If $c = \frac {1}{2}$ and $f(x) = 2x -x^2$ , then interval of $x$ in which $LMVT$, is applicable, is 

 $(i)$  $f (x)$ is continuous and defined for all real numbers

$(ii)$ $f '(-5) = 0 \,; \,f '(2)$ is not defined and $f '(4)  = 0$

$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f (x)$

$(iv)$ $f ''(2)$ is undefined, but $f ''(x)$ is negative everywhere else.

$(v)$ the signs of  $f '(x)$ is given below

Possible graph of $y = f (x)$ is

In which of the following functions Rolle’s theorem is applicable ?

Consider the function $f(x) = {e^{ - 2x}}$ $sin\, 2x$ over the interval $\left( {0,{\pi \over 2}} \right)$. A real number $c \in \left( {0,{\pi \over 2}} \right)\,,$ as guaranteed by Rolle’s theorem, such that $f'\,(c) = 0$ is

If the function  $f(x) =  - 4{e^{\left( {\frac{{1 - x}}{2}} \right)}} + 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}$ and $g(x)=f^{-1}(x) \,;$ then the value of $g'(-\frac{7}{6})$ equals

If the Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1 ]$ for the point $c = \frac{1}{2}$ , then the value of $2a + b$ is