Given If f(x) is continuous at x = 0, find the value of k.

Question

Given $straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator 1 minus cos space 2 straight x over denominator straight x squared end fraction comma space straight x not equal to 0 end cell row cell space space space space space space space space space space space space space straight k space space space space space space comma space straight x equals 0 end cell end table close$
If f(x) is continuous at x = 0, find the value of k.

Solution

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 space 2 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 space space space space space space straight k 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 below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 below fraction numerator 1 minus cos space 2 straight x over denominator straight x squared end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator 2 space sin squared straight x over denominator straight x squared end fraction equals 2 Lt with straight x rightwards arrow 0 below fraction numerator sin squared straight x over denominator straight x squared end fraction space space space space space space space space space space space space space space space equals 2 Lt with straight x rightwards arrow 0 below open parentheses fraction numerator sin space straight x over denominator straight x end fraction close parentheses squared equals 2 left parenthesis 1 right parenthesis squared equals 2 cross times 1 equals 2 Since space straight f left parenthesis straight x right parenthesis space space is space continous space at space straight x equals 0 therefore space space space space 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 rightwards double arrow space straight k equals 2.

For Daily Free Study Material Join wiredfaculty

Continuity And Differentiability

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location