Question

Find the value of k so that the function f is continuous at the indicated point
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell space straight k space straight x squared comma space straight x greater or equal than 1 end cell row cell space 4 space space space space comma space straight x less than 1 end cell end table close at space straight x equals 1$

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Solution

Here space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell space straight k space straight x squared comma space straight x greater or equal than 1 end cell row cell space 4 space space space space comma space straight x less than 1 end cell end table close space Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow 1 to the power of minus below left parenthesis straight x right parenthesis equals 4 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 straight k space straight x squared 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 left square bracket Put space straight x equals 1 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 1 to the power of plus right square bracket space space space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 1 to the power of plus below left curly bracket straight k left parenthesis 1 plus straight h right parenthesis squared right curly bracket equals Lt with straight x rightwards arrow straight h below left curly bracket straight k left parenthesis 1 plus straight h squared plus 2 straight h right parenthesis right curly bracket space space space space space space space space space space space space space space space space space equals straight k left parenthesis 1 plus 0 plus 0 right parenthesis equals straight k Since space straight f space is space continous space at space straight x equals 1 therefore 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 plus below straight f left parenthesis straight x right parenthesis equals 4 therefore space space space space space space space space space space space straight k space equals 4

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