Question

Find the value of k so that the function f is continuous at the indicated point
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight k left parenthesis straight x squared minus 2 straight x space right parenthesis comma space if space straight x less than 0 end cell row cell space space cos space straight x space space space space space comma space if space straight x greater or equal than 0 end cell end table close at space straight x equals 0$

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 left parenthesis straight x squared minus 2 straight x space right parenthesis comma space if space straight x less than 0 end cell row cell space space cos space straight x space space space space space comma space if space straight x greater or equal than 0 end cell end table close 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 curly bracket straight k left parenthesis straight x squared minus 2 straight x right parenthesis right curly bracket 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 straight k left curly bracket left parenthesis 0 minus straight h right parenthesis squared minus 2 left parenthesis 0 minus straight h right parenthesis right curly bracket equals straight k Lt with straight h rightwards arrow 0 below left parenthesis straight h squared plus 2 straight h right parenthesis space space space space space space space space space space space space space space space space equals straight k left parenthesis 0 plus 0 right parenthesis equals 0 Also space straight f left parenthesis 0 right parenthesis equals cos space 0 equals 1 therefore space Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis not equal to straight f left parenthesis 0 right parenthesis therefore space straight f left parenthesis straight x right parenthesis space is space not space continous space for space any space value space of space straight k.

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