The value of ‘c’ in Rolle’s theorem for  on [-1,1]

The correct answer is: 0

The given function is
f(x) = log(x² + 2) – log3
Statement of Rolle’s theorem: When a function is continuous on the interval [a, b], differentiable interval (a, b) and follows f(a) = f(b) then f’(c)=0.
The given interval is [-1, 1]
f(-1) = log[(-1)²+2]-log3
=log3-log3
f(1) = log[(1)² + 2] – log3
= log3 – log3
= 0
f(-1) = f(1)
Let c be such that f’(c) = 0


f'(c) = 0

c = 0
So, the value of c is 0

For such questions, we should know the properties of Rolle's theorem.