Determine the value of the constant k so that the functionis continous at x=1.

Question

Determine the value of the constant k so that the function
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x squared minus 3 straight x plus 2 over denominator straight x minus 1 end fraction comma space if space straight x not equal to 1 end cell row cell space space space space space space space space space space straight k space space space space space space space space space comma space if space straight x equals 1 end cell end table close$
is continous at x=1.

Solution

$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x squared minus 3 straight x plus 2 over denominator straight x minus 1 end fraction comma space if space straight x not equal to 1 end cell row cell space space space space space space space space space space straight k space space space space space space space space space comma space if space straight x equals 1 end cell end table close space Lt with straight x rightwards arrow 1 below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 below fraction numerator straight x squared minus 3 straight x plus 2 over denominator straight x minus 1 end fraction equals Lt with straight x rightwards arrow 1 below fraction numerator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x minus 2 right parenthesis over denominator straight x minus 1 end fraction space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow below left parenthesis straight x minus 2 right parenthesis equals 1 minus 2 equals negative 1 Since space straight f left parenthesis straight x right parenthesis space is space continous space at space straight x equals 1 therefore space straight f left parenthesis 1 right parenthesis equals space Lt with straight x rightwards arrow 1 below straight f left parenthesis straight x right parenthesis space space space space space space rightwards double arrow space space space straight k equals negative 1.$

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