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 Rolle's theorem holds for the function $f(x)=x^{3}-a x^{2}+b x-4, x \in[1,2]$ with $f ^{\prime}\left(\frac{4}{3}\right)=0,$ then ordered pair $( a , b )$ is equal to

Suppose that $f (0) = - 3$ and $f ' (x) \le 5$ for all values of $x$. Then the largest value which $f (2)$ can attain is

If from mean value theorem, $f'({x_1}) = {{f(b) - f(a)} \over {b - a}}$, then

The value of $c$ in the Lagrange's mean value theorem for the function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}-4 \mathrm{x}^{2}+8 \mathrm{x}+11$ when $\mathrm{x} \in[0,1]$ is

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