Learn Rolle’s theorem example in calculus, practice problems for finding the value of number 'c' which satisfies conclusion of Rolle's Theorem for a function.
Let \('f'\) be a function defined in [a, b] such that
A \(f(x)=2x^3+7x^2-9x\) in [0, 1]
B \(f(x)=sinx\) in \([0, \pi]\)
C \(f(x)=\dfrac {x-1}{x-4}\) in [2, 5]
D \(f(x)=x^2+5x+6\) in [–3, –2]
\(f'(x)=0\; \text{in } [a, b]\),
A c= 3
B c=\(\dfrac {3}{2}\)
C c= 2
D c= 5
A c= 2
B c= 1.7
C c= \(\dfrac {4}{3}\)
D c= –5
A c= \(\dfrac {\pi}{4}\)
B c= \(-\dfrac {\pi}{8}\)
C c= \(\dfrac {\pi}{6}\)
D c= \(\dfrac {\pi}{3}\)
then \(f(a) = f(b)\) is not possible for distinct values of 'a' and 'b' as it will violate Rolle's theorem.
by intermediate value theorem,
\(f(x)=0\), will have only one root and it will lie in (p, q)
A exactly one real root
B exactly two real roots
C at least two real roots
D at least four real roots
Function which does not satisfy Rolle's theorem will satisfy one of the following conditions
(1) Discontinuous in [a, b]
(2) Non differentiable in (a, b)
(3) \(f(a)\neq f(b)\)
A \(f\) is discontinuous at \(x=0\)
B \(f\) is non differentiable at \(x=0\)
C \(f\) is discontinuous at \(x=\dfrac {1}{2}\)
D \(f\) is discontinuous at \(x=\dfrac {-1}{2}\)