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 in space left parenthesis 0 comma space straight pi right parenthesis space such space that space straight f apostrophe left parenthesis straight c right parenthesis space equals space 0.
Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space 4 space cos space straight c equals 0 comma space straight i. straight e. comma space cos space straight c equals 0
therefore space straight c equals straight pi over 2 comma space fraction numerator 3 straight pi over denominator 2 end fraction comma space fraction numerator 5 straight pi over denominator 2 end fraction comma... space space space rightwards double arrow space straight c equals straight pi over 2 space lies space in space left parenthesis 0 comma space straight pi right parenthesis
Rolle apostrophe straight s space theorem space is space verified.
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602. If space space straight f open parentheses straight x close parentheses equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin open parentheses straight a plus 1 close parentheses plus 2 sinx over denominator straight x end fraction comma space straight x less than 0 end cell row cell 2 space space space space space space space space space space space space space space space space space space space space space space comma space x equals 0 end cell row cell fraction numerator square root of 1 plus b x end root minus 1 over denominator straight x end fraction space space space space space space comma space straight x greater than 0 end cell end table close
is continuous at x = 0, then find the values of a and b.
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603. if space cos space left parenthesis straight a plus straight y right parenthesis space equals space cos space straight y space then space prove space that space dy over dx space equals fraction numerator cos squared left parenthesis straight a plus straight y right parenthesis over denominator sin space straight a end fraction
Hence space show space that space sin space straight a space fraction numerator straight d squared straight y over denominator dx squared end fraction plus sin space 2 space left parenthesis straight a plus straight y right parenthesis dy over dx equals 0
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604. Find space dy over dx space if space straight y equals sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x minus 4 square root of 1 minus 4 straight x squared end root over denominator 5 end fraction close square brackets
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605.

Show that the function straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x minus 3 close vertical bar comma space straight x element of bold R bold comma is  continuous but not differentiable at x=3. 

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606.

Determine the value of the constant ‘k’ so that the function  is continuous at x = 0.

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 Multiple Choice QuestionsLong Answer Type

607.

For what value of k is the following function continuous at x = 2?

f ( x )  =2x + 1     ;   x<2    k            ;   x = 2        3x - 1  ;   x>1       


608.

Differentiate the following function w.r.t. x:

y = sinxx + sin-1x


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609.

Find dydx if (x2 + y2)2 = xy.


610.

If y =3 cos ( log x ) + 4 sin ( log x ), then show that x2 d2ydx2 + x dydx + y = 0


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