Determine if f is definend by from Mathematics Continuity and D

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Advertisement

Continuity and Differentiability

Long Answer Type

31. 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 square root of 1 plus straight x minus end root square root of 1 minus straight x end root over denominator sin space straight x end fraction comma space if space space straight x not equal to 0 end cell row cell space space space space space space space space space space space space 1 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 end table close

125 Views

Short Answer Type

32.

Determine if f is defined by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x squared space sin 1 over straight x comma space straight x not equal to 0 end cell row cell space space 0 space space space space space comma space straight x equals 0 end cell end table close$

74 Views

33.

Determine if f define by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space straight x over denominator straight x end fraction plus space cos space straight x comma space straight x not equal to 0 end cell row cell space space space space space space space space space space 2 space space space space space comma space straight x equals 0 end cell end table close at space straight x equals 0$

95 Views

Long Answer Type

Advertisement

34.
Determine if f is definend by
$straight f left parenthesis straight x 0 right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell xsin 1 over straight x comma space straight x not equal to 0 end cell row cell space space 0 space space space space comma space straight x equals 0 end cell end table close$

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell xsin 1 over straight x comma space straight x not equal to 0 end cell row cell space space 0 space space space space comma space straight x equals 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 space Lt with straight x rightwards arrow 0 to the power of minus below straight x space sin 1 over straight x space space space space space space left square bracket Put space straight x equals 0 minus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 to the power of minus right square bracket space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 below left parenthesis negative straight h right parenthesis sin open parentheses fraction numerator 1 over denominator 0 minus straight h end fraction close parentheses equals space Lt with straight x rightwards arrow 0 below left parenthesis negative straight h right parenthesis sin open parentheses negative 1 over straight h close parentheses space space space space space space space space equals Lt with straight x rightwards arrow 0 below left parenthesis negative straight h right parenthesis open parentheses negative sin 1 over straight h close parentheses equals Lt with straight x rightwards arrow below space straight h space sin 1 over straight h equals 0 space space space space space space space space space space space space space space space space space space space space open square brackets table row cell because Lt with straight x rightwards arrow 0 below straight h equals 0 space and space sin 1 over straight h space is space bounded space as space open vertical bar sin 1 over straight h close vertical bar less or equal than 1 end cell row cell therefore Lt with straight x rightwards arrow 0 below straight h space sin 1 over straight h equals 0 space as space we space know space that end cell row cell because Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis space straight g left parenthesis straight x right parenthesis equals 0 space if space Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis equals 0 space and space straight g left parenthesis straight x right parenthesis space is space bounded end cell end table close square brackets

Agian 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 plus below straight x space sin 1 over straight x space space left square bracket Put space straight x equals 0 plus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 to the power of plus right square bracket space space space space space space equals Lt with straight x rightwards arrow 0 below left parenthesis 0 plus straight h right parenthesis. sin open parentheses fraction numerator 1 over denominator 0 plus straight h end fraction close parentheses equals Lt with straight x rightwards arrow 0 below straight h space sin 1 over straight h equals 0 space space left square bracket As space expalained space above right square bracket Also space straight f left parenthesis 0 right parenthesis equals 0 therefore 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

⇒ f is continuous at x = 0.

76 Views

Advertisement

Advertisement

Short Answer Type

35. Find all points of continuity of f, where

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space straight x over denominator straight x end fraction comma space straight x less than 0 end cell row cell space straight x plus 1 space comma space straight x greater or equal than 0 end cell end table close at space straight x equals 0

79 Views

36. Discuss the continuity of 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 4 straight x over denominator straight x squared end fraction comma space straight x not equal to 0 end cell row cell space space space space space space space 8 space space space space comma space straight x equals 0 end cell end table close at space straight x equals 0

83 Views

Long Answer Type

37.

If space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell open vertical bar straight x close vertical bar sin 1 over straight x comma space if space straight x not equal to 0 end cell row cell space space space space space space space 0 comma space space space space space if space straight x equals 0 end cell end table close

then discuss continuity of f(x) at x = 0.

74 Views

Short Answer Type

38. Discuss the continuity at x = 0 of the function

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell open vertical bar straight x close vertical bar space cos 1 over straight x comma space straight x equals 0 end cell row cell space space space space space space 0 space space space space comma space straight x equals 0 end cell end table close

104 Views

Advertisement

39. Show 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 e to the power of 1 over straight x end exponent minus 1 over denominator straight e to the power of begin display style 1 over straight x plus 1 end style end exponent end fraction comma space straight x not equal to 0 end cell row cell space space space space space space 0 space space space space space space space comma space straight x equals 0 end cell end table close

is discontinous at x=0.

75 Views

40.

Let space 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 end cell end table if space straight x less than 2 end cell row cell table row cell 2 space if space straight x equals 2 end cell row cell straight x squared space if space straight x greater than 2 end cell end table end cell end table close

Show that 2 is a removable discontinuity of f.

83 Views

3
4
5

Advertisement