The value of $c$ for the Lagrange's mean value theorem for $f(x)=\sqrt{x^2-x}, x \in[1,4]$ is

$f:[1,3] \rightarrow R$ is a function defined as $f(x)=x^3+a x^2+b x$. If $f(1)-f(3)=0$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$, then $a-b$ is equal to

Consider the quadratic equation $ax^2+bx+c=0$,where $2a+3b+6c=0$ and let $g(x)=\frac{ax^3}{3}+\frac{bx^2}{2}+cx$.
Statement-$I$ : The given quadratic equation $ax^2+bx+c=0$ has at least one root in $(0,1)$.
Statement-$II$ : Rolle's theorem is applicable to $g(x)$ on $[0,1]$.
Then

If a function $f$ is differentiable on $R$ such that $f^{\prime}(x) \leq 4$ for all $x \in R$; and if $f(2)=-6$ and $f(6)=8$,then the value of $f(4)$ belongs to the interval

If $f$ is a differentiable function such that $f(2x + 1) = f(1 - 2x)$ for all $x \in R$,then the minimum number of roots of the equation $f'(x) = 0$ in $x \in (-5, 10)$,given that $f(2) = f(5) = f(10)$,is:

Let $f(x)$ be continuous on $[0,6]$ and differentiable on $(0,6)$. Let $f(0)=12$ and $f(6)=-4$. If $g(x)=\frac{f(x)}{x+1}$, then for some Lagrange's constant $c \in(0,6)$, $g^{\prime}(c)=$