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Continuity And Differentiability

Question
CBSEENMA12034513

For what value of k is the function

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space 5 straight x over denominator 3 straight x end fraction comma space if space straight x not equal to 0 end cell row cell space space space space space space straight k space space space space space space comma space if space straight x equals 0 end cell end table close
continous space 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 fraction numerator sin space 5 straight x over denominator 3 straight x end fraction comma space if space straight x not equal to 0 end cell row cell space space space space space space straight k space space space space space space comma space if space straight x equals 0 end cell end table close space
space Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 below fraction numerator sin space 5 straight x over denominator 3 straight x end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator sin space 5 straight x over denominator 5 straight x end fraction cross times 5 over 3 equals 5 over 3 space Lt with straight x rightwards arrow 0 below fraction numerator sin space 5 straight x over denominator 5 straight x end fraction equals 5 over 3 cross times 1 equals 5 over 3
Since space straight f left parenthesis straight x right parenthesis space is space continous space at space straight x equals 0
therefore space straight f left parenthesis 0 right parenthesis equals Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis
rightwards double arrow space space space space space straight k equals 5 over 3

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