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Continuity and Differentiability

Short Answer Type

61. find the value of the k so that the given function is continuous at the indicated point:

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight k space cos space straight x over denominator straight pi minus 2 straight x end fraction comma space if space straight x not equal to straight pi over 2 end cell row cell space space space space space space space 3 space space space space space space comma space if space straight x equals straight pi over 2 end cell end table close at space straight x equals straight pi over 2

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62. Find the value of a and b so that the function f given by

straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell 1 space space space comma space space space space space straight x less or equal than 3 end cell row cell ax plus straight b comma space space space space 3 less than straight x less than 5 end cell row cell 7 space space space comma space space space space straight x greater or equal than 5 end cell end table close

is continous at x=3 and x=5

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63. If the function

straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell space space space 3 ax plus straight b space space if space straight x greater than 1 end cell row cell space space space space space space 11 space space space space space space space space if space straight x equals 1 end cell row cell space space 5 ax minus 2 straight b space space if space straight x less than 1. end cell end table close

is continous at x=1, find the values of a and b.

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64. For what value of k is the following function continuous at x = 2?

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell table row cell 2 straight x plus 1 comma space space space space straight x less than 2 end cell row cell space space space straight k space space space space comma space space space space space space straight x equals 2 end cell end table end cell row cell space 3 straight x minus 1 comma space space space space space straight x greater than 2 end cell end table close

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Long Answer Type

65. Given that

straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell fraction numerator 1 minus cos space 4 straight x over denominator straight x squared end fraction space space space space space space space space space space comma space if space straight x less than 0 end cell row cell space space space space space space straight a space space space space space space space space space space space space space space space space space space space space space space space comma space if space straight x equals 0 end cell row cell fraction numerator square root of straight x over denominator square root of 16 plus square root of straight x end root minus 4 end fraction space space comma space if space straight x greater than 0 end cell end table close

If f(x) is continous at x=0, find the value of a.

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66. Determine the value of a, b_y c for which the functions
$straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell fraction numerator sin left parenthesis straight a plus 1 right parenthesis straight x plus sin space straight x over denominator straight x end fraction space space space comma space straight x less than 0 end cell row cell space space space space space space space space space space space space space space space space space space space space straight c space space space space space space space space space space space space space space space comma space straight x equals 0 end cell row cell fraction numerator square root of straight x plus bx squared end root minus square root of straight x over denominator bx to the power of begin display style 3 over 2 end style end exponent end fraction space space space space space space space comma space straight x greater than 0 end cell end table close$
is continous at x=0.

We space have straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell fraction numerator sin left parenthesis straight a plus 1 right parenthesis straight x plus sin space straight x over denominator straight x end fraction space space space comma space straight x less than 0 end cell row cell space space space space space space space space space space space space space space space space space space space space straight c space space space space space space space space space space space space space space space comma space straight x equals 0 end cell row cell fraction numerator square root of straight x plus bx squared end root minus square root of straight x over denominator bx to the power of begin display style begin inline style bevelled 3 over 2 end style end style end exponent end fraction space space space space space space space comma space straight x greater than 0 end cell end table close space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator sin left parenthesis straight a plus 1 right parenthesis straight x plus sin space straight x over denominator straight x end fraction equals Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator sin left parenthesis straight a plus 1 right parenthesis straight x over denominator straight x end fraction plus space Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator sin space straight x over denominator straight x end fraction space space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator sin left parenthesis straight a plus 1 right parenthesis straight x over denominator left parenthesis straight a plus 1 right parenthesis straight x end fraction. left parenthesis straight a plus 1 right parenthesis plus space Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator sin space straight x over denominator straight x end fraction equals 1. left parenthesis straight a plus 1 right parenthesis plus 1 equals straight a plus 2 space Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator square root of straight x plus bx squared end root minus square root of straight x over denominator bx to the power of begin display style bevelled 3 over 2 end style end exponent end fraction space space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator square root of straight x plus bx squared end root minus square root of straight x over denominator bx to the power of bevelled 3 over 2 end exponent end fraction cross times fraction numerator square root of straight x plus bx squared end root plus square root of straight x over denominator square root of straight x plus bx squared end root plus square root of straight x end fraction space space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator left parenthesis straight x plus bx squared right parenthesis minus straight x over denominator bx to the power of bevelled 3 over 2 end exponent open parentheses square root of straight x plus bx squared end root minus square root of straight x close parentheses end fraction equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator bx squared over denominator bx to the power of bevelled 3 over 2 end exponent open parentheses square root of straight x plus bx squared end root minus square root of straight x close parentheses end fraction space space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 below fraction numerator bx squared over denominator bx squared open parentheses square root of 1 plus bx end root plus 1 close parentheses end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator 1 over denominator square root of straight x plus bx plus 1 end root end fraction equals fraction numerator 1 over denominator square root of straight x plus 0 end root plus 1 end fraction equals 1 half.

Also f(0) = c
Since f is continuous at x = 0

therefore space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 0 right parenthesis rightwards double arrow space straight a plus 2 equals 1 half equals straight c space rightwards double arrow space straight a equals negative 3 over 2 comma straight c equals 1 half

b can be any real number except 0.

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Short Answer Type

67. Is the function defined by f(x) = | x |, a continuous function ?

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68. Discuss the continuity of the function f given by f(x) = x³ + x ² –1.

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69. Discuss the continuity of the function f defined by

straight f left parenthesis straight x right parenthesis equals 1 over straight x comma space straight x not equal to 0

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70. Examine the following functions for continuity :
f(x)=x-5

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Long Answer Type

66. Determine the value of a, by c for which the functionsis continous at x=0.

Short Answer Type