Let $f$ be differentiable for all $x$. If $f(1) = -2$ and $f'(x) \ge 2$ for $x \in [1, 6]$,then

Let $f(x) = \begin{cases} x^\alpha \ln x, & x > 0 \\ 0, & x = 0 \end{cases}$. Rolle's theorem is applicable to $f$ for $x \in [0, 1]$ if $\alpha = $

Let $f(1) = -2$ and $f'(x) \ge 4.2$ for $1 \le x \le 6$. The possible value of $f(6)$ lies in the interval

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

For a polynomial $g(x)$ with real coefficients,let $m_g$ denote the number of distinct real roots of $g(x)$. Suppose $S$ is the set of polynomials with real coefficients defined by $S = \{(x^2-1)^2(a_0+a_1x+a_2x^2+a_3x^3) : a_0, a_1, a_2, a_3 \in \mathbb{R}\}$. For a polynomial $f$,let $f'$ and $f''$ denote its first and second order derivatives,respectively. Then the minimum possible value of $(m_f + m_{f'})$,where $f \in S$,is

If the function $f(x)=a x^3+b x^2+11 x-6$ satisfies the conditions of Rolle's theorem in $[1,3]$ and $f^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0$,then $a+b=$