If $c$ is a point at which Rolle's theorem holds for the function, $f(\mathrm{x})=\log _{\mathrm{e}}\left(\frac{\mathrm{x}^{2}+\alpha}{7 \mathrm{x}}\right)$ in the interval $[3,4],$ where $\alpha \in \mathrm{R},$ then $f^{\prime \prime}(\mathrm{c})$ is equal to

If the function  $f(x) =  – 4{e^{\left( {\frac{{1 – x}}{2}} \right)}} + 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}$ and $g(x)=f^{-1}(x) \,;$ then the value of $g'(-\frac{7}{6})$ equals

Let $f(x) = (x-4)(x-5)(x-6)(x-7)$ then –

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

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