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

Let $f$ be any function defined on $R$ and let it satisfy the condition

$|f( x )-f( y )| \leq\left|( x - y )^{2}\right|, \forall( x , y ) \in R$ If $f(0)=1,$ then

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 $c \in[0,1]$

Let $f :[2,4] \rightarrow R$ be a differentiable function such that $\left(x \log _e x\right) f^{\prime}(x)+\left(\log _e x\right) f(x)+f(x) \geq 1$, $x \in[2,4]$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$.

Consider the following two statements:

$(A): f(x) \leq 1$, for all $x \in[2,4]$

$(B)$ : $f(x) \geq \frac{1}{8}$, for all $x \in[2,4]$

Then,

The function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfy the conditions of Rolle's theorem in $[1, 3]. $ The values of  $a $ and $ b $ are

In the mean value theorem, $f(b) - f(a) = (b - a)f'(c)$if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of  $c$  is