The function is not defined at x = 0. Find the value of/(x) so

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Continuity and Differentiability

Long Answer Type

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41. The function $straight f left parenthesis straight x right parenthesis equals fraction numerator log left parenthesis 1 plus ax right parenthesis minus log left parenthesis 1 minus bx right parenthesis over denominator straight x end fraction$ is not defined at x = 0. Find the value of/(x) so that f is continuous at x = 0.

Here space straight f left parenthesis straight x right parenthesis equals fraction numerator log left parenthesis 1 plus ax right parenthesis minus log left parenthesis 1 minus bx right parenthesis over denominator straight x end fraction 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 log left parenthesis 1 plus ax right parenthesis minus log left parenthesis 1 minus bx right parenthesis over denominator straight x end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator log left parenthesis 1 plus ax right parenthesis over denominator straight x end fraction minus bold Lt with bold x bold rightwards arrow bold 0 below fraction numerator bold log bold left parenthesis bold 1 bold minus bold bx bold right parenthesis over denominator bold x end fraction space space space space space space space space space space space space space equals straight a space Lt with straight x rightwards arrow 0 below 1 over ax log left parenthesis 1 plus ax right parenthesis minus straight b. Lt with straight x rightwards arrow 0 below 1 over bx log left parenthesis 1 minus bx right parenthesis space space space space space space space space space space space space space equals straight a space Lt with straight x rightwards arrow 0 below log left parenthesis 1 plus ax right parenthesis to the power of 1 over ax end exponent plus straight b. Lt with straight x rightwards arrow 0 below log left parenthesis 1 minus bx right parenthesis to the power of negative 1 over bx end exponent space space space space space space space space space space space space space equals straight a space log space straight e plus straight b space log space straight e equals straight a plus straight b

For f(x) to be continuous at x = 0, we have

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 space space space space straight i. straight e. comma space straight f left parenthesis 0 right parenthesis equals straight a plus straight b therefore space function space must space be space defined space as space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator log left parenthesis 1 plus ax right parenthesis minus log left parenthesis 1 minus bx right parenthesis over denominator straight x 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 straight a plus straight b comma space space space space space space space space space space space space space space space straight x equals 0 end cell end table close

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Short Answer Type

42. Discuss continuity of f(x) at x = 0, if

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator log left parenthesis 1 plus ax right parenthesis minus log left parenthesis 1 minus bx right parenthesis over denominator straight x 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 space space space space space straight a plus straight b space space space space space space space space space space space space space space space space space space space comma space straight x equals 0 end cell end table close

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43. Discuss continuity of f(x) at x = 0, if

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell 1 over straight x plus fraction numerator log left parenthesis 1 minus straight x right parenthesis over denominator straight x squared end fraction comma space space if space straight x not equal to 0 end cell row cell space space space space space space space space minus 1 half space space space space space space space space space space space space space comma space if space straight x equals 0 end cell end table close

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44. Discuss continuity of f(x) at x = 0, if

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight e to the power of straight x minus 1 minus straight x over denominator straight x squared end fraction comma space if space straight x not equal to end cell row cell space space space space space space 1 half space space space space space space space space space comma space if space straight x equals 0 end cell end table close

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45. 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 straight x squared minus 25 over denominator straight x minus 5 end fraction comma space if space straight x not equal to 5 end cell row cell space space space space space space space straight k space space space space space space space comma space if space straight x equals 5 end cell end table close

is continous at x=5.

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46. Determine 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 straight x squared minus 36 over denominator straight x minus 6 end fraction comma space if space straight x not equal to 6 end cell row cell space space space space space space space straight k space space space space space space space comma space if space straight x equals 6 end cell end table close

is continuous at x = 6.

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47. 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 straight x squared minus 3 straight x plus 2 over denominator straight x minus 1 end fraction comma space if space straight x not equal to 1 end cell row cell space space space space space space space space space space straight k space space space space space space space space space comma space if space straight x equals 1 end cell end table close

is continous at x=1.

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48. Determine the value of 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 straight x squared minus 9 over denominator straight x minus 3 end fraction comma space straight x not equal to 1 end cell row cell space space space space space space straight k space space space space space comma space straight x equals 3 end cell end table close

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49. 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 ?

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50. 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

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