Question

Find k so that $straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space kx over denominator straight x end fraction space comma space straight x not equal to 0 end cell row cell space 4 plus straight x space space space space space comma space straight x equals 0 end cell end table close$

is continuous at x = 0.

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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 kx over denominator straight x end fraction space comma space straight x not equal to 0 end cell row cell space 4 plus straight x space 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 open parentheses fraction numerator sin space kx over denominator kx end fraction cross times straight k close parentheses equals straight k space Lt with straight x rightwards arrow 0 below fraction numerator sin space kx over denominator kx end fraction equals straight k cross times 1 equals straight k

Also f(0)=Value of (4+x) at x=0

therefore space Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 0 right parenthesis therefore space space space space space space space straight k equals 4

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