Question

Find the value of k so that the function f is continuous at the indicated point
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight k space straight x squared comma space straight x less or equal than 2 end cell row cell 3 space space space space comma space straight x greater than 2 end cell end table close at space straight x equals 2$

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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 straight k space straight x squared comma space straight x less or equal than 2 end cell row cell 3 space space space space comma space straight x greater than 2 end cell end table close space space space Lt with straight x rightwards arrow 2 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of minus 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 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 2 minus straight h comma 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 equals Lt with straight h rightwards arrow 0 below left curly bracket straight k left parenthesis 2 minus straight h squared right parenthesis right curly bracket equals straight k space Lt with straight h rightwards arrow 0 below left parenthesis straight h squared minus 4 straight h plus 4 right parenthesis equals straight k left parenthesis 0 minus 0 plus 4 right parenthesis equals 4 straight k space space space Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of plus below 3 equals 3 Since space straight f left parenthesis straight x right parenthesis space is space continous space at space straight x equals 2 therefore space Lt with straight x rightwards arrow 2 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis straight x right parenthesis rightwards double arrow space 4 straight k equals 3 space space space rightwards double arrow space straight k equals 3 over 4

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