Show that the function f given by f(x) = | x | + | x –1 |,x �
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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
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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.
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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.

    Here space straight f left parenthesis straight x right parenthesis equals open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar 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 left parenthesis open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar right parenthesis 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 space space left square bracket Put space straight x equals 0 minus straight h comma space straight h greater than 0 right square bracket space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 0 minus straight h close vertical bar plus open vertical bar 0 minus straight h minus 1 close vertical bar right parenthesis equals Lt with straight h rightwards arrow 0 below open curly brackets straight h plus left parenthesis 1 plus straight h right parenthesis close curly brackets equals 0 plus 1 plus 0 equals 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 left parenthesis open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar right parenthesis 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 space space left square bracket Put space straight x equals 0 plus straight h comma space straight h greater than 0 right square bracket space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 0 plus straight h close vertical bar plus open vertical bar 0 plus straight h minus 1 close vertical bar right parenthesis equals Lt with straight h rightwards arrow 0 below open curly brackets open vertical bar straight h close vertical bar plus open vertical bar negative left parenthesis 1 minus straight h right parenthesis close vertical bar close curly brackets space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis straight h plus 1 minus straight h right parenthesis equals Lt with straight h rightwards arrow 0 below left parenthesis straight h plus 1 minus straight h right parenthesis equals Lt with straight h rightwards arrow 0 below 1 equals 1 Also space straight f left parenthesis 0 right parenthesis space equals open vertical bar 0 close vertical bar plus open vertical bar 0 plus 1 close vertical bar equals 0 plus 1 equals 1 therefore 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 plus below straight f left parenthesis straight x right parenthesis equals straight F left parenthesis 0 right parenthesis therefore straight f space is space continous space at space straight x equals 0. Again space Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of minus below left parenthesis open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar right parenthesis 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 1 minus straight h comma space straight h greater than 0 right square bracket 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 equals Lt with straight h rightwards arrow 0 below open vertical bar 1 minus straight h close vertical bar plus open vertical bar 1 minus straight h minus 1 close vertical bar equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 1 minus straight h close vertical bar plus open vertical bar negative straight h close vertical bar right parenthesis 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 equals Lt with straight h rightwards arrow 0 below left parenthesis 1 minus straight h plus straight h right parenthesis equals Lt with straight h rightwards arrow 0 below 1 equals 1 space Lt with straight x rightwards arrow 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of plus below left parenthesis open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar right parenthesis space space space left square bracket Put space straight x equals 1 plus straight h comma space straight h greater than 0 right square bracket space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 1 plus straight h close vertical bar plus open vertical bar 1 plus straight h minus 1 close vertical bar right parenthesis equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 1 plus straight h close vertical bar plus open vertical bar straight h close vertical bar right parenthesis space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis 1 plus straight h plus straight h right parenthesis equals 1 plus 0 plus 0 equals 1 straight f left parenthesis 1 right parenthesis equals open vertical bar 1 close vertical bar plus open vertical bar 1 minus 1 close vertical bar equals 1 plus 0 equals 1 therefore Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 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.
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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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