$(i)$  $f (x)$ 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 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

Let $y = f (x)$ and $y = g (x)$ be two differentiable function in $[0,2]$ such that  $f(0) = 3,$ $f(2) = 5$ , $g (0) = 1$ and $g(2) = 2$. If there exist atlellst one $c \in \left( {0,2} \right)$ such that $f'(c)=kg'(c)$,then $k$ must be

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

The value of $\left[ {\frac{{\log \left( {\frac{x}{e}} \right)}}{{x - \,e}}} \right]\,\forall x\, > \,e$ is equal to (where [.] denotes greatest integer function)

If $f(x) = \cos x,0 \le x \le {\pi \over 2}$, then the real number $ ‘c’ $ of the mean value theorem is