+91-8076753736 support@wiredfaculty.com
logo
logo

Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12034456
Wired Faculty App

Prove that the function
straight f left parenthesis straight x right parenthesis open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x over denominator open vertical bar straight x close vertical bar plus 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 straight k space space space space space space space comma space straight x equals 0 end cell end table close
remains discontinuous at x = 0, regardless of the choice of k.

Solution
Here space straight f left parenthesis straight x right parenthesis open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x over denominator open vertical bar straight x close vertical bar plus 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 straight k space space space 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 Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator straight x over denominator open vertical bar straight x close vertical bar plus 2 straight x squared end fraction 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 to the power of minus right square bracket space space space space space space space equals Lt with straight h rightwards arrow 0 below fraction numerator 0 minus straight h over denominator open vertical bar 0 minus straight h close vertical bar plus 2 left parenthesis 0 minus straight h right parenthesis squared end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator negative straight h over denominator straight h plus 2 straight h squared end fraction equals Lt with straight h rightwards arrow 0 below minus fraction numerator straight h over denominator 1 plus 2 straight h end fraction equals negative fraction numerator 1 over denominator 1 plus 0 end fraction equals negative 1 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 straight x over denominator open vertical bar straight x close vertical bar plus 2 straight x squared end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 space space space space space space space space equals Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar 0 plus straight h close vertical bar over denominator open vertical bar 0 plus straight h close vertical bar plus 2 left parenthesis 0 plus straight h right parenthesis squared end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator straight h over denominator straight h plus 2 straight h squared end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator 1 over denominator 1 plus 2 straight h end fraction equals fraction numerator 1 over denominator 1 plus 0 end fraction equals 1 therefore Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis not equal to Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis rightwards double arrow space Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis space does space not space exist
⇒ f(x) is discontinuous whatever k may be.

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation