Question

In the following , find the value of the constant k so that the given function is continuous at the indicated point:
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight k space straight x plus 5 space space comma space if space straight x less or equal than 2 end cell row cell straight x minus 1 space space space space space comma space if space straight x greater than 2 end cell end table close at space straight x equals 2.$

Solution

Her 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 plus 5 space space comma space if space straight x less or equal than 2 end cell row cell straight x minus 1 space space space space space comma space if space straight x greater than 2 end cell end table close space Lt with straight x rightwards arrow 2 to the power of minus below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow 2 to the power of minus below left parenthesis kx plus 5 right parenthesis space space space space space space space space left square bracket Put space straight x equals 2 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 curly bracket straight k left parenthesis 2 minus straight h right parenthesis plus 5 right curly bracket equals straight k left parenthesis 2 minus 0 right parenthesis plus 5 equals 2 straight k plus 5 space Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow 2 to the power of plus below left parenthesis straight x minus 1 right parenthesis space space space space space space space space space space left square bracket Put space straight x equals 2 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 space equals Lt with straight h rightwards arrow 0 below left parenthesis 2 plus straight h minus straight h right parenthesis equals 2 plus 0 minus 1 equals 1 space space 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 space space space space rightwards double arrow space 2 straight k plus 5 equals 1 rightwards double arrow space 2 straight k equals negative 4 space rightwards double arrow straight k equals negative 2

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