Q. 1025.0( 1 Vote )

# State True

False

Rolle’s Theorem states that, Let f : [a, b] R be continuous on [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers.Then there exists some c in (a, b) such that f’(c) = 0.

We have, f(x) = |x – 1| in [0, 2].

Since, polynomial and modulus functions are continuous everywhere f(x) is continuous.

Now, x-1=0

x=1

We need to check if f(x) is differentiable at x=1 or not.

We have,

Lf’(1) =

=

= ( f(x) = 1-x, if x< 1)

=

=

Rf’(1) =

=

= ( f(x) = x-1, if 1< x)

=

Lf’(1) ≠ Rf’(1)

f(x) is not differentiable at x=1.

Hence, rolle’s theorem is not applicable on f(x) since it is not differentiable at x=1 (0,2).

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

Verify the Rolle’Mathematics - Exemplar

The value of c inMathematics - Exemplar

For the function Mathematics - Exemplar

Verify the Rolle’Mathematics - Exemplar

Verify the Rolle’Mathematics - Exemplar

Discuss theRD Sharma - Volume 1

Using Rolle’s theMathematics - Exemplar

Verify the Rolle’Mathematics - Exemplar

State True Mathematics - Exemplar

Discuss the appliMathematics - Exemplar