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Continuity and Differentiability

Long Answer Type

111. Find all points of discontinuity of f, where f is defined by

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator open vertical bar straight x close vertical bar over denominator straight x end fraction comma space space space space if space straight x not equal to 0 end cell row cell 0 comma space space space space space space space space space if space straight x equals 0 end cell end table close

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112. Find all points of discontinuity of f, where f is defined by

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x over denominator open vertical bar straight x close vertical bar end fraction comma space if space straight x less than 0 end cell row cell space minus 1 comma space space if space straight x greater or equal than 0 end cell end table close

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113.

Find all points of discontinuity of f, where f is defined by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell space straight x plus 1 comma space if space straight x greater or equal than 1 end cell row cell straight x squared plus 1 comma space if space straight x less than 1 end cell end table close$

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114.

Find all points of discontinuity of f, where f is defined by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x cubed minus 3 comma space if space straight x less or equal than 2 end cell row cell straight x squared plus 1 comma space if space straight x greater than 2 end cell end table close$

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115.

Find all points of discontinuity of f, where f is defined by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x to the power of 10 minus 1 comma space space if space straight x less or equal than 1 end cell row cell straight x squared comma space space space space space space space space space space space if space straight x greater than 1 end cell end table close$

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116. Is the function defined by

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x plus 5 comma space if space straight x less or equal than 1 end cell row cell straight x minus 5 comma space if space straight x greater than 1 end cell end table close

a continuous function?

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117. Discuss the continuity of the function f, where/is defined by

straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell 3 comma space space if space 0 less or equal than straight x less or equal than 1 space space end cell row cell 4 comma space space if space 1 less than straight x less than 3 end cell row cell 5 comma space if space 3 less or equal than straight x less or equal than 1 space end cell end table close

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Short Answer Type

118. Discuss the continuity of the function f, where f is defined by

straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell 2 straight x comma end cell row cell 0 comma end cell row cell 4 straight x comma end cell end table close table row cell if space straight x less than 0 space space space space space space end cell row cell if space 0 less or equal than straight x less or equal than 1 end cell row cell if space straight x greater than 1 space space space space space space end cell end table

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Long Answer Type

119. Discuss the continuity of the function f , where fis defined by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell negative 2 comma space space space space if space straight x less or equal than negative 1 end cell row cell 2 straight x comma space space if minus 1 less than straight x less or equal than 1 end cell row cell 2 space space space space space if space straight x greater than 1 end cell end table close$

$straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell negative 2 comma space space space space if space straight x less or equal than negative 1 end cell row cell 2 straight x comma space space if minus 1 less than straight x less or equal than 1 end cell row cell 2 space space space space space if space straight x greater than 1 end cell end table close$
Function f is defined at all points of the real line.
When x < – 1, we have f(x) = – 2; which is constant and so it is continuous.
$At space straight x equals negative 1 space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of minus below left parenthesis negative 2 right parenthesis equals negative 2 space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of plus below left parenthesis 2 straight x right parenthesis equals 2 left parenthesis negative 1 right parenthesis equals negative 2 Also space space space space space space space space straight f left parenthesis negative 1 right parenthesis equals negative 2 therefore space stack space space space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis negative 1 right parenthesis$

∴ f is continuous at x = – 1
In the interval –1 < x < 1, we have f(x) = 2 x, which being a linear polynomial, is continuous.
$At space straight x equals negative 1 space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of minus below left parenthesis 2 straight x right parenthesis equals 2 left parenthesis 1 right parenthesis equals 2 space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of plus below left parenthesis 2 right parenthesis equals 2 Also space space space space space space space space straight f left parenthesis 1 right parenthesis equals 2 left parenthesis 1 right parenthesis equals 2 therefore space stack space space space space space space space space space space Lt space with straight x rightwards arrow negative 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 1 right parenthesis$
∴ f is continuous at x = 1
When x > 1, we have f (x) = 2. which is constant and so it is continuous.

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Short Answer Type

120. Show that the function of given by

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x comma space space space space if space straight x greater or equal than 1 end cell row cell straight x squared comma space if space straight x less than 1 end cell end table close

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