Prove that the functionremains discontinuous at x = 0, regardles
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    Continuity and Differentiability

    Multiple Choice QuestionsShort Answer Type

    21.

    Examine the  continuity of 
    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 minus straight a close vertical bar over denominator straight x minus straight a end fraction space space space space space space straight x not equal to straight a end cell row cell space space space space space space space space 1 space space space space space space space space space space space straight x equals straight a end cell end table close space at space space straight x equals straight a

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    22. Examine the continuity of f (x) at x = 0.
    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 2 open vertical bar straight x close vertical bar end fraction comma space straight x not equal to 0 end cell row cell space space space space space 1 half comma space straight x equals 0 end cell end table close
    79 Views
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    23. Examine the continuity of f (x) at x = 0.
    If space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x minus open vertical bar straight x close vertical bar comma space straight x not equal to 0 end cell row cell space space space space space space 2 space space space comma space straight x equals 0 end cell end table close
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    24.  Test the continuity of the function f(x) at the origin :
    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 straight x not equal to 0 end cell row cell space space space 1 space space space space comma space space space space straight x equals 0 end cell end table close
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    Multiple Choice QuestionsLong Answer Type

    25. Test the continuity of the function f(x) at the originstraight 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 straight x not equal to 0 end cell row cell space space space 0 space space space space comma space space space space straight x equals 0 end cell end table close
    then show that f(x) is discontinuous at x = 0.
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    26. Prove that the function
    straight f left parenthesis straight x right parenthesis 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 plus 2 straight x squared end fraction comma space straight x not equal to 0 end cell row cell space space space space space space space space space straight k space space space space space space space comma space straight x equals 0 end cell end table close
    remains discontinuous at x = 0, regardless of the choice of k.

    Here space straight f left parenthesis straight x right parenthesis 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 plus 2 straight x squared end fraction comma space straight x not equal to 0 end cell row cell space space space space space space space space space straight k space space space space space space space comma space straight x equals 0 end cell end table close space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator straight x over denominator open vertical bar straight x close vertical bar plus 2 straight x squared end fraction space space space space space left square bracket Put space straight x equals 0 minus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 to the power of minus right square bracket space space space space space space space equals Lt with straight h rightwards arrow 0 below fraction numerator 0 minus straight h over denominator open vertical bar 0 minus straight h close vertical bar plus 2 left parenthesis 0 minus straight h right parenthesis squared end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator negative straight h over denominator straight h plus 2 straight h squared end fraction equals Lt with straight h rightwards arrow 0 below minus fraction numerator straight h over denominator 1 plus 2 straight h end fraction equals negative fraction numerator 1 over denominator 1 plus 0 end fraction equals negative 1 space Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator straight x over denominator open vertical bar straight x close vertical bar plus 2 straight x squared end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket Put space straight x equals 0 plus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 to the power of plus right square bracket space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar 0 plus straight h close vertical bar over denominator open vertical bar 0 plus straight h close vertical bar plus 2 left parenthesis 0 plus straight h right parenthesis squared end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator straight h over denominator straight h plus 2 straight h squared end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator 1 over denominator 1 plus 2 straight h end fraction equals fraction numerator 1 over denominator 1 plus 0 end fraction equals 1 therefore Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis not equal to Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis rightwards double arrow space Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis space does space not space exist
    ⇒ f(x) is discontinuous whatever k may be.
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    27. Show that the function f given by f(x) = | x | + | x –1 |,x ∈ R is continuous both at x =.0 and x = 1.
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    28.

    Discuss continuity of the function f given by

    f(x) = | x – 1| + | x – 2 ] at x = 1 and x = 2.

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    Multiple Choice QuestionsShort Answer Type

    29. Discuss the continuity or otherwise of the function f(x) defined by
    straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin 3 straight x over denominator straight x end fraction comma space straight x not equal to 0 end cell row cell space space space space 1 space space space space space space comma space straight x equals 0 end cell end table close space at space straight x equals 0
    87 Views
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    30. Discuss the continuity or otherwise of the function f(x) defined by
    straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space 2 straight x over denominator sin space 3 straight x end fraction comma space straight x not equal to 0 end cell row cell space space space space space space 2 space space space space space comma space straight x equals 0 end cell end table close at space straight x equals 0
    88 Views
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