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

In  $[0, 1]$ Lagrange's mean value theorem is $ NOT$  applicable to

 $(i)$  $f (x)$ 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

Which of the following function can satisfy Rolle's theorem ?

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $ [-1, 1] $ because

If the equation

${a_n}{x^{n - 1}} + \,{a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1}x = 0,\,{a_1} \ne 0,n\, \geqslant \,2,$

has a positive root $x= \alpha ,$ then the equation 

$n{a_n}{x^{n - 1}} + \,(n - 1){a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1} = 0$

has a positive root which is