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Continuity And Differentiability

Question
CBSEENMA12034607
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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 plus 1 comma space if space straight x less or equal than straight pi end cell row cell cos space straight x space space comma space if space straight x greater than straight pi end cell end table close at space straight x equals straight pi

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 plus 1 comma space if space straight x less or equal than straight pi end cell row cell cos space straight x space space comma space if space straight x greater than straight pi end cell end table close space Lt with straight x rightwards arrow straight pi to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow straight pi to the power of minus below left parenthesis kx plus 1 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 straight pi minus 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 straight pi to the power of minus right square bracket space 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 left curly bracket straight k left parenthesis straight pi minus straight h right parenthesis plus 1 right curly bracket equals straight k left parenthesis straight pi minus 0 right parenthesis equals 1 equals straight k space straight pi plus 1 space Lt with straight x rightwards arrow straight pi to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow straight pi to the power of plus below space cos space straight x 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 straight pi 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 straight pi to the power of plus 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 cos left parenthesis straight pi plus straight h right parenthesis equals Lt with straight h rightwards arrow 0 below left parenthesis cos space straight pi space cos space straight h minus sin space straight pi space sin space straight h right parenthesis space space space space space space space space space space space space space space space space space equals cos space straight pi.1 minus sin space straight pi.0 equals cos equals negative 1 Since space straight f space is space continous space at space straight x equals straight pi therefore space space Lt with straight x rightwards arrow straight pi to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow straight pi to the power of plus below straight f left parenthesis straight x right parenthesis therefore space straight k space ππ equals negative 1 space rightwards double arrow straight k space straight pi equals negative 2 space space rightwards double arrow space straight k equals negative 2 over straight pi

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