Question

Verify the truth of Rolle's Theorem for the function
$straight f left parenthesis straight x right parenthesis equals space sin space 3 straight x comma space straight x element of open square brackets 0 comma space straight pi over 3 close square brackets$

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

Here space straight f left parenthesis straight x right parenthesis equals sin space 3 straight x left parenthesis straight i right parenthesis space Since space sine space curve space is space continuous space therefore space straight f left parenthesis straight x right parenthesis space is space continuous space function space in space open square brackets 0 comma space straight pi over 3 close square brackets left parenthesis ii right parenthesis space straight f apostrophe left parenthesis straight x right parenthesis equals 3 space cos space 3 straight x comma space which space exists space in space open parentheses 0 comma space straight pi over 3 close parentheses therefore space straight f left parenthesis straight x right parenthesis space is space derivable space in space open parentheses 0 comma space straight pi over 3 close parentheses left parenthesis iii right parenthesis straight f left parenthesis 0 right parenthesis equals sin space 0 equals 0 comma space straight f open parentheses straight pi over 3 close parentheses equals sin space straight pi equals 0 therefore space straight f left parenthesis 0 right parenthesis equals straight f open parentheses straight pi over 3 close parentheses therefore space straight f left parenthesis straight x right parenthesis space satisfies space all space the space conditions space of space Rolle apostrophe straight s space theorem. space therefore space there space must space exist space at space least space one space value space straight c space of space straight x space such space that space straight f apostrophe left parenthesis straight c right parenthesis space equals space 0 space where straight c element of space open parentheses 0 comma space straight pi over 3 close parentheses. Now space straight f apostrophe left parenthesis straight c right parenthesis equals 3 space cos space 3 straight c therefore space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space 3 space cos space 3 straight c equals 0 space straight i. straight i. comma space cos space 3 space straight c equals 0 therefore space space space 3 space straight c equals straight pi over 2 comma space fraction numerator 3 straight pi over denominator 2 end fraction comma space fraction numerator 5 straight pi over denominator 2 end fraction comma... therefore space space space space space space straight c equals straight pi over 6 comma space fraction numerator 3 straight pi over denominator 2 end fraction comma space fraction numerator 5 straight pi over denominator 6 end fraction comma... space Now space straight c equals straight pi over 6 space lies space in space open parentheses 0 comma space straight pi over 3 close parentheses

∴ Rolle's Theorem is verified.

For Daily Free Study Material Join wiredfaculty

Continuity And Differentiability

Verify the truth of Rolle's Theorem for the function
$straight f left parenthesis straight x right parenthesis equals space sin space 3 straight x comma space straight x element of open square brackets 0 comma space straight pi over 3 close square brackets$

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location