Question

find the value of the k so that the given function is continuous at the indicated point:
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight k space cos space straight x over denominator straight pi minus 2 straight x end fraction comma space if space straight x not equal to straight pi over 2 end cell row cell space space space space space space space 3 space space space space space space comma space if space straight x equals straight pi over 2 end cell end table close at space straight x equals straight pi over 2$

Solution

Here space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight k space cos space straight x over denominator straight pi minus 2 straight x end fraction comma space if space straight x not equal to straight pi over 2 end cell row cell space space space space space space space 3 space space space space space space comma space if space straight x equals straight pi over 2 end cell end table close space Lt with straight x rightwards arrow begin inline style straight pi over 2 end style below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow begin inline style straight pi over 2 end style below fraction numerator straight k space cos space straight x over denominator straight pi minus 2 straight x end fraction space space space space space space left square bracket Put space straight x equals begin inline style straight pi over 2 end style 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 begin inline style straight pi over 2 end style right square bracket space space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 below fraction numerator straight k space cos open parentheses straight pi over 2 plus straight h close parentheses over denominator straight pi minus 2 left parenthesis straight pi over 2 minus straight h right parenthesis end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator negative straight k space sin space straight h over denominator negative 2 straight h end fraction equals straight k over 2 space Lt with straight h rightwards arrow 0 below fraction numerator sin space straight h over denominator straight h end fraction equals straight k over 2 cross times 1 equals straight k over 2 space Also space straight f open parentheses begin inline style straight pi over 2 end style close parentheses equals 3 Since space straight f space is space continous space at space straight x equals begin inline style straight pi over 2 end style therefore space Lt with straight x rightwards arrow begin inline style straight pi over 2 end style below straight f left parenthesis straight x right parenthesis equals straight f open parentheses begin inline style straight pi over 2 end style close parentheses space space space space rightwards double arrow space space straight k over 2 equals 3 space space space space space space rightwards double arrow straight k equals 6

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