Given $f (x) =4\,\, - \,\,{\left( {\frac{1}{2}\, - \,x} \right)^{2/3}}\,$ $g (x) = \left\{ \begin{array}{l}\frac{{\tan \,\,[x]}}{x}\,\,\,\,,\,\,x \ne \,0\\1\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x\, = \,0\end{array} \right.$

$h (x) = \{x\}$   $k (x) = {5^{{{\log }_2}(x\, + \,3)}}$then in $[0, 1]$ Lagranges Mean Value Theorem is $NOT$ applicable to

If the function $f(x) = 2x^2 + 3x + 5$ satisfies $LMVT$ at $x = 3$ on the closed interval $[1, a]$ then the value of $a$ is equal to

Examine the applicability of Mean Value Theorem:

$(i)$ $f(x)=[x]$ for $x \in[5,9]$

$(ii)$ $f(x)=[x]$ for $x \in[-2,2]$

$(iii)$ $f(x)=x^{2}-1$ for $x \in[1,2]$

The function $f(x) = {(x - 3)^2}$ satisfies all the conditions of mean value theorem in $[3, 4].$ A point on $y = {(x - 3)^2}$, where the tangent is parallel to the chord joining $ (3, 0)$  and $(4, 1)$  is

If for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ 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