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

lf Rolle's theorem holds for the function $f(x) =2x^3 + bx^2 + cx, x \in [-1, 1],$  at the point $x = \frac {1}{2},$ then $2b+ c$ equals

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 the functions $f ( x )=\frac{ x ^3}{3}+2 bx +\frac{a x^2}{2}$ and $g(x)=\frac{x^3}{3}+a x+b x^2, a \neq 2 b$ have a common extreme point, then $a+2 b+7$ is equal to

Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq$ 0 , and $f(x)=g(x) \sin x$.

$STATEMENT$ $-1: \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.and

$STATEMENT$ $-2: f^{\prime}(0)=g(0)$.

From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$   $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $