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

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f\left( 0 \right) = 2 = g\left( 1 \right)\;,\;\;g\left( 0 \right) = 0,$ and $f\left( 1 \right) = 6,$ then for some $c \in \left] {0,1} \right[$  . .

Verify Mean Value Theorem for the function $f(x)=x^{2}$ in the interval $[2,4]$

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

For every pair of continuous functions $f, g:[0,1] \rightarrow R$ such that $\max \{f(x): x \in[0,1]\}=\max \{g(x): x \in[0,1]\}$, the correct statement$(s)$ is (are) :

$(A)$ $(f(c))^2+3 f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$

$(B)$ $(f(c))^2+f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$

$(C)$ $(f(c))^2+3 f(c)=(g(c))^2+g(c)$ for some $c \in[0,1]$

$(D)$ $(f(c))^2=(g(c))^2$ for some $c \in[0,1]$

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