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Continuity And Differentiability

Question
CBSEENMA12034514

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 sin space 2 straight x over denominator 5 straight x end fraction comma space if space straight x not equal to 0 end cell row cell space space space space space space space straight k space space space space space comma space if space straight x equals 0 end cell end table close
is space continous space at space straight x equals 0

Solution
straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space 2 straight x over denominator 5 straight x end fraction comma space if space straight x not equal to 0 end cell row cell space space space space space space space straight k space space space space space comma space if space straight x equals 0 end cell end table close
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 2 straight x over denominator 5 straight x end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator sin space 2 straight x over denominator 2 straight x end fraction cross times 2 over 5 equals 2 over 5 cross times Lt with straight x rightwards arrow below fraction numerator sin space 2 straight x over denominator 2 straight x end fraction equals 2 over 5 cross times 1 equals 2 over 5.
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 straight x right parenthesis equals Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis space space space space space space space space space space space space space space space space space space space rightwards double arrow space straight k equals 2 over 5

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