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

Consider a quadratic equation $ax^2 + bx + c = 0,$ where $2a + 3b + 6c = 0$ and let $g(x) = a\frac{{{x^3}}}{3} + b\frac{{{x^2}}}{2} + cx.$

Statement $1:$ The quadratic equation has at least one root in the interval $(0, 1).$

Statement $2:$ The Rolle's theorem is applicable to function $g(x)$ on the interval $[0, 1 ].$

Verify Rolle's Theorem for the function $f(x)=x^{2}+2 x-8, x \in[-4,2]$

For the function $f(x) = {e^x},a = 0,b = 1$, the value of $ c$ in mean value theorem will be

From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$   $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $