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Continuity And Differentiability

Question
CBSEENMA12035760

Determine the value of the constant ‘k’ so that the functionstraight f left parenthesis straight x right parenthesis space equals space open curly brackets table attributes columnalign left end attributes row cell fraction numerator Kx over denominator vertical line straight x vertical line end fraction comma space If space straight x space less than 0 end cell row cell 3 comma space if space straight x greater or equal than space 0 end cell end table close  is continuous at x = 0.

Solution
straight f left parenthesis straight x right parenthesis space equals space open curly brackets table attributes columnalign left end attributes row cell fraction numerator Kx over denominator vertical line straight x vertical line end fraction comma space If space straight x space less than 0 end cell row cell 3 comma space if space straight x greater or equal than space 0 end cell end table close
stack lim space straight f left parenthesis straight x right parenthesis with straight x space rightwards arrow 0 below space equals stack space lim with straight x rightwards arrow 0 below space straight f left parenthesis space straight x right parenthesis space equals space straight f left parenthesis 0 right parenthesis
rightwards double arrow space limit as straight x space rightwards arrow 0 of space fraction numerator negative kx over denominator straight x end fraction space equals space stack lim space 3 with straight x rightwards arrow 0 below space equals space 3
rightwards double arrow negative straight k space equals 3
rightwards double arrow straight k equals negative 3

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