Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12034520

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 1 minus cos 2 straight x over denominator 2 straight x squared end fraction comma space straight x not equal to 0 end cell row cell space space space space space space space space space space space space space space straight k space space space space comma space straight x equals 0 end cell end table close
is continuous at x = 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 1 minus cos 2 straight x over denominator 2 straight x squared end fraction comma space straight x not equal to 0 end cell row cell space space space space space space space space space space space space space space straight k space space space space comma 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 1 minus cos 2 straight x over denominator 2 straight x squared end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator 2 sin squared straight x over denominator 2 straight x squared end fraction equals Lt with straight x rightwards arrow 0 below open parentheses fraction numerator sin space straight x over denominator straight x end fraction close parentheses squared equals left parenthesis 1 right parenthesis squared equals 1
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 space space space rightwards double arrow space straight k equals 1

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